Introductory notes in topology

Stephen Semmes

Rice University

Contents

1 Topological spaces 5

1.1 Neighborhoods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Other topologies on R 6

3 Closed sets 6

3.1 Interiors of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Metric spaces 8

4.1 Other metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

5 The real numbers 9

5.1 Additional properties . . . . . . . . . . . . . . . . . . . . . . . . . 105.2 Diameters of bounded subsets of metric spaces . . . . . . . . . . 10

6 The extended real numbers 11

7 Relatively open sets 11

7.1 Additional remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 12

8 Convergent sequences 12

8.1 Monotone sequences . . . . . . . . . . . . . . . . . . . . . . . . . 138.2 Cauchy sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

9 The local countability condition 14

9.1 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159.2 Sequentially closed sets . . . . . . . . . . . . . . . . . . . . . . . 15

10 Local bases 16

11 Nets 16

11.1 Sub-limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

12 Uniqueness of limits 17

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13 Regularity 18

13.1 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

14 An example 19

14.1 Topologies and subspaces . . . . . . . . . . . . . . . . . . . . . . 20

15 Countable sets 21

15.1 The axiom of choice . . . . . . . . . . . . . . . . . . . . . . . . . 2215.2 Strong limit points . . . . . . . . . . . . . . . . . . . . . . . . . . 22

16 Bases 22

16.1 Sub-bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2316.2 Totally bounded sets . . . . . . . . . . . . . . . . . . . . . . . . . 23

17 More examples 24

18 Stronger topologies 24

18.1 Completely Hausdorff spaces . . . . . . . . . . . . . . . . . . . . 25

19 Normality 25

19.1 Some remarks about subspaces . . . . . . . . . . . . . . . . . . . 2619.2 Another separation condition . . . . . . . . . . . . . . . . . . . . 27

20 Continuous mappings 27

20.1 Simple examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 2820.2 Sequentially continuous mappings . . . . . . . . . . . . . . . . . . 28

21 The product topology 29

21.1 Countable products . . . . . . . . . . . . . . . . . . . . . . . . . . 3021.2 Arbitrary products . . . . . . . . . . . . . . . . . . . . . . . . . . 31

22 Subsets of metric spaces 31

22.1 The Baire category theorem . . . . . . . . . . . . . . . . . . . . . 3222.2 Sequences of open sets . . . . . . . . . . . . . . . . . . . . . . . . 32

23 Open sets in R 33

23.1 Collections of open sets . . . . . . . . . . . . . . . . . . . . . . . 33

24 Compactness 34

24.1 A class of examples . . . . . . . . . . . . . . . . . . . . . . . . . . 34

25 Properties of compact sets 35

25.1 Disjoint compact sets . . . . . . . . . . . . . . . . . . . . . . . . . 3525.2 The limit point property . . . . . . . . . . . . . . . . . . . . . . . 36

26 Lindelof ’s theorem 37

26.1 The Lindelof property . . . . . . . . . . . . . . . . . . . . . . . . 3726.2 Applications to metric spaces . . . . . . . . . . . . . . . . . . . . 38

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27 Continuity and compactness 38

27.1 The extreme value theorem . . . . . . . . . . . . . . . . . . . . . 3927.2 Semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

28 Characterizations of compactness 40

28.1 Countable and sequential compactness . . . . . . . . . . . . . . . 4128.2 Some variants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

29 Products of compact sets 42

29.1 Sequences of subsequences . . . . . . . . . . . . . . . . . . . . . . 4329.2 Compactness of closed intervals . . . . . . . . . . . . . . . . . . . 43

30 Filters 44

30.1 Nets and filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4530.2 Mappings and filters . . . . . . . . . . . . . . . . . . . . . . . . . 46

31 Refinements 47

31.1 Another characterization of compactness . . . . . . . . . . . . . . 47

32 Ultrafilters 48

32.1 Connections with set theory . . . . . . . . . . . . . . . . . . . . . 4932.2 Tychonoff’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 49

33 Continuous real-valued functions 50

34 Compositions and inverses 51

34.1 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

35 Local compactness 52

36 Localized separation conditions 53

37 σ-Compactness 53

38 Topological manifolds 54

38.1 Unions of bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

39 σ-Compactness and normality 55

40 Separating points 56

40.1 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

41 Urysohn’s lemma 57

41.1 Complete regularity . . . . . . . . . . . . . . . . . . . . . . . . . 5841.2 R \ {0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

42 Countable bases 58

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43 One-point compactification 59

43.1 Supports of continuous functions . . . . . . . . . . . . . . . . . . 59

44 Connectedness 60

44.1 Connected topological spaces . . . . . . . . . . . . . . . . . . . . 6044.2 Connected sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6144.3 Other properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 6244.4 Pathwise connectedness . . . . . . . . . . . . . . . . . . . . . . . 6244.5 Connected components . . . . . . . . . . . . . . . . . . . . . . . . 6344.6 Local connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 6444.7 Local pathwise connectedness . . . . . . . . . . . . . . . . . . . . 65

45 A little set theory 66

45.1 Mappings and their properties . . . . . . . . . . . . . . . . . . . . 6645.2 One-to-one correspondences . . . . . . . . . . . . . . . . . . . . . 6645.3 One-to-one mappings . . . . . . . . . . . . . . . . . . . . . . . . . 6745.4 Some basic properties . . . . . . . . . . . . . . . . . . . . . . . . 6745.5 Bases of topological spaces . . . . . . . . . . . . . . . . . . . . . . 6845.6 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6845.7 Properties of exponentials . . . . . . . . . . . . . . . . . . . . . . 6945.8 Countable dense sets . . . . . . . . . . . . . . . . . . . . . . . . . 6945.9 Additional properties of exponentials . . . . . . . . . . . . . . . . 69

46 Some more set theory 70

46.1 Zorn’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7046.2 Hausdorff’s maximality principle . . . . . . . . . . . . . . . . . . 7146.3 Choice functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 7146.4 Comparing sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7246.5 Well-ordered sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 7346.6 Comparing well-ordered sets . . . . . . . . . . . . . . . . . . . . . 7446.7 Well-ordered subsets . . . . . . . . . . . . . . . . . . . . . . . . . 74

47 Some additional topics 75

47.1 Products of finite sets . . . . . . . . . . . . . . . . . . . . . . . . 7547.2 The Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7647.3 Quotient spaces and mappings . . . . . . . . . . . . . . . . . . . 7747.4 Homotopic paths . . . . . . . . . . . . . . . . . . . . . . . . . . . 7947.5 The fundamental group . . . . . . . . . . . . . . . . . . . . . . . 8047.6 Topological groups . . . . . . . . . . . . . . . . . . . . . . . . . . 8247.7 Topological vector spaces . . . . . . . . . . . . . . . . . . . . . . 8347.8 Uniform spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8447.9 The supremum norm . . . . . . . . . . . . . . . . . . . . . . . . . 8547.10Uniform continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 8647.11The supremum metric . . . . . . . . . . . . . . . . . . . . . . . . 8847.12An approximation theorem . . . . . . . . . . . . . . . . . . . . . 8947.13Infinitely many variables . . . . . . . . . . . . . . . . . . . . . . . 90

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Appendix

A Additional homework assignments 91

A.1 Limit points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91A.2 Dense open sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91A.3 Combining topologies . . . . . . . . . . . . . . . . . . . . . . . . 91A.4 Complements of countable sets . . . . . . . . . . . . . . . . . . . 92A.5 Combining topologies again . . . . . . . . . . . . . . . . . . . . . 92A.6 Combining topologies on R . . . . . . . . . . . . . . . . . . . . . 92A.7 Connections with product topologies . . . . . . . . . . . . . . . . 92

References 93

1 Topological spaces

A topology on a set X is a collection τ of subsets of X, the open subsets of Xwith respect to the topology, such that the empty set ∅ and X itself are opensets, the intersection of finitely many open sets is an open set, and the union ofany family of open sets is an open set. For any set X, the indiscrete topologyhas ∅, X as the only open sets, and the discrete topology is defined by sayingthat all subsets of X are open. On the real line R, a set U ⊆ R is open inthe standard topology if for every x ∈ U there are real numbers a, b such thata < x < b and (a, b) ⊆ U , where (a, b) is the open interval consisting of all realnumbers r such that a < r < b.

Let (X, τ) be a topological space. We say that X satisfies the first separation

condition if for every x, y ∈ X with x 6= y there is an open set V ⊆ X such thaty ∈ V and x 6∈ V . To be more precise, x and y may be chosen independentlyof each other, and so the first separation condition actually means that thereare a pair of open subsets V and V of X such that y ∈ V and x 6∈ V as well asx ∈ V and y 6∈ V . We say that X satisfies the second separation condition, orequivalently that X is a Hausdorff topological space, if for every x, y ∈ X withx 6= y there are open subsets U , V of X such that x ∈ U , y ∈ V , and U , V aredisjoint, i.e., U ∩ V = ∅. For the first homework assignment, discuss the casewhere X is a topological space with only finitely many elements which satisfiesthe first separation condition, and give an example of a topological space thatsatisfies the first separation condition but not the second.

1.1 Neighborhoods

The definition of a topological space may seem a bit strange at first, and cer-tainly quite abstract. It is sometimes helpful to look at a topological spaceX in terms of neighborhoods, where a neighborhood of a point x ∈ X can bedefined as any open set U ⊆ X such that x ∈ U . Alternatively, one may takethe neighborhoods of x in X to be the subsets of X that contain an open setcontaining x. At any rate, the neighborhoods of x in X are supposed to contain

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all elements of X which are sufficiently close to x. The definition of a topologyis a convenient way to make this precise, which can then be used for relatednotions of convergence, continuity, etc. In practice, one might start with a classof neighborhoods of elements of a set X, and use this to get a topology on X.Specifically, a set U ⊆ X would be defined to be an open set in this situationif for every x ∈ U there is a neighborhood of x in X which is contained in U .For example, the standard topology on the real line is essentially defined in thisway, using open intervals as a basic class of neighborhoods in R.

2 Other topologies on R

In addition to the standard topology on the real line R, let us consider a coupleof “exotic topologies” τ−, τ+, defined as follows. We say that U ⊆ R is an openset with respect to the topology τ− if for every x ∈ U there is a real numbera < x such that (a, x] ⊆ U , where the interval (a, x] consists of the real numberst which satisfy a < t ≤ x. Similarly, U ⊆ R is an open set with respect to thetopology τ+ if for every x ∈ U there is a real number b > x such that [x, b) ⊆ U ,where [x, b) consists of the real numbers t which satisfy x ≤ t < b. One cancheck that these are topologies on R, and that U ⊆ R is an open set in thestandard topology if and only if it is open with respect to τ− and τ+.

If (X, τ) is a topological space and E ⊆ X, then we say that E is dense inX if for every nonempty open set U ⊆ X, E ∩ U 6= ∅. Equivalently, E is densein X if for every x ∈ X and every open set U ⊆ X such that x ∈ U , there is apoint y ∈ E which approximates x in the sense that y ∈ U . Observe that X isautomatically dense in itself, and if X is equipped with the discrete topology,then X is the only dense set in X. One can check that the set Q of rationalnumbers is dense in the real line with respect to the standard topology, andalso with respect to the topologies τ−, τ+ described in the previous paragraph.This uses the fact that for every pair of real numbers a, b with a < b there is arational number r such that a < r < b.

3 Closed sets

Let (X, τ) be a topological space, and let E ⊆ X and p ∈ X be given. We saythat p is adherent to E in X, or equivalently that p is an adherent point of Ein X, if for every open set U ⊆ X such that p ∈ U there is an x ∈ E withx ∈ U . We say that p is a limit point of E in X if for every open set U ⊆ Xsuch that p ∈ U there is a y ∈ E with y 6= p and y ∈ U . Every element ofE is automatically adherent to E, and every limit point of E is automaticallyadherent to E as well. An adherent point of E in X which is not an element ofE is automatically a limit point of E, and a limit point of E may or may notbe an element of E.

The complement of a subset A of X is denoted X \A and defined to be theset of z ∈ X such that z 6∈ A. Thus the complement of the complement of A is

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equal to A. The complement of the union of a family of subsets of X is equalto the intersection of the corresponding family of complements of the sets, andsimilarly the complement of the intersection of a family of subsets of X is equalto the union of the corresponding family of complements of the sets. We saythat A ⊆ X is closed if the complement of A in X is an open set. It followsfrom the definition of a topology that ∅, X are closed subsets of X, the unionof finitely many closed subsets of X is a closed set, and the intersection of anyfamily of closed subsets of X is a closed set.

The closure of E ⊆ X is denoted E and defined to be the set of points in Xthat are adherent to E, which is the same as the union of E and the set of limitpoints of E in X. Equivalently, z ∈ X is not in the closure of E if there is anopen set U ⊆ X such that z ∈ U and E ∩ U = ∅, in which case U is containedin the complement of the closure of E in X. It follows that X \ E is equal tothe union of the open subsets of X contained in X \ E, so that X \ E is alsoan open set in X, which implies that E is a closed set in X. If E is a closedsubset of X, then it is easy to see that E = E, because every element of thecomplement of E is contained in the open set X \ E which does not intersectE by definition, and hence is not adherent to E. Note that a subset E of X isdense in X if and only if E = X.

3.1 Interiors of sets

Let X be a topological space. The interior of a set A ⊆ X is denoted IntA,and defined to be the set of x ∈ A for which there is an open set U ⊆ X suchthat x ∈ U and U ⊆ A. If U ⊆ X is any open set such that U ⊆ A, then everyelement of U is contained in A, so that

U ⊆ IntA.

Thus IntA may be described equivalently as the union of all of the open subsetsof X that are contained in A. In particular, it follows that IntA is automaticallyan open subset of X too. It is easy to check that the complement of the interiorof A ⊆ X in X is equal to the closure of the complement of A in X, which is tosay that

X \ IntA = X \ A.

This is essentially the same as saying that

X \ E = Int(X \ E)

for every E ⊆ X, which was also mentioned earlier, in effect.If A1, A2 are subsets of X, then

Int(A1 ∩ A2) = (Int A1) ∩ (IntA2).

This follows from the definition of the interior of a set. Similarly,

(Int A1) ∪ (IntA2) ⊆ Int(A1 ∪ A2).

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Thus,E1 ∪ E2 = E1 ∪ E2

andE1 ∩ E2 ⊆ E1 ∩ E2

for every E1, E2 ⊆ X.

4 Metric spaces

By a metric space we mean a set M together with a function d(x, y) defined forx, y ∈ M such that (i) d(x, y) is a nonnegative real number for every x, y ∈ Mwhich is equal to 0 if and only if x = y, (ii) d(x, y) = d(y, x) for every x, y, z ∈ M ,and (iii) the triangle inequality holds, which is to say that

d(x, z) ≤ d(x, y) + d(y, z)

for every x, y, z ∈ M . In this event we say that d(x, y) defines a metric on M .For every x ∈ M and positive real number r, we put

B(x, r) = {y ∈ M : d(x, y) < r},

the open ball centered at x with radius r in M . We say that U ⊆ M is an openset if for every x ∈ U there is an r > 0 such that B(x, r) ⊆ U . One can checkthat this defines a topology on M . Specifically, if U1, . . . , Un are open subsetsof M in this sense and x is an element of the intersection U1 ∩ · · · ∩ Un, thenfor each i = 1, . . . , n, x is an element of Ui, and there is an ri > 0 such thatB(x, ri) ⊆ Ui. If r is equal to the minimum of r1, . . . , rn, then r > 0 and B(x, r)is contained in U1 ∩ · · · ∩Un. This shows that U1 ∩ · · · ∩Un is an open set in M ,and the fact that arbitrary unions of open subsets of M are open sets is evensimpler. For every x ∈ M and r > 0, one can check that the open ball B(x, r)is an open set, because for every y ∈ B(x, r) we have that

B(y, t) ⊆ B(x, r)

with t = r − d(x, y) > 0, by the triangle inequality.On any set M , the discrete metric is defined by putting d(x, y) equal to 0

when x = y and to 1 when x 6= y. One can check that this is a metric, and thatthe corresponding topology on M is the discrete topology.

4.1 Other metrics

Let (M,d(x, y)) be a metric space. For every positive real number ρ, ρ d(x, y)defines a metric on M which determines the same topology on M as d(x, y)does, i.e., the same class of open subsets of M . To be more precise, the openball in M centered at a point w ∈ M and with radius t > 0 with respect to theinitial metric d(x, y) is the same as the open ball in M centered at w with radiusρ t with respect to the new metric ρ d(x, y). Similarly, for every positive real

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number τ , dτ (x, y) = min(d(x, y), τ) defines a metric on M which determinesthe same topology as d(x, y). In order to check that dτ (x, y) satisfied the triangleinequality, observe that

min(a + b, τ) ≤ min(a, τ) + min(b, τ)

for every a, b ≥ 0. The open ball in M centered at a point z ∈ M and withradius r > 0 with respect to d(x, y) is the same as the open ball in M centeredat z with radius r with respect to dτ (x, y) when r ≤ τ . When r > τ , any openball in M with radius r with respect to dτ (x, y) contains every element of M .However, the equivalence of balls with small radius is sufficient to imply thatthe two metrics determine the same class of open subsets of M , i.e., the sametopology. As another example along these lines, consider d′(x, y) =

√d(x, y).

For any real numbers a, b ≥ 0,

a2 + b2 ≤ a2 + 2 a b + b2 = (a + b)2,

and one can use this to show that d′(x, y) satisfies the triangle inequality andhence is a metric on M . In this case, the open ball centered at a point p ∈ Mand with radius r > 0 with respect to d(x, y) is the same as the open ballcentered at p with radius

√r with respect to d′(x, y), and this implies that the

associated topologies are the same.

5 The real numbers

If x is a real number, then the absolute value |x| of x is equal to x when x ≥ 0and to −x when x ≤ 0. Hence |x| ≥ 0 for every x ∈ R, and |x| = 0 if and onlyif x = 0. One can check that

|x + y| ≤ |x| + |y|

for every x, y ∈ R, and therefore that |x − y| defines a metric on R, known asthe standard metric. The topology determined by the standard metric is thestandard topology. To be more precise, the open ball in R centered at x ∈ R

with radius r > 0 is the same as the open interval (x − r, x + r).If A ⊆ R and b ∈ R, then we say that b is an upper bound for A if a ≤ b for

every a ∈ A. We say that c ∈ R is the least upper bound or supremum of A,denoted supA, if c is an upper bound for A and if c ≤ b for every upper boundb of A. It follows directly from the definition that the supremum of A is uniqueif it exists, i.e., if c1, c2 ∈ R both satisfy the requirements of suprema of A, thenc1 ≤ c2, c2 ≤ c1, and thus c1 = c2. The completeness property of the real linestates that every nonempty set of real numbers which has an upper bound hasa supremum. The rational numbers are not complete in this sense, because theset of t ∈ Q such that t ≥ 0 and t2 ≤ 2 is a nonempty subset of Q with anupper bound that does not have a supremum in Q, and in fact the supremumin the real line is equal to

√2.

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Similarly, w ∈ R is a lower bound for B ⊆ R if w ≤ x for every x ∈ B,and z ∈ R is the greatest lower bound or infimum of B, denoted inf B, if z isa lower bound of B and every lower bound w ∈ R of B satisfies w ≤ z. Asin the previous situation, the infimum is automatically unique if it exists. If Bis a nonempty set of real numbers with a lower bound, then the completenessproperty for the real numbers implies that B has an infimum, which can beobtained as the supremum of the set of lower bounds of B. Alternatively, theinfimum of B is equal to the negative of the supremum of

−B = {x ∈ R : −x ∈ B}.

5.1 Additional properties

The completeness property of the real numbers can be used to show that everya ∈ R with a > 0 has a positive square root. Specifically, let A be the set ofr ∈ R such that r ≥ 0 and r2 ≤ a. Clearly 0 ∈ A, and thus A 6= ∅. If a ≤ 1,then r ≤ 1 for every r ∈ A, while if a ≥ 1, then r ≤ a for every a ∈ A. HenceA is a nonempty set of real numbers with an upper bound, and thus there is ab ∈ R which is the least upper bound of A. If b2 < a, then one can show thatthere is an ǫ > 0 such that (b+ǫ)2 = b2 +2 b ǫ+ǫ2 < a. Similarly, if b2 > a, thenone can show that there is a δ > 0, δ < b, such that (b−δ)2 = b2−2 b δ+δ2 > a.In both cases one can get a contradiction with the fact that b = supA, and itfollows that b2 = a. It is easy to show directly that a positive square root of ais unique.

If r, t ∈ R and r ≤ t, then the closed interval [r, t] consists of the x ∈ R

such that r ≤ x ≤ t. Suppose that Il = [rl, tl], l = 1, 2, . . ., is a sequence ofclosed intervals in the real line such that Il ⊆ Il+1 for every l ≥ 1. Equivalently,rl ≤ rl+1 ≤ tl+1 ≤ tl for every l ≥ 1, which implies that rl ≤ rn ≤ tn ≤ tl when1 ≤ l ≤ n. Using completeness one can show that there is an x ∈ R such thatx ∈ Il for every l ≥ 1. To be more precise, the intersection

⋂∞l=1 Il of all of the

Il’s is the same as the closed interval [r, t], where r is the supremum of the setof rl’s and t is the infimum of the set of tl’s, l ≥ 1.

5.2 Diameters of bounded subsets of metric spaces

Let (M,d(x, y)) be a metric space. A set E ⊆ M is said to be bounded if thereis a point p ∈ M and a positive real number r such that d(x, p) ≤ r for everyx ∈ E. In this case, d(x, q) ≤ d(p, q) + r for every q ∈ M and x ∈ E, by thetriangle inequality. Equivalently, E ⊆ M is bounded if the set of real numbersof the form d(x, y), x, y ∈ E, is bounded. If E ⊆ M is bounded and nonempty,then the diameter of E is denoted diam E and defined by

diam E = sup{d(x, y) : x, y ∈ E}.Suppose that E1, E2 are nonempty subsets of M such that E1 ⊆ E2. If E2 isbounded, then E1 is bounded too, and

diam E1 ≤ diam E2.

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If E ⊆ M is bounded and nonempty, then the closure E of E is bounded, and

diam E = diam E.

For if x, y ∈ E and ǫ > 0 is arbitrary, then there are x′, y′ ∈ E such thatd(x, x′), d(y, y′) < ǫ. Hence d(x, y) ≤ d(x′, y′) + 2 ǫ, by the triangle inequality,which implies that d(x, y) ≤ diam E +2 ǫ. Because this holds for every x, y ∈ E,we may conclude that E is bounded and diamE ≤ diam E + 2 ǫ, which impliesin turn that diamE ≤ diam E since ǫ > 0 is arbitrary. The opposite inequalityholds automatically, and therefore the diameters of E and E are equal.

6 The extended real numbers

By definition, the extended real numbers consist of the real numbers togetherwith two additional elements, −∞ and +∞, such that

−∞ < x < +∞

for every x ∈ R. The notions of upper and lower bounds make sense for setsof extended real numbers, with +∞ as an upper bound and −∞ as a lowebound for any set of extended real numbers. Using the completeness propertyfor the real numbers, one can check that every nonempty set of extended realnumbers has a unique supremum and infimum, defined in the same way as forsets of real numbers. In particular, if E is a nonempty set of real numberswhich does not have an upper bound in R, then the supremum of E is equal to+∞ in the extended real numbers. Similarly, the infimum of a nonempty set ofreal numbers without a finite lower bound is equal to −∞ in the extended realnumbers.

A set U of extended real numbers is an open set if for every x ∈ U eitherx ∈ R and there are real numbers a, b such that a < x < b and (a, b) ⊆ U ,or x = −∞ and there is a real number b such that the set [−∞, b) of extendedreal numbers y < b is contained in U , or x = +∞ and there is a real numbera such that the set (a,+∞] of extended real numbers z > a is contained in U .This defines a topology on the set of extended real numbers, for which the realline R is a dense open set. This topology agrees with the one on R in the sensethat E ⊆ R is open in R with the standard topology if and only if E is openin the extended real numbers. If E ⊆ R, then +∞ is a limit point of E in theextended real numbers if and only if E does not have an upper bound in R, andsimilarly −∞ is a limit point of E in the extended real numbers if and only ifE does not have a lower bound in R.

7 Relatively open sets

Suppose that (X, τ) is a topological space and that Y ⊆ X. There is a naturaltopology on Y induced by the one on X, in which E ⊆ Y is open in Y whenthere is an open set U ⊆ X such that E = U ∩ Y . In this event we say that E

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is relatively open in Y . If Y happens to be an open set in X, then E ⊆ Y isrelatively open in Y if and only if E is an open set in X, but otherwise relativelyopen subsets of Y may not be open in X. Following the general definition ofclosed subsets of a topological space, E ⊆ Y is relatively closed in Y if Y \ Eis relatively open in Y . Hence E is relatively closed in Y when Y \ E = U ∩ Yfor some open set U in X, which implies that E = (X \ U) ∩ Y and X \ Uis closed in X. Conversely, if E = A ∩ Y for some closed set A in X, thenY \ E = (X \ A) ∩ Y and X \ A is open in X, and E is relatively closed in Y .In particular, if Y is closed in X, then E ⊆ Y is relatively closed in Y if andonly if E is closed in X. For any Y , the relative closure of E ⊆ Y , which is theclosure of E in Y as a topological space itself, is equal to the intersection of Ywith the closure of E in X.

If (M,d(x, y)) is a metric space and Y ⊆ M , then the restriction of d(x, y) tox, y ∈ Y defines a metric on Y . The topology on Y associated to the restrictionof the metric to Y is the same as the topology on Y induced from the topologyon X determined by the metric. Indeed, if U ⊆ X is open in X, then U ∩ Yis open in Y with respect to the restriction of the metric to Y . Conversely, theopen ball in Y with center y ∈ Y and radius r > 0 is automatically equal to theintersection of Y with the open ball in X with center y and radius r. Therefore,every open ball in Y is relatively open with respect to the topology induced fromthe one on X. If E ⊆ Y is an open set with respect to the topology determinedby the restriction of the metric to Y , then E is equal to a union of open balls inY , the open balls in Y that are contained in E. Each of these open balls is theintersection of Y with an open ball in X, and hence their union is the same asthe intersection of Y with the union of the corresponding open balls in X, anopen set in X.

7.1 Additional remarks

Let X be a topological space. If U ⊆ X is an open set, then for every Y ⊆ X,U ∩Y is relatively open in Y . Conversely, if U ⊆ X and U ∩Y is relatively openin Y for every Y ⊆ X, then U is an open set in X, since we can take Y = X.Similarly, E ⊆ X is a closed set if and only if E ∩Y is relatively closed in Y forevery Y ⊆ X. As a refinement of this type of question, one can look for classesof subsets Y of X such that open and closed subsets of X are characterized byintersections with Y being relatively open or closed in Y , respectively, for everyY in the particular class of subsets of X. For example, the class of open subsetsY of X has this property. Actually, any class of open subsets of X whose unionis equal to X has this property. Any class of subsets of X for which the unionof their interiors is equal to X has this property for the same reasons.

8 Convergent sequences

A sequence {xj}∞j=1 of elements of a topological space X converges to x ∈ Xif for every open set U ⊆ X with x ∈ U there is a positive integer L such that

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xj ∈ U for every j ≥ L. If X is equipped with the indiscrete topology, thenevery sequence {xj}∞j=1 of elements of X converges to every x ∈ X. If X isequipped with the discrete topology and {xj}∞j=1 is a sequence of elements ofX that converges to x ∈ X, then there is an L ≥ 1 such that xj = x for everyj ≥ L, because {x} is an open set in X. If X is any topological space, and ifthere is an L ≥ 1 such that xj = x for every j ≥ L, then {xj}∞j=1 converges tox. In the real line, the sequence {xj}∞j=1 defined by xj = 1/j for each j ≥ 1converges to 0 with respect to the standard topology.

Let X be a topological space, let x be an element of X, and consider thesequence {xj}∞j=1 such that xj = x for every j. This sequence automaticallyconverges to x in X, as in the previous paragraph. If X satisfies the firstseparation condition, y ∈ X, and y 6= x, then there is an open set V ⊆ X suchthat y ∈ V and x 6∈ V , and it follows that {xj}∞j=1 does not converge to y in X.In the other direction, if y ∈ X, and if every open set V ⊆ X with y ∈ V alsocontains x, then {xj}∞j=1 converges to y.

Suppose that X is a topological space, E ⊆ X, and {xj}∞j=1 is a sequence ofelements of E which converges to x ∈ X. Under these conditions, x is adherentto E, and hence an element of the closure E of E, and an element of E if Eis a closed set. If xj 6= x for every j, then x is a limit point of E. Note thatthe first separation condition is equivalent to the statement that subsets of Xwith exactly one element are closed. More precisely, for each x ∈ X, the firstseparation condition implies that X \ {x} can be expressed as the union of afamily of open subsets of X, so that X \ {x} is an open set in X too.

8.1 Monotone sequences

A sequence {xl}∞l=1 of real numbers is said to be monotone increasing if xl ≤ xn

for every n ≥ l ≥ 1. Similarly, a sequence {yl}∞l=1 of real numbers is said to bemonotone decreasing if yl ≤ yn when n ≥ l ≥ 1. Every monotone increasing ordecreasing sequence of real numbers which is bounded in the sense that the termsof the sequence are contained in a bounded set in R converges. Specifically, onecan check that a monotone increasing sequence of real numbers which is boundedfrom above converges to the supremum of the set of terms in the sequence, anda monotone decreasing sequence of real numbers which is bounded from belowconverges to the infimum of the terms in the sequence. In particular, this usesthe completeness property of the real numbers.

Now consider the real line equipped with the exotic topologies τ+, τ−, whereU ⊆ R is an open set with respect to τ+ if and only if for every x ∈ U there isa t ∈ R with t > x such that [x, t) ⊆ U , and V ⊆ R is an open set with respectto τ− if and only if for every y ∈ V there is an r ∈ R with r < y such that(r, y] ⊆ V . A monotone increasing sequence of real numbers which is boundedfrom above converges to the supremum of the terms in the sequence with respectto the topology τ−, but a monotone decreasing sequence converges with respectto τ− if and only if it is eventually constant. In the same way, a monotonedecreasing sequence of real numbers which is bounded from below converges tothe infimum of the terms in the sequence with respect to the topology τ+, but

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a monotone increasing sequence converges with respect to τ+ if and only if itis eventually constant. To be more precise, a sequence is said to be eventuallyconstant if all but finitely many terms in the sequence are the same.

8.2 Cauchy sequences

Let (M,d(x, y)) be a metric space. A sequence {xl}∞l=1 of elements of M is saidto be a Cauchy sequence if for every ǫ > 0 there is an L ≥ 1 such that

d(xl, xn) < ǫ for every l, n ≥ L.

If {xl}∞l=1 converges to a point x ∈ M , then {xl}∞l=1 is a Cauchy sequence,because for every ǫ > 0 there is an L such that

d(xl, x) <ǫ

2when l ≥ L,

and henced(xl, xn) ≤ d(xl, x) + d(xn, x) <

ǫ

2+

ǫ

2= ǫ

when l, n ≥ L. A metric space in which every Cauchy sequence converges issaid to be complete.

One can use the completeness property of the real numbers in terms ofordering to show that the real line equipped with the standard metric |x − y|is complete as a metric space. For suppose that {xl}∞l=1 is a Cauchy sequenceof real numbers, and consider El = {xn : n ≥ l} for each l. Because {xl}∞l=1 isa Cauchy sequence, El is a bounded set of real numbers for every l ≥ 1, andhence has an infimum rl and a supremum tl. The fact that {xl}∞l=1 is a Cauchysequence implies that tl − rl converges to 0 as a sequence of real numbers. Onecan check that the supremum of the rl’s is equal to the infimum of the tl’s, andthat {xl}∞l=1 converges to their common value.

9 The local countability condition

A topological space X satisfies the local countability condition at a point p ∈ Xif there is a sequence {Uℓ(p)}∞ℓ=1 of open subsets of X such that p ∈ Uℓ(p) andUℓ+1(p) ⊆ Uℓ(p) for every ℓ ≥ 1, and for every open set U ⊆ X with p ∈ Uthere is an ℓ ≥ 1 such that Uℓ(p) ⊆ U . It is easy to arrange for the Uℓ(p)’sto be decreasing, by replacing Uℓ(p) with U1(p) ∩ · · · ∩ Uℓ(p) if necessary, andthis ensures that Uℓ(p) ⊆ U implies Uk(p) ⊆ U when k ≥ ℓ. The topology of ametric space satisfies the local countability condition at every point, because onecan take Uℓ(p) to be the open ball centered at p with radius 1/ℓ. The real lineequipped with the topology τ+, in which U ⊆ R is open if and only if for everyp ∈ U there is a t ∈ R with t > p such that [p, t) ⊆ U , also satisfies the localcountability condition at every point, because one can take Uℓ(p) = [p, p+1/ℓ).Similarly, the real line equipped with the topology τ−, in which U ⊆ R is openif and only if for every p ∈ U there is an r ∈ R, r < t, such that (r, t] ⊆ U ,satisfies the local countability condition at every point, with Uℓ(p) = (p−1/ℓ, p].

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Suppose that X is a topological space that satisfies the local countabilitycondition at a point p ∈ X, and let {Uℓ(p)}∞ℓ=1 be a sequence of open sets as inthe previous paragraph. If p is adherent to E ⊆ X, then for every ℓ ≥ 1, thereis an xℓ ∈ E such that xℓ ∈ Uℓ(p). It is easy to see that {xℓ}∞ℓ=1 converges top in X under these conditions. If p is a limit point of E, then one can choosexℓ such that xℓ 6= p for each ℓ ≥ 1 too. Remember that there are analogousstatements in the other direction that work in any topological space.

9.1 Subsequences

Let {xi}∞i=1 be a sequence of elements of some set X. A subsequence of {xi}∞i=1

is a sequence of the form {xin}∞n=1, where i1 < i2 < · · · is a strictly increasing

sequence of positive integers. Suppose now that X is a topological space. LetE1 be the set of x ∈ X for which there is a subsequence {xin

}∞n=1 of {xi}∞i=1

such that {xin}∞n=1 converges to x in X. Let E2 be the set of x ∈ X with the

property that for every open set U ⊆ X with x ∈ U there are infinitely manypositive integers l such that xl ∈ U . It follows directly from the definitionsthat E1 ⊆ E2. Conversely, if X satisfies the local countability condition at eachpoint, then one can check that E2 ⊆ E1 and hence E1 = E2. In any topologicalspace X, it is easy to see that E2 is a closed set. Therefore, if X satisfies thelocal countability condition at each point, then E1 is a closed set in X too.

9.2 Sequentially closed sets

Let (X, τ) be a topological space. A set E ⊆ X is said to be sequentially closed

if for every sequence {xi}∞i=1 of elements of E and every x ∈ X such that {xi}∞i=1

converges to x in X we have that x ∈ E. Closed subsets of X in the usual senseare automatically sequentially closed. Conversely, sequentially closed subsets ofX are closed in the usual sense when X satisfies the local countability conditionat each point. More generally, if Y ⊆ X satisfies the local countability conditionat each of its elements with respect to the topology induced by τ on X and ifE is sequentially closed in X, then E ∩ Y is relatively closed in Y .

It is easy to see that the intersection of any family of sequentially closedsubsets of X is sequentially closed. If E1, E2 are sequentially closed subsetsof X, then their union E1 ∪ E2 is sequentially closed, because any sequence ofelements of E1∪E2 has a subsequence whose terms are contained in E1 or in E2.Of course a subsequence of a convergent sequence converges to the same limit.It follows that the collection τ of complements of sequentially closed subsetsof X defines a topology on X that contains τ . However, it is not clear that asequence in X which converges with respect to τ also converges with respect toτ .

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10 Local bases

A local base for the topology of a topological space X at a point p ∈ X is acollection B(p) of open subsets of X such that p ∈ V for every V ∈ B(p), andfor every open set U ⊆ X with p ∈ U there is a V ∈ B(p) which satisfies V ⊆ U .The collection of all open subsets of X that contain p as an element is a localbase for the topology at p. In a metric space, the open balls centered at p forma local base for the topology at p. The exotic topologies τ+, τ− on the realline are characterized by the local bases B+(p), B−(p), where B+(p) consistsof the intervals [p, t), t > p, and B−(p) consists of the intervals (r, p], r < p.In any topological space, the topology is determined by choices of local basesat all of the elements of the space. A topological space X satisfies the localcountability condition at a point p ∈ X if and only if there is a local base B(p)for the topology at p whose elements can be enumerated by a sequence. Asnoted previously, a sequence of open subsets of X that form a local base for thetopology of X at p can be modified to get a decreasing sequence of open setswhich form a local base for the topology at p using the intersection of the firstn open sets in the original sequence for each n.

Let B(p) be a local base for the topology of X at p, and let E ⊆ X be given.It is easy to see that p is adherent to E if and only if for every U ∈ B(p) thereis a q ∈ E such that q ∈ U . Similarly, p is a limit point of E if and only if forevery U ∈ B(p) there is a q ∈ E such that q ∈ U and q 6= p.

11 Nets

A partial order on a set A is a binary relation ≺ such that (i) x ≺ x for everyx ∈ A, (ii) x ≺ y and y ≺ x imply x = y for every x, y ∈ A, and (iii) x ≺ yand y ≺ z imply x ≺ z for every x, y, z ∈ A. A partially-ordered set (A,≺) isa directed system if for every collection a1, . . . , an of finitely many elements ofA there is an a ∈ A such that ai ≺ a for i = 1, . . . , n. For example, the set ofpositive integers is a directed system with respect to the usual ordering. If Xis a topological space, p ∈ X, and B(p) is a local base for the topology of X atp, then B(p) becomes a directed system with the ordering ≺ defined by U ≺ Vwhen V ⊆ U for every U, V ∈ B(p). The ordering is the reverse of inclusion,because smaller open sets represent greater precision at p. If U1, . . . , Un ∈ B(p),then their intersection U1 ∩ · · · ∩ Un is an open set in X that contains p, andhence there is a V ∈ B(p) such that V ⊆ U1 ∩ · · · ∩Un. This shows that B(p) isa directed system with respect to the ordering ≺.

If (A,≺) is a directed system and X is any set, then a net {xa}a∈A indexedby A with values in X assigns to every a ∈ A an element xa of X . If A is the setof positive integers with the usual ordering, then this is the same as a sequenceof elements of X. If X is a topological space, {xa}a∈A is a net of elements ofX along the directed system A, and x ∈ X, then {xa}a∈A is said to converge

to x if for every open set U ⊆ X there is a b ∈ A such that xa ∈ U for everya ∈ A which satisfies b ≺ a. This is the same as the definition of convergence

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of a sequence when A is the set of positive integers with the usual ordering. IfE ⊆ X and {xa}a∈A is a net of elements of E which converges to x ∈ X, then xis adherent to E. Conversely, every adherent point of E is the limit of a net ofelements of E. Similarly, if {xa}a∈A is a net of elements of E which convergesto x ∈ X and xa 6= x for every a ∈ A, then x is a limit point of E, and everylimit point of E is the limit of a convergent net of elements of E \ {x}.

11.1 Sub-limits

Let X be a topological space, let (A,≺) be a directed system, and let {xa}a∈A

be a net of elements of X indexed by A. Let E be the set of x ∈ X with theproperty that for every a ∈ A and open set U ⊆ X with x ∈ U there is a b ∈ Asuch that a ≺ b and xb ∈ U . It is easy to see from the definition that E is aclosed set in X. Fix an x ∈ E. Let Ax be the set of ordered pairs (a, U) wherea ∈ A and U ⊆ X is an open set with x ∈ U . It would be sufficient for thepresent purposes to use only open sets U in a local base for the topology of Xat x. If (a, U), (b, V ) ∈ Ax, then put (a, U) ≺x (b, V ) when a ≺ b and V ⊆ U .One can check that this defines a partial ordering on Ax which makes Ax intoa directed system. For (a, U) ∈ Ax, let x(a,U) be an element of X of the formxb, where b ∈ A, a ≺ b, and xb ∈ U . By construction, {x(a,U)}(a,U)∈Ax

is a netof elements of X indexed by Ax that converges to x.

12 Uniqueness of limits

Let X be a Hausdorff topological space, let (A,≺) be a directed system, andlet {xa}a∈A be a net of elements of X which converges to x, y ∈ X. If x 6= y,then the Hausdorff property implies that there are open subsets U , V of X suchthat x ∈ U , y ∈ V , and U ∩ V = ∅. Because of convergence, there are b, b′ ∈ Asuch that xa ∈ U when b ≺ a and xa ∈ V when b′ ≺ a. If c ∈ A and b, b′ ≺ c,then xa ∈ U ∩ V when c ≺ a, a contradiction. This shows that the limit of aconvergent net in a Hausdorff space is unique, and in particular the limit of aconvergent sequence is unique.

Conversely, suppose that X is a topological space which is not Hausdorff.Hence there are x, y ∈ X such that x 6= y and for every pair of open subsets U ,V of X with x ∈ U and y ∈ V , we have that U ∩ V 6= ∅. Let A be the set ofordered pairs (U, V ) of open subsets of X with x ∈ U and y ∈ V . There is anatural partial ordering ≺ on A, where (U, V ) ≺ (U ′, V ′) for (U, V ), (U ′, V ′) ∈ Awhen U ′ ⊆ U and V ′ ⊆ V . If (U1, V1), . . . , (Un, Vn) are finitely many elementsof A, and if we put U = U1 ∩ · · · ∩ Un, V = V1 ∩ · · · ∩ Vn, then (U, V ) ∈ A and(Ui, Vi) ≺ (U, V ) for i = 1, . . . , n. This shows that (A,≺) is a directed system.For each (U, V ) ∈ A, let x(U,V ) be an element of U ∩V . This defines a net alongA with values in X. By construction, this net converges to x and to y. It followsthat a topological space is Hausdorff if every convergent net in the space has aunique limit.

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13 Regularity

Let (M,d(x, y)) be a metric space, and let x, y be elements of M , with x 6= y.If r = d(x, y), then B(y, r) is an open set in M that contains y but not x, andhence the topology on M determined by the metric satisfies the first separationcondition. If t = d(x, y)/2, then B(x, t), B(y, t) are disjoint open sets in Mwhich contain x, y, respectively, and it follows that M is a Hausdorff topologicalspace. For every x ∈ M and r ≥ 0, let Vr(x) be the set of w ∈ M such thatd(w, x) > r. If z ∈ Vr(x) and t = d(z, x) − r, then t is a positive real number,and one can use the triangle inequality to check that B(z, t) is contained inVr(x). It follows that Vr(x) is an open set in M . For x ∈ M and r ≥ 0, theclosed ball with center x and radius r is denoted B(x, r) and defined to be theset of w ∈ M with d(w, x) ≤ r. This is the same as the complement of Vr(x)in M , and hence B(x, r) is a closed set in M . One can also show that B(x, r)is a closed set by verifying directly that every limit point of B(x, r) in M is anelement of B(x, r).

A topological space X satisfies the third separation condition, or is regular,if it satisfies the first separation condition, and if for every x ∈ X and closedset E ⊆ X such that x 6∈ E there are open subsets U , V of X such that x ∈ U ,E ⊆ V , and U ∩ V = ∅. The first separation condition is equivalent to thestatement that for every y ∈ X, the set consisting of y and no other elementsis closed in X, and therefore a regular topological space is Hausdorff becauseone can take E = {y} when y 6= x. Suppose that (M,d(x, y)) is a metric space,x ∈ M , and E ⊆ M is a closed set such that x ∈ M \ E. Because M \ E is anopen set, there is an r > 0 such that B(x, r) ⊆ M \E. It follows that B(x, r/2),Vr/2(x) are disjoint open subsets of M that contain x, E, respectively, and thusM is a regular topological space.

13.1 Subspaces

Let X be a topological space, and let Y be a subset of X, equipped with thetopology induced by the one on X. If X satisfies the first separation condition,then it is easy to see that Y does too. Namely, if p and q are distinct elementsof Y , then p and q are also distinct elements of X, and hence there is an openset V in X such that q ∈ V and p 6∈ V . This implies that V ∩ Y is a relativelyopen set in Y that contains q and not p, as desired. Similarly, if X is Hausdorff,then Y is Hausdorff too. Suppose now that X is regular, and let us check thatY is regular as well. Of course, X satisfies the first separation condition in thiscase, and hence Y does too. Let x be an element of Y , and let E be a relativelyclosed subset of Y that does not contain x. Thus there is a closed set E1 in Xsuch that E = E1 ∩Y , and x 6∈ E1, since x ∈ Y \E. This implies that there areopen sets U1, V1 in X such that x ∈ U1, E1 ⊆ V1, and U1 ∩ V1 = ∅, because Xis regular. Therefore U1 ∩ Y and V1 ∩ Y are relatively open subsets of Y suchthat x ∈ U1 ∩ Y , E ⊆ V1 ∩ Y , and (U1 ∩ Y ) ∩ (V1 ∩ Y ) = ∅, as desired.

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14 An example

Let V be the space of real-valued functions on the set Z+ of positive integers.This is basically the same as the space of sequences of real numbers, but forthe sake of notational simplicity it is convenient to express the elements of V asfunctions. If ρ is a positive real-valued function on Z+ and f ∈ V, then put

Nρ(f) = {g ∈ V : |f(l) − g(l)| < ρ(l) for every l ∈ Z+},

the neighborhood of f associated to ρ in V. Let us say that U ⊆ V is an openset if for every f ∈ U there is a positive real-valued function ρ on Z+ such thatNρ(f) ⊆ U . Clearly the empty set and V are open subsets of V in this sense.Suppose that U1, . . . , Un are open subsets of V, and that f ∈ U1 ∩ · · · ∩Un. Foreach i = 1, . . . , n, there is a positive real-valued function ρi on Z+ such thatNρi

(f) ⊆ Ui. Ifρ(l) = min(ρ1(l), . . . , ρn(l))

for l ∈ Z+, then ρ is a positive real-valued function on Z+, and

Nρ(f) = Nρ1(f) ∩ · · · ∩ Nρn

(f) ⊆ U1 ∩ · · · ∩ Un.

Hence U1 ∩ · · · ∩ Un is an open set in V. It is easy to see from the definitionthat a union of open subsets of V is an open set, and therefore that we have atopology on V.

If f ∈ V, ρ is a positive real-valued function on Z+, and g ∈ Nρ(f), then

ρ′(l) = ρ(l) − |f(l) − g(l)| > 0

for every l ∈ Z+ and Nρ′(g) ⊆ Nρ(f), which implies that Nρ(f) is an open setin V. If f1, f2 are distinct elements of V, which means that f1(l) 6= f2(l) forsome l, then one can choose a positive real-valued function ρ on Z+ such that2 ρ(l) < |f1(l) − f2(l)| for some l, and hence Nρ(f1) ∩ Nρ(f2) = ∅. Thus V isHausdorff, and one can also show that V is regular. However, V does not satisfythe local countability condition at any point. For suppose that f ∈ V and thatU1, U2, . . . is a sequence of open sets in V that contain f and form a local basefor the topology of V at f . Let ρ1, ρ2, . . . be a sequence of positive real-valuedfunctions on Z+ such that Nρi

(f) ⊆ Ui for each i. Put

ρ(n) =min(ρ1(n), . . . , ρn(n))

n.

Thus ρ is a positive real-valued function on Z+, ρ(l) < ρi(l) when l ≥ i, andNρ(f) does not contain Nρi

(f) ⊆ Ui for any i, a contradiction.If {fj}∞j=1 is a sequence of elements of V, f ∈ V, {fj(l)}∞j=1 converges to f(l)

as a sequence of real numbers in the standard topology on R for every l ∈ Z+,and there is an n ≥ 1 such that fj(l) = f(l) when j, l ≥ n, then it is easy to seethat {fj}∞j=1 converges to f in the topology defined on V. Conversely, suppose

19

that {fj}∞j=1 is a sequence of elements of V which converges to f ∈ V in thistopology. Let ρ be the nonnegative real-valued function on Z+ defined by

ρ(n) =sup{min(|fj(l) − f(l)|, 1) : j, l ≥ n}

2.

By construction, ρ is monotone decreasing on Z+. If ρ(n) > 0 for every n ∈ Z+,then there is an n0 ≥ 1 such that fj ∈ Nρ(f) when j ≥ n0, by convergence inV. This implies that

|fj(l) − f(l)| < ρ(l) ≤ ρ(n0)

when j, l ≥ n0, contradicting the definition of ρ(n0). Therefore ρ(n) = 0 forsome n, which implies that fj(l) = f(l) when j, l ≥ n. Also, {fj(l)}∞j=1 convergesto f(l) as a sequence of real numbers in the standard topology on R for everyl ∈ Z+, and so we are back in the situation described at the beginning of theparagraph. If V0 is the space of f ∈ V such that f(l) 6= 0 for only finitely manyl, then V0 is closed in V. Indeed, if f(l) 6= 0 for infinitely many l, then we canchoose a positive function ρ on Z+ such that ρ(l) ≤ |f(l)| when f(l) 6= 0, andevery h ∈ Nρ(f) satisfies h(l) 6= 0 when f(l) 6= 0, which is to say that Nρ(f) iscontained in the complement of V0 in V. The functions on Z+ with values inthe rational numbers Q are dense in V, and similarly the Q-valued functions inV0 are dense in V0.

14.1 Topologies and subspaces

Let X be a set, and suppose that X1,X2, . . . is a sequence of subsets of X suchthat Xi ⊆ Xi+1 for each i and X =

⋃∞i=1 Xi. Suppose too that each Xi is

equipped with a topology τi, and that these topologies are compatible in thesense that the topology on Xi induced by τi+1 on Xi+1 is the same as τi. Itfollows that the topology on Xi induced by τl on Xl is the same as τi whenl > i. Let us say that U ⊆ X is an open set in X if U ∩Xi is an open set in Xi

with respect to τi for every i ≥ 1. It is easy to check that this defines a topologyτ on X, using the fact that each τi is a topology on Xi. If Ui ⊆ Xi is an openset with respect to τi, then the compatibility condition implies that there is anopen set Ui+1 ⊆ Xi+1 with respect to ti+1 such that Ui = Ui+1∩Xi. Repeatingthe process, for every l > i there is an open set Ul ⊆ Xl with respect to τl suchthat Ul−1 = Ul ∩ Xl−1. Hence Ul = Un ∩ Xl when n ≥ l ≥ i. If U =

⋃∞l=i Ul,

then Un = U ∩ Xn for every n ≥ i. In particular, U is an open set in X, and itfollows that τi is the same as the topology induced on Xi by τ .

As a variant of this construction, suppose also that X is a vector space overthe real numbers, and that each Xi is a linear subspace of X. Under suitableconditions, one might first say that a convex set U ⊆ X is an open set in X ifU ∩ Xi is an open set in Xi for every i ≥ 1. One can then use the convex opensets to generate a topology on X in the usual way, i.e., where V ⊆ X is an openset if for every x ∈ V there is a convex open set U ⊆ X such that x ∈ U andU ⊆ V .

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15 Countable sets

Let A be a set, and suppose that there is a sequence {xj}∞j=1 of elements of Ain which every element of A appears at least once. It may be that A has onlyfinitely many elements, and that some elements are repeated infinitely often inthe sequence. At any rate, one can systematically go through the sequence andskip terms that are repeated. If A has infinitely many elements, then this willleave a sequence {yl}∞l=1 of elements of A in which every element of A occursexactly once. In this event we say that A is countably infinite. If B ⊆ A hasinfinitely elements, then we can also go through the sequence and just keep theterms in B, to get a sequence {zn}∞n=1 of elements of B in which every elementof B appears exactly once. It follows that B is countably infinite too.

If one adds an element to a countably-infinite set, then the resulting set iscountably-infinite, because one can get an enumeration of the elements in thenew set by putting the new element in the first term and sliding the other termsover one step. Similarly, the union of a countably-infinite set with a set with onlyfinitely many elements is countably infinite. One can enumerate the elements ofa union of two countably-infinite sets by a sequence by using the terms indexedby even integers for one set and the terms indexed by odd integers for the otherset. If A1, A2, . . ., is a sequence of sets with only finitely many elements, thenone can make a sequence listing the elements of the union A =

⋃∞j=1 Aj by first

listing the elements of A1, then the elements of A2, etc., and hence A has onlyfinitely many elements or is countably infinite. The Cartesian product Z+×Z+,consisting of ordered pairs of positive integers, can be expressed as a union of asequence of finite subsets, and is therefore countable infinite. Using this one cancheck that the union of a sequence of countably-infinite sets is countably-infinite.Basically, one can list the elements of the union by a doubly-indexed sequence,and convert that into a sequence of the usual type using an enumeration ofZ+ × Z+.

The set Z+ of positive integers can be enumerated trivially with the sequencexj = j and hence is countably infinite. The set Z of all integers can be enu-merated by the sequence y2l = l, y2l+1 = −l and is also countably infinite. Forevery positive integer n, the set of integer multiples of 1/n is basically the sameas the set of all integers and therefore countably infinite, and consequently theset Q of rational numbers is countably infinite, being the union of these setsover all n ∈ Z+. Let B be the set of functions f on Z+ such that f(l) = 0 or1 for every l ∈ Z+. Suppose that {fj}∞j=1 is a sequence of functions in B. Iff is the function on Z+ such that f(l) = 1 when fl(l) = 0 and f(l) = 0 whenfl(l) = 1, then f ∈ B and f is different from fj for every positive integer j. Itfollows that B is uncountable, which means that it is neither finite nor countablyinfinite. One can use this to show that the real line is uncountable. Specifically,every element of B corresponds to a real number in the interval [0, 1], using theelement of B as a binary expansion. Every element of [0, 1] occurs in this wayat least once and at most twice. If there were a sequence listing all of the realnumbers in [0, 1], then B would be countably infinite as well, a contradiction.

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15.1 The axiom of choice

Let A be a set, and suppose that f is a function on A which assigns to everya ∈ A a set f(a) ⊆ A. Let B be the set of a ∈ A such that a ∈ B when a 6∈ f(a)and a 6∈ B when a ∈ f(a). By construction, B 6= f(a) for every a ∈ A, sincea is an element of exactly one of B, f(a) for every a ∈ A. This generalizes theuncountability of the set of all binary sequences.

The axiom of choice can be formulated as the statement that for any setA there is a mapping from the collection of all nonempty subsets of A into Awhich associates to every B ⊆ A with B 6= ∅ an element b of A such that b ∈ B.As another common formulation, suppose that I is a nonempty set and that forevery i ∈ I, Ai is a nonempty set. That is to say, {Ai}i∈I is a nonempty familyof nonempty sets. The axiom of choice asserts that there is a function f on Iwith values in

⋃i∈I Ai such that f(i) ∈ Ai for every i ∈ I. The generalized

Cartesian product∏

i∈I Ai of the Ai’s is customarily defined as the set of suchfunctions f , and the axiom of choice is then the claim that this generalizedCartesian product is nonempty when I and each Ai, i ∈ I, are nonempty.

15.2 Strong limit points

Let X be a topological space, and suppose that E ⊆ X and p ∈ X. Let ussay that p is a strong limit point of E if for every open set U ⊆ X with p ∈ Uthere are infinitely many elements of E in U . In particular, there is a q ∈ E ∩Usuch that q 6= p under these conditions, which implies that a strong limit pointis automatically a limit point in the usual sense. Conversely, if X satisfies thefirst separation condition, then one can check that an ordinary limit point isautomatically a strong limit point. In any topological space X, the set of stronglimit points of a set E ⊆ X is a closed set.

Similarly, a point p ∈ X is a condensation point of a set E ⊆ X if for everyopen set U ⊆ X with p ∈ U , E ∩U is uncountable. Hence a condensation pointof E is a strong limit point. Again one can check that the set of condensationpoints of E is a closed set in X.

16 Bases

Let X be a topological space. A collection B of open subsets of X is said tobe a base for the topology of X if every open set V ⊆ X can be expressed as aunion of open subsets of X in B. Equivalently, B is a base for the topology of Xif for every open set V ⊆ X and every x ∈ V there is a U ∈ B such that x ∈ Uand U ⊆ V . This is the same as saying that for every p ∈ X, the collectionB(p) of U ∈ B such that p ∈ U is a local base for the topology of X at p. Forexample, the collection of open intervals (a, b) with a, b ∈ R, a < b, is a basefor the standard topology of the real line. If X is any set, then the collection ofone-element subsets {x}, x ∈ X, is a base for the discrete topology on X. If Mis a metric space, then the collection of open balls in M with arbitrary centersand radii form a base for the topology associated to the metric.

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The existence of a base for the topology with only finitely or countably manyelements is a very interesting condition on a topological space. For instance,the collection of open intervals (a, b) with a, b ∈ Q, a < b, forms a countablebase for the standard topology on the real line. More generally, suppose that(M,d(x, y)) is a metric space and that {xj}∞j=1 is a sequence of points in M suchthat the set E consisting of all of the xj ’s, j ∈ Z+, is dense in M . In this eventthe family B(xj , 1/l), j, l ∈ Z+, of open balls forms a base for the topologyof M , which can be rearranged into a ordinary sequence using the countabilityof Z+ × Z+. Conversely, let X be any topological space with a base B for itstopology. For every U ∈ B with U 6= ∅, pick a point x(U) ∈ U , and let E bethe set of x(U), U ∈ B, U 6= ∅. Because B is a base for the topology of X, E isdense in X. If B has only finitely or countably many elements, then E has onlyfinitely or countably many elements.

16.1 Sub-bases

Let X be a topological space, and let B0 be a collection of open subsets ofX. Consider the collection B1 of subsets of X of the form U1 ∩ · · · ∩ Un, whereU1, . . . , Un are finitely many elements of B0. We say that B0 is a sub-base for thetopology of X if B1 is a base for the topology of X. For example, the collection ofopen half-lines (−∞, a) and (b,+∞) for a, b ∈ R is a sub-base for the standardtopology on the real line, and it suffices to take a, b to be rational numbers. IfB0 is any collection of subsets of a set X such that

⋃{U : U ∈ B0} = X, thenit is easy to see that B0 is a sub-base for a topology on X, in which the opensubsets of X are unions of elements of the corresponding collection B1.

Note that the collection of finite subsets of the set Z+ of positive integer iscountably infinite. Indeed, every finite subset of Z+ is a subset of {1, . . . , n} forsome n ∈ Z+, and of course there are only finitely many subsets of {1, . . . , n}for each n ∈ Z+. This shows that the collection of finite subsets of Z+ can beexpressed as the union of a sequence of finite sets, and hence is countable, andit follows that the collection of finite subsets of any countable set is countabletoo. If B0 is a collection of only finitely or countably many subsets of a set Xand B1 is obtained from B0 as in the previous paragraph, then the precedingobservation implies that B1 also has only finitely or countably many elements.If a topological space X has a sub-base B0 for its topology with only finitely orcountably many elements, it therefore has a base B1 for its topology with onlyfinitely or countably many elements as well.

16.2 Totally bounded sets

Let (M,d(x, y)) be a metric space. A set E ⊆ M is totally bounded if for everyǫ > 0 there is a finite set Aǫ ⊆ M such that

E ⊆⋃

x∈Aǫ

B(x, ǫ).(16.1)

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Thus a totally bounded set is bounded, since the union of finitely many boundedsubsets of M is bounded. Suppose that Aǫ ⊆ M satisfies (16.1). For every

x ∈ Aǫ with B(x, ǫ) ∩ E 6= ∅, choose y(x) ∈ E such that d(y(x), x) < ǫ. Let Aǫ

be the set of y(x)’s obtained in this way. By construction, Aǫ ⊆ E and

E ⊆⋃

y∈Aǫ

B(y, 2 ǫ).(16.2)

Clearly the number of elements of Aǫ is less than or equal to the number ofelements of Aǫ when Aǫ has only finitely many elements. Therefore, withoutloss of generality, we may suppose that Aǫ ⊆ E in the definition of a totallybounded set. If M is totally bounded, then M is separable, in the sense thatthere is a dense set A ⊆ M with only finitely or countably many elements, i.e.,A =

⋃∞n=1 A1/n.

17 More examples

The exotic topologies τ+, τ− on the real line can be characterized by the basesB+, B− for their topologies, respectively, where B+ consists of the intervals ofthe form [a, b) and B− consists of the intervals of the form (a, b], with a, b ∈ R

and a < b. It has been mentioned previously that the rational numbers aredense in the real line with respect to each of these exotic topologies, and thateach of these topologies has a countable local base for the topology at everypoint. If B′

+ is any base for the topology of R equipped with the topology τ+,then for every x ∈ R there is an element U(x) of B′

+ which contains x and iscontained in [x,+∞). If x, y ∈ R and x < y, then x ∈ U(x) \ U(y), and henceU(x) 6= U(y). It follows that any base for the topology of R equipped with τ+

has to be uncountable, using the fact that the real line is uncountable. Similarly,any base for the R equipped with τ− has to be uncountable. Consequently, thereare no metrics on R for which the associated topologies are τ+ or τ−, since ametric space that contains a countable dense set has a countable base for itstopology.

Each interval [a, b) in R is an open set with respect to the topology τ+, andit is also closed, because its complement (−∞, a) ∪ [b,+∞) in the real line isan open set in the topology τ+. If x ∈ R, E ⊆ R is closed with respect to thetopology τ+, and x 6∈ E, then there is a t ∈ R such that [x, t) ⊆ R \ E. Hence(R, τ+) is a regular topological space, since [x, t), (−∞, x)∪ [t,+∞) are disjointopen sets in the topology that contain x, E, respectively. For the same reasons,(R, τ−) is a regular topological space.

18 Stronger topologies

Suppose that τ1, τ2 are topologies on a set X such that τ2 contains τ1. Thus theopen subsets of X with respect to τ1 are also open subsets of X with respect

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to τ2. In this event one may say that τ2 is a stronger topology on X thanτ1. If p ∈ X is a limit point of E ⊆ X with respect to τ2, then p is also alimit point of E with respect to τ1. A sequence or net of elements of X thatconverges with respect to τ2 automatically converges with respect to τ1, andwith the same limit. If X satisfies the first separation condition with respect toτ1, then X automatically satisfies the first separation condition with respect toτ2. The analogous statement works for the Hausdorff property too. The sametype of argument does not work for regularity, since a stronger topology hasmore closed sets as well.

18.1 Completely Hausdorff spaces

A topological space X is said to be completely Hausdorff if for every pair ofelements p, q of X with p 6= q there are open subsets U , V of X such thatp ∈ U , q ∈ V , and U ∩ V = ∅. This obviously implies that X is Hausdorff,since U ⊆ U , V ⊆ V . If τ1, τ2 are topologies on X such that X is completelyHausdorff with respect to τ1 and τ1 ⊆ τ2, then it is easy to see that X is alsocompletely Hausdorff with respect to τ2. It is also easy to check that subspacesof completely Hausdorff spaces are completely Hausdorff with respect to theinduced topology.

A topological space X is regular if and only if it satisfies the first separationcondition, and for each point p ∈ X and open set U ⊆ X with p ∈ U there is anopen set W ⊆ X such that p ∈ W and W ⊆ U . In this formulation of regularity,X\U corresponds to the closed subset of X that does not contain p in the earlierformulation. Using this more local description of regularity, it is easy to see thatregular topological spaces are completely Hausdorff. Thus a completely Haus-dorff topological space is also said to satisfy the separation condition numbertwo-and-a-half, between the second and third separation conditions.

19 Normality

A topological space X satisfies the fourth separation condition, or is normal, ifit satisfies the first separation condition, and if for every pair E1, E2 of disjointclosed subsets of X there are disjoint open subsets U1, U2 of X such that E1 ⊆ U1

and E2 ⊆ U2. A normal topological space X is Hausdorff and regular, since thefirst separation condition implies that subsets of X with one element are closed.

If (M,d(x, y)) is a metric space, then M is normal with respect to the topol-ogy associated to the metric. For let a pair E1, E2 of disjoint closed subsetsof M be given. For every x ∈ E1, choose a positive real number r1(x) suchthat B(x, r1(x)) ⊆ M \ E2, which we can do because x ∈ M \ E2 and M \ E2

is an open set. Similarly, for every y ∈ E2, choose an r2(y) > 0 such thatB(y, r2(y)) ⊆ M \ E1. Let U1 be the union of the open balls B(x, r1(x)/2),x ∈ E1, and let U2 be the union of the open balls B(y, r2(y)/2), y ∈ E2. Thus U1,U2 are open subsets of M which contain E1, E2, respectively, by construction.Suppose for the sake of a contradiction that there is a point z ∈ U1 ∩U2. Hence

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there are x ∈ E1 and y ∈ E2 such that d(x, z) < r1(x)/2 and d(y, z) < r2(y)/2.By the triangle inequality,

d(x, y) <r1(x) + r2(y)

2.(19.1)

However, r1(x) and r2(y) are each less than or equal to d(x, y), because of theway that they were chosen and the fact that x ∈ E1, y ∈ E2. This contradicts(19.1), and it follows that U1, U2 are disjoint, as desired.

Now consider the real line equipped with the exotic topology τ+, in whichU ⊆ R is an open set if and only if for every x ∈ U there is a t ∈ R, t > x, suchthat [x, t) ⊆ U . Let E1, E2 be disjoint closed subsets of R with respect to thetopology τ+. For every x ∈ E1, there is a t1(x) > x such that [x, t1(x)) ⊆ R\E2,since R \ E2 is an open set with respect to τ+ which contains x. Similarly, forevery y ∈ E2 there is a t2(y) > y such that [y, t2(y)) ⊆ R \ E1. Let U1 bethe union of the intervals [x, t1(x)), x ∈ E1, and let U2 be the union of theintervals [y, t2(y)), y ∈ E2. By construction, U1, U2 are open subsets of R withrespect to τ+ which contain E1, E2, respectively. Suppose for the sake of acontradiction that there is a point z ∈ U1 ∩ U2. Hence there are x ∈ E1 andy ∈ E2 such that z is an element of both [x, t1(x)) and [y, t2(y)). Equivalently,x, y ≤ z and z < t1(x), t2(y). If x ≤ y, then y ∈ [x, t1(x)), contradicting the waythat t1(x) was chosen, since y ∈ E2. For the same reasons, y ≤ x is impossible,which implies that U1∩U2 = ∅. This shows that (R, τ+) is a normal topologicalspace, and (R, τ−) is a normal topological space as well, where τ− is the exotictopology generated by intervals of the form (a, b].

19.1 Some remarks about subspaces

Let X be a topological space, let Y be a subset of X, and let A1, A2 be disjointrelatively closed subsets of Y . Because A1, A2 are relatively closed in Y , thereare closed subsets E1, E2 of X such that A1 = E1 ∩ Y , A2 = E2 ∩ Y . However,one may not be able to take E1, E2 to be disjoint in X, even though A1, A2

are disjoint in Y . For example, if X is the real line with the standard topology,Y = R \ {0}, A1 = (−∞, 0), and A2 = (0,+∞), then A1, A2 are relativelyclosed subsets of R \ {0}, and E1 = (−∞, 0], E2 = [0,+∞) are the only closedsubsets of R whose intersections with R \ {0} are equal to A1, A2, respectively.

This indicates that the naive argument for showing that a subspace of anormal topological space is also normal with respect to the induced topologydoes not work. Of course, there is no problem in the example mentioned in theprevious paragraph, since A1, A2 are already open sets as well as being relativelyclosed in R\{0}. We also know that R\{0} is normal with respect to the inducedtopology, because the induced topology is the same as the topology induced bythe restriction of the standard metric on R to R \ {0}. Similarly, any subspaceY of a metric space (X, d(x, y)) is normal with respect to the induced topology,because the induced topology on Y is the same as the topology determined bythe restriction of the metric d(x, y) to x, y ∈ Y .

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19.2 Another separation condition

A pair of subsets A, B of a topological space X are said to be separated if

A ∩ B = A ∩ B = ∅.(19.2)

We say that X satisfies the fifth separation condition, or is completely normal, ifX satisfies the first separation condition, and if for every pair A, B of separatedsubsets of X there are disjoint open subsets U , V of X such that A ⊆ Uand B ⊆ V . This implies that X is normal, since disjoint closed sets areautomatically separated. One can check that metric spaces satisfy the fifthseparation condition, using essentially the same argument as before. Similarly,the real line equipped with either of the exotic topologies τ+, τ− also satisfiesthe fifth separation condition, for basically the same reasons as before.

If A,B ⊆ Y ⊆ X, then it is easy to see that A, B are separated as subsets ofX if and only if they are separated as subsets of Y , with respect to the topologyon Y induced by the given topology on X. This is because a point p ∈ Y isadherent to A as a subset of X if and only if p is adherent to A as a subsetof Y , with respect to the induced topology. If X satisfies the fifth separationcondition, then it follows that Y does too, since one can apply the separationcondition on X directly to separated subsets of Y .

20 Continuous mappings

Suppose that X, Y are topological spaces and that f : X → Y is a functiondefined on X with values in Y , which is the same as a mapping from X to Y .We say that f is continuous at a point p ∈ X if for every open set V ⊆ Ywith f(p) ∈ V there is an open set U ⊆ X such that p ∈ U and f(x) ∈ V forevery x ∈ U . If f is continuous at p, (A,≺) is a directed system, and {xa}a∈A

is a net that converges to p in X, then the net {f(x)}a∈A converges to f(p) inY . Conversely, if f is not continuous at p, then there is an open set V ⊆ Ysuch that f(p) ∈ V and for every open set U ⊆ X with p ∈ U there is a pointx(U) ∈ U such that f(x(U)) 6∈ V . The x(U)’s form a net indexed by the opensubsets of X that contain p ordered by reverse inclusion, and this net convergesto p by construction. The f(x(U))’s form a net in Y indexed in the same waythat does not converge to f(p). Therefore f is continuous at p if f maps everyconvergent net in X with limit p to a convergent net in Y with limit f(p).

We say that f is a continuous mapping from X to Y if f is continuous atevery point p ∈ X. If f : X → Y is continuous and W ⊆ Y is an open set, then

f−1(W ) = {x ∈ X : f(x) ∈ W}is an open set in X, because for every x ∈ f−1(W ) there is an open set U inX such that x ∈ U and U ⊆ f−1(W ). It is easy to see from the definitionof continuity that every mapping with this property is continuous. For anymapping f : X → Y , one can check that

X \ f−1(A) = f−1(Y \ A)

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for every A ⊆ Y , and hence that f is continuous if and only if f−1(A) is closedin X for every closed set A ⊆ Y .

20.1 Simple examples

Here are some simple examples of continuous mappings f from a topologicalspace X into a topological space Y .

(1) If X is equipped with the discrete topology, then any mapping f fromX into any topological space Y is continuous.

(2) If Y is equipped with the indiscrete topology, then any mapping f fromany topological space X into Y is continuous.

(3) If X = Y is the real line with the standard topology, then continuity isequivalent to the familiar definition in terms of ǫ’s and δ’s. Similarly, if X andY are metric spaces, then continuity is equivalent to an analogous definition interms of ǫ’s and δ’s.

(4) Constant mappings between arbitrary topological spaces are continuous.(5) If X = Y with the same topology, then the identity mapping f(x) = x

is continuous.(6) If X = Y as sets, and τ1, τ2 are topologies on X such that τ2 ⊆ τ1, then

the identity mapping is continuous as a mapping from (X, τ1) to (X, τ2).(7) If X ⊆ Y and X is equipped with the topology induced from the one on

Y , then the standard embedding f(x) = x is continuous as a mapping from Xinto Y .

(8) If X is equipped with the indiscrete topology and Y satisfies the firstseparation condition, then every continuous mapping from X into Y is constant.

(9) A mapping f : X → Y is said to be locally constant if for every p ∈ Xthere is an open set U ⊆ X such that p ∈ U and f is constant on U . Locallyconstant mappings are always continuous.

(10) If Y is equipped with the discrete topology, then every continuous map-ping from X into Y is locally constant.

20.2 Sequentially continuous mappings

Let X, Y be topological spaces, and let f be a mapping from X to Y . We saythat f is sequentially continuous at a point p ∈ X if for every sequence {xi}∞i=1

of elements of X which converges to p in X, {f(xi)}∞i=1 converges to f(p) as asequence in Y . If f is continuous at p in the usual sense, then f is sequentiallycontinuous at p, because sequences are special cases of nets. Conversely, sequen-tial continuity at p implies continuity at p when the domain satisfies the localcountability condition at p. Similarly, f : X → Y is sequentially continuous if itis sequentially continuous at every point p ∈ X. Hence continuous mappings aresequentially continuous. If X satisfies the local countability condition at eachpoint, then every sequentially continuous mapping f : X → Y is continuous.

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21 The product topology

Suppose that (X1, τ1), (X2, τ2) are topological spaces. Consider the Cartesianproduct X1 × X2, consisting of all ordered pairs (x1, x2) with x1 ∈ X1 andx2 ∈ X2. The product topology τ1 × τ2 on X1 × X2 can be defined by sayingthat W ⊆ X1 ×X2 is an open set if and only if for every (x1, x2) ∈ W there areopen sets U1 ⊆ X1, U2 ⊆ X2 such that x1 ∈ U1, x2 ∈ U2, and U1 × U2 ⊆ W .Equivalently, If U ⊆ X1 and V ⊆ X2 are open sets, then U × V is an open setin X1 × X2, and products of open sets like these form a base for the producttopology on X1×X2. If (A,≺) is a directed system and {(x1,a, x2,a)}a∈A is a netin X1×X2, then this net converges to (x1, x2) ∈ X1×X2 in the product topologyif and only if the nets {x1,a}a∈A, {x2,a}a∈A converge to x1, x2, respectively, inX1, X2.

Let p1 : X1 × X2 → X1, p2 : X1 × X2 be the two coordinate projectionsassociated to the Cartesian product, which is to say that

p1(x1, x2) = x1, p2(x1, x2) = x2

for every (x1, x2) ∈ X1 ×X2. It is easy to see from the definition of the producttopology that p1, p2 are automatically continuous as mappings from X1 × X2

with the product topology to X1, X2 with their given topologies, respectively.In general, if X, Y are topological spaces and f is a mapping from X to Y , thenf : X → Y is an open mapping if

f(U) = {y ∈ Y : y = f(x) for some x ∈ U}is an open set in Y for every open set U ⊆ X. One can check that the coordinatemappings p1, p2 are automatically open mappings too.

If E1 ⊆ X1 and E2 ⊆ X2 are closed sets, then E1 × E2 is a closed set inX1 × X2 with respect to the product topology. This is because

(X1 × X2) \ (E1 × E2) = ((X1 \ E1) × X2) ∪ (X1 × (X2 \ E2)),

which is the union of two open subsets of X1 × X2 with respect to the prod-uct topology. If X1 and X2 both satisfy the first, second, or third separationcondition, then one can check that X1 × X2 satisfies the same condition withrespect to the product topology. If B1, B2 are bases for the topologies of X1,X2, respectively, then

B = {U1 × U2 : U1 ∈ B1, U2 ∈ B2}is a base for the product topology on X1×X2. In particular, if B1, B2 have onlyfinitely or countably many elements, then there are only finitely or countablymany elements in B as well. There are analogous statements for local bases forthe topologies of X1, X2 at individual elements, and for the local countabilitycondition. If the topologies on X1, X2 are determined by metrics d1(x1, y1),d2(x2, y2), then it is easy to see that

d(x, y) = max(d1(x1, y2), d2(x2, y2)), x = (x1, y1), y = (y1, y2) ∈ X1 × X2,

is a metric on X1 × X2 that determines the product topology on X1 × X2.

29

21.1 Countable products

If X1,X2, . . ., is a sequence of sets, then their Cartesian product X =∏∞

i=1 Xi

can be defined as the set of sequences x = {xi}∞i=1 such that xi ∈ Xi forevery i ≥ 1. Suppose that each Xi is equipped with a topology τi too. Thecorresponding product topology τ on X is defined by saying that U ⊆ X is anopen set if for every x ∈ U there are open sets Ui ⊆ Xi, i ≥ 1, such that xi ∈ Ui

for every i ≥ 1,∏∞

i=1 Ui ⊆ U , and Ui = Xi for all but finitely many i. Anothertopology τ ′ on X can be defined in the same way but without the restrictionthat Ui = Xi for all but finitely many i, which means that the open subsets ofX with respect to τ ′ are much smaller in general than for the product topology.We shall call this the strong product topology on X, to distinguish it from theordinary product topology.

With respect to both the ordinary and strong product topologies on X, thecoordinate projections pi : X → Xi defined by pi(x) = xi are continuous openmappings for each i. The product topology on X may be described as theweakest topology on X for which these coordinate mappings are continuous foreach i. If Ei ⊆ Xi is a closed set for each i, then E =

∏∞i=1 Ei is a closed set in X

with respect to the product topology, and hence also with respect to the strongproduct topology, because X \E can be expressed as the union of open sets, andis therefore open. If Xi satisfies the first, second, or third separation conditionfor each i, then one can check that X also satisfies the same condition, withrespect to both the ordinary and strong product topologies. If Xi is equippedwith the discrete topology for each i, then the strong product topology on X isthe discrete topology too.

Let Bi be a base for the topology of Xi for each i, and let B(n) be thecollection of open subsets of X with respect to the product topology of the form∏∞

i=1 Ui, where Ui ∈ Bi when i ≤ n and Ui = Xi when i > n. It is easy to seethat

B =

∞⋃

n=1

B(n)

is a base for the product topology on X. If Bi has only finitely or countablymany elements for each i, then B(n) has only finitely or countably many ele-ments for each n, and hence B has only finitely or countably many elements aswell. As before, there are analogous statements for local bases and the localcountability condition. Suppose that each Xi is equipped with a metric di(·, ·)that determines the topology τi. Without loss of generality, we may supposealso that di(p, q) ≤ 1/i for every p, q ∈ Xi and i ≥ 1, since otherwise we canreplace di(p, q) with min(di(p, q), 1/i). In this case, one can check that

d(x, y) = sup{di(xi, yi) : i ≥ 1}

is a metric on X that determines the product topology on X.

30

21.2 Arbitrary products

Let I be a set, and suppose that for every i ∈ I we have a set Xi. Let Xbe the generalized Cartesian product

∏i∈I Xi of the Xi’s, which is to say the

set of functions f defined in I with values in⋃

i∈I Xi such that f(i) ∈ Xi forevery i ∈ I. Suppose that each Xi is equipped with a topology τi. The producttopology on X is defined by saying that U ⊆ X is an open set if for everyf ∈ U there are open sets Ui ⊆ Xi, i ∈ I, such that f(i) ∈ Ui for every i ∈ I,∏

i∈I Ui ⊆ U , and Ui = X for all but finitely many i ∈ I. The condition thatf(i) ∈ Ui for every i ∈ I is equivalent to f ∈ ∏

i∈I Ui. One can get anothertopology with much smaller open subsets in general by allowing the Ui’s to bearbitrary open subsets of the Xi’s, without the requirement that Ui = Xi forall but finitely many i. As in the previous section, we shall call this the strongproduct topology on X.

The coordinate projections pi : X → Xi are defined by pi(f) = f(i) foreach i ∈ I, and are continuous open mappings with respect to the ordinaryand strong product topologies on X. The product topology on X is again theweakest topology for which these coordinate mappings are all continuous. IfEi ⊆ Xi is a closed set for each i ∈ I, then E =

∏i∈I Ei is a closed set in

X with respect to the product topology, and hence with respect to the strongproduct topology, because X \E can be expressed as the union of open set, andis therefore open. As before, one can check that X satisfies the first, second, orthird separation condition when Xi satisfies the same condition for each i ∈ I.If Xi is equipped with the discrete topology for each i ∈ I, then the strongproduct topology on X is again the discrete topology.

22 Subsets of metric spaces

Let (M,d(x, y)) be a metric space. If A ⊆ M and r > 0, then put

Ar =⋃

x∈A

B(x, r),

which is the same as the set of y ∈ M for which there is an x ∈ A such thatd(x, y) < r. By construction,

A ⊆ Ar

and Ar is an open set in M for every r > 0, being a union of open sets. Observethat

Ar ⊆ At

when r < t. Furthermore, ⋂

r>0

Ar = A.

Specifically, y ∈ M is an element of the closure A of A if and only if y is adherentto A, which means that for every r > 0 there is an x ∈ A such that d(x, y) < r,and which is the same as saying that y ∈ Ar for every r > 0. We can just as well

31

restrict our attention to r > 0 of the form 1/n, where n is a positive integer, toget that

∞⋂

n=1

A1/n = A.

In particular, every closed set in M can be expressed as the intersection of asequence of open sets. Equivalently, every open set in M can be expressed asthe union of a sequence of closed sets. If X is a topological space which satisfiesthe first separation condition, p ∈ X, and X also satisfies the local countabilitycondition at p, then {p} can be expressed as the intersection of a sequence ofopen subsets of X.

22.1 The Baire category theorem

Let (M,d(x, y)) be a complete metric space, and let U1, U2, . . . be a sequenceof dense open subsets of M . The Baire category theorem states that the inter-section

⋂∞i=1 Ui is dense in M under these conditions. To see this, let x0 ∈ M

and r0 > 0 be given, and consider the closed ball B0 = B(x0, r0). BecauseU1 is a dense open set in M , there is an x1 ∈ U1 and an r1 > 0 such thatB1 = B(x1, r1) ⊆ B0 ∩ U1 and r1 ≤ 1. Similarly, there is an x2 ∈ U2

and an r2 > 0 such that B2 = B(x2, r2) ⊆ B1 ∩ U2 and r2 ≤ 1/2. In gen-eral, for every positive integer i there is an xi ∈ Ui and an ri > 0 such thatBi = B(xi, ri) ⊆ Bi−1 ∩ Ui and ri ≤ 1/i. By construction, {xi}∞i=1 is a Cauchysequence in M and therefore converges to some point x ∈ M . Since xi ∈ Bl forevery i ≥ l and Bl is a closed ball, we get that x ∈ Bl for every l ≥ 0. Hencex ∈ B0 and x ∈ Ul for every l ≥ 1, as desired. It follows from the Baire categorytheorem that the set of rational numbers is not the intersection of a sequenceof open subsets of the real line.

22.2 Sequences of open sets

Let B be the set of all binary sequences, and remember that there is a one-to-one correspondence between B and the real line. There is also a one-to-onecorrespondence between B and the set of sequences of integers, or even the setof sequences of elements of B.

If X is a topological space with a countable base for its topology, then thecollection of open subsets of X has cardinal number less than or equal to thatof B. If X is any topological space for which the collection of open sets hascardinality less than or equal to that of B, then the collection of sequences ofopen subsets of X also has cardinality less than or equal to the cardinality of B.Hence the collection of subsets of X which can be expressed as the intersection ofa sequence of open subsets of X has cardinality less than or equal to that of B inthis case. However, the collection of all subsets of X may have larger cardinality.This happens when X is the real line, for instance, since the cardinality of thecollection of all subsets of a set is always strictly larger than the cardinality ofthe set itself.

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23 Open sets in R

Let U be a nonempty open set in the real line, with respect to the standardtopology. For every x ∈ U , there are a, b ∈ R such that a < x < b and(a, b) ⊆ U , since U is an open set. Moreover, there are extended real numbersa′, b′ with a′ < x < b′ which satisfy the following three conditions:

(a′, b′) ⊆ U ;(23.1)

a′ = −∞ or a′ 6∈ U ;(23.2)

b′ = +∞ or b′ 6∈ U.(23.3)

Specifically, a′ is the infimum of the a < x such that (a, x) ⊆ U , and b′ is thesupremum of the b > x such that (x, b) ⊆ U . Any two open intervals satisfying(23.1), (23.2), and (23.3) are necessarily the same interval or are disjoint subsetsof the real line. For any such open interval, one can pick a rational number in theinterval, and different intervals are then associated to different rational numbersbecause of the disjointness of the intervals. It follows that every nonempty openset in the real line can be expressed as the union of a finite or countably-infinitecollection of disjoint open intervals, where the intervals may be bounded orunbounded, depending on the situation.

One can consider similar matters for the real line equipped with the exotictopologies τ+, τ−. If U ⊆ R is an open set with respect to τ+ and x ∈ U , thenthere is a b > x such that [x, b) ⊆ U . By taking the supremum of these b’s,it follows that there is an extended real number b′ > x such that [x, b′) ⊆ Uand either b′ = +∞ or b′ 6∈ U . By taking the infimum of the a ≤ x such that[a, x] ⊆ U , one gets an extended real number a′ ≤ x such that either a′ = −∞and (−∞, x] ⊆ U , or −∞ < a′ < x, (a′, x] ⊆ U , and a′ 6∈ U , or −∞ < a′ ≤ x,[a′, x] ⊆ U , and (c, x] 6⊆ U when c < a′. There are analogous statements for theexotic topology τ− based on intervals of the form (a, b].

23.1 Collections of open sets

If X is a topological space with a countable base for its topology, then thecardinality of the collection of open subsets of X is less than or equal to thecardinality of the real line, because every open set in X can be described by asequence of elements of a countable base for the topology of X. In particular,the cardinality of the collection of open subsets of the real line is less than orequal to the cardinality of R. Of course the cardinality of R is less than orequal to the cardinality of the set of all open subsets of R, since R \ {x} isan open set in R for every x ∈ R. One can also show that the cardinality ofthe collection of all open subsets of R is less than or equal to the cardinality ofR using the fact that every open set in R is the union of a sequence of openintervals. Remember that the cardinality of the real line is the same as thatof the set B of binary sequences. Also, the cardinality of B × B is equal tothe cardinality of B. It follows that the cardinality of the collection of all openintervals in R is less than or equal to the cardinality of B. Hence the cardinality

33

of the collection of all open subsets of R is less than or equal to the cardinalityof the set of sequences of elements of B, and therefore less than or equal to thecardinality of B. A similar argument can be applied to the real line equippedwith the exotic topologies τ+ or τ−.

24 Compactness

Let X be a topological space and let E ⊆ X be given. By an open covering ofE we mean a family {Ui}i∈I of open subsets of X such that

E ⊆⋃

i∈I

Ui.

We say that E is compact in X if for every open covering {Ui}i∈I of E in Xthere is an finite subcovering. This means that there are i1, . . . , in ∈ I such that

E ⊆ Ui1 ∪ · · · ∪ Uin.

If E has only finitely many elements, then E is automatically compact. A famoustheorem states that every closed interval [a, b] in the real line is compact, withrespect to the standard topology on R. We shall not give the usual proof here,but instead we shall see later how this can be obtained as a consequence ofgeneral results about compactness.

If Y ⊆ X and E ⊆ Y , then one can also consider compactness of E relative toY , using the topology induced from the one on X. A nice feature of compactnessis that E is compact relative to X if and only if E is compact relative to Y ,basically because every open covering of E relative to X can be converted to anopen covering relative to Y and vice-versa. Suppose that τ1, τ2 are topologieson X such that the open subsets of X with respect to τ1 are also open withrespect to τ2. If E ⊆ X is compact with respect to τ2, then E is compact withrespect to τ1 as well, because every open covering of E with respect to τ1 is anopen covering with respect to τ2. If X is equipped with the discrete topologyand E ⊆ X is compact, then E has only finitely many elements.

24.1 A class of examples

Let X be a topological space, and let {xl}∞l=1 be a sequence of elements of Xthat converges to x ∈ X. Let K be the set consisting of the xl’s, l ≥ 1, and x.It is easy to see that K is a compact set in X. For suppose that {Ui}i∈I is anarbitrary covering of K by open subsets of X. Since x ∈ K, there is an i0 ∈ Isuch that x ∈ Ui0 . Because {xl}∞l=1 converges to x in X, there is an L ≥ 1 suchthat xl ∈ Ui0 when l ≥ L. If we choose i1, . . . , iL−1 ∈ I such that xl ∈ Uil

forl = 1, . . . , L − 1, then K ⊆ Ui0 ∪ Ui1 ∪ · · · ∪ UiL−1

, as desired. Suppose nowthat X satisfies the local countability condition at every point. In this case, aset E ⊆ X is closed in X if and only if for every sequence {xl}∞l=1 of elementsof E which converges to a point x ∈ X we have that x ∈ E. It follows from

34

the previous remarks that E ⊆ X is a closed set in X if and only if E ∩ K isrelatively closed in K for every compact set K ⊆ X. Consequently, U ⊆ X isan open set in X if and only if U ∩K is relatively open in K for every compactset K ⊆ X.

25 Properties of compact sets

Let X be a topological space, let K1, K2 be compact subsets of X, and let usshow that their union K1 ∪K2 is a compact set in X. Let {Ui}i∈I be any opencovering of K1 ∪ K2 in X. Hence {Ui}i∈I is an open covering of K1 and K2

individually, and it follows that there are finite subsets I1, I2 of I such that

K1 ⊆⋃

i∈I1

Ui, K2 ⊆⋃

i∈I2

Ui.

Therefore I1 ∪ I2 ⊆ I has only finitely many elements and

K1 ∪ K2 ⊆⋃

i∈I1∪I2

Ui.

Thus there is a finite subcovering of K1 ∪ K2 from the open covering {Ui}i∈I ,and it follows that K1 ∪ K2 is compact.

If K ⊆ X is compact and E ⊆ X is closed, then E ∩ K is compact. For if{Ui}i∈I is any open covering of E ∩K in X, then the Ui’s together with X \Eform an open covering of K, which implies that K is contained in the union offinitely many Ui’s and X\E. Clearly E∩K can be covered by the same collectionof finitely many Ui’s, and therefore E ∩ K is compact. If X is Hausdorff andK ⊆ X is compact, then K is a closed set in X. To see this, let p ∈ X \ Kbe given, and for every q ∈ K let U(p, q), V (p, q) be disjoint open subsets of Xsuch that p ∈ U(p, q) and q ∈ V (p, q). Since K is compact, there are finitelymany elements q1, . . . , qn of K such that K is contained in the union V (p) ofV (p, q1), . . . , V (p, qn). If U(p) is the intersection of U(p, q1), . . . , U(p, qn), thenU(p) is an open set in X such that p ∈ U(p) and U(p)∩V (p) = ∅. In particular,U(p) ⊆ X \ K, so that X \ K =

⋃p∈K U(p) is an open set in X, and hence

K is closed. By contrast, if X is equipped with the indiscrete topology, thenevery subset of X is compact. Similarly, if X is an infinite set equipped withthe topology in which W ⊆ X is an open set exactly when W is the empty setor X \ W has only finitely many elements, then X satisfies the first separationcondition, and one can check that every subset of X is again compact.

25.1 Disjoint compact sets

Let X be a Hausdorff topological space. If p ∈ X, K ⊆ X is compact, andp ∈ X \ K, then there are open subsets U(p), V (p) of X such that p ∈ U(p),K ⊆ V (p), and U(p) ∩ V (p) = ∅, as before. Thus X satisfies the analogueof the third separation condition for compact sets instead of closed sets. If

35

H ⊆ X is another compact set and H ∩ K = ∅, then it follows that there arefinitely many elements p1, . . . , pr of H such that H is contained in the unionU of U(p1), . . . , U(pr). If V is the intersection of V (p1), . . . , V (pr), then U , Vare open subsets of X such that H ⊆ U , K ⊆ V , and U ∩ V = ∅. Hence Xsatisfies the analogue of the fourth separation condition for compact sets insteadof closed sets. In particular, a compact Hausdorff topological space X is normal,because closed subsets of X are compact. Similarly, if X is a regular topologicalspace, H ⊆ X is compact, E ⊆ X is closed, and H ∩ E = ∅, then H, E arecontained in disjoint open subsets of X. If X is completely Hausdorff, thendisjoint compact subsets of X are contained in open sets with disjoint closures,by analogous arguments.

25.2 The limit point property

If X is a topological space, then E ⊆ X has the limit point property if everyinfinite set A ⊆ E has a limit point in E. Similarly, E ⊆ X has the strong limit

point property if every infinite set A ⊆ E has a strong limit point in E. Thestrong limit point property implies the limit point property, and the converseholds when X satisfies the first separation condition, since every limit point isthen a strong limit point. These two conditions are quite similar to compactness.For instance, if E has only finitely many elements, then E automatically hasthe strong limit point property. If E ⊆ Y ⊆ X, then E satisfies one of theseconditions relative to Y if and only if it satisfies the same condition relativeto X. If E1, E2 ⊆ X satisfy the limit point property or the strong limit pointproperty, then E1 ∪ E2 has the same feature. For if A ⊆ E1 ∪ E2 has infinitelymany elements, then at least one of A∩E1, A∩E2 has infinitely many elements,and hence a limit point or strong limit point in E1 or E2. The intersection ofa set which satisfies either of these versions of the limit point property with aclosed set in X also satisfies the same version of the limit point property, sincea closed set contains all of its limit points and therefore all limit points of itssubsets. If K ⊆ X is compact, then K satisfies the strong limit point property.For if A ⊆ K is an infinite set without a strong limit point in K, then for everyp ∈ K there is an open set U(p) ⊆ X such that p ∈ U(p) and A has onlyfinitely many elements in U(p). Compactness implies that there are finitelymany elements p1, . . . , pn of K such that K ⊆ U(p1) ∪ · · · ∪ U(pn). Hence Ahas only finitely many elements, a contradiction. As a partial converse to thisstatement, suppose that E ⊆ X has the strong limit point property and thatV1, V2, . . . is a sequence of open subsets of X such that E ⊆ ⋃∞

i=1 Vi. If for everyn ≥ 1 there is an xn ∈ E \ ⋃n

i=1 Vi, then the set A ⊆ E of xn’s has infinitelymany elements. A strong limit point p ∈ E of A would be in Vl for some l,which would imply that xn ∈ Vl for some n > l, a contradiction. It follows thatE ⊆ ⋃n

i=1 Vi for some n.

36

26 Lindelof ’s theorem

Let X be a topological space, let B be a base for the topology of X, and let{Ui}i∈I be any family of open subsets of X. Put

Bi = {V ∈ B : V ⊆ Ui}

for each i ∈ I, so that ⋃

V ∈Bi

V = Ui.

If we put

B′ =⋃

i∈I

Bi,

then we get that ⋃

V ∈B′

V =⋃

i∈I

( ⋃

V ∈Bi

V)

=⋃

i∈I

Ui.

If V ∈ B′, then V ∈ Bi for some i ∈ I, and we let i(V ) be an element of I withthis property. Thus V ⊆ Ui(V ) for each V ∈ B′, and we let I1 be the set of i(V )with V ∈ B′ chosen in this way. It follows that

⋃

V ∈B′

V ⊆⋃

V ∈B′

Ui(V ) =⋃

i∈I1

Ui,

so that ⋃

i∈I

Ui =⋃

V ∈B′

V ⊆⋃

i∈I1

Ui.

This implies that ⋃

i∈I1

Ui =⋃

i∈I

Ui,

because the left side is automatically contained in the right side, since I1 ⊆ I.If B has only finitely or countably many elements, then B′ ⊆ B has only finitelyor countably many elements, and it is easy to see that I1 has only finitely orcountably many elements as well.

26.1 The Lindelof property

A subset E of a topological space X is said to have the Lindelof property if everyopen covering of E in X can be reduced to a finite or countable subcovering.Thus compact sets have the Lindelof property, by definition. The Lindelof prop-erty is often defined initially for topological spaces, and subsets of topologicalspaces are treated as topological spaces themselves, using the induced topology.This is equivalent to defining the Lindelof property directly for subsets, as be-fore, by an argument like the one for compact sets. If a subset E of a topologicalspace X has both the Lindelof and strong limit point properties, then it is easy

37

to see that E is compact, because the strong limit point property implies thatevery countable open covering of E has a finite subcovering.

Let X be a topological space, and let B be a base for the topology of X. IfY is any subset of X, then the collection BY of the sets of the form U ∩ Y withU ∈ B is a base for the topology on Y induced from the topology on X. Inparticular, if B has only finitely or countably many elements, then BY has onlyfinitely or countably many elements as well. If B has only finitely or countablymany elements, then every subset of X has the Lindelof property, by the earlierargument. In this case, it follows that every subset of X with the strong limitpoint property is compact.

26.2 Applications to metric spaces

Let (M,d(x, y)) be a metric space. If E ⊆ M is compact, then it is easy tosee that E is totally bounded, by considering the covering of E by open ballsof radius ǫ centered at elements of E. Similarly, if E has the Lindelof property,then for every ǫ > 0, E can be covered by finitely or countably many openballs of radius ǫ centered at elements of E. In particular, if M has the Lindelofproperty, then M is separable, since one can apply the previous statement toǫ = 1/n for each n ∈ Z+. Conversely, if M is separable, then there is a base forthe topology of M with only finitely or countably many elements, which impliesthat M has the Lindelof property, as before.

If E ⊆ M has the limit point property, then E is totally bounded. Otherwise,there is an ǫ > 0 such that E cannot be covered by finitely many balls with radiusǫ. This implies that there is an infinite sequence x1, x2, x3, . . . of elements of Esuch that d(xj , xl) ≥ ǫ when j < l. The set A ⊆ E of xj ’s has infinitely manyelements and no limit points, which gives a contradiction. If M has the limitpoint property, then M is totally bounded, which implies that M is separable,and hence that M has the Lindelof property. It follows that M is compact,using the hypothesis that M have the limit point property again. If E ⊆ M hasthe limit point property, then E is compact, since one can apply the previousargument to E, considered as a metric space itself, using the restriction of themetric d(x, y) to x, y ∈ E.

27 Continuity and compactness

Suppose that X and Y are topological spaces, and that f is a continuous map-ping from X to Y . If E ⊆ X is compact, then

f(E) = {y ∈ Y : y = f(x) for some x ∈ E}

is compact in Y . To see this, let {Vi}i∈I be any open covering of f(E) in Y .For every i ∈ I, put

Ui = f−1(Vi) = {x ∈ X : f(x) ∈ Vi}.

38

Because f : X → Y is continuous, Ui is an open set in X for every i ∈ I. Thecompactness of E in X implies that there are finitely many indices i1, . . . , in ∈ Isuch that

E ⊆ Ui1 ∪ · · · ∪ Uin.

This implies in turn that

f(E) ⊆ Vi1 ∪ · · · ∪ Vin.

This uses the observation that

f(f−1(A)) ⊆ A

for every A ⊆ Y . Hence we get a finite subcovering of f(E) from the initialopen covering {Vi}i∈I in Y , and it follows that f(E) is compact in Y .

27.1 The extreme value theorem

Suppose that f is a continuous real-valued function on a topological space X,and that E is a compact subset of X. Thus f(E) is a compact set in R, as before.In particular, f(E) is a closed set in R, because the real line is Hausdorff withrespect to the standard topology. We also know that compact subsets of metricspaces are totally bounded, so that f(E) is totally bounded with respect to thestandard metric on R. This implies that f(E) is a bounded set in R, which canalso be seen by covering f(E) by the open intervals (−n, n) for n ∈ Z+, andreducing to a finite subcovering to get that f(E) ⊆ (−n, n) for some n ∈ Z+.If E 6= ∅, so that f(E) 6= ∅, then f(E) has a supremum and infimum in R, bythe completeness property of the real line. Because f(E) is also closed, one cancheck that the supremum and infimum of f(E) are contained in f(E), which isto say that the maximum and minimum of f on E are attained.

27.2 Semicontinuity

A real-valued function f on a topological space X is said to be upper semi-

continuous if f−1((−∞, b)) is an open set in X for every b ∈ R. Similarly,f : X → R is said to be lower semicontinuous if f−1((a,+∞)) is an open set inX for every a ∈ R. Thus continuous real-valued functions are both upper andlower semicontinuous, since (−∞, b) and (a,+∞) are open subsets of R for ev-ery a, b ∈ R. Conversely, if f : X → R is both upper and lower semicontinuous,then it is easy to see that f is continuous, because

f−1((a, b)) = f−1((a,+∞)) ∩ f−1((−∞, b))

is an open set in X for every a, b ∈ R with a < b.Suppose that f : X → R is upper semicontinuous and that E ⊆ X is

compact. Of course,

X =

∞⋃

n=1

f−1((−∞, n)),

39

and one can use the compactness of E to get that E ⊆ f−1((−∞, n)) for somen, so that f(E) has an upper bound in R. If E 6= ∅, then it follows that f(E)has a supremum in R, and we put t = sup f(E). Suppose for the sake of acontradiction that t 6∈ f(E), so that

E ⊆∞⋃

n=1

f−1((−∞, t − 1/n)).

This implies that E ⊆ f−1((−∞, t − 1/n)) for some n, by compactness. Hencet−1/n is an upper bound for f(E), contradicting the fact that t is the supremumof f(E). This shows that f attains its maximum on E when f : X → R

is upper semicontinuous and E ⊆ X is nonempty and compact. Similarly, fattains its minimum on E when f : X → R is lower semicontinuous and E ⊆ Xis nonempty and compact. This can be derived from the previous argumentapplied to −f , because f : X → R is lower semicontinuous if and only if −f isupper semicontinuous.

28 Characterizations of compactness

Let X be a topological space, and suppose that K ⊆ X is compact. If {Ei}i∈I

is a family of closed subsets of X such that

K ∩ Ei1 ∩ Ei2 ∩ · · · ∩ Ein6= ∅(28.1)

for every collection i1, . . . , in of finitely many indices in I, then

K ∩⋂

i∈I

Ei 6= ∅.(28.2)

Otherwise, the complements X \ Ei of the Ei’s in X form an open covering ofK with no finite subcovering. Conversely, if K has this property, then one cancheck that K is compact, by considering the complements of the sets in an opencovering of K.

Suppose that K ⊆ X and that {Ei}i∈I is a family of closed subsets of Xfor which (28.1) holds for any i1, . . . , in ∈ I. Let A be the collection of finitenonempty subsets of I, ordered by inclusion. For every A ∈ A, let xA be anelement of K ∩⋂

i∈A Ei, which exists, by hypothesis. Thus {xA}A∈A is a net ofelements of K indexed by A as a directed system. If there is an x ∈ K such thatfor every open set U ⊆ X with x ∈ U and every A ∈ A there is an A′ ∈ A suchthat A ⊆ A′ and xA′ ∈ U , then x ∈ Ei for every i ∈ I, because x is adherent toEi for each i ∈ I, and hence (28.2) holds.

Conversely, let K be a compact set in X, let (A,≺) be a directed system,and let {xa}a∈A be a net of elements of K. For every a ∈ A, let Ea be theclosure in X of the set of xb with b ∈ A and a ≺ b. If a1, . . . , an ∈ A, thenthere is an a ∈ A such that ai ≺ a for i = 1, . . . , n, and thus Ea ⊆ Eai

for eachi. Hence K ∩ Ea1

∩ · · · ∩ Eancontains K ∩ Ea and is nonempty in particular.

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Because K is compact, there is an x ∈ K such that x ∈ Ea for every a ∈ A, andthus for every open set U ⊆ X with x ∈ U and every a ∈ A there is a b ∈ Asuch that a ≺ b and xb ∈ U , as desired.

28.1 Countable and sequential compactness

A subset K of a topological space X is said to be countably compact if everycountable open covering of K in X can be reduced to a finite subcovering. Inanalogy with the previous characterization of compact sets in terms of familiesof closed subsets of X, countable compactness can be characterized in the sameway in terms of countable families of closed subsets of X. More precisely, K ⊆ Xis countably compact if for every sequence U1, U2, U3, . . . of open subsets of Xsuch that K ⊆ ⋃∞

j=1 Uj , we have that K ⊆ ⋃nj=1 Uj for some positive integer

n. We may as well restrict our attention to the case where Uj ⊆ Uj+1 for each

j, since this can be arranged by replacing Uj with⋃j

l=1 Ul for each j. This isequivalent to saying that if E1, E2, E3, . . . is a sequence of closed subsets of Xsuch that Ej+1 ⊆ Ej and K ∩ Ej 6= ∅ for each j, then K ∩ ⋂∞

j=1 Ej 6= ∅, bytaking complements in X, as before.

If K ⊆ X has the strong limit point property, then we have seen that K iscountably compact. Conversely, suppose that K is countably compact, and letx1, x2, x3, . . . be a sequence of elements of K. Let El be the closure in X of theset of xj with j ≥ l for each positive integer l. By construction, El is a closedsubset of X such that El+1 ⊆ El and K ∩ El 6= ∅ for each l. Thus countablecompactness of K implies that there is an element p of K ∩ ⋂∞

l=1 El, as in theprevious paragraph. If U ⊆ X is an open set with p ∈ U , then for each l ∈ Z+

there is a j ≥ l such that xj ∈ U , because p ∈ El. In particular, if the xj ’sare distinct elements of K, then this implies that p is a strong limit point ofthe set of xj ’s. It follows that K has the strong limit point property when K iscountable compact.

A set K ⊆ X is said to be sequentially compact if for every sequence {xi}∞i=1

of elements of K there is a subsequence {xil}∞l=1 of {xi}∞i=1 and an x ∈ K

such that {xil}∞l=1 converges to x. If K is sequentially compact, then it is easy

to see that K has the strong limit point property. More precisely, if A ⊆ Khas infinitely many elements, then one can apply sequential compactness to asequence of distinct elements of A. This leads to a limit in K of a convergentsubsequence of this sequence, which is a strong limit point of A. Conversely, ifX satisfies the local countability condition at every point, and if K ⊆ X hasthe strong limit point property, then K is sequentially compact. To see this, let{xi}∞i=1 be a sequence of elements of K. If there is an x ∈ K such that x = xi

for infinitely many positive integers i, then {xi}∞i=1 has a constant subsequencethat converges to x trivially. Otherwise, the set A ⊆ K consisting of the xi’shas infinitely many elements, and there is a strong limit point x ∈ K of A.Using the local countability condition for X at x, one can show that there is asubsequence of {xi}∞i=1 that converges to x, as desired.

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28.2 Some variants

Let (M,d(x, y)) be a metric space. If {xi}∞i=1 is a Cauchy sequence in M witha convergent subsequence {xil

}∞l=1, then {xi}∞i=1 converges to the same limit.Hence every Cauchy sequence in a sequentially compact set in M converges.

Let us say that E ⊆ M has property (∗) if for every set A ⊆ E with infinitelymany elements and every ǫ > 0 there is a bounded set B ⊆ A with infinitelymany elements such that diam B < ǫ. Equivalently, E has property (∗) if forevery sequence of elements of E and every ǫ > 0 there is a subsequence of thesequence whose terms are contained in a bounded set with diameter less than ǫ.If E is totally bounded, then E has property (∗), and the converse holds by thesame argument used to show that a set with the limit point property is totallybounded. If every sequence in E has a subsequence which is a Cauchy sequence,then E has property (∗). Conversely, suppose that E ⊆ M has property (∗),and that a sequence of elements of E is given. Using property (∗), one canget a subsequence of the sequence whose terms are contained in a bounded setwith diameter less than 1. By repeating the process, one can get a subsequenceof the subsequence whose terms are contained in a bounded set with diameterless than 1/2. For each positive integer n, one can apply property (∗) to get asubsequence of the initial sequence which is also a subsequence of the previoussubsequence when n ≥ 2 whose terms are contained in a bounded set withdiameter less than 1/n. The sequence whose nth term is the nth term of thenth subsequence obtained in this way is a subsequence of the initial sequenceand is a Cauchy sequence. Therefore, E ⊆ M is totally bounded if and only ifevery sequence of elements of E has a subsequence which is a Cauchy sequence.If M is complete, then it follows that E ⊆ M is sequentially compact if andonly if E is closed in M and totally bounded. One can also show more directlythat a subset of a complete metric space is compact when it is closed and totallybounded, using an argument like the traditional proof of compactness of closedintervals in the real line.

29 Products of compact sets

Let X be a topological space, and suppose that B is a collection of open subsetsof X which forms a base for the topology of X. If K ⊆ X is compact withrespect to B, in the sense that every covering of K by elements of B can bereduced to a finite subcovering, then K is compact in the usual sense. For let{Ui}i∈I be an arbitrary open covering of K in X, and let B′ be the set of V ∈ Bfor which there is an i ∈ I such that V ⊆ Ui. Because B is a base for thetopology of X, each Ui is equal to the union of the V ∈ B such that V ⊆ Ui.Therefore K ⊆ ⋃

i∈I Ui =⋃

V ∈B′ V , which is to say that the V ’s in B′ coverK. By hypothesis, there are V1, . . . , Vn ∈ B′ which cover K. The definition ofB′ implies that there are i1, . . . , in ∈ I such that Vl ⊆ Uil

for l = 1, . . . , n, andhence that Ui1 , . . . , Uin

cover K. This shows that K is compact. There is ananalogous statement for the Lindelof property, and this argument is very similar

42

to the one used to show that every E ⊆ X has the Lindelof property when thereis a countable base for the topology of X.

Now let X1, X2 be topological spaces, and consider the Cartesian productX1 × X2 with the product topology. If K1, K2 are compact subsets of X1, X2,respectively, then K1 ×K2 is compact in X1 ×X2. To show this, it is enough toshow that every covering {Ui × Vi}i∈I of K1 × K2 by products of open sets inX1, X2 can be reduced to a finite subcovering, by the remarks in the precedingparagraph. For every x ∈ K1, let I(x) be the set of i ∈ I such that x ∈ Ui.Since {x} × K2 ⊆ K1 × K2, {Vi}i∈I(x) is an open covering of K2 in X2. Thecompactness of K2 in X2 implies that there is a finite set I0(x) ⊆ I(x) such thatK2 is covered by Vi, i ∈ I0(x). If U(x) is the intersection of Ui, i ∈ I0(x), thenU(x) is an open set in X1 which contains x. The compactness of K1 implies thatthere are x1, . . . , xn ∈ K1 such that K1 is covered by the U(xi)’s, 1 ≤ i ≤ n.Because U(x)×K2 is covered by Ui × Vi, i ∈ I0(x), for every x ∈ K1, it followsthat K1×K2 is covered by the Ui×Vi’s with i ∈ I0(x1)∪· · ·∪I0(xn), as desired.

29.1 Sequences of subsequences

Let X1,X2, . . . be a sequence of topological spaces, and let X =∏∞

i=1 Xi betheir Cartesian product. Suppose that Ei ⊆ Xi, i ≥ 1, are sequentially com-pact, and consider their Cartesian product E =

∏∞i=1 Ei. We would like to

show that E is sequentially compact in X when X is equipped with the producttopology. It follows from the definition of the product topology that a sequenceof elements of X converges to an element of X if and only if the correspondingsequence of coordinates in Xi to converges to the ith coordinate of the limit forevery i ≥ 1. Suppose that a sequence of elements of E is given. Because E1

is sequentially compact, there is a subsequence of this sequence whose coordi-nates in X1 converge to an element of E1. Similarly, because E2 is sequentiallycompact, there is a subsequence of the subsequence for which the coordinatesin X2 converge to an element of E2. Proceeding in this way, we get a sequenceof subsequences of the initial sequence, where each new subsequence is a subse-quence of the preceding subsequence, and where the coordinates in Xn of thenth subsequence converge to an element of En for every n ≥ 1. If we form asequence in which the nth term is equal to the nth term of the nth subsequence,then this sequence is a subsequence of the initial sequence which converges inX to an element of E. Of course, the analogous argument for finite products issimpler, and in particular the last step would not be needed in that case.

29.2 Compactness of closed intervals

Let B be the set of all binary sequences, i.e., the set of sequences x = {xi}∞i=1

such that xi = 0 or 1 for each i. Equivalently, B is the Cartesian product∏∞i=1 Bi, with Bi = {0, 1} for each i. Using the product topology, based on the

discrete topology on the Bi’s, B becomes a Hausdorff topological space which issequentially compact, since the Bi’s are sequentially compact trivially. It followsthat B is a compact space, using the fact that the topology on B is determined

43

by a metric, or by observing that B has a countable base for its topology andtherefore has the Lindelof property.

Consider the real-valued function f on B defined by f(x) =∑∞

i=1 xi 2−i,which associates to each binary sequence a real number in the usual way. Onecan check that f : B → R is a continuous mapping with respect to the producttopology on B. It follows that the closed unit interval [0, 1] is a compact set inthe real line, since f(B) = [0, 1]. Of course, one can also show this more directly.

30 Filters

Let X be a set. A filter on X is a nonempty collection F of nonempty subsetsof X such that

A ∩ B ∈ F for every A,B ∈ F(30.1)

andif A ∈ F and A ⊆ E ⊆ X, then E ∈ F .(30.2)

In particular, this implies that X ∈ F for every filter F on X. Conversely,F = {X} is a filter on any nonempty set X.

Similarly, if X is an infinite set, then

F = {A ⊆ X : X \ A has only finitely many elements}(30.3)

is a filter on X. If X is any set and p ∈ X, then it is easy to see that

Fp = {A ⊆ X : p ∈ A}(30.4)

is a filter on X. More generally, if A is a nonempty subset of any set X, then

FA = {B ⊆ X : A ⊆ B}(30.5)

is a filter on A.Suppose now that X is a topological space. A filter F on X is said to

converge to a point p ∈ X if for every open set U in X such that p ∈ U , we havethat U ∈ F . Thus for instance the filter Fp in (30.4) automatically convergesto p, for any topological space X. Conversely, if a filter F on X converges top ∈ X and X is equipped with the discrete topology, then U = {p} is an openset in X that contains p, and hence {p} ∈ F . This implies that

Fp ⊆ F ,(30.6)

because of (30.2). In the other direction, if A ∈ F , then A∩{p} ∈ F , by (30.1),so that A ∩ {p} 6= ∅, and hence p ∈ A. Thus F ⊆ Fp, which implies that

F = Fp(30.7)

under these conditions.

44

Let X be any topological space again, let p be an element of X, and put

F(p) = {A ⊆ X : p ∈ U ⊆ A for some open set U ⊆ X}.(30.8)

It is easy to see that this is a filter on X, because the intersection of two opensubsets of X is also an open set in X. Note that F(p) converges to p byconstruction, and that a filter F on X converges to p if and only if F(p) ⊆ F .

If X is a Hausdorff topological space, then the limit of a convergent filteron X is unique. To see this, suppose for the sake of a contradiction that F is afilter on X that converges to p, q ∈ X, and that p 6= q. If X is Hausdorff, thenthere are disjoint open subsets U , V of X such that p ∈ U and q ∈ V . Theconvergence of F to p and to q implies that U and V are both elements of F ,and hence that U ∩ V ∈ F . This is a contradiction, since U ∩ V = ∅, while theelements of F are supposed to be nonempty subsets of X.

Conversely, if X is not Hausdorff, then there is a filter on X that convergesto two different elements of X. More precisely, if X is not Hausdorff, then thereare elements p, q of X such that U ∩ V 6= ∅ for every pair of open subsets U , Vof X with p ∈ U , q ∈ V . In this case, one can check that

F = {A ⊆ X : U ∩ V ⊆ A for some open sets U, V ⊆ X(30.9)

such that p ∈ U and q ∈ V }

is a filter on X that converges to both p and q, as desired.If F is a filter on a topological space X that converges to a point p ∈ X,

and if E ∈ F , then p is adherent to E, so that p ∈ E. To see this, let U be anopen set in X such that p ∈ U . Because F converges to p, we have that U ∈ F ,and hence E ∩U ∈ F , by (30.2). This implies that E ∩U 6= ∅, by the definitionof a filter, as desired.

Conversely, if p ∈ X is adherent to a set E ⊆ X, then there is a filter F onX that converges to p and contains E as an element. By hypothesis, E ∩U 6= ∅for every open set U ⊆ X such that p ∈ U , and we can take

F = {A ⊆ X : E ∩ U ⊆ A for some open set U ⊆ X with p ∈ U}.(30.10)

It is not difficult to verify that this has the required properties.

30.1 Nets and filters

Let A be a nonempty set, and let ≺ be a partial ordering on A such that (A,≺)is a directed system. Also let {xa}a∈A be a net of elements of a set X, and put

Ea = {xb : b ∈ A, a ≺ b}(30.11)

for each a ∈ A. Thus

F = {E ⊆ X : Ea ⊆ E for some a ∈ A}(30.12)

is a nonempty collection of nonempty subsets of X, and we would like to checkthat F is a filter on X. Let E,E′ ∈ F be given, and let a, a′ be elements of

45

A such that Ea ⊆ E and Ea′ ⊆ E′. Because A is a directed system, there is ab ∈ A such that a, a′ ≺ b, which implies that

Eb ⊆ Ea ∩ Ea′ ⊆ E ∩ E′.(30.13)

Hence E ∩ E′ ∈ F , so that F satisfies (30.1). By construction, F also satisfies(30.2), and is therefore a filter.

Suppose now that X is a topological space. If {xa}a∈A converges to a pointx ∈ X, then for each open set U in X with x ∈ U there is an a ∈ A such thatEa ⊆ U . This implies that U ∈ F , so that F also converges to x. Conversely,if F converges to x, then U ∈ F for every open set U ⊆ X with x ∈ U . Thisimplies that Ea ⊆ U for some a ∈ A, and hence that {xa}a∈A converges to x.

In the other direction, let F be a filter on a set X, and let us considersome nets associated to F , as follows. Let ≺ be the partial ordering on Fcorresponding to reversed inclusion, so that E1 ≺ E2 when E1, E2 ∈ F andE2 ⊆ E1. If E,E′ ∈ F , then E ∩E′ ∈ F and E,E′ ≺ E ∩E′, which shows thatF is a directed system with respect to ≺. If x(E) ∈ E for each E ∈ F , then{x(E)}E∈F is a net of elements of X indexed by F as a directed system. If Xis a topological space and F converges to a point x in X, then it is easy to seethat each of these nets associated to F also converges to x.

Conversely, suppose that F does not converge to x in X. This means thatthere is an open set U in X such that x ∈ U and U 6∈ F . If E ∈ F , then E 6⊆ U ,since otherwise U would be an element of F , by the definition of a filter. Ifx(E) ∈ E \ U for each E ∈ F , then {x(E)}E∈F is a net associated to F as inthe previous paragraph that does not converge to x.

30.2 Mappings and filters

Let X and Y be sets, let f be a mapping from X into Y , and let F be a filteron X. Under these conditions, it is easy to see that

f∗(F) = {E ⊆ Y : f−1(E) ∈ F}(30.14)

is a filter on Y .Now let X, Y be topological spaces. If F converges to p ∈ X and f is

continuous at p, then we would like to check that f∗(F) converges to f(p). LetV be an open set in Y that contains f(p) as an element. If f is continuous atp, then there is an open set U in X such that p ∈ U and U ⊆ f−1(V ). If Fconverges to p in X, then U ∈ F , and hence V ∈ f∗(F), as desired.

Conversely, let F(p) be the filter on X generated by open sets U ⊆ X suchthat p ∈ U , as in (30.8). Suppose that f∗(F(p)) converges to f(p) in Y , and letus check that f is continuous at p. If V is an open set in Y that contains f(p),then V ∈ f∗(F(p)) by hypothesis, and hence f−1(V ) ∈ F(p). This implies thatthere is an open set U in X such that p ∈ U ⊆ f−1(V ), as desired.

46

31 Refinements

Let X be a set, and let F be a filter on X. A filter F ′ is said to be a refinement

of F whenF ⊆ F ′.(31.1)

Suppose now that X is a topological space, F is a filter on X, p ∈ X, and

p ∈ E(31.2)

for each E ∈ F . This last is equivalent to the condition that

E ∩ U 6= ∅(31.3)

for every open set U ⊆ X such that p ∈ U . It is easy to see that

F ′ = {A ⊆ X : E ∩ U ⊆ A for some E ∈ F and(31.4)

open set U ⊆ X such that p ∈ U}

is a filter on X, because F is a filter and the intersection of two open subsetsof X is also open. If A ∈ F , then A ∈ F ′, since we can take E = A and U = Xin (31.4), and hence F ′ is a refinement of F . Similarly, if U is an open set in Xthat contains p, then U ∈ F ′, since we can take E = X in (31.4), and thus F ′

converges to p in X.Conversely, suppose that F is a filter on X, and that F ′ is a filter on X

which is a refinement of F and which converges to a point p ∈ X. The latterimplies that U ∈ F ′ for every open set U in X that contains p. If E ∈ F , thenE ∈ F ′, because F ′ is a refinement of F , and hence

E ∩ U ∈ F ′.(31.5)

In particular, E ∩ U 6= ∅ for every E ∈ F and open set U ⊆ X that contains p.It follows that p ∈ E for each E ∈ F , as in the previous paragraph.

31.1 Another characterization of compactness

Let X be a topological space, and suppose that K ⊆ X is compact. Also let Fbe a filter on X such that K ∈ F , and consider the collection of closed sets Ein X, with E ∈ F . If E1, . . . , En are finitely many elements of F , then

E1 ∩ · · · ∩ En ∩ K ∈ F ,(31.6)

and in particular this set is not empty. It follows that

E1 ∩ · · · ∩ En ∩ K 6= ∅,(31.7)

and hence that ( ⋂

E∈F

E)∩ K 6= ∅,(31.8)

47

by an earlier characterization of compactness. This implies that there is filterF ′ on X which is a refinement of F and which converges to a point p ∈ K, bythe previous discussion of refinements.

Conversely, suppose that K is a subset of X with the property that any filterF on X with K ∈ F has a refinement that converges to an element of K, andlet us show that K is compact. Let {Ei}i∈I be a family of closed subsets of Xsuch that

Ei1 ∩ · · · ∩ Ein∩ K 6= ∅(31.9)

for every collection i1, . . . , in of finitely many elements of I. If

F = {A ⊆ X : Ei1 ∩ · · · ∩ Ein∩ K ⊆ A for some i1, . . . , in ∈ I},(31.10)

then it is easy to check that F is a filter on X that contains K as an element.By hypothesis, there is a refinement F ′ of F that converges to an element pof K. This implies that p ∈ E for each E ∈ F , by the previous discussion ofrefinements again. Of course, Ei ∈ F for each i ∈ I, by construction, and sop ∈ Ei for each i ∈ I, because Ei is a closed set in X for every i ∈ I. Thus

p ∈( ⋂

i∈I

Ei

)∩ K,(31.11)

and hence ( ⋂

i∈I

Ei

)∩ K 6= ∅.(31.12)

This implies that K is a compact set in X, by an earlier characterization ofcompactness.

32 Ultrafilters

Let X be a set, and let F be a filter on X. We say that F is an ultrafilter on Xif it is maximal in the sense that if F ′ is a filter on X which is a refinement of F ,then F ′ = F . For example, if p ∈ X, then it is easy to see that the collection Fp

of subsets of X that contain p as an element is an ultrafilter on X. Ultrafiltersof this form are said to be principal ultrafilters on X.

Suppose that F is a filter on X, A ⊆ X, and

A ∩ E 6= ∅(32.1)

for every E ∈ F . In this case, it is easy to see that

FA = {B ⊆ X : A ∩ E ⊆ B for some E ∈ F}(32.2)

is a filter on X which is a refinement of F . If F is an ultrafilter on X, then itfollows that FA = F under these conditions. Note that A ∈ FA automatically,so that FA = F implies that A ∈ F . Conversely, if A ∈ F , then FA = F .

If F is any filter on X, A ⊆ X, and there is an E ∈ F such that A ∩E = ∅,then E ⊆ X \ A, and therefore X \ A ∈ F . This shows that either A ∩ E 6= ∅

48

for every E ∈ F , or X \ A ∈ F . If F is an ultrafilter, then we have seen thatA ∈ F in the first case, as in the preceding paragraph.

Now suppose that F ,F ′ are filters on X, and that F ′ is a refinement of F .If A ∈ F ′ and E ∈ F , then A ∩ E ∈ F ′, and thus A ∩ E 6= ∅. If F has theproperty that A ∈ F whenever A ⊆ X satisfies A ∩ E 6= ∅ for every E ∈ F ,then it follows that F = F ′, and hence that F is an ultrafilter on X.

Let F be a filter on X, and suppose that for every A ⊆ X, either A ∈ F orX \A ∈ F . If A ⊆ X satisfies A∩E 6= ∅ for every E ∈ F , then we cannot haveX \A ∈ F , and so we conclude that A ∈ F . Thus F satisfies the criterion in theprevious paragraph, so that F is an ultrafilter on X. To summarize, a filter Fon X is an ultrafilter if and only if for each A ⊆ X, either A ∈ F or X \A ∈ F .

32.1 Connections with set theory

Using the axiom of choice, through the Hausdorff maximality principle or Zorn’slemma, one can show that every filter F on a set X has a refinement U which isan ultrafilter on X. More precisely, the collection of all filters on X is a partially-ordered set with respect to inclusion, which is the ordering corresponding torefinement of filters. Let C be a nonempty chain of filters on X, which is to saya nonempty collection of filters on X such that for every F1,F2 ∈ C, either F2

is a refinement of F1, or the other way around. Let F be the union of the filtersin C, which means that E ⊆ X is an element of F if and only if E ∈ F for someF ∈ C. It is not difficult to check that F is a filter on X under these conditions.The main point is that if E1, E2 ∈ F , then E1 ∩ E2 ∈ F too. To see this, letF1, F2 be elements of C that contain E1, E2 as elements, respectively. BecauseC is a chain of filters on X, either F1 ⊆ F2 or F2 ⊆ F1. This implies that E1

and E2 are both contained in F1, or that they are both contained in F2. HenceE1 ∩ E2 is an element of F1 or F2, by the definition of a filter. In both cases,it follows that E1 ∩ E2 ∈ F , as desired. Of course, F is a refinement of everyelement of C, in addition to being a filter on X. This permits one to apply theHausdorff maximality principle or Zorn’s lemma to the partially-ordered set offilters on X that are refinements of a given filter F on X to show that F has arefinement which is an ultrafilter on X.

If X is a topological space and K ⊆ X, then we have seen that K is compactif and only if every filter F on X with K ∈ F has a refinement which convergesto an element of K. If U is an ultrafilter on X, K ⊆ X is compact, and K ∈ U ,then it follows that U converges to an element of K. Conversely, suppose thatK ⊆ X and every ultrafilter U on X with K ∈ U converges to an element of K.If F is a filter on X with K ∈ F , then there is an ultrafilter U on X which isa refinement of F , as in the previous paragraph. Of course, K ∈ U , so that Uconverges to an element of K by hypothesis, and hence K is compact.

32.2 Tychonoff’s theorem

Let X and Y be sets, and let f be a mapping from X into Y . If F is an ultrafilteron X, then f∗(F) is an ultrafilter on Y . To see this, it suffices to check that for

49

each A ⊆ Y , either A ∈ f∗(F) or Y \ A ∈ f∗(F). This is the same as sayingthat either f−1(A) ∈ F or f−1(Y \A) = X \ f−1(A) ∈ F , which holds becauseF is an ultrafilter on X.

Now let I be a nonempty set, and suppose that Xi is a topological space foreach i ∈ I. Also let Ki be a conpact subset of Xi for each i ∈ I. Under theseconditions, Tychonoff’s theorem states that K =

∏i∈I Ki is a compact subset of

X =∏

i∈I Xi with respect to the product topology. To prove this, it is enoughto show that if U is an ultrafilter on X such that K ∈ U , then U converges onX to an element of K. Let pi : X → Xi be the standard coordinate projectionfor each i ∈ I, which sends elements of X to their ith coordinates in Xi. Theargument in the preceding paragraph implies that (pi)∗(U) is an ultrafilter onXi for each i ∈ I. It is also easy to see that Ki ∈ (pi)∗(U) for each i ∈ I,because K ∈ U . It follows that (pi)∗(U) converges in Xi to an element xi ofKi for each i ∈ I, because Ki is a compact subset of Xi for each i. If x is theelement of K such that pi(x) = xi for every i ∈ I, then one can check that Uconverges to x on X, as desired.

33 Continuous real-valued functions

If X, Y are topological spaces, then the space of continuous mappings from Xto Y is denoted C(X,Y ). In the special case where Y is the real line with thestandard topology, we may use the abbreviation C(X) instead of C(X,R).

Suppose that X is a topological space, p ∈ X, and f1, f2 are real-valuedfunctions on X which are continuous at p, and let us check that the sum f1 +f2

is continuous at p too. Let ǫ > 0 be given, and let U1, U2 be open subsets of Xsuch that p ∈ U1, U2,

|f1(x) − f1(p)| <ǫ

2

for every x ∈ U1, and

|f2(x) − f2(y)| <ǫ

2

for every x ∈ U2. By the triangle inequality for the absolute value function,

|(f1(x) + f2(x)) − (f1(p) + f2(p))| ≤ |f1(x) − f1(p)| + |f2(x) − f2(p)| < ǫ

for every x ∈ U1 ∩U2. It follows that f1 + f2 is continuous at p, since U1 ∩U2 isan open set in X that contains p. Similarly, one can show that the product f1 f2

is continuous at p. It is a bit simpler to start with the special cases where f1 orf2 is a constant function, and where f1(p) = 0 or f2(p) = 0. The general casecan then be derived from these two special cases using the previous assertion forsums. Of course, these arguments are analogous to more classical versions forfunctions on the real line. As a consequence, C(X) is a commutative ring withrespect to pointwise addition and multiplication, since the sum and product oftwo continuous real-valued functions on X is a continuous function on X.

One can also look at continuity of sums and products in terms of compo-sitions, which is discussed next. More precisely, the continuity of real-valued

50

functions f1, f2 on X is equivalent to the continuity of the combined mapping

F (x) = (f1(x), f2(x)),

as a mapping from X into R2 = R×R, where R2 is equipped with the producttopology associated to the standard topology on R. Addition and multiplica-tion on R correspond to mappings from R2 into R, and one can verify thatthese mappings are continuous with respect to the standard topology on R andthe associated product topology on R2, by classical arguments. The sum andproduct of f1, f2 on X are the same as the compositions of F with these twomappings from R2 into R, so that their continuity can be derived from thecontinuity of compositions of continuous mappings.

If f is a real-valued function on X such that f(x) 6= 0 for every x ∈ X, then1/f(x) defines a real-valued function on X as well. If f is continuous at a pointp ∈ X, then one can show that 1/f is continuous at p too, using arguments likethose for functions on the real line, as before. Alternatively, t 7→ 1/t definesa continuous mapping from R \ {0} into itself, with respect to the topologyinduced on R \ {0} by the standard topology on R, as in the classical situation.This permits the continuity of 1/f on X to be derived from the continuityof compositions of continuous mappings, since 1/f can be expressed as thecomposition of f with t 7→ 1/t on R \ {0}.

34 Compositions and inverses

Suppose that X1, X2, and X3 are topological spaces, f1 is a mapping from X1

to X2, and f2 is a mapping from X2 to X3. As usual, the composition f2 ◦ f1

is the mapping from X1 to X3 given by (f2 ◦ f1)(x) = f2(f1(x)). If p ∈ X1,f1 is continuous at p, and f2 is continuous at f1(p), then it is easy to see thatf2 ◦ f1 is continuous at p. Consequently, f1 ∈ C(X1,X2) and f2 ∈ C(X2,X3)imply that f2 ◦ f1 ∈ C(X1,X3), which one can also check by showing that(f2 ◦ f1)

−1(W ) = f−11 (f−1

2 (W )) is an open set in X1 whenever W is an openset in X3.

Let X, Y be sets, and let IX , IY be the identity mappings on X and Y ,which is to say that IX(x) = x for every x ∈ X and IY (y) = y for every y ∈ Y .A mapping f : X → Y is invertible if there is a mapping f−1 : Y → X suchthat f−1 ◦ f = IX and f ◦ f−1 = IY . Because f−1 ◦ f = IX , f is a one-to-

one mapping from X to Y , or equivalently f : X → Y is an injection, whichmeans that f(x) = f(x′) implies x = x′ for every x, x′ ∈ X. Similarly, becausef ◦f−1 = IY , f maps X onto Y , or equivalently f : X → Y is a surjection, whichmeans that for every y ∈ Y there is an x ∈ X such that f(x) = y. Conversely,if f is a one-to-one mapping from X onto Y , which is the same as saying that fis a one-to-one correspondence from X onto Y or that f : X → Y is a bijection,then f is invertible, and f−1 : Y → X can be defined by f−1(y) = x whenf(x) = y for x ∈ X, y ∈ Y . Observe in particular that the inverse f−1 of aninvertible mapping f is unique. If X1, X2, and X3 are sets and f1 : X1 → X2,f2 → X3 are invertible mappings, then one can check that f2 ◦ f1 : X1 → X3 is

51

invertible, and that (f2 ◦f1)−1 = f−1

1 ◦f−12 . If X, Y are topological spaces, then

a homeomorphism from X onto Y is an invertible mapping f : X → Y whichis continuous and whose inverse f−1 : Y → X is continuous. If X1, X2, X3

are topological spaces and f1 : X1 → X2, f2 : X2 → X3 are homeomorphisms,then f2 ◦ f1 : X1 → X3 is a homeomorphism too. Of course the inverse of ahomeomorphism is a homeomorphism as well.

34.1 Compact spaces

Let X, Y be topological spaces, and suppose that f is a one-to-one continuousmapping from X onto Y . If X is compact and Y is Hausdorff, then the inversemapping f−1 is also continuous as a mapping from Y onto X, which is to saythat f is a homeomorphism from X onto Y . To see this, it will be convenientto let the inverse mapping be denoted g. It suffices to show that g−1(U) isan open set in Y for every open set U in X. Equivalently, it is enough toshow that g−1(E) is a closed set in Y for every closed set E in X. Of course,g−1(E) = f(E) in this situation. If X is compact, then every closed set E ⊆ Xis also compact, by previous results. This implies that f(E) is compact in Ywhen E ⊆ X is closed, because f is continuous. If Y is Hausdorff, then compactsubsets of Y are closed, and hence g−1(E) = f(E) is a closed set in Y for everyclosed set E ⊆ X, as desired. It is easy to give examples where f−1 : Y → X isnot continuous when X is not required to be compact or Y is not required tobe Hausdorff.

35 Local compactness

A topological space X is said to be locally compact if for every p ∈ X thereis an open set U ⊆ X such that p ∈ U and U is contained in a compact setin X. Suppose that X is locally compact and that K ⊆ X is compact. Forevery p ∈ K, there is an open set U(p) ⊆ X such that p ∈ U(p) and U(p)is contained in a compact set. Because K is compact, there are finitely manyelements p1, . . . , pn of K such that K is contained in U =

⋃ni=1 U(pi). Of course

U is an open set in X, and U is contained in a compact set since the union offinitely many compact sets is a compact set.

If X is Hausdorff, then X is locally compact if and only if for every p ∈ Xthere is an open set U ⊆ X such that p ∈ U and U is compact. Suppose that Xis a locally compact Hausdorff topological space, p ∈ X, and U ⊆ X is an openset such that p ∈ U and U is compact. Observe that ∂U = U \ U is a compactset in X, since it is the intersection of the compact set U with the closed setX \ U . Because p ∈ U ⊆ X \ ∂U and X is Hausdorff, there are disjoint opensubsets V , W of X such that p ∈ V and ∂U ⊆ W . Hence U1 = U ∩V is an openset in X, p ∈ U , and U1 ⊆ U , which implies that U1 is compact. We also havethat U1 ⊆ X \W and hence U1 ⊆ X \W , since W is an open set. It follows thatU1 ⊆ U , since U1 ⊆ U and ∂U ⊆ W , by construction. If U0 is any open set inX such that p ∈ U0, there is an open set U ⊆ X such that p ∈ U , U ⊆ U0, and

52

U is compact, i.e., the intersection of U0 with any open set that contains p andhas compact closure. By the previous argument, there is an open set U1 ⊆ Xsuch that p ∈ U1, U1 ⊆ U ⊆ U0, and U1 is compact. In particular, a locallycompact Hausdorff topological space is regular, since one can take U0 = X \ Ewhen E ⊆ X is closed and p 6∈ E, and then U1, X \U1 are disjoint open subsetsof X that contain p, E, respectively.

36 Localized separation conditions

Let X be a topological space. The Hausdorff property is equivalent to askingthat for every p, q ∈ X with p 6= q there is an open set U ⊆ X such that p ∈ Uand q is not an element of the closure of U . Similarly, X is regular if and onlyif X satisfies the first separation condition and for every p ∈ X and open setV ⊆ X with p ∈ V there is an open set W ⊆ X such that p ∈ W and the closureof W is contained in V . Also, normality is the same as saying that X satisfiesthe first separation condition and for every A, V ⊆ X with A closed, V open,and A ⊆ V there is an open set W ⊆ X such that A ⊆ W and W ⊆ V . If Xis regular and K, V are subsets of X with K compact, V open, and K ⊆ V ,then there is an open set W ⊆ X such that K ⊆ W and W ⊆ V , by the usualcovering arguments.

Suppose now that X is a locally compact Hausdorff topological space. Inthis case, X is regular, and every compact set in X is contained in an openset with compact closure. Using the observation at the end of the precedingparagraph, it follows that if K ⊆ X is compact and V ⊆ X is an open set withK ⊆ V , then there is an open set W ⊆ X such that K ⊆ W , W is compact, andW ⊆ V . If K1, K2 are disjoint compact subsets of X, then there are disjointopen sets V1, V2 ⊆ X such that Ki ⊆ Vi for i = 1, 2, as in any Hausdorff space.Hence there are open sets W1,W2 ⊆ X such that Ki ⊆ Wi for i = 1, 2 and W1,W2 are disjoint compact subsets of X.

37 σ-Compactness

Let X be a topological space. A set E ⊆ X is said to be σ-compact if thereis a sequence of compact subsets K1,K2,K3, . . . of X such that E =

⋃∞i=1 Ki.

If K ⊆ X is compact, then K is automatically σ-compact, using Ki = K forevery i ≥ 1. It is easy to see that σ-compact sets have the Lindelof property,and that the union of a sequence of σ-compact sets is σ-compact as well. If X isσ-compact, X =

⋃∞i=1 Ki for some compact sets Ki, and E ⊆ X is closed, then

E is σ-compact, because E ∩ Ki is compact for each i and E =⋃∞

i=1(E ∩ Ki).Similarly, if X is σ-compact and E is the union of a sequence of closed sets,then E is σ-compact. If the topology on X is determined by a metric, X is σ-compact, and U ⊆ X is an open set, then it follows that U is σ-compact, sinceevery open set in a metric space can be expressed as the union of a sequence ofclosed sets.

53

Suppose now that X is a locally compact topological space which is alsoσ-compact, so that X =

⋃∞i=1 Ki for some sequence of compact sets Ki, i ∈ Z+.

Local compactness implies that for each i ≥ 1 there is an open set Ui ⊆ Xand a compact set Ki ⊆ X such that Ki ⊆ Ui ⊆ Ki. Put Vl =

⋃li=1 Ui and

Kl =⋃l

i=1 Ki for every l ≥ 1, so that the Vl’s are open subsets of X, the Kl’s

are compact subsets of X, Vl ⊆ Vl+1, Kl ⊆ Kl+1, and Vl ⊆ Kl for every l ≥ 1,and

X =

∞⋃

l=1

Vl =

∞⋃

l=1

Kl.

The Vl’s form an open covering of X, and therefore every compact set in X iscontained in Vn for some positive integer n, depending on the compact set. Ifone assumes that X is Hausdorff too, then one can take Ki = Ui for each i. Inthis case, one can make some additional adjustments to the construction to geta sequence of open subsets W1,W2,W3, . . . of X such that X =

⋃∞j=1 Wj and

Wj is a compact set contained in Wj+1 for every j ≥ 1. Note that a topologicalspace is σ-compact when it is locally compact and has the Lindelof property.

Of course, the real line R is σ-compact with respect to the standard topology,since closed and bounded intervals are compact. It follows that R has theLindelof property, which one could also get from the fact that the open intervals(a, b) with a < b being rational numbers is a countable base for the topologyof R. If the set of irrational numbers were σ-compact, then it would be theunion of a sequence of closed subsets of R in particular. This would imply thatthe set of rational numbers is the intersection of a sequence of open subsetsof R, contradicting the Baire category theorem. However, the set of irrationalnumbers has the Lindelof property, and in fact every subset of the real line hasthe Lindelof property, since there is a countable base for the topology of R.

38 Topological manifolds

Let n be a positive integer, and let Rn be the space of n-tuples of real numbers,which is the same as the Cartesian product of n copies of the real line. This leadsto a natural topology on Rn, which is the product topology corresponding tothe standard topology on the real line. This topology is also the one determinedby the standard Euclidean metric on Rn. Note that the product of n closedintervals in R determines a compact subset of Rn, which implies that Rn islocally compact and σ-compact. One can also get a countable base for thestandard topology on Rn, using such a base for the standard topology on R, orthe fact that Rn is a separable metric space.

A topological space X is said to be locally Euclidean with dimension nif for every p ∈ X there is an open set U ⊆ X such that p ∈ U and U ishomeomorphic to an open set in Rn, with respect to the standard topology onRn, as in the previous paragraph. In this case, one might also say that X is ann-dimensional topological manifold, although one often asks that X be Hausdorffas a topological space too. Another condition that is frequently included is that

54

there be a base for the topology of X with only finitely or countably manyelements. If X is locally Euclidean, then X is locally compact, because Rn islocally compact. Remember that X has the Lindelof property when there isa base for the topology of X with only finitely or countably many elements,and that a locally compact topological space with the Lindelof property is σ-compact.

Of course, Rn is locally Euclidean with dimension n with respect to thestandard topology, as is any open subset of Rn with respect to the correspondinginduced topology. It is well known that the unit sphere in Rn+1 with respectto the standard Euclidean metric is locally Euclidean with dimension n too.

38.1 Unions of bases

Let X be a topological space, and suppose that {Ui}i∈I is a collection of opensubsets of X such that X =

⋃i∈I Ui. If for every i ∈ I there is a collection Bi of

open subsets of Ui which form a base for the topology of Ui, then B =⋃

i∈I Bi

is a base for the topology of X. For let p ∈ X and an open set V ⊆ X withp ∈ V be given. By hypothesis, p ∈ Ui for some i ∈ I, and hence there is anopen set W ∈ Bi such that p ∈ W and W ⊆ Ui ∩ V , since Bi is a base for thetopology of Ui. In particular, W ∈ B, p ∈ W , and W ⊆ V , as desired.

Suppose now that X is locally Euclidean with dimension n. If an open setU ⊆ X is homeomorphic to an open set in Rn, then there is a collection BU ofcountably many open subsets of U that forms a base for the induced topologyon U . This follows from the fact that there is a countable base for the topologyof Rn, as mentioned earlier. If X has the Lindelof property, then X can becovered by a family of finitely or countably many such open subsets U . Thisimplies that there is a countable base for the topology of X, by the remarks inthe previous paragraph.

39 σ-Compactness and normality

Let X be a locally compact Hausdorff topological space which is σ-compact, andlet W1,W2, . . . be a sequence of open subsets of X such that Wj is a compactset contained in Wj+1 for every j ≥ 1 and X =

⋃∞j=1 Wj . Suppose that A, B

are disjoint closed subsets of X, and put Aj = A ∩ Wj , Bj = B ∩ Wj for everypositive integer j. Hence Aj , Bj are disjoint compact subsets of Wj+1 with

Aj ⊆ A ⊆ X \ B and Bj ⊆ B ⊆ X \ A

for each j. Because A1, B1 are disjoint compact subsets of X and X is Hausdorff,there are disjoint open subsets U1, V1 of X such that A1 ⊆ U1, B1 ⊆ V1. Wemay as well suppose that U1 ⊆ W2∩(X \B) and V1 ⊆ W2∩(X \A), by replacingU1, V1 with their intersections with W2 ∩ (X \ B), W2 ∩ (X \ A), respectively.By shrinking U1, V1 further, using the hypothesis that X is a locally compactHausdorff topological space, we may suppose that U1, V1 are disjoint compactsubsets of X which satisfy U1 ⊆ W2 ∩ (X \ B), V1 ⊆ W2 ∩ (X \ B). In general,

55

we would like to choose disjoint open subsets Uj , Vj of X recursively in such away that Aj ⊆ Uj , Bj ⊆ Vj , Uj−1 ⊆ Uj and Vj−1 ⊆ Vj when j ≥ 2,

Uj ⊆ Wj+1 ∩ (X \ B), Vj ⊆ Wj+1 ∩ (X \ A),

and Uj , Vj are disjoint compact sets for every positive integer j. If j ≥ 2 andwe have chosen Uj−1, Vj−1 in this manner, then

Aj = Aj ∪ Uj−1, Bj = Bj ∪ Vj−1

are disjoint compact subsets of X such that

Aj ⊆ Wj+1 ∩ (X \ B), Bj ⊆ Wj+1 ∩ (X \ A),

which permits us to repeat the process. It follows that U =⋃∞

j=1 Uj , V =⋃∞j=1 Vj are disjoint open subsets of X which contain A, B, respectively. There-

fore X is normal.

40 Separating points

Let X be a topological space, and let C(X) be the space of continuous real-valued functions on X, as usual. We say that C(X) separates points in X iffor every p, q ∈ X with p 6= q there is an f ∈ C(X) such that f(p) 6= f(q). IfC(X) separates points in X, then X has to be a Hausdorff topological space.For suppose that p, q ∈ X and f(p) 6= f(q). If f(p) < f(q), then there is anr ∈ R such that f(p) < r < f(q). In this case, U = {x ∈ X : f(x) < r} andV = {x ∈ X : f(x) > r} are disjoint open subsets of X containing p and q,respectively. The case where f(p) > f(q) can be handled similarly, and onecan refine the argument a bit to get that X is completely Hausdorff. If C(X)separates points on X and Y ⊆ X, then it is easy to see that C(Y ) separatespoints on Y with respect to the induced topology. If τ1, τ2 are topologies onX such that continuous real-valued functions on X with respect to τ1 separatepoints and τ1 ⊆ τ2, then continuous real-valued functions on X with respect toτ2 separate points as well.

40.1 Some examples

On the real line, the continuous function f(x) = x is sufficient to separatepoints. Similarly, if j and n are positive integers with j ≤ n, then the jthstandard coordinate function on Rn sends x = (x1, . . . , xn) to xj . These arecontinuous real-valued functions on Rn with respect to the standard topologyon Rn, which separate points in Rn.

Let n be a positive integer again, and let X be a Hausdorff topological spacewhich is locally Euclidean with dimension n. Also let W ⊆ X be an open setwhich is homeomorphic to an open set W ′ ⊆ Rn, and let φ′ be a continuousreal-valued function on Rn for which the set of x ∈ Rn with φ′(x) 6= 0 is

56

contained in a compact set K ′ with K ′ ⊆ W ′. We can restrict φ′ to W ′ anduse the homeomorphism between W and W ′ to get a continuous real-valuedfunction φ on W corresponding to φ′ and a compact set K ⊆ W correspondingto K ′ such that K contains every x ∈ W with φ(x) 6= 0. If we put φ(x) = 0when x ∈ X \W , then one can check that φ is a continuous real-valued functionon X. In particular, one can use functions like these to separate points in X.

If (M,d(x, y)) is a metric space, then one can use the triangle inequality tocheck that fp(x) = d(x, p) is a continuous function on M for every p ∈ M . It iseasy to see that these functions separate points in M .

If X is any set, then the indicator function 1A associated to a subset A ofX is defined on X by putting 1A(x) equal to 1 when x ∈ X, and to 0 whenx ∈ X \ A. If X is a topological space, then it is easy to see that the indicatorfunction 1A associated to A ⊆ X is continuous if and only if A is both open andclosed in X. A topological space X is said to be totally separated if for everypair of points p, q ∈ X with p 6= q, there is a set A ⊆ X which is both open andclosed such that p ∈ A and q ∈ X \A, in which case the corresponding indicatorfunctions separate points in X.

41 Urysohn’s lemma

Let X be a topological space. If f is a continuous real-valued function on Xand a, c are real numbers such that a < c, then the sets of x ∈ X such thatf(x) ≤ a, f(x) ≥ c are disjoint closed subsets of X. If moreover b ∈ R anda < b < c, then the sets of x ∈ X such that f(x) < b, f(x) > b are disjoint opensubsets of X which contain the previous closed sets, respectively.

Conversely, suppose that X is normal, and that E0, E1 are disjoint closedsubsets of X. Let {ri}∞i=1 be an enumeration of the rational numbers r such that0 < r < 1, i.e., ri ∈ Q∩ (0, 1) for every i ≥ 1, and every element of Q∩ (0, 1) isequal to ri for exactly one i. Because X is normal, there is an open set Ur1

⊆ Xsuch that E0 ⊆ Ur1

and Ur1⊆ X \ E1. Similarly, there is an open set Ur2

⊆ Xsuch that E0 ⊆ Ur2

, Ur2⊆ X\E1, and Ur1

⊆ Ur2if r1 < r2, Ur2

⊆ Ur1if r1 > r2.

Continuing in this way, we get an open set Ur ⊆ X for every r ∈ Q∩ (0, 1) suchthat E0 ⊆ Ur, Ur ⊆ X \ E1, and Ur ⊆ Ut when r < t. Consider the real-valuedfunction defined on X by putting f(x) equal to the infimum of r ∈ Q ∩ (0, 1)such that x ∈ Ur when there is such an r, and f(x) = 1 otherwise. Equivalently,f(x) is equal to the supremum of r ∈ Q ∩ (0, 1) such that x ∈ X \ Ur whenthere is such an r, and f(x) = 0 otherwise. By construction, f(x) = 0 whenx ∈ E0, f(x) = 1 when x ∈ E1, and 0 ≤ f(x) ≤ 1 for every x ∈ X. One cancheck that f is a continuous function on X, using the fact that Ur, X \ Ur areopen subsets of X for every r ∈ Q ∩ (0, 1). If X is a locally compact Hausdorfftopological space, K ⊆ X is compact, W ⊆ X is an open set, and K ⊆ W , thenone can use analogous arguments to show that there is a continuous real-valuedfunction h on X such that h(x) = 1 when x ∈ K, h(x) = 0 when x ∈ X \ W ,and 0 ≤ h(x) ≤ 1 for every x ∈ X.

57

41.1 Complete regularity

A topological space X is said to be completely regular if it satisfies the firstseparation condition, and if for every point p ∈ X and closed set E ⊆ X withp 6∈ E there is a continuous function f : X → R such that f(p) 6= 0 and f(x) = 0for every x ∈ E. Of course, one may as well ask that f(p) = 1, by multiplyingf by a nonzero real number, if necessary. One can also ask that 0 ≤ f(x) ≤ 1for every x ∈ X, by taking the maximum of f(x) and 0 and then the minimumof the result and 1. Urysohn’s lemma implies that normal topological spacesare completely regular, and it is easy to see that completely regular spaces areregular. Thus completely regular spaces are said to satisfy separation conditionnumber three-and-a-half, and are also known as Tychonoff spaces.

It is easy to check that subspaces of completely regular spaces are completelyregular as well. In particular, subspaces of normal spaces are completely regular.Locally compact Hausdorff topological spaces are completely regular, by thevariant of Urysohn’s lemma for them mentioned previously. If X is completelyregular, K ⊆ X is compact, E ⊆ X is closed, and E ∩ K = ∅, then there isa continuous function f : X → [0, 1] such that f(x) = 0 for every x ∈ E andf(y) = 1 for every y ∈ K. Indeed, one can use compactness of K and completeregularity to cover K by finitely many open sets that can be separated from Eby suitable continuous functions, and then combine these functions to separateK from E. If X is a topological space where continuous real-valued functionsseparate points, then one can use similar arguments to show that X satisfiesthe analogue of complete regularity in which E is also required to be compact.One can then repeat the process to separate a pair of disjoint compact subsetsof X by a continuous real-valued function on X.

41.2 R \ {0}Consider the real line with the standard topology. The standard topology on R

is determined by the standard metric, and in particular R is a normal topologicalspace. The set R \ {0} of nonzero real numbers is a topological space too, withthe topology induced by the standard topology on the real line. This topologyis also determined by the restriction of the standard metric to R\{0}, and thusR \ {0} is normal. As subsets of the real line, the sets of positive and negativereal numbers are open sets. As subsets of the set of nonzero real numbers, thesesets are both open and closed. The function equal to 0 on the negative realnumbers and equal to 1 on the positive real numbers is a continuous functionon the set of nonzero real numbers. Of course, this function is not the restrictionto the nonzero real numbers of a continuous function on the real line.

42 Countable bases

Let X be a topological space, and suppose that B is a base for the topology of Xwith only finitely or countably many elements. This is a very nice property for atopological space to have, as we have seen. Suppose that X is also regular, and

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let U be an open set in X. If p ∈ U , then there is an open set V (p) in X suchthat p ∈ V (p) and V (p) ⊆ U , by regularity. Thus U =

⋃{V (p) : p ∈ U}, and itfollows that there is a set A ⊆ U with only finitely or countably many elementssuch that U =

⋃{V (p) : p ∈ A}, since B has only finitely or countably manyelements. This implies that U =

⋃{V (p) : p ∈ A}, so that every open set inX can be expressed as the union of only finitely or countably many closed setsunder these conditions. Equivalently, every closed set in X can be expressed asthe intersection of only finitely or countably many open sets. If X is a locallycompact Hausdorff space, then the same argument shows that every open set inX is σ-compact.

A famous theorem of Urysohn states that if X is normal and has a countablebase for its topology, then there is a metric on X that determines the sametopology on X. This was extended to regular spaces with countable bases byTychonoff.

43 One-point compactification

If X is a locally compact Hausdorff topological space which is not compact,then the one-point compactification of X is the topological space X∗ defined asfollows. As a set, X∗ consists of the elements of X together with an additionalpoint p∗, the point at infinity. The open subsets of X are also considered tobe open subsets of X∗, and conversely they are the only open subsets of X∗

contained in X, i.e., which do not contain p∗. If V ⊆ X∗ and p∗ ∈ V , then Vis considered to be an open set in X∗ if and only if V ∩ X is an open set in Xand there is a compact set K ⊆ X such that X \ K ⊆ V ∩ X. It is easy tosee that this defines a topology on X∗. Specifically, the intersection of finitelymany open subsets of X∗ that contain p∗ is an open set in X∗ because the unionof finitely many compact subsets of X is a compact set in X. By construction,X is a dense open set in X∗. One can check that X∗ is Hausdorff using thehypothesis that X is locally compact. If {Vi}i∈I is any open covering of X∗,then there is an i∗ ∈ I such that p∗ ∈ Vi∗ , and thus there is a compact setK ⊆ X such that X \ K ⊆ Vi∗ ∩ X. Since {Vi ∩ X}i∈I is an open covering ofX, and hence of K, there is a collection i1, . . . , in of finitely many elements of Isuch that K ⊆ ⋃n

l=1(Vil∩X). It follows that X∗ ⊆ Vi∗ ∪

⋃nl=1 Vil

, and thereforeX∗ is compact.

43.1 Supports of continuous functions

Let X be a topological space, and let C(X) be the space of continuous real-valued functions on X, as usual. The support of f ∈ C(X) is denoted supp fand defined to be the closure of the set of x ∈ X such that f(x) 6= 0. Iff1, f2 ∈ C(X), then

supp f1 + f2 ⊆ (supp f1) ∪ (supp f2)

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andsupp f1 f2 ⊆ (supp f1) ∩ (supp f2).

Let us specialize to the situation where X is a locally compact Hausdorff topo-logical space. The space of continuous real-valued functions on X with compactsupport is denoted Ccom(X). If f1, f1 ∈ Ccom(X), then f1 + f2 ∈ Ccom(X), bythe previous observation. Similarly, if f1, f2 ∈ C(X) and at least one of f1, f2

has compact support in X, then the product f1 f2 also has compact support.Let us say that f ∈ C(X) vanishes at infinity if for every ǫ > 0 there is a com-pact set K ⊆ X such that |f(x)| < ǫ when x ∈ X \ K. A continuous functionwith compact support automatically vanishes at infinity. The space of contin-uous real-valued functions on X which vanish at infinity is denoted C0(X). Iff1, f2 ∈ C0(X), then it is not too difficult to show that f1 + f2 ∈ C0(X). Iff1 ∈ C0(X), and if f2 ∈ C(X) is bounded, which is to say that there is anA ≥ 0 such that |f2(x)| ≤ A for every x ∈ X, then f1 f2 ∈ C0(X). Of courseCcom(X) = C0(X) = C(X) when X is compact. Suppose that X is not compact,and let X∗ be the one-point compactification of X. If f ∈ C0(X), then f extendsto a continuous function on X∗ which is equal to 0 at the point p∗ at infinity,and conversely the restriction of any such function on X∗ to X lies in C0(X).In the same way, continuous functions on X with compact support correspondto continuous functions on X∗ which are equal to 0 on a neighborhood of p∗.

44 Connectedness

44.1 Connected topological spaces

A topological space X is not connected if there are nonempty disjoint opensubsets U1, U2 of X such that

U1 ∪ U2 = X.

Equivalently, X is not connected if there are nonempty disjoint closed subsetsE1, E2 of X such that

E1 ∪ E1 = X.

This is also the same as having a set E ⊆ X which is both open and closedsuch that E 6= ∅,X. Otherwise, if X does not have any of these equivalentproperties, then X is connected.

For example, if X is equipped with the discrete topology and X has at leasttwo elements, then X is not connected. If X is equipped with the indiscretetopology, then X is connected. If X is an infinite set with the topology in whichU ⊆ X is an open set when U = ∅ or X \U has only finite many elements, thenX is connected, because

U1 ∩ U2 6= ∅for any nonempty open subsets U1, U2 of X.

The real line with the standard topology is connected. For suppose that E1,E2 are nonempty disjoint closed subsets of R whose union is R, and let x1, x2

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be elements of E1, E2, respectively. We may as well suppose that x1 < x2, sinceotherwise we can relabel the indices. One can check that

sup(E1 ∩ [x1, x2]) ∈ E1 ∩ E2,

contradicting the disjointness of E1 and E2.

44.2 Connected sets

Let X and Y be topological spaces, and suppose that f : X → Y is a continuousmapping such that f(X) = Y . If Y is not connected, then there are disjointnonempty open subsets V1, V2 of Y such that V1∪V2 = Y . This implies that X isnot connected, because f−1(V1), f−1(V2) are disjoint nonempty open subsets ofX whose union is equal to X. This is the same as saying that the connectednessof X implies the connectedness of Y under these conditions.

A set E in a topological space X is said to be connected if it is connectedas a topological space itself, using the topology induced from the one on X. Amore direct characterization can be given as follows. A pair of subsets A, B ofX are said to be separated if

A ∩ B = A ∩ B = ∅,

where A and B are the closures of A and B in X. One can check that E ⊆ Xis not connected if and only if E is the union of a pair of nonempty separatedsubsets of X. This uses the fact that the closure of A ⊆ E relative to E is equalto the intersection of E with the closure of A relative to X, because a pointp ∈ E is adherent to A as a subset of E if and only if p is adherent to A as asubset of X. In particular, a pair of sets A,B ⊆ E is separated relative to Eif and only if they are separated relative to X. It is easy to see that a pair ofdisjoint closed subsets of a topological space are separated, as well as a pair ofdisjoint open sets. Conversely, if a topological space is the union of a pair ofseparated sets, then the two sets are both open and closed.

Of course, one can also define connectedness of a subset E of a topologicalspace X directly in terms of separated sets as in the previous paragraph, whichis to say that E ⊆ X is not connected if there are nonempty separated setsA,B ⊆ X such that A ∪ B = E. If E ⊆ Y ⊆ X, then one can check that Eis connected as a subset of Y , with the topology induced from the one on X,if and only if E is connected as a subset of X. This is because A,B ⊆ Y areseparated relative to Y if and only if A, B are separated relative to X, sincetheir closures relative to Y are the same as the intersection of their closuresrelative to X with Y , as before. In particular, E is connected as a subset of Xif and only if E is connected as a subset of itself, with the topology induced bythe one on X.

If X and Y are topological spaces, f : X → Y is continuous, and E isa connected subset of X, then f(E) is connected in Y . This can be derivedfrom the analogous statement with E = X mentioned earlier, or shown directlyusing the characterization of connected subsets of topological spaces in terms of

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separated sets. As an extension of the connectedness of the real line, one canshow that a set E ⊆ R is connected with respect to the standard topology on R

if and only if for every x, y ∈ E with x < y, the interval [x, y] is contained in E.The “only if” part of this statement is easy to check, while the “if” part is verysimilar to the connectedness of R. Now let a, b be real numbers such that a < b,and let f be a continuous real-valued function on [a, b]. One can also think of fas a continuous function on the real line, by putting f(x) = f(a) when x < a,and f(x) = f(b) when x > b. If t is a real number such that f(a) < t < f(b) orf(b) < t < f(a), then the intermediate value theorem implies that there is anx ∈ (a, b) such that f(x) = t. This follows from the connectedness of f([a, b]),using the connectedness of [a, b] and the continuity of f .

44.3 Other properties

Let X be a topological space. If E1, E2 are connected subsets of X and E1∩E2 isnot empty, then E1∪E2 is also connected. For suppose that there are nonemptyseparated sets A, B of X such that

A ∪ B = E1 ∪ E2.

It is easy to see that

A1 = A ∩ E1, B1 = B ∩ E1

andA2 = A ∩ E2, B2 = B ∩ E2

are pairs of separated sets in X. The remaining point is that A1, B1 6= ∅ orA2, B2 6= ∅.

Similarly, if E is a connected set in X, then one can show that the closureE of E in X is connected. Otherwise, there are nonempty separated sets A, Bin X such that A∪B = E. If A0 = A∩E and B0 = B ∩E, then A0 and B0 areseparated sets in X too. Again the remaining point is to check that A0, B0 6= ∅under these conditions.

If E ⊆ X has the property that for every p, q ∈ E there is a connected setE(p, q) ⊆ E such that p, q ∈ E(p, q), then E is connected. As usual, if E is notconnected, then E is the union of a pair of nonempty separated sets A, B in X.Let p, q be elements of A, B, respectively. It is easy to see that

A ∩ E(p, q), B ∩ E(p, q)

are nonempty separated sets whose union is E(p, q), contradicting the connect-edness of E(p, q).

44.4 Pathwise connectedness

A set E in a topological space X is said to be pathwise connected if for everyp, q ∈ E there are real numbers a, b with a ≤ b and a continuous mapping

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f : [a, b] → X such that f(a) = p, f(b) = q, and f([a, b]) ⊆ E. Here [a, b] is theclosed interval in the real line with endpoints a, b, equipped with the topologyinduced by the standard topology on R. In particular, [a, b] is connected, andso

E(p, q) = f([a, b])

is connected. It follows from a previous observation that E is connected whenE is pathwise connected.

There are famous examples of closed sets in the plane that are connectedbut not pathwise connected. These examples also show that the closure of apathwise-connected set may not be pathwise connected. Specifically,

E = {(x, y) ∈ R2 : x > 0, y = sin(1/x)}

is pathwise connected, while

E = E ∪ {(0, y) ∈ R2 : −1 ≤ y ≤ 1}

is not, because there is no continuous path in E that connects any point in Eto any point of the form (0, y), −1 ≤ y ≤ 1. Instead of the sine function, similarexamples could be based on any nonconstant continuous periodic real-valuedfunction on the real line.

Let a, b, and c be real numbers with a ≤ b ≤ c, and let φ, ψ be continuousmappings from the closed intervals [a, b], [b, c] into a topological space X, re-spectively. If φ(b) = ψ(b), then it is easy to see that the function on [a, c] equalto φ on [a, b] and to ψ on [b, c] is also continuous as a mapping from [a, c] intoX. In particular, the union of two pathwise-connected subsets of a topologicalspace is pathwise connected when they have a point in common, just as forconnected sets.

Remember that a set E ⊆ Rn is said to be convex if for every x, y ∈ E, theline segment in Rn connecting x and y is contained in E. Thus convex sets arealways pathwise connected. Of course, a ball in Rn with respect to the standardEuclidean metric is convex. Subsets of Rn which are “rectangular boxes” in thesense that the are Cartesian products of intervals are also convex.

44.5 Connected components

Remember that a binary relation x ∼ y on a set X is said to be an equivalence

relation if it satisfies the following three conditions. First, this relation shouldbe reflexive, in the sense that x ∼ x for every x ∈ X. Second, this relationshould be symmetric, which is to say that x ∼ y if and only if y ∼ x for everyx, y ∈ X. Third, this relation should be transitive, so that x ∼ y and y ∼ zimply that x ∼ z for every x, y, z ∈ X. If x ∼ y is an equivalence relation on Xand x ∈ X, then put

E(x) = {w ∈ X : w ∼ x}.It is easy to see that E(x) = E(y) when x ∼ y, and that

E(x) ∩ E(y) = ∅

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when x 6∼ y. Of course, x ∈ E(x) for each x ∈ X, by reflexivity, so that

⋃

x∈X

E(x) = X.

Subsets of X of the form E(x) for some x ∈ X are known as equivalence classesassociated to the equivalence relation x ∼ y. A collection of pairwise-disjointsubsets of X whose union is equal to X is called a partition of X. Thus thecollection of equivalence classes corresponding to an equivalence relation on Xis a partition of X, and conversely it is easy to see that any partition of Xdetermines an equivalence relation for which the subsets of X in the partitionare the equivalence classes.

Now let X be a topological space, and consider the binary relation definedon X by saying that x ∼c y when there is a connected set E ⊆ X such thatx, y ∈ E. One can check that this is an equivalence relation on X, usingproperties of connected sets discussed earlier. The equivalence classes in Xcorresponding to this equivalence relation are connected subsets of X, known asthe connected components of X. Equivalently, the connected component E(x)of X containing a point x ∈ X is equal to the union of all of the connectedsubsets of X that contain x as an element. Connected components of X aremaximal connected subsets of X, and they are automatically closed subsets ofX, because the closure of a connected set in X is also connected.

Similarly, another binary relation on X mat be defined by saying that x ∼p ywhen x, y ∈ X can be connected by a continuous path in X. This is also anequivalence relation on X, and the equivalence classes associated to this relationare known as pathwise-connected components of X. It is easy to see that theseare pathwise-connected subsets of X, which are maximal with respect to thisproperty. Every pathwise-connected component of X is contained in a connectedcomponent of X, but the connected components of X may be larger than thepathwise-connected components. Each connected component of X is in fact theunion of the pathwise-connected components that it contains.

44.6 Local connectedness

If X is a topological space, then there are three basic types of local connectednessconditions that one might consider. In the first condition, we ask that for eachpoint p ∈ X there be an open set V ⊆ X such that p ∈ V and every elementq of V is contained in a connected subset of X that also contains p. This isequivalent to asking that the connected components of X be open sets. Thesecond condition is stronger, and asks that for each p ∈ X and open set U ⊆ Xwith p ∈ U there be an open set V in X such that p ∈ V ⊆ U , and each elementq of V is contained in a connected subset of U that also contains p. This is thesame as saying that every open subset of X satisfies the first condition, as atopological space itself, with respect to the topology induced from the one on X.The third condition is simpler, and asks that for each p ∈ X and open set U ⊆ Xwith p ∈ U there be a connected open set W in X such that p ∈ W ⊆ U . This

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obviously implies the second condition, and it is easy to see that the conversealso holds, by taking W to be the connected component of U that contains p.Thus one normally defines local connectedness in terms of either the second orthird conditions, which are equivalent.

44.7 Local pathwise connectedness

As before, there are three basic types of local pathwise connectedness conditionsfor a topological space X that one might consider. The first is that for each pointp ∈ X there be an open set V ⊆ X such that p ∈ V and every element q of V canbe connected to p by a continuous path in X. This is equivalent to asking thatthe pathwise-connected components of X be open sets. Note that this impliesthat the pathwise-connected components of X are also closed sets, since thecomplement of any pathwise-connected component in X is equal to the unionof the all of the other pathwise-connected components, which are open sets inthis case. This condition also implies that the pathwise-connected componentsof X are the same as the connected components of X. More precisely, if E is aconnected component of X, then E is automatically the union of the pathwise-connected components of X that it contains. If E were to contain more than onepathwise-connected component, then it would have to be disconnected, since thepathwise-connected components are pairwise disjoint and open in this situation.

The second local pathwise connectedness condition asks that for each p ∈ Xand open set U ⊆ X there be an open set V in X such that p ∈ V ⊆ U , and everyelement of V can be connected to p by a continuous path in U . This is the sameas saying that every open set U in X satisfies the first condition described inthe previous paragraph, as a topological space itself, with the topology inducedby the one on X. The third condition asks that for each p ∈ X and open setU ⊆ X that contains p there be a pathwise-connected open set W in X suchthat p ∈ W ⊆ U . This is equivalent to the second condition, since the thirdcondition obviously implies the second condition, and one can get the converseby taking W to be the pathwise-connected component of U that contains p.

Thus one normally says that X is locally pathwise connected if it satisfieseither of the second or third conditions, described in the previous paragraph. IfX is locally pathwise connected, then every open set U ⊆ X is too, as before.This implies that the pathwise-connected components of U are the same asthe connected components of U , as in the discussion of the first condition. Inparticular, if U is connected, then it follows that U is pathwise connected inthis situation. Note that Rn is locally pathwise connected with respect to thestandard topology for each positive integer n, and hence connected open subsetsof Rn are also pathwise connected.

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45 A little set theory

45.1 Mappings and their properties

Let A, B be sets, and let f be a function on A with values in B, also known as amapping from A into B. We say that f is one-to-one or injective if f(a) = f(a′)implies that a = a′ for every a, a′ ∈ A. We say that f maps A onto B, or thatf is a surjection, if for every b ∈ B there is an a ∈ A such that f(a) = b. If f isa one-to-one mapping of A onto B, then we say that f is a bijection, or that fis a one-to-one correspondence between A and B. In general, f(A) denotes theset of b ∈ B for which there is an a ∈ A such that f(a) = b.

Suppose that A, B, C are sets and f : A → B and g : B → C are mappingsbetween them. In this case, the composition g ◦ f of f and g is the mappingfrom A to C defined by

(g ◦ f)(a) = g(f(a))

for every a ∈ A. It is easy to see that the composition of two injections is aninjection, and that the composition of two surjections is a surjection. Hencethe composition of two bijections is a bijection as well. If f : A → B is abijection, then the inverse mapping f−1 : B → A is defined by f−1(b) = awhen f(a) = b. The inverse mapping is a bijection whose inverse is f . Byconstruction, f−1 ◦ f is the identity mapping on A, and f ◦ f−1 is the identitymapping on B. Conversely, if there is a mapping f−1 : B → A such that f−1 ◦fis the identity mapping on A and f ◦ f−1 is the identity mapping on B, thenf is a bijection, and f−1 is the same as the inverse mapping just described. Iff : A → B and g : B → C are both bijections, then it is easy to see that

(g ◦ f)−1 = f−1 ◦ g−1.

45.2 One-to-one correspondences

Let A, B be sets. If there is a one-to-one correspondence from A onto B, thenwe may express this formally by

#A = #B.

Of course, #A = #A automatically, because the identity mapping on A is aone-to-one correspondence of A to itself. Also, #A = #B is equivalent to#B = #A, because the inverse of a bijection is a bijection. Similarly, #A = #Band #B = #C imply #A = #C for any sets A, B, and C, since the compositionof two bijections is a bijection. Note that A is a finite set with exactly n elementsfor some positive integer n if and only if #A = #{1, . . . , n}. A set A is countablyinfinite if and only if #A = #Z+, where Z+ denotes the set of positive integers.

If B is the set of infinite binary sequences and [0, 1] is the unit interval inthe real line, then #B = #[0, 1]. To see this, let f be the standard mappingfrom B onto [0, 1], which sends a binary sequence b = {bi}∞i=1 to the real numberf(b) =

∑∞i=1 bi 2−i. For each positive integer n, let An be the finite set of b ∈ B

such that bi = 0 for every i ≥ n or bi = 1 for every i ≥ n. If A =⋃∞

n=1 An, then

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A is countably infinite. Let E be the set of rational numbers r ∈ [0, 1] such thatr is an integer multiple of 2−l for some nonnegative integer l. The restrictionof f to B \ A is a one-to-one mapping onto [0, 1] \ E. The restriction of f toA maps onto E but is not one-to-one. However, A and E are both countablyinfinite, and so there is a one-to-one correspondence between them. Combiningthis with the restriction of f to B\A leads to a one-to-one correspondence fromB onto [0, 1], as desired.

45.3 One-to-one mappings

If there is a one-to-one mapping from a set A into a set B, then this may beexpressed formally by #A ≤ #B. Thus #A = #B implies #A ≤ #B and#B ≤ #A. If A, B, C are sets such that #A ≤ #B and #B ≤ #C, then#A ≤ #C, since the composition of two injections is also an injection.

Suppose that A, B are nonempty sets, and that f is a one-to-one mappingfrom A into B. Fix α ∈ A, and let g : B → A be the mapping defined byg(b) = a when a ∈ A and b = f(a) and g(b) = α when b ∈ B \f(A). Clearly g isa surjection. Hence #A ≤ #B implies that there is a mapping from B onto A.Conversely, suppose that there is a mapping g from B onto A. Assuming theaxiom of choice, there is a mapping f : A → B such that g ◦ f is the identitymapping on A. Therefore f is one-to-one.

If A, B are sets such that #A ≤ #B and #B ≤ #A, then #A = #B.Without loss of generality, we may suppose that B ⊆ A, since #B ≤ #A meansthat there is a one-to-one correspondence between B and a subset of A. In thiscase, #A ≤ #B implies that there is an injective mapping f : A → A such thatf(A) ⊆ B. Let A1, A2, . . . be the sequence of subsets of A defined by A1 = f(A)and Al+1 = f(Al) for l ≥ 1. One can use induction to show that Al+1 ⊆ Al foreach l. If C =

⋂∞l=1 Al, then it is easy to check also that f(C) = C. Because

C ⊆ B, it suffices to show that there is a one-to-one correspondence from A \Conto B \ C. Put E1 = A \ f(A1) and El = Al \ Al+1 when l ≥ 1, and observethat f(El) = El+1 for each l. By construction, the El’s are pairwise disjoint,and

⋃∞l=1 El = A \ C. For each x ∈ E1, let O(x) be the orbit of x with respect

to f , which is to say the set consisting of x, f(x), f(f(x)), etc. These orbitsare pairwise disjoint, and their union is A \ C. We would like to show thatthere is a one-to-one mapping from each orbit O(x) onto O(x) ∩ B. All butpossibly the initial element x of O(x) is automatically contained in B, sincef(A) ⊆ B. It follows that either the identity mapping or a simple shift is aone-to-one mapping of O(x) onto O(x) ∩ B, as desired.

45.4 Some basic properties

Let A, B, C be sets, with B ⊆ A and A ∩ C = ∅. If B is countably infiniteand C has only finitely or countably many elements, then B ∪ C is countablyinfinite. One can use this to show that #(A ∪ C) = #A. For example, one can

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apply this to check that

#(0, 1) = #[0, 1) = #(0, 1] = #[0, 1].

Also, f(x) = 1/2 + x/(1 + 2 |x|) is a one-to-one mapping from the real line R

onto the open interval (0, 1), and hence #R = #(0, 1). Remember that the setQ of rational numbers is countable. As another example, one can check that#(R \ Q) = #R.

In general, if A, C are sets, A is infinite, and #C ≤ #A, then there is anargument based on the axiom of choice which implies that #(A ∪ C) = #A.Another argument based on the axiom of choice implies that #A ≤ #B or#B ≤ #A for any pair of sets A, B.

45.5 Bases of topological spaces

Let (X, τ) be a topological space which has a countable base for its topology.Thus there is a sequence U1, U2, . . . of open subsets of X such that every openset in X can be expressed as a union of some collection of the Uj ’s. Of course,the union of any collection of the Uj ’s is an open set in X, since each Uj is open.Let B be the set of all binary sequences, and consider the mapping φ : B → τwhich sends a binary sequence b = {bj}∞j=1 to the open set in X that is theunion of the Uj ’s such that bj = 1. If bj = 0 for each j, then we interpret φ(b)as being the empty set. Because the Uj ’s form a base for the topology of X, φmaps B onto the collection τ of all open subsets of X. Similarly, let ψ : τ → Bbe the mapping that sends an open set V in X to the binary sequence which isequal to 1 for each positive integer j such that Uj ⊆ V and equal to 0 otherwise.Because the Uj ’s form a base for the topology of X, each open set V in X isequal to the union of the Uj ’s such that Uj ⊆ V . Equivalently, φ(ψ(V )) = Vfor each open set V in X, which implies that ψ is a one-to-one mapping of τinto B. Consequently, #τ ≤ #B.

Suppose now that (X, τ) is the real line with the standard topology. Foreach real number t, the set (−∞, t) of r ∈ R with r < t is an open set inR. This defines a one-to-one mapping of R into τ . Hence #R ≤ #τ , andtherefore #τ = #R, since the open intervals in R with rational endpoints forma countable base for the topology.

45.6 Exponentials

Let A, B, be sets. The set of all functions on A with values in B is sometimesdenoted BA. This is the same as the Cartesian product of a family of copiesof B, with one copy of B for each element of A. This may also be denoted 2A

when B is the set consisting of exactly the two elements 0 and 1. The set ofall subsets of A may be identified with 2A, where a set E ⊆ A corresponds tothe function on A which is equal to 1 at elements of E and to 0 at elementsof A \ E. If A and B are finite sets, then BA is a finite set. In this case, thenumber of elements of BA is the number of elements of B to the power whichis the number of elements of A.

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45.7 Properties of exponentials

Let A, B, and C be sets. Remember that the Cartesian product A×B consistsof the ordered pairs (a, b) with a ∈ A and b ∈ B. There is a natural one-to-onecorrespondence between the set CA×B of mappings from A×B into C and theset (CB)A of mappings from A into the set of mappings from B into C. For iff(a, b) is a function on A×B with values in C, then for each a ∈ A we can viewfa(b) = f(a, b) as a function on B with values in C. Equivalently, a 7→ fa is amapping from A into CB . Thus we get a mapping from CA×B into (CB)A. Itis easy to see that this is a one-to-one correspondence, as desired. If A and Bare countably infinite, then A × B is countably infinite, and

#(CB)A = #CA×B = #CZ+ .

By definition, the set B of binary sequences is the same as 2Z+ . Consider

the set ZZ+

+ of all sequences of positive integers. One can check that

#B = #2Z+ ≤ #ZZ+

+ ≤ #BZ+ ,

since 2 ≤ #Z+ ≤ #B. However, the remarks in the previous paragraph imply

that #BZ+ = #(2Z+)Z+ = #2Z+×Z+ = #2Z+ = #B. Therefore #ZZ+

+ = #B.

45.8 Countable dense sets

Suppose that X is a Hausdorff topological space which satisfies the local count-ability condition at each point and contains a countable dense set E. Underthese conditions, for each x ∈ X there is a sequence {xl}∞l=1 of elements of Ethat converges to x. Because X is Hausdorff, the limit of any convergent se-quence of elements of X is unique. Let B be the set of all binary sequences, andlet C be the set of convergent sequences of elements of E. Thus C ⊆ EZ+ , andhence #C ≤ #EZ+ = #B. There is a natural mapping from C into X, whichsends a convergent sequence to its limit in X. By hypothesis, this is a mappingfrom C onto X. Therefore #X ≤ #C ≤ #B.

If X is the real line with the standard topology, then X satisfies the condi-tions described in the previous paragraph, and #X = #B.

45.9 Additional properties of exponentials

Let A, B, and C be sets, with A ∩ B = ∅. There is a natural one-to-onecorrespondence between the set CA∪B of mappings from A ∪ B into C and theCartesian product CA × CB of the set CA of mappings from A into C withthe set CB of mappings from B into C. Basically, a mapping from A ∪ B intoC is the same as a combination of a mapping from A into C and a mappingfrom B into C. If A and B are countable sets, then A ∪ B is countable, and itfollows that #CA = #CB = #CA∪B . Thus #CA × CB = #CZ+ , and hence#CZ+ × CZ+ = #CZ+ .

In particular, the set B of all binary sequences is the same as CZ+ withC = {0, 1}. Therefore #B × B = #B.

69

46 Some more set theory

46.1 Zorn’s lemma

Let (A,≺) be a partially-ordered set. If C ⊆ A, then the restriction of theordering ≺ to C also defines a partial ordering on C. If the restriction of ≺ to Cis a linear ordering on C, then C is said to be a chain in A. The term “chain”may also be used to refer to any linearly ordered set, not just in the contextof subsets of a fixed partially-ordered set. An element b of A is said to be anupper bound for C ⊆ A if x ≺ b for every x ∈ C. An element a of A is said tobe maximal in A if for every c ∈ A with a ≺ c, we have that a = c. If a ∈ A isan upper bound for A, then a is automatically maximal in A. If A is linearlyordered, then every maximal element of A is also an upper bound for A, butthis is not normally the case for partially-ordered sets.

Suppose that A is a partially-ordered set, and that every chain C in A hasan upper bound in A. Under these conditions, Zorn’s lemma asserts that A hasa maximal element.

If A has only finitely many elements, then every chain C in A has onlyfinitely many elements, and hence a maximal element which is an upper boundfor C. In this case, Zorn’s lemma can be understood directly, as follows. Let a1

be an element of A. If a1 is a maximal element of A, then there is nothing to do,and we stop. Otherwise, if a1 is not a maximal element of A, then there is ana2 ∈ A such that a1 ≺ a2 and a1 6= a2. If a2 is maximal in A, then we stop, andotherwise we can repeat the process. If A has only finitely many elements, thenthe process has to stop in finitely many steps, leading to a maximal element ofA.

Suppose now that A is countably infinite, so that there is a sequence {xj}∞j=1

of elements of A in which every element of A occurs exactly once. If x1 isa maximal element of A, then we stop. Otherwise, put j1 = 1, and let j2be the smallest positive integer such that j2 > 1 and x1 ≺ xj2 . If xj2 ismaximal, then we stop, and otherwise we repeat the process. In general, ifj1 = 1 < j2 < · · · < jn have been chosen, and xjn

is maximal in A, then westop. Otherwise, we let jn+1 be the smallest positive integer such that jn+1 > jn

and xjn≺ xjn+1

. If the process stops after finitely many steps, then we get amaximal element of A. Otherwise, we get a subsequence {xjl

}∞l=1 of {xj}∞j=1

such that xjl≺ xjl=1

for each l. Thus the set C consisting of the xjl’s is a chain

in A. Suppose for the sake of a contradiction that b ∈ A is an upper bound forC. Let k be the unique positive integer such that b = xk. Note that xjl

≺ xk

for every l ∈ Z+, because b is an upper bound for C. This implies that k = jp

for some p, by construction. Hence b = xk = xjp≺ xjp+1

, which implies in turnthat b = xjp+1

, because xjp+1≺ b. This is a contradiction, since jp < jp+1, so

that xjp6= xjp+1

.

70

46.2 Hausdorff’s maximality principle

Let (A,≺) be a partially ordered set again. The Hausdorff maximality principle

states that there is a maximal chain C1 in A. More precisely, this means thatC1 is a chain in A, and if C is a chain in A such that C1 ⊆ C, then C1 = C.It is easy to see that the Hausdorff maximality principle implies Zorn’s lemma,because an upper bound for a maximal chain in A is a maximal element of A.

Conversely, one can show that Zorn’s lemma implies the Hausdorff maximal-ity principle, by considering the collection E of chains in A as a partially orderedset with respect to inclusion. Thus a chain L in E is a collection of chains in A,with the property that if C,C ′ ∈ L, then either C ⊆ C ′ or C ′ ⊆ C. In this case,one can check that the union

⋃C∈L C of the elements of L is also a chain in A,

which is an upper bound for L as an element of E . This shows that E satisfiesthe hypothesis of Zorn’s lemma, and the conclusion of Zorn’s lemma applied toE is exactly the same as the existence of a maximal chain in A.

If A is a partially ordered set with only finitely many elements, then it is easyto get a maximal chain in A, by taking a chain in A with a maximal numberof elements. Suppose now that A is countably infinite, and let {xj}∞j=1 be asequence of elements of A in which every element of A occurs exactly once.Put j1 = 1, and let j2 be the smallest positive integer greater than 1 such thatx1 ≺ xj1 or xj1 ≺ x1, if there is one. In general, if 1 = j1 < j2 < · · · < jn havebeen chosen such that {xj1 , . . . , xjn

} is a chain in A, then we let jn+1 be thesmallest positive integer such that jn+1 > jn and {xj1 , . . . , xjn

, xjn+1} is also a

chain in A, if there is one. It is easy to see that this process leads to a maximalchain in A, whether it stops after finitely many steps, or continues indefinitely.

46.3 Choice functions

Let I be a nonempty set, and suppose that for each i ∈ I, Ei is a nonemptyset. Consider the collection P of ordered pairs (A, f), where A is a nonemptysubset of I, and f is a function from A into

⋃i∈A Ei such that f(i) ∈ Ei for

each i ∈ A. If (A, f), (B, g) ∈ P, then put (A, f) ≺ (B, g) when A ⊆ B andf(i) = g(i) for every i ∈ A, so that f is equal to the restriction of g to B. Itis easy to see that this defines a partial ordering on P. We would like to useZorn’s lemma or the Hausdorff maximality principle to show that there is anelement (A, f) of P with A = I.

Suppose that L is a chain in P, and consider

AL =⋃

{A : (A, f) ∈ L}.

If (A, f), (B, g) ∈ L, then (A, f) ≺ (B, g) or (B, g) ≺ (A, f), because L is achain in P. Hence f(i) = g(i) for every i ∈ A ∩ B. If i ∈ AL, then i ∈ A forsome (A, f) ∈ L, and we put fL(i) = f(i). This does not depend on the choiceof (A, f) ∈ L such that i ∈ A, by the previous observation. Equivalently, thegraph of fL as a subset of the Cartesian product of AL and

⋃i∈AL

Ei is theunion of the graphs of the functions f with (A, f) ∈ L. At any rate, this defines

71

a function fL on AL with values in⋃

i∈ALEi such that f〉 ∈ Ei for each i ∈ AL.

Thus (AL, fL) ∈ P, and (A, f) ≺ (AL, fL) for every (A, f) ∈ L by construction.This shows that every chain L in P has a upper bound, so that Zorn’s

lemma implies that there is a maximal element (A0, f0) in P. Alternatively, theHausdorff maximality principle implies that there is a maximal chain L0 in P,and if (A0, f0) = (AL0

, fL0) is as in the previous paragraph, then it is easy to

see that (A0, f0) is maximal in P. If A0 6= I, then let i1 be an element of I \A0,and put A1 = A0 ∪ {i1}. Also put f1(i) = f0(i) for every i ∈ A0, and let f1(i1)be an element of Ei1 . This defines a function f1 on A1 with values in

⋃i∈A1

Ei

such that f1(i) ∈ Ei for each i ∈ A1, so that (A1, f1) ∈ P. By construction,(A0, f0) ≺ (A1, f1) and (A0, f0) 6= (A1, f1), contradicting maximality of (A0, f0).Thus A0 = I, as desired.

This shows that the axiom of choice follows from Zorn’s lemma or the Haus-dorff maximality principle. Conversely, there are well-known arguments bywhich Zorn’s lemma or the Hausdorff maximality principle may be derived fromthe axiom of choice.

46.4 Comparing sets

Let A, B be sets, and let P be the collection of ordered pairs (A1, f1) such thatA1 ⊆ A and f1 is a one-to-one mapping from A1 into B. If (A1, f1), (A2, f2) ∈ P,then put (A1, f1) ≺ (A2, f2) when A1 ⊆ A2 and f1 is equal to the restriction off2 to A1. It is easy to see that this defines a partial ordering on P.

Let L be a chain in P, and put

AL =⋃

{A1 : (A1, f1) ∈ L}.

If (A1, f1), (A2, f2) ∈ L, then (A1, f1) ≺ (A2, f2) or (A2, f2) ≺ (A1, f1), andhence f1(x) = f2(x) for every x ∈ A1 ∩ A2. Thus we can define fL : AL → Bby fL(x) = f1(x) when (A1, f1) ∈ L and x ∈ A1, as in the previous situation.If x, y ∈ AL, then there are (A1, f1), (A2, f2) ∈ L such that x ∈ A1 and y ∈ A2.Again (A1, f1) ≺ (A2, f2) or (A2, f2) ≺ (A1, f1), because L is a chain in P. Inthe first case, x, y ∈ A2 and fL(x) = f2(x), fL(y) = f2(y). If fL(x) = fL(y),then f2(x) = f2(y), and hence x = y, since f2 : A2 → B is one-to-one. In thesecond case, x, y ∈ A1 and fL(x) = f1(x), fL(y) = f1(y), and the injectivityof f1 : A1 → B implies that x = y when fL(x) = fL(y). This shows thatfL : AL → B is injective, so that (AL, fL) ∈ P. Of course, we also have that(A1, f1) ≺ (AL, fL) for every (A1, f1) ∈ L, by construction.

Thus every chain L in P has an upper bound, so that Zorn’s lemma impliesthat there is a maximal element (A0, f0) in P. Alternatively, the Hausforffmaximality principle implies that there is a maximal chain L0 in P, and if(A0, f0) = (AL0

, fL0) is as in the previous pargraph, then it is easy to see that

(A0, f0) is maximal in P. If A0 6= A and f0(A0) 6= B, then let a be an elementof A \ A0, and let b be an element of B \ f0(A0). Put A′

0 = A0 ∪ {a}, anddefine f ′

0 : A′0 → B by f ′

0(x) = f0(x) when x ∈ A0 and f ′0(a) = b. Thus f ′

0 is aone-to-one mapping from A′

0 into B, and hence (A′0, f

′0) ∈ P. This contradicts

72

the maximality of (A0, f0), since (A0, f0) ≺ (A′0, f

′0) and (A0, f0) 6= (A′

0, f′0) by

construction.It follows that either A0 = A or f0(A0) = B. In the first case, f0 is a one-to-

one mapping from A into B, so that #A ≤ #B. In the second case, the inverseof f0 defines a one-to-one mapping from B into A, and hence #B ≤ #A.

46.5 Well-ordered sets

A linearly-ordered set (A,≺) is said to be well-ordered if for every nonemptysubset B of A there is a b ∈ B such that b ≺ c for each c ∈ B. Of course, everynonempty finite subset of a linearly ordered set has a minimal element, so thatlinearly-ordered sets with only finitely many elements are well-ordered. The setZ+ of positive integers with the standard ordering is also well-ordered.

The well-ordering principle states that any set admits a well-ordering. Thereare well-known argumnents by which this may be derived from the axiom ofchoice. Conversely, if A is equipped with a well-ordering, then one can get amapping from the set of nonempty subsets of A into A that associates to eachnon-empty set B ⊆ A an element of B by taking the minimal element of B withrespect to the ordering. Thus the well-ordering principle implies the axiom ofchoice.

Suppose that I is a partially-ordered set, and that for each i ∈ I, Ai is apartially-ordered set. Suppose also that Ai ∩ Aj = ∅ when i 6= j, which caneasily be arranged by replacing Ai with Ai × {i} if necessary. One can define apartial ordering on A =

⋃i∈I Ai in the following way. If x, y ∈ A are elements

of Ai for the same i ∈ I, then one can use the ordering on Ai to order x and y inA. Otherwise, if x ∈ Ai and y ∈ Aj for some i, j ∈ I with i 6= j, then x precedesy in the ordering on A if and only if i precedes j in the ordering on I. It iseasy to see that this defines a partial ordering on A, using the correspondingproperties of the partial orderings on I and each Ai. Similarly, this defines alinear ordering on A when I and each Ai are linearly-ordered. If I and each Ai

are well-ordered, then A is well-ordered too. For if B is a nonempty subset ofA, then

BI = {i ∈ I : Ai ∩ B 6= ∅}is a nonempty subset of I, and hence has a minimal element i0. The minimalelement of B in A is then the same as the minimal element of B ∩ Ai0 as asubset of Ai0 .

If (E1,≺1) and (E2,≺2) are partially-ordered sets, then the lexicographic

ordering ≺ on E = E1×E2 is defined as follows. If x1, y1 ∈ E1 and x2, y2 ∈ E2,then

(x1, x2) ≺ (y1, y2)

when either x1 ≺1 y1 and x1 6= y1, or x1 = y1 and x2 ≺2 y2. It is well-knownand easy to check that this defines a partial ordering on E, which is a linearordering when ≺1, ≺2 are linear orderings on E1, E2, respectively. Similarly, Eis well-ordered by ≺ when E1, E2 are well-ordered by ≺1, ≺2, respectively. This

73

can also be seen as a special case of the discussion in the previous paragraph,with I = E1 and Ai = {i} × E2 for each i ∈ E1.

46.6 Comparing well-ordered sets

Let (A,≺) be a well-ordered set, and let α be an order automorphism on A,which is to say a one-to-one mapping from A onto itself such that α(x) ≺ α(y)for every x, y ∈ A with x ≺ y. If α is not the identity mapping on A, thenthe set of x ∈ A such that α(x) 6= x is nonempty, and hence this set has aminimal element x0. However, it is not difficult to show that α(x0) = x0 underthese conditions, from which one may conclude that α is the identity mappingon A. This implies that an order isomorphism between two well-ordered sets isnecessarily unique.

A subset I of A is said to be an (order) ideal in A if for each x ∈ A andy ∈ I with x ≺ y we have that x ∈ I. If I, I ′ are ideals in A and α is anorder isomorphism from I onto I ′, then one can check that α is the identitymapping, and hence that I = I ′, by basically the same type of argument asbefore. Of course, A is an ideal in itself, and so we get in particular that anorder isomorphism α from A onto an ideal I ′ in A is the identity mapping, andthat I ′ = A in this case. This would not work without the hypothesis that I ′

be an ideal in A, since there are plenty of order-preserving one-to-one mappingsfrom the set Z+ of positive integers into itself.

If a is any element of A, then it is easy to see that

Ia = {x ∈ A : x ≺ a, x 6= a}is an ideal in A. Conversely, if I is an ideal in A, I 6= A, and a is the minimalelement of A \ I, then I = Ia. It follows that if I, I ′ are any two ideals in A,then either I ⊆ I ′ or I ′ ⊆ I.

Suppose now that A and B are well-ordered sets. A well-known theoremstates that either A is isomorphic to an ideal in B, or that B is isomorphicto an ideal in A. This includes the possibility that A is isomorphic to B, andotherwise A or B is isomorphic to a proper ideal in the other set. Note that theideal and the isomorphism in the conclusion of this theorem are unique, by thepreceding discussion. The basic idea of the proof of the theorem is quite simple,since the minimal element of A should correspond to the minimal element of B,and so on.

46.7 Well-ordered subsets

Let (A,≺) be a well-ordered set, let I be an ideal in A, and let φ be a one-to-oneorder-preserving mapping from I into A. Note that φ(I) is not asked to be anideal in A here. Let us check that

x ≺ φ(x)

for every x ∈ I under these conditions. Otherwise, there is a minimal elementy of I such that φ(y) ≺ y and φ(y) 6= y. Let Iy be the set of x ∈ A such that

74

x ≺ y and x 6= y, as before. Of course, Iy ⊆ I, because I is an ideal in A. Theminimality of y implies that x ≺ φ(x) for every x ∈ Iy. Note that φ(x) ≺ φ(y)and φ(x) 6= φ(y) when x ∈ Iy, because φ is order-preserving and one-to-oneby hypothesis. Thus x ≺ φ(x) ≺ φ(y) for every x ∈ Iy. If φ(y) = x for somex ∈ Iy, then we would have that x = φ(x) = φ(y), contradicting the fact thatφ(x) 6= φ(y). Hence φ(y) 6∈ Iy, which says exactly that y ≺ φ(y), contradictingthe hypothesis that φ(y) ≺ y and φ(y) 6= y.

If φ(I) is also asked to be an ideal in A, then we can apply the previousargument to φ and to its inverse, to get that φ(x) = x for every x ∈ I andthus φ(I) = I. This case was discussed earlier, using a simpler but analogousargument.

Let B be a subset of A, which can also be considered as a well-ordered set,using the restriction of the ordering ≺ from A to B. Also let φ be a one-to-oneorder-preserving mapping from A onto an ideal in B. More precisely, φ(A) issupposed to be an ideal in B as a well-ordered set, but φ(A) may not be anideal in A. If φ(A) 6= B, then there is a b ∈ B \ φ(A), and φ(a) ≺ b for everya ∈ A, because φ(A) is an ideal in B. In particular, b ≺ φ(b), since b is also anelement of A. We also know that x ≺ φ(x) for every x ∈ A, as before, so thatb ≺ φ(b). This implies that φ(b) = b, contradicting the fact that b 6∈ φ(A). Itfollows that φ(A) = B when φ is a one-to-one order-preserving mapping from Aonto an ideal in B ⊆ A. This extends the case where B = A that was mentionedbefore.

If B is any well-ordered set, then either B is isomorphic to an ideal in A,or A is isomorphic to an ideal in B, as discussed previously. If B is a subsetof A, with the induced ordering, and A is isomorphic to an ideal in B, then itfollows that A is ismorphic to B, as in the preceding paragraph. Thus we mayconclude that every subset B of A is isomorphic to an ideal in A.

47 Some additional topics

47.1 Products of finite sets

Let X1,X2, . . . be a sequence of finite sets, and let X =∏∞

j=1 Xj be theirCartesian product. More precisely, suppose that each Xj is equipped withthe discrete topology, and let X be equipped with the corresponding producttopology. We may as well suppose also that each Xj has at least two elements,since X = ∅ when Xj = ∅ for any j, and the Xj ’s with only one element donot really affect the topology on X. We have seen before that X is compact,because it is sequentially compact and has a countable base for its topology.Let us now give another proof of this fact.

Let {Uα}α∈A be an arbitrary open covering of X, and suppose for the sakeof a contradiction that X is not covered by finitely many Uα’s. Of course, Xcan be identified with the union of the sets

{x1} ×∞∏

j=2

Xj(47.1)

75

with x1 ∈ X1. If each of these sets can be covered by only finitely many Uα’s,then X can be covered by only finitely many Uα’s, since X1 has only finitelymany elements. Thus there is an x1 ∈ X1 such that (47.1) cannot be coveredby finitely many Uα’s. Similarly, there is an x2 ∈ X2 such that

{x1} × {x2} ×∞∏

j=3

Xj(47.2)

cannot be covered by finitely many Uα’s. Continuing in this way, we get asequence x1, x2.x3, . . . of elements of X1,X2,X3, . . ., respectively, such that

{x1} × {x2} × · · · × {xn} ×∞∏

j=n+1

Xj(47.3)

cannot be covered by finitely many Uα’s for any positive integer n. Let x be theelement of X corresponding to the sequence x1, x2, x3, . . .. Because {Uα}α∈A

covers X, there is an α0 ∈ A such that x ∈ Uα0. We also have that (47.3) is

contained in Uα0for some n, since Uα0

is an open set in X with respect to theproduct topology. This contradicts the hypothesis that (47.3) cannot be coveredby finitely many Uα’s for any n, as desired.

In particular, one can apply this argument with Xj = {0, 1} for each j, toget that the space B of all binary sequences is compact with respect to theproduct topology. As before, the function that assigns to each binary sequencea real number in the usual way defines a continuous mapping from B onto [0, 1],and hence the compactness of B then implies that [0, 1] is compact. Of course,the preceding proof of the compactness of X and thus B is very similar to thestandard proof of the compactness of [0, 1], and so it is not surprising that onecan use it to recover the compactness of [0, 1]. One can also show that B ishomeomorphic to Cantor’s middle-thirds set, which is reviewed next.

47.2 The Cantor set

Put E0 = [0, 1], E1 = [0, 1/3] ∪ [2/3, 1], and

E2 = [0, 1/9] ∪ [2/9, 1/3] ∪ [2/3, 7/9] ∪ [8/9, 1].(47.4)

In general, if n is a nonnegative integer, then En is a closed subset of [0, 1]consisting of 2n pairwise-disjoint closed intervals of length 3−n. To go from En

to En+1, one removes the open middle third of each of the intervals in En, asin the first two steps. Thus En+1 ⊆ En for each n, and

E =∞⋂

n=0

En(47.5)

is Cantor’s middle-thirds set. This is a closed subset of [0, 1], which does notcontain any interval in the real line of positive length.

76

Let An be the set consisting of the endpoints of the 2n intervals in En foreach n ≥ 0. Thus An has 2n+1 elements, An ⊆ En, and An ⊆ An+1 for each n.This implies that

A =

∞⋃

n=1

An(47.6)

is a countable set contained in the Cantor set E, and one can check that A isdense in E. More precisely, if x ∈ En for some n ≥ 0, and if I is the intervalin En of length 3−n that contains x, then the distance from x to the nearestendpoint of I is less than or equal to one-half of 3−n.

As usual, every element of [0, 1] can be represented by a binary expansion,and sometimes by two binary expansions. Similarly, every element of [0, 1] has aternary expansion, which is to say an expansion base 3, consisting of 0’s, 1’s, and2’s. It is well known that the Cantor set consists exactly of elements of [0, 1] withternary expansions consisting of only 0’s and 2’s. Some elements of the Cantorset also have a ternary expansion that includes a 1, but ternary expensions thatconsist of only 0’s and 2’s are unique when they exist. Every ternary expansiondetermines an element of [0, 1], and thus every ternary expansion with only 0’sand 2’s determines an element of the Cantor set.

If b = {bj}∞j=1 is a binary sequence, so that bj = 0 or 1 for each j, then

f(b) =

∞∑

j=1

2 bj 3−j .(47.7)

is the element of [0, 1] corresponding to the ternary expansion whose jth termis 2 bj . Of course, these ternary expansions contain only 0’s and 2’s, and everyternary expansion with only 0’s and 2’s comes from a unique binary sequence bin this way. Thus f defines a one-to-one mapping from the set B of all binarysequences onto the Cantor set E, as in the previous paragraph. One can checkthat f is a homeomorphism from B as a product of a sequence of copies of {0, 1}with the product topology associated to the discrete topology on {0, 1} onto theCantor set with the topology induced by the standard topology on the real line.

47.3 Quotient spaces and mappings

Let X be a set, and let ∼ be an equivalence relation on X. Also let X/ ∼ bethe set of equivalence classes in X determined by ∼, and let π be the naturalquotient mapping from X onto X/ ∼, which sends each point x ∈ X to theequivalence class in X that contains x. If X is a topological space, then thequotient topology is defined on X/ ∼ by saying that a set W ⊆ X/ ∼ is an openset if and only if π−1(W ) is an open set in X. It is easy to see that this definesa topology on X, using the fact that

π−1( ⋃

α∈A

Eα

)=

⋃

α∈A

π−1(Eα)(47.8)

77

andπ−1

( ⋂

α∈A

Eα

)=

⋂

α∈A

π−1(Eα)(47.9)

for any family {Eα}α∈A of subsets of X/ ∼.Similarly, let X and Y be topological spaces, and let f be a mapping from

X onto Y . We say that f is a quotient mapping if V ⊆ Y is an open set if andonly if f−1(V ) is an open set in X. Equivalently, f is a quotient mapping if fis continuous and V ⊆ Y is an open set whenever f−1(V ) is an open set in X.Note that a one-to-one quotient mapping is the same as a homeomorphism.

If X is a topological space, ∼ is an equivalence relation on X, and π isthe canonical quotient mapping from X onto the corresponding quotient spaceX/ ∼, then π is a quotient mapping with respect to the quotient topology onX/ ∼, by construction. Conversely, let X and Y be topological spaces, and letf be a mapping from X onto Y . Consider the relation on X defined by x ∼ x′

when f(x) = f(x′). It is easy to see that this defines an equivalence relationon X, for which the equivalence classes are the subsets of X of the form f−1(y)for some y ∈ Y . This permits one to identify X/ ∼ with Y , and π with f , sothat f is a quotient mapping if and only if the topology on Y corresponds tothe quotient topology on X/ ∼.

Let X and Y be topological spaces again, and remember that

X \ f−1(E) = f−1(Y \ E)(47.10)

for any mapping f : X → Y and subset E of Y . Using this, one can check thata mapping f from X onto Y is a quotient mapping if and only if it has theproperty that E ⊆ Y is a closed set if and only if f−1(E) is a closed set in X.As before, this is equivalent to asking that f be continuous, and that E ⊆ Y bea closed set whenever f−1(E) is a closed set in X.

Observe that f(f−1(V )) ⊆ V for every mapping f : X → Y and subsetV of Y , and that f(f−1(V )) = V when f maps X onto Y . This implies thata continuous mapping from X onto Y is a quotient mapping if it sends opensubsets of X to open subsets of Y , or if it sends closed subsets of X to closedsubsets of Y .

Suppose that f is a continuous mapping from X onto Y , that X is compact,and that Y is Hausdorff. Thus closed subsets of X are also compact, andhence are sent to compact subsets of Y , because f is continuous. We alsoknow that ompact subsets of Y are closed sets, because Y is Hausdorff. Thisshows that f maps closed subsets of X to closed subsets of Y under theseconditions, so that f is a quotient mapping, as in the previous paragraph. Thiswas mentioned previously in the special case where f is also one-to-one, so thatf is a homeomorphism.

Remember that a topological space Y satisfies the first separation conditionif and only if {y} is a closed set in Y for every y ∈ Y . If f : X → Y is a quotientmapping, then it follows that Y satisfies the first separation condition if andonly if f−1(y) is a closed set in X for every y ∈ Y .

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47.4 Homotopic paths

Let X be a topological space, and let p, q be elements of X. Also let f0, f1

be continuous mappings from [0, 1] into X such that f0(0) = f1(0) = p andf0(1) = f1(1) = q. We say that f0 and f1 are homotopic as continuous pathsin X from p to q if there is a continuous mapping F from [0, 1] × [0, 1] into Xsuch that

F (0, t) = p and F (1, t) = q(47.11)

for every t ∈ [0, 1], and

F (r, 0) = f0(r) and F (r, 1) = f1(r)(47.12)

for every r ∈ [0, 1]. Thus ft(r) = F (r, t) is a continuous path in X from p to q asa function of r ∈ [0, 1] for each t ∈ [0, 1], which reduces to the given paths f0(r),f1(r) when t = 0, 1. The function F (r, t) defines a homotopy between f0(r) andf1(r), and the continuity of this function on [0, 1] × [0, 1] basically means thatwe have a continuous family of paths from p to q in X.

If X = Rn with the standard topology for some positive integer n, then

F (r, t) = (1 − t) f0(r) + t f1(t)(47.13)

defines a homotopy between f0(r) and f1(r). Hence every pair of paths in Rn

with the same endpoints are homotopic. Similarly, if X is a convex set in Rn,with the topology induced by the standard topology on Rn, then every pair ofpaths in X with the same endpoints are homotopic, using the same homotopy(47.13). However, if X = R2 \ {(0, 0)}, with the toppology induced by thestandard topology on R2, then it can be shown that different paths in X withthe same endpoints may not be homotopic in X. Basically, this is becausedifferent paths in X may wrap around (0, 0) in different ways, even if (0, 0) isnot included in X.

As another class of examples, let X be any topological space again, and letf be a continuous mapping of [0, 1] into X. Also let a be a continuous mappingfrom [0, 1] into [0, 1] such that a(0) = 0 and a(1) = 1. Thus a maps [0, 1] onto[0, 1], by the intermediate value theorem. In this case,

F (r, t) = f((1 − t) r + t a(r))(47.14)

defines a homotopy between f(r) and f(a(r)). In effect, this uses the convexityof [0, 1], instead of the convexity of X. This shows that any continuous path fin X is homotopic to any continuous reparameterization of f .

Let X be any topological space, and let p, q be elements of X, as before. If fis a continuous mapping from [0, 1] into X such that f(0) = p and f(1) = q, thenf is obviously homotopic to itself, because F (r, t) = f(r) defines a homotopybetween f and itself. Similarly, if f0, f1 are continuous mappings from [0, 1] intoX such that f0(0) = f1(0) = p, f0(1) = f1(1) = q, and F (r, t) is a homotopybetween f0 and f1, then F (r, 1 − t) defines a homotopy between f1 and f0.Thus the property of being homotopic is symmetric in the two paths. If f2 is

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another continuous mapping from [0, 1] into X such that f2(0) = p, f2(1) = q,and f2 is homotopic to f1, then it is not too difficult to show that f0 is alsohomotopic to f2. Basically, one simply continues a homotopy from f0 to f1 witha homotopy from f1 to f2, using a change of variables to get a homotopy from f0

to f2 defined on [0, 1]× [0, 1]. This implies that the property of being homotopicdefines an equivalence relation on the set of continuous paths in X from p to q.The corresponding equivalence classes are known as homotopy classes.

47.5 The fundamental group

Let X be a topological space, and let p be an element of X. A continuousmapping f : [0, 1] → X such that f(0) = f(1) = p is called a continuous loopbased at p. If g : [0, 1] → X is another continuous loop based at p, then we cancombine f and g to get a continuous loop f · g based at p. More precisely, thiscan be defined by

(f · g)(r) = f(2r) when 0 ≤ r ≤ 1/2(47.15)

= g(2r − 1) when 1/2 ≤ r ≤ 1.

If f0, f1, g0, and g1 are continuous loops based at p such that f0 is homotopicto f1 and g0 is homotopic to g1, then it is easy to check that f0 ·g0 is homotopicto f1 · g1, by combining homotopies between f0 and f1 and between g0 and g1

in the same way that individual loops are combined to get a homotopy betweenf0 · g0 and f1 · g1.

If f is a continuous loop in X based at p, then let [f ] be the homotopyclass of continuous loops based at p that contains f , which is the collection ofall continuous loops in X based at p that are homotopic to f . If g is anothercontinuous loop in X based at p, then put

[f ] · [g] = [f · g].(47.16)

If f ′, g′ are continuous loops in X based at p that are homotopic to f , g,respectively, then f ′ · g′ is homotopic to f · g, as in the preceding paragraph,and hence

[f ′ · g′] = [f · g].(47.17)

This implies that [f ] · [g] is a well-defined operation on homotopy classes ofcontinuous loops in X based at p, since it does not depend on the choice ofrepresentatives of the homotopy classes [f ] and [g].

If f , g, and h are continuous loops in X based at p, then (f · g) · h andf · (g · h) are also continuous loops in X based at p, which are not normallyquite the same. However, it is easy to see that (f · g) · h and f · (g · h) arereparameterizations of each other, and hence they are homotopic to each other.This implies that

([f ] · [g]) · [h] = [f ] · ([g] · [h]),(47.18)

so that the corresponding operation on homotopy classes of continuous loops inX based at p is associative.

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Let cp : [0, 1] → X be the constant loop based at p, defined by cp(r) = p foreach r ∈ [0, 1]. If f is any continuous loop in X based at p, then it is easy tosee that cp · f and f · cp can both be given as reparameterizations of f . Thuscp · f and f · cp are both homotopic to f , so that

[cp] · [f ] = [f ] · [cp] = [f ].(47.19)

This shows that [cp] plays the role of the identity element for this operation onhomotopy classes of continuous loops in X based at p.

If f is a continuous loop in X based at p, then let f be the continuous loopin X based at p defined by

f(r) = f(1 − r).(47.20)

If f ′ is another continuous loop in X based at p which is homotopic to f , thenit is easy to see that f ′ is homotopic to f , by replacing r with 1 − r in thehomotopy between f and f ′. Thus the homotopy class [f ] depends only on the

homotopy class [f ], and is denoted [f ]−1. One can also check that f · f and f ·fare both homotopic to cp as continuous loops in X based at p, so that

[f ]−1 · [f ] = [f ] · [f ]−1 = [cp].(47.21)

It follows that the collection of homotopy classes of continuous loops in Xbased at p forms a group with respect to the operation discussed in the previousparagraphs, known as the fundamental group of X at p. If q is another elementof X and ap,q is a continuous path in X from p to q, then one can show that thefundamental group of X at p is isomorphic to the fundamental group of X at q.The basic idea is to map continuous loops in X based at p to continuous loopsin X based at q, using ap,q to go back and forth between p and q. It is easyto see that this mapping sends homotopic loops at p to homotopic loops at q,and thus leads to a mapping from homotopy classes of loops at p to homotopyclasses of loops at q. One can also check that this mapping is compatible with theoperations on homotopy classes of loops discussed in the previous paragraphs, sothat it defines a homomorphism from the fundamental group of X at p into thefundamental group of X at q. There is an analogous mapping from continuousloops in X at q to continuous loops in X at p, for which the correspondingmapping between homotopy classes is the inverse of the preceding one. If X ispathwise connected, then the fundamental group of X at p is isomorphic to thefundamental group of X at q for every p, q ∈ X.

If X = Rn with the standard topology for some positive integer n, then thefundamental group of X is trivial at every point, because any two continuousloops based at the same point are homotopic to each other, as before. Similarly,if X is a convex set in Rn equipped with the topology induced by the standardtopology on Rn, then the fundamental group of X is trivial at every point. IfX = R2 \ {(0, 0)}, then it can be shown that the fundamental group of X atany point is isomorphic to the group Z of integers with respect to addition.The isomorphism is given by the winding number of a loop in R2 \ {(0, 0)},which counts the number of times that such a loop wraps around the origin,also taking orientations into account.

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47.6 Topological groups

Let G be a group. Thus G is a set equipped with a binary operation, which canbe expressed multiplicatively by x y for x, y ∈ G. This operation is supposed tosatisfy the associative law

(x y) z = x (y z)(47.22)

for every x, y, z ∈ G, and there should also be an identity element e in G suchthat

e x = x e = x(47.23)

for every x ∈ G. Each x ∈ G is supposed to have an inverse x−1 ∈ G thatsatisfies

x x−1 = x−1 x = e,(47.24)

and which is uniquely determined by x.In order for G to be a topological group, G should also be equipped with

a topology for which the group operations are continuous. More precisely, thismeans that

(x, y) 7→ x y(47.25)

should be continuous as a mapping from G × G into G, using the producttopology on G × G associated to the given topology on G. Similarly,

x 7→ x−1(47.26)

should be continuous as a mapping from G to itself. This should actually be ahomeomorphism from G onto itself, because (x−1)−1 = x for each x ∈ G, whichimplies that this mapping is its own inverse.

If a, b ∈ G, thenLa(x) = a x, Rb(x) = x b(47.27)

define mappings from G into itself, which are the left and right translationmappings corresponding to a, b, respectively. Continuity of multiplication inthe group implies that La and Rb are continuous mappings on G for each a, b ∈G. Note that La and Rb are invertible as mappings on G for each a, b ∈ G,with inverses given by La−1 , Rb−1 , respectively. Thus La and Rb are actuallyhomeomorphisms from G onto itself for each a, b ∈ G, since La−1 and Rb−1 arealso continuous.

If a, b ∈ G and A,B ⊆ G, then put

aB = La(B), A b = Rb(A),(47.28)

andAB = {x y : x ∈ A, y ∈ B}.(47.29)

If A and B are open subsets of G, then aB and Ab are also open sets in Gfor every a, b ∈ G, because La and Rb are homeomorphisms on G, as in the

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previous paragraph. Similarly, if A and B are closed subsets of G, then aB andAb are closed sets for every a, b ∈ G. Observe that

AB =⋃

a∈A

aB =⋃

b∈B

Ab(47.30)

for every A,B ⊆ G. If A or B is an open set in G, then it follows that AB isan open set in G, because it is a union of open subsets of G. Let us also put

A−1 = {a−1 : a ∈ A}.(47.31)

If A is an open or closed set in G, then A−1 has the same property, becausex 7→ x−1 is a homeomorphism on G.

It is customary to also ask that {e} be a closed set in a topological groupG. This implies that {a} is a closed set in G for every a ∈ G, and hence that Gsatisfies the first separation condition, because of continuity of translations, asbefore. One can also show that G is Hausdorff and even regular, as follows.

Suppose that E is a closed subset of G that does not contain e. Thus G \Eis an open set that contains e. This implies that there are open subsets U , Vof G containing e such that

U V −1 ⊆ G \ E,(47.32)

because of the continuity of the group operations. Equivalently, this means thatU V −1 ∩ E = ∅, which implies that

U ∩ E V = ∅.(47.33)

Note that E ⊆ E V , since e ∈ V , and that E V is an open set, because Vis open. If instead E1 is a closed set in G that does not contain a particularelement x ∈ G, then one can apply the preceding argument to E = x−1 E1. IfU , V are as before, then x U is an open set in G that contains x, and which isdisjoint from the open set x E V = E1 V that contains V . This shows that G isregular, and hence Hausdorff, as desired.

Of course, any group is a topological group with respect to the discretetopology. The real line is also a topological group with respect to addition andthe standard topology.

If G is a topological group, and if G satisfies the local countability conditionat e, then G satisifies the local countability condition at each point, because ofcontinuity of translations. A well-known theorem states that the topology on Gis determined by a metric under these conditions.

47.7 Topological vector spaces

Let V be a vector space over the real numbers. Thus V is set equipped withoperations of addition and scalar multiplication that satisfy certain conditions.In order for V to be a topological vector space, V should also be equipped witha topology for which the vector space operations are continuous. More precisely,

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this means that addition should be continuous as a mapping from V × V intoV , using the product topology on V × V associated to the given topology onV . Similarly, scalar multiplication should be continuous as a mapping fromR×V into V , using the product topology on R×V associated to the standardtopology on R and the given topology on V .

Of course, a vector space V is an abelian group with respect to addition.The additive inverse −v of a vector v ∈ V is equal to −1 ∈ R times v in thesense of scalar multiplication. Thus the continuity of v 7→ −v follows from thecontinuity of scalar multiplication in this case. As for topological groups, it iscustomary to ask that {0} be a closed set in a topological vector space V , where0 is the additive identity element in V . This implies that V is Hausdorff andeven regular, as before.

A nonnegative real-valued function ‖v‖ on a vector space V over the realnumbers is said to be a norm if ‖v‖ = 0 if and only if v = 0,

‖t v‖ = |t| ‖v‖(47.34)

for every t ∈ R and v ∈ V , and

‖v + w‖ ≤ ‖v‖ + ‖w‖(47.35)

for every v, w ∈ V . Here |t| is the absolute value of a real number t. In thiscase, it is easy to see that

d(v, w) = ‖v − w‖(47.36)

defines a metric on V . One can also check that V is a topological vector spacewith respect to the topology determined by this metric.

If n is a positive integer, then Rn is a topological vector space, where thevector space operations are defined coordinatewise in the usual way, and Rn isequipped with the standard topology. The standard topology on Rn is the sameas the topology associated to the standard Euclidean metric, which is the metricassociate to the standard Euclidean norm on Rn as in the previous paragraph.More generally, any product of copies of R is a topological vector space, wherethe vector space operations are again defined coordinatewise, and one uses theproduct topology associated to the standard topology on R. However, if thespace is the product of infinitely many copies of R, then one can show that thistopology is not determined by a norm.

47.8 Uniform spaces

If X is a topological space, then many of the same notions related to continuityand convergence can be defined as on a metric space. However, this does notinclude Cauchy sequences or uniform continuity. Indeed, one can give examplesof metrics d1(x, y) and d2(x, y) on a set X that determine the same topologyon X, but not the same Cauchy sequences or uniformly continuous mappings.Of course, this is closely related to the fact that a homeomorphism betweenmetric spaces may not be uniformly continuous, and that its inverse may notbe uniformly continuous.

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To deal with this, there is an abstract notion of a uniform structure on a setX. This involves subsets of the product X × X that contain the diagonal ∆,which is the set of (x, y) ∈ X ×X with x = y. Basically, these subsets of X ×Xare supposed to describe uniform families of neighborhoods of elements of X,but we shall not go into the details here. If X is equipped with a metric d(x, y),for instance, then one can consider subsets of X × X containing all (x, y) withX, y ∈ X and d(x, y) < ǫ for some ǫ > 0.

Topological groups provide another interesting special case of this, whichincludes topological vector spaces in particular. Let G be a topological group,and let U be an open set in G that contains the identity element e. This leads toan interesting subset of G×G that contains the diagonal, consisting of all (x, y)with x, y ∈ G such that x y−1 ∈ U . Alternatively, one could look at the set of all(x, y) with x, y ∈ G such that y−1 x ∈ U . This is the same as the previous setwhen G is abelian, but otherwise the difference could be significant. One couldalso consider the set of (x, y) with x, y ∈ G such that (x y−1)−1 = y x−1 ∈ U ,which amounts to replacing U with U−1. This is not too serious, because of thecontinuity of x 7→ x−1 on G. Similarly, the condition (y−1 x)−1 = x−1 y ∈ U isthe same as y−1 x ∈ U−1.

47.9 The supremum norm

Let X be a (nonempty) topological space, and let Cb(X) be the space of boundedcontinuous real-valued functions on X. More precisely, a real-valued function fon X is said to be bounded on X if f(X) is a bounded subset of the real line,so that |f(x)| ≤ A for some nonnegative real number A and every x ∈ X. If Xis compact, then f(X) is a compact set in R for every continuous real-valuedfunction f on X, and hence f(X) is automatically bounded in R. It is easy tosee that Cb(X) is a vector space over the real numbers with respect to pointwiseaddition and scalar multiplication, which may be considered as a linear subspaceof the vector space C(X) of all continuous real-valued functions on X. If C(X)separates points on X, then Cb(X) separates points on X too, because one canget take maxima and minima of arbitrary continuous real-valued functions on Xwith constants to get bounded continuous functions on X. Of course, constantfunctions on any topological space are bounded and continuous.

If f ∈ Cb(X), then the supremum norm of f on X is defined by

‖f‖ = supx∈X

|f(x)|.(47.37)

It is easy to check that this is indeed a norm on Cb(X) of X as a vector spaceover the real numbers, which determines the corresponding supremum metric

d(f, g) = ‖f − g‖(47.38)

on Cb(X). As usual, a sequence {fj}∞j=1 of bounded continuous real-valuedfunctions on X converges to another bounded continuous real-valued functionf on X with respect to the supremum metric if and only if {fj}∞j=1 converges

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to f uniformly on X. If a sequence {fj}∞j=1 of bounded continuous real-valuedfunctions on X converges to a real-valued function f uniformly on X, then f isbounded and continuous on X, by standard arguments. Similarly, one can showthat Cb(X) is complete as a metric space with respect to the supremum metric,in the sense that every Cauchy sequence in Cb(X) converges to an element ofCb(X). If {fj}∞j=1 is a Cauchy sequence in Cb(X) with respect to the supremummetric, then {fj(x)}∞j=1 is a Cauchy sequence in R for each x ∈ X, which impliesthat {fj(x)}∞j=1 converges in R for each x ∈ X, because the real line is completewith respect to the standard metric. This shows that {fj}∞j=1 converges to a real-valued function f poitwise on X, and one can then use the Cauchy conditionwith respect to the supremum metric to check that {fj}∞j=1 converges to funiformly on X, which implies that f is bounded and continuous on X, asbefore.

Suppose now that X is a locally compact Hausdorff topological space whichis not compact. Remember that a continuous real-valued function f on X issaid to vanish at infinity on X if for each ǫ > 0 there is a compact set Kǫ ⊆ Xsuch that |f(x)| < ǫ for every x ∈ X \Kǫ. This implies that f is bounded on X,since any continuous function on X is bounded on compact subsets of X. Thusthe space C0(X) of continuous real-valued functions on X that vanish at infinityis contained in Cb(X), and in fact C0(X) is a linear subspace of Cb(X). It is nottoo difficult to show that C0(X) is also closed with respect to the supremummetric on Cb(X).

Remember that the support of a continuous real-valued function f on X isdefined to be the closure of the set of x ∈ X such that f(x) 6= 0, and that thespace of continuous real-valued functions with compact support on X is denotedCcom(X). This is a linear subspace of C0(X), and it is not too difficult to showthat Ccom(X) is dense in C0(X) with respect to the supremum metric. To seethis, let f ∈ C0(X) and ǫ > 0 be given, and let Kǫ be a compact set in X suchthat |f(x)| < ǫ for every x ∈ X \ Kǫ, as in the preceding paragraph. UsingUrysohn’s lemma, one can get a continuous real-valued function θǫ on X suchthat θǫ(x) = 1 for every x ∈ Kǫ, θǫ(x) = 1 for every x ∈ X, and θǫ has compactsupport on X. Thus θǫ f has compact support on X, and

|f(x) − θǫ(x) f(x)| = (1 − θǫ(x)) |f(x)| < ǫ(47.39)

for every x ∈ X, as desired.

47.10 Uniform continuity

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces. As usual, a mapping f fromM into N is said to be uniformly continuous if for every ǫ > 0 there is a δ > 0such that

ρ(f(x), f(y)) < ǫ(47.40)

for every x, y ∈ M such that d(x, y) < δ. Of course, every uniformly continuousmapping from M into N is continuous, and conversely it is well known thatevery continuous mapping from M into N is uniformly continuous when M

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is compact. As mentioned earlier, uniform continuity of mappings betweentopological spaces does not make sense without some additional structure, butthere is a special case where a type of uniform continuity condition is natural,that we shall described next.

Let X, Y be topological spaces, and let (N, ρ(u, v)) be a metric space again.Also let f(x, y) be a function on X × Y with values in N . In particular, Ncould be the real line with the standard metric, so that f(x, y) is a real-valuedfunction on X ×Y . Let us say that f(x, y) is continuous in x at a point x0 ∈ Xuniformly in y if for every ǫ > 0 there is an open set U of X such that x0 ∈ Usuch that

ρ(f(x, y), f(x0, y)) < ǫ(47.41)

for every x ∈ U and y ∈ Y . If this condition holds for every x0 ∈ X, then wesay that f(x, y) is continuous in x uniformly in y.

If f(x, y) is continuous in x at x0 ∈ X uniformly in y, then f(x, y) is obviouslycontinuous as a function of x at x0 for each y ∈ Y . In the other direction, letus show that f(x, y) is continuous in x at x0 uniformly in y when f(x, y) iscontinuous as a function on X × Y with the product topology at (x0, y) forevery y ∈ Y and Y is compact. This implies that f(x, y) is continuous in xuniformly in y when f(x, y) is continuous on X ×Y with respect to the producttopology and Y is compact.

Let ǫ > 0 be given, and for each y ∈ Y , let W (y) be an open set in X × Ysuch that (x0, y) ∈ W (y) and

ρ(f(x, z), f(x0, y)) <ǫ

2(47.42)

for every (x, z) ∈ W (y). By definition of the product topology on X × Y ,there are open subsets U(y), V (y) of X, Y , respectively, such that x0 ∈ U(y),y ∈ V (y), and

U(y) × V (y) ⊆ W (y).(47.43)

If x ∈ U(y) and z ∈ V (y), then we can apply (47.42) to (x, z) and to (x0, z) toget that

ρ(f(x, z), f(x0, z)) ≤ ρ(f(x, z), f(x0, y)) + ρ(f(x0, y), f(x0, z))(47.44)

<ǫ

2+

ǫ

2= ǫ.

If Y is compact, then there are finitely many elements y1, . . . , yn of Y such that

Y =

n⋃

j=1

V (yj).(47.45)

In this case,

U =

n⋂

j=1

U(yj)(47.46)

is an open set in X that contains x0, and it is easy to see that (47.44) holds forevery x ∈ U and z ∈ Y , as desired.

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47.11 The supremum metric

Let Y be a topological space, and let (N, ρ(u, v)) be a metric space. Rememberthat a subset of N is said to be bounded if it is contained in a ball of finiteradius, and that a mapping f : Y → N is bounded when f(Y ) is a bounded setin N . Let Cb(Y,N) be the space of bounded continuous mappings from Y intoN . If Y is compact and f : Y → N is continuous, then f(Y ) is compact andhence bounded in Y , and Cb(Y,N) reduces to the space C(Y,N) of all continuousmappings from Y into N . If N is the real line with the standard metric, thenCb(Y,N) is the same as the vector space Cb(Y ) of bounded continuous real-valuedfunctions on Y , discussed earlier.

If f, g ∈ Cb(Y,N), then put

θ(f, g) = supy∈Y

ρ(f(y), g(y)).(47.47)

One can check that this defines a metric on Cb(Y,N), known as the supremummetric. If N = R with the standard metric, then this is the same as the metricassociated to the supremum norm on Cb(Y ). As before, a sequence {fj}∞j=1 ofbounded continuous mappings from Y into N converges to another boundedcontinuous mapping f from Y into N with respect to the supremum metric ifand only if {fj}∞j=1 converges to f uniformly on Y . If {fj}∞j=1 is a sequenceof bounded continuous mappings from Y into N that converges uniformly toa mapping f : Y → N , then f is also bounded and continuous, by standardarguments. If N is complete, then one can check that Cb(Y,N) is completewith respect to the supremum metric as well. More precisely, completenessof N implies that a Cauchy sequence {fj}∞j=1 in Cb(Y,N) with respect to thesupremum metric converges pointwise to a mapping f : Y → N , and one canthen show that the Cauchy condition with respect to the supremum metricimplies that {fj}∞j=1 converges to f uniformly on Y .

Let X be another topological space, and suppose that Φ is a continuous map-ping from X into Cb(Y,N), with respect to the supremum metric on Cb(Y,N). Ifx ∈ X, then Φ(x) is a bounded continuous mapping from Y into N , which canthen be evaluated at y ∈ Y to get an element Φ(x)(y) of N . Thus Φ determinesa mapping φ from X × Y into N , where

φ(x, y) = Φ(x)(y)(47.48)

for every x ∈ X and y ∈ Y . The hypothesis that Φ map X into Cb(Y,N) meansthat φ(x, y) is a bounded continuous function of y ∈ Y for each x ∈ X. Thecondition that Φ be a continuous mapping from X into Cb(Y,N) with respectto the supremum metric is then equivalent to asking that φ(x, y) be continuousin x uniformly in y.

If φ : X×Y → N has the properties that φ(x0, y) is continuous as a functionof y ∈ Y for some x0 ∈ X, and that φ(x, y) is continuous as a function of x at x0

uniformly in y, then it is easy to see that φ is continuous as a function on X×Ywith respect to the product topology at (x0, y) for each y ∈ Y . In particular, ifΦ is a continuous mapping from X into Cb(Y,N), as in the preceding paragraph,

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then the corresponding function φ(x, y) is continuous as a mapping from X ×Ywith the product topology into N . Conversely, if φ is a continuous mappingfrom X × Y into N and Y is compact, then φ(x, y) is bounded and continuousas a function of y ∈ Y for each x ∈ X, and hence φ(x, y) determines a mappingΦ from X into Cb(Y,N) as in (47.48). We have seen before that φ(x, y) is alsocontinuous in x uniformly in y under these conditions, so that Φ is continuouswith respect to the supremum metric on Cb(Y,N).

47.12 An approximation theorem

Let X and Y be compact Hausdorff topological spaces, and let f(x, y) be acontinuous real-valued function on X×Y , with respect to the product topology.We would like to show that f(x, y) can be approximated by finite sums ofproducts of continuous functions on X and on Y , uniformly on X × Y .

Let ǫ > 0 be given. Because f(x, y) is continuous on X × Y and Y iscompact, we have seen previously that f(x, y) is continuous in x uniformly iny. This implies that for each x ∈ X there is an open set U1(x) ⊆ X such thatx ∈ U1(x) and

|f(w, y) − f(x, y)| < ǫ(47.49)

for every w ∈ U1(x) and y ∈ Y . Using Urysohn’s lemma, for each x ∈ X we canget a continuous nonnegative real-valued function φx on X such that φx(x) > 0and φx(w) = 0 when w ∈ X \ U1(x). Put

U2(x) = {w ∈ W : φx(w) > 0},(47.50)

so that U2(x) is an open set in X that contains x and is contained in U1(x).Because X is compact, there are finitely many elements x1, . . . , xn of X such

that

X =n⋃

l=1

U2(xl).(47.51)

Thusn∑

l=1

φxl(w) > 0(47.52)

for each w ∈ X, and we put

ψj(w) = φxj(w)

( n∑

l=1

φxl(w)

)−1

(47.53)

for j = 1, . . . , n. This is a nonnegative real-valued continuous function on X foreach j, which satisfies ψj(w) = 0 when w ∈ X \ U2(xj) and

n∑

j=1

ψj(w) = 1(47.54)

for every w ∈ X.

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It follows that

f(w, y) −n∑

j=1

ψj(w) f(xj , y) =

n∑

j=1

ψj(w) (f(w, y) − f(xj , y))(47.55)

for every w ∈ X and y ∈ Y , and hence

∣∣∣∣f(w, y) −n∑

j=1

ψj(w) f(xj , y)

∣∣∣∣ ≤n∑

j=1

ψj(w) |f(w, y) − f(xj , y)|.(47.56)

If ψj(w) 6= 0, then w ∈ U2(xj) ⊆ U1(xj), so that (47.49) holds with x = xj forevery y ∈ Y . Plugging this into (47.56), we get that

∣∣∣∣f(w, y) −n∑

j=1

ψj(w) f(xj , y)

∣∣∣∣ < ǫn∑

j=1

ψj(w) = ǫ(47.57)

for every w ∈ X and y ∈ Y , as desired.

47.13 Infinitely many variables

Let I be an infinite set, and suppose that Xi is a topological space for eachi ∈ I. Let X be the Cartesian product

∏i∈I Xi, equipped with the product

topology, and let xi ∈ Xi be the ith component of x ∈ X for each i ∈ I. Alsolet (N, ρ(u, v)) be a metric space, and let f be a continuous mapping from Xinto N . We would like to approximate f by functions of finitely many variables.

If x ∈ X and ǫ > 0, then there is an open set U(x) in X such that x ∈ U(x)and

ρ(f(x), f(y)) <ǫ

2(47.58)

for every y ∈ U(x). More precisely, we may as well take U(x) to be of the form

U(x) =∏

i∈I

Ui(x),(47.59)

where Ui(x) is an open subset of Xi that contains xi for each i ∈ I, and

I(x) = {i ∈ I : Ui(x) 6= Xi}(47.60)

is a finite subset of I, because of the way that the product topology on X isdefined. If w, y ∈ U(x), then we can apply (47.58) twice to get that

ρ(f(w), f(y)) ≤ ρ(f(w), f(x)) + ρ(f(x), f(y)) <ǫ

2+

ǫ

2= ǫ.(47.61)

In particular, this holds when w ∈ U(x), y ∈ X, and wi = yi for each i ∈ I(x),since y ∈ U(x) as well in this case.

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Suppose now that Xi is compact for each i ∈ I, so that X is also compactwith respect to the product topology, by Tychonoff’s theorem. This impliesthat there is a finite set A ⊆ X such that

X =⋃

x∈A

U(x),(47.62)

since the U(x)’s form an open covering of X. Put

I1 =⋃

x∈A

I(x),(47.63)

which is a finite subset of I. If w, y ∈ X and wi = yi for every i ∈ I1, then(47.61) implies that

ρ(f(w), f(y)) < ǫ,(47.64)

because w ∈ U(x) for some x ∈ A, for which I(x) ⊆ I1.

A Additional homework assignments

Note that the first homework assignment was mentioned in Section 1.

A.1 Limit points

Let X be a topological space, and for E ⊆ X, put

E′ = {p ∈ X : p is a limit point of E}.

Discuss the question of whether E′ is closed, with examples or general results,as appropriate, e.g., how might this be related to the separation conditions?

A.2 Dense open sets

Let (X, τ) be a topological space. Suppose that τ ′ is the collection of opensubsets U of X with respect to τ such that either U = ∅ or U is dense in X withrespect to τ . For the homework, please show the following statements. First, τ ′

also satisfies the requirements of a topology on X. Second, X is automaticallynot Hausdorff with respect to τ ′ when X has at least two elements. Third,suppose that X satisfies the first separation condition with respect to τ . Ifevery element of X is a limit point of X with respect to τ , then X satisfies thefirst separation condition with respect to τ ′ too.

A.3 Combining topologies

Let X be a set, and suppose that τ1, τ2 are topologies on X. Let τ = τ1 ∨ τ2 bethe collection of subsets W of X such that for every x ∈ W there are subsetsU , V of X with x ∈ U , x ∈ V , U ∈ τ1, V ∈ τ2, and U ∩ V ⊆ W . Show thefollowing statements. First, τ is a topology on X. Second, τ1, τ2 ⊆ τ . Third,

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if τ is any other topology on X such that τ1, τ2 ⊆ τ , then τ ⊆ τ . Fourth, ifd1(x, y), d2(x, y) are metrics on X for which the associated topologies are τ1,τ2, respectively, then d(x, y) = max(d1(x, y), d2(x, y)) defines a metric on X forwhich the associated topology is τ .

A.4 Complements of countable sets

Let X be a set. Let us say that U ⊆ X is an open set if U = ∅ or X \U has onlyfinitely or countably many elements. Show that this defines a topology on X.Of course, this is analogous to the situation mentioned previously in which theopen sets are the empty set and the sets whose complements have only finitelymany elements. Discuss some of the properties of this topology. For instance,under what conditions is X a Hausdorff space? Which sets E ⊆ X are dense inX? When does a sequence of elements of X converge?

A.5 Combining topologies again

Let X be a set equipped with a topology τ1, and let τ2 be the topology on Xdescribed in Appendix A.4 in which a set V ⊆ X is an open set if V = ∅ or X \Vhas only finitely or countable many elements. Consider the topology τ = τ1∨ τ2

on X as defined in Appendix A.3. What happens when X has only finitely orcountably many elements? In general, when is a point p ∈ X a limit point of aset E ⊆ X with respect to τ? When is a set E ⊆ X dense in X with respectto τ? Which sequences in X converge with respect to τ? Does (X, τ) satisfythe first separation condition? Is (X, τ) Hausdorff when (X, τ1) is Hausdorff?If (X, τ1) satisfies the local countability condition at a point p ∈ X, then underwhat conditions does (X, τ) satisfy the local countability condition at p?

A.6 Combining topologies on R

Let τ1 be the standard topology on the real line, and let τ2 be the topology onR defined in Appendix A.4. Thus U ⊆ R is an open set with respect to τ2 ifU = ∅ or R\U has only finitely or countably many elements. Let τ = τ1∨τ2 bethe topology on R obtained by combining these two topologies as in AppendixA.3. Show that (R, τ) is not a regular topological space, so that there is a pointp ∈ R and a closed set E ⊆ R with respect to τ such that p 6∈ E and U ∩V 6= ∅for every pair of open subsets U , V of R with respect to τ which satisfy p ∈ Uand E ⊆ V .

A.7 Connections with product topologies

This is not so much another homework assignment, as some remarks aboutconnections between product topologies and the way of combining topologiesdescribed in Appendix A.3. Let (X1, τ1) and (X2, τ2) be topological spaces.Also let τ ′

1 be the collection of subsets of X1 × X2 of the form U1 × X2, whereU1 is an open set in X1 with respect to τ1. Similarly, let τ ′

2 be the collection of

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subsets of X1 × X2 of the form X1 × U2, where U2 is an open set in X2 withrespect to τ2. It is easy to see that τ ′

1 and τ ′2 are topologies on X1 × X2. More

precisely, τ ′1 is the same as the product topology on X1 × X2 associated to τ1

on X1 and the indiscrete topology on X2, and τ ′2 is the same as the product

topology on X1 ×X2 associated to the indiscrete topology on X1 and τ2 on X2.The product topology on X1 × X2 associated to τ1 on X1 and τ2 on X2 is thesame as the combination τ ′

1 ∨ τ ′2 of τ ′

1 and τ ′2 on X1 ×X2 as before. Conversely,

suppose now that τ1, τ2 are topologies on the same set X, and let τ = τ1 ∨ τ2

be their combination, as before. We can also consider the product topology onX×X defined using τ1 on the first factor of X and τ2 on the second factor of X.If we identify X with the “diagonal” in X × X, consisting of the ordered pairs(x, x) with x ∈ X, then the combined topology τ = τ1 ∨ τ2 on X correspondsexactly to the topology induced on the diagonal by the product topology onX × X associated to τ1 and τ2 on the two factors of X.

References

[1] J. Baum, Elements of Point Set Topology, Dover, 1991.

[2] T. Gamelin and R. Greene, Introduction to Topology, 2nd edition, Dover,1999.

[3] M. Gemignani, Elementary Topology, Dover, 1990.

[4] P. Halmos, Naive Set Theory, Springer-Verlag, 1974.

[5] H. Herrlich, Axiom of Choice, Lecture Notes in Mathematics 1876,Springer-Verlag, 2006.

[6] J. Hocking and G. Young, Topology, 2nd edition, Dover, 1988.

[7] I. James, Introduction to Uniform Spaces, Cambridge University Press,1990.

[8] I. James, Topologies and Uniformities, Springer-Verlag, 1999.

[9] D. Kahn, Topology, Dover, 1995.

[10] I. Kaplansky, Set Theory and Metric Spaces, 2nd edition, Chelsea, 1977.

[11] R. Kasriel, Undergraduate Topology, Dover, 2009.

[12] J. Kelley, General Topology, Springer-Verlag, 1975.

[13] J. Kelley, I. Namioka, et al., Linear Topological Spaces, Springer-Verlag,1976.

[14] S. Krantz, The Elements of Advanced Mathematics, 2nd edition, Chapman& Hall / CRC, 2002.

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[15] S. Krantz, Real Analysis and Foundations, 2nd edition, Chapman & Hall/ CRC, 2005.

[16] S. Krantz, A Guide to Real Variables, Mathematical Association of Amer-ica, 2009.

[17] S. Krantz, A Guide to Topology, Mathematical Association of America,2009.

[18] S. Krantz, Essentials of Topology with Applications, CRC Press, 2010.

[19] J. Lee, Introduction to Topological Manifolds, Springer-Verlag, 2000.

[20] G. McCarty, Topology: An Introduction with Application to Topological

Groups, 2nd edition, Dover, 1988.

[21] B. Mendelson, Introduction to Topology, 3rd edition, Dover, 1990.

[22] G. Moore, Zermelo’s Axiom of Choice, Springer-Verlag, 1982.

[23] W. Page, Topological Uniform Structures, Dover, 1988.

[24] M. Rudin, Lectures on Set Theoretic Topology, American MathematicalSociety, 1975.

[25] W. Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill,1976.

[26] L. Steen and J. Seebach, Jr., Counterexamples in Topology, Dover, 1995.

[27] G. Whyburn, Topological Analysis, 2nd revised edition, Princeton Univer-sity Press, 1964.

[28] S. Willard, General Topology, Dover, 2004.

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