September 2020

INTRODUCTION TO TOPOLOGY(SOME ADDITIONAL BASIC EXERCISES)

MICHAEL MEGRELISHVILI

Abstract. We provide some additional exercises in the course Topology-8822205(Bar-Ilan University). We are going to update this file several times (during thecurrent semester).

Contents

1. Metric spaces 12. Topological spaces 53. Topological products 84. Compactness 95. Quotients 116. Hints and solutions 12References 25

1. Metric spaces

Exercise 1.1. Give an example of a metric space (X, d) containing two balls suchthat

B(a1, r1) ( B(a2, r2) but r2 < r1.

Hint: think about metric subspaces of R.

Exercise 1.2. Let (X, d) be a metric space and

0 < r + d(a, b) < R.

Prove that B(b, r) ⊆ B(a,R). Conclude that the ball B(a,R) is an open subset of X.

Exercise 1.3. Let X be a set. Define

d∆(x, y) :=

{0 for x = y

1 for x 6= y

Show that (X, d∆) is an ultrametric space and describe the balls and spheres accordingto their radii.

Date: September 14, 2020.1

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Exercise 1.4. Find non-isometric metric spaces X, Y such that there exist isometricembeddings f : X ↪→ Y , g : Y ↪→ X.

Exercise 1.5. Let (V, || · ||) be a normed space. Show that every translation

Tz : V → V, Tz(x) = z + x

is an isometry. Conclude that all open balls B(a, r) with the same radius r (anda ∈ V ) are isometric.

Exercise 1.6. Give geometric descriptions of B[v, r], B(v, r), S(v, r) in R2 for v =(1, 2), r = 3 with respect to the following metrics: (a) Euclidean d; (b) d1; (c) dmax.

Exercise 1.7.

(1) Consider the normed spaces (C[0, 1], || · ||max) and (C[0, 1], || · ||1). Give anintuitive description of the ”closed ball” Bmax[θ, 3], B1[θ, 3] with radius r = 3and the center in the zero function θ : [0, 1]→ R, x 7→ 0.

(2) Explain theoretically why B1[θ, 3] ⊂ Bmax[θ, 3].

Exercise 1.8. Let (X, d) be a pseudometric space. Prove that the following conditionsare equivalent:

(1) (X, d) is a metric space.(2) Every finite subset F ⊂ X is closed.(3) For every converging sequence f : N → X, xn := f(n) the limit lim

n→∞xn is

unique.(4) (X, d) satisfies the Hausdorff property.(5) ∩{B(a, r)| r > 0} = {a}.

Exercise 1.9. (p-adic metric on Z)For every given prime p define on the set Z of all integers the following metric

dp(x, y) :=

{0 for x = y1pk

for k = k(x, y) = max{i : pi|(x− y)}Show that

(1) Is an ultrametric and diam(Z, dp) = 1.Hint: Observe that

k(x, z) ≥ min{k(x, y), k(y, z)}.(2) Every translation Tz : Z→ Z, Tz(x) = z + x is an isometry.(3) lim

n→∞pn = 0. Conclude: (Z, dp) does not contain isolated points.

(4) Every ball B(0, r) (center is 0) is a subgroup of (Z,+). Every ball B(a, r) isclopen.

(5) For every ball B(a, r) and every b ∈ B(a, r) we have B(a, r) = B(b, r).That is, every point inside a ball is its center ! Try to generalize to everyultrametric space.Remark: see the schematic pictures of (Z, d3) and (Z, d7) Figures 1 and 2.

(6) Describe the ”next stage” in the picture for (Z, d3).

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Figure 1. p=3

Figure 2. p=7

Exercise 1.10. (Cantor cube) Let

X = {0, 1}N = {x = (x1, x2, · · · )| xk ∈ {0, 1}}

be the set of all binary sequences. Define d : X ×X → [0,∞) as

d(x, y) :=

{0 for x = y1k

for k = k(x, y) = min{i : xi 6= yi}

Show that

(1) d is an ultrametric and diam(X, d) = 1.Hint: Observe that

k(x, z) ≥ min{k(x, y), k(y, z)}

(2) Give an example of a converging sequence with distinct members (hence, isnot eventually constant).

(3) Every ball B(θ, r) (center is the zero-sequence θ := (0, 0, · · · )) is a subgroupof the group ({0, 1}N,+) (where the sum + of sequences is ”modulo 2”).

Exercise 1.11. (Hilbert space l2)

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Define

l2 := {x = (x1, x2, · · · )| xi ∈ R∞∑i=0

x2i <∞}

||x|| :=

√√√√ ∞∑i=0

x2i

Then l2 is a normed space and 〈x, y〉 :=∑∞

i=0 xiyi is an inner (scalar) product.Show that

(1) There exists a sequence in l2 which has no limit but converges coordinate-wise(in contrast to the Euclidean space Rn).

(2) There exist isometric embeddings Rn ↪→ l2, (N, d∆) ↪→ l2.

Recall that d∆(x, y) :=

{0 for x = y

1 for x 6= y

(3) Find a linear isometric embedding f : l2 → l2 which is not onto.(4) ∗ There exists a metric space M with 4 elements which cannot be embedded

isometrically into l2 (and hence into any Rn). Show also that 4 is the minimumpossible natural number with above property.Hint: when d(x, z) = d(x, y) + d(y, z) in R2, Rn ?

Exercise 1.12. Let (M,d) be a metric space. Show that for every closed subsetA ⊆M there exists a sequence On of open subsets in M such that A = ∩nOn.

Exercise 1.13. Show that the p-adic metric space (Z, dp) (see Exercise 1.9) is notcomplete. What about the Cantor cube (from Exercise 1.10) ?

Exercise 1.14. Show that the normed space (C[−1, 1], || · ||1) defined by

||f ||1 =

∫ 1

−1

|f(x)|dx

is not a Banach space.

Exercise 1.15. Give an example of a linear map f : E1 → E2 between two normedspaces which is not continuous.

Exercise 1.16. (Cantor Set)Let C be the set of real numbers that are sums of series of the form

∞∑k=1

ak3k

where k ∈ {0, 2}.

In other words, C consists of the real numbers that have the form 0.a1a2...ak... withoutthe digit 1 in the number system with base 3. Prove that

(1) C is contained in [0, 1].(2) C does not meet (1

3, 2

3).

(3) C does not meet (3s+13k, 3s+2

3k) for every s, k ∈ N.

(4) Find a geometric description of C(Hint: removing from [0, 1] countably many open intervals).

Remark: See for example [1], [5] or/and wikipedia.

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Totally bounded metric spaces. A metric space (M,d) is said to be totallybounded (or, precompact) if for every ε > 0 there exists a finite subset Aε of Xsuch that Aε is ε-dense in (M,d). A subset Y of M is said to be totally bounded ifthe metric subspace (Y, dY ) is totally bounded.

Exercise 1.17. Let (M,d) be a metric space.

(1) Every finite subset is totally bounded;(2) Every totally bounded subset is bounded;(3) Finite union of totally bounded subsets is totally bounded;(4) If X is a totally bounded subset of M then every subset Y of X is also totally

bounded.(5) The closure cl(Y ) of a totally bounded subset Y is also totally bounded;

Exercise 1.18.

(1) Let (X, d) be a complete metric space and Y ⊂ X is a closed subset. Then themetric subspace (Y, dY ) is also complete.

(2) Let (Y, dY ) is a metric subspace of (X, d). Show that if (Y, dY ) is completethen Y is closed in X.

2. Topological spaces

Exercise 2.1. Prove that for every pseudometric space (X, d) the pair (X, top(d)) isa topological space.

Exercise 2.2. Prove that any intersection ∩iτi of topologies τi on the same set X isa topology. Show that it is not true, in general, for unions.

Exercise 2.3. Let (X, τ) be a topological space. Show that the following conditionsare equivalent:

(1) (X, τ) ∈ T1;(2) Every singleton {a} is closed;(3) Every finite subset F ⊂ X is closed;(4) τcofin ⊂ τ .

Exercise 2.4. On Z define the following family τ≤ of subsets

τ≤ := {∅,Z, Ok : k ∈ Z} where Ok := {x ∈ Z : x ≤ k}.Show that:

(1) (Z, τ≤) is a connected topological space with property T0.(2) It is not (pseudo)metrizable.(3) Every continuous map (Z, τ≤)→ R into the reals is constant.

Exercise 2.5. On Z describe the smallest topology τcofin with the following property:every singleton {a} is closed in Z. Show that (Z, τcofin) is connected, not Hausdorffand not (pseudo)metrizable.

Exercise 2.6. (generalized subspace topology) Let (X, τ) be a topological space andf : Y → X be a function. Define

τf,Y := {f−1(O) : O ∈ τ}.

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Prove that (Y, τf,Y ) is a topological space.Remark: In the case of a subset Y ⊂ X and the inclusion f = in : Y → X we get thesubspace topology on Y .

Exercise 2.7. (heredity of the continuity)Let f : X → Y be a continuous map between topological spaces and X1 ⊂ X, Y1 ⊂ Y

be subspaces such that f(X1) ⊂ Y1. Show the continuity of the following induced map

f∗ : X1 → Y1, x 7→ f(x).

Conclude that, in particular, the following induced maps are continuous:

(1) Y → Z, y 7→ f(y)(2) Y → f(Y ), y 7→ f(y)

are also continuous.Conclude also that if f : X → Z is a homeomorphism then also the full restriction

f ∗Y : Y → f(Y ) is a homeomorphism.

Exercise 2.8. Show that for any topological space Y the following maps are alwayscontinuous:

(1) (X, τdiscr)→ Y .(2) Y → (X, τtriv).

Exercise 2.9. Show that the following conditions are equivalent:

(1) X is not connected.(2) There exists a clopen subset A ⊂ X such that ∅ 6= A 6= X.(3) There exists a continuous map f : X → R such that f(X) is a two point set.

Exercise 2.10. Show (by giving corresponding examples) that the separation proper-ties Ti, i ∈ {0, 1, 2, 3} are not, in general, preserved by continuous onto maps.

Exercise 2.11. Which topological properties from the following list are hereditary:

(1) connectedness(2) Ti, i ∈ {0, 1, 2, 3}(3) compactness(4) B2

(5) B1

(6) Metrizability.

Remark: a property is said to be hereditary if every topological subspace inherits it.

Exercise 2.12. Let X = R ∪ {p}, where p /∈ R. Define

τ := {O ⊂ X| p ∈ O ⇒ X \O is countable}.Show that

(1) (X, τ) is a Hausdorff topological space.(2) There exists A ⊂ X such that scl(A) 6= cl(A).(3) (X, τ) is not metrizable.(4) (X, τ) is a normal topological space. That is, (X, τ) ∈ T4.

Exercise 2.13. Let C1, C2 be closed subsets of X and f : X → Y be a function suchthat X = C1∪C2 and the restrictions fC1 : C1 → Y and fC1 : C1 → Y are continuous(on subspaces C1 and C2, respectively). Show that f is continuous. Conclude that

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this remains true for FINITELY many closed subspaces (but not for infinitely many).Show also that it is not true in general if C1, or C2 is not closed.

Exercise 2.14. Let f1, f2 be two continuous functions X → Y , where Y is a Haus-dorff space, A be a dense subset in X such that f1(a) = f2(a) for every a ∈ A. Showthat f1(x) = f2(x) for every x ∈ X.

Exercise 2.15. (H. Furstenberg’s proof of the infinitude of primes)Define a topology τF on the integers Z, by declaring a subset U ⊆ Z to be an open

set if and only if it is either the empty set, ∅ or it is a union of arithmetic sequencesS(a, b) := aZ + b (for a 6= 0). In other words, U is open if and only if every x ∈ Uadmits some non-zero integer a such that S(a, x) ⊆ U . Prove:

(1) (Z, τF ) is a topological space.(2) Finite nonempty subset of this topological space is not open.(3) S(a, b) are clopen.(4) * Using (1,2,3) conclude that there are infinitely many prime numbers.

Hint: use the equality Z \ {−1, 1} = ∪{S(p, 0) : p is prime}.

Exercise 2.16. Prove or disprove:

(1) Let A ⊂ B ⊂ X. If A is dense in B and B is dense in X then A is dense inX.

(2) Finite intersection ∩ni=1Ai of dense subsets Ai in X is dense in X.(3) Finite intersection ∩ni=1Oi of open dense subsets Oi in X is dense in X.(4) Countable intersection ∩i∈NOi of open dense subsets Oi in X is dense in X.

Exercise 2.17. Let ∂(A) = A \ int(A) be the boundary of a subset A ⊂ X in atopological space X. Prove the following formula

(∂(A))c = int(Ac) ∪ int(A).

Exercise 2.18. Let A be a subset of a topological space X. Define its indicatorfunction

ξA : X → {0, 1}, ξ(x) =

{1 for x ∈ A0 for x /∈ A

Show that the points of discontinuity of this function is just the boundary ∂(A) of A.Conclude that ξA is continuous if and only if A is clopen.

Exercise 2.19. (Sorgenfrey line)Let X = R. Define τs as follows:

τs := {O ⊂ R : x ∈ O ⇒ ∃ε > 0 [x, x+ ε) ⊆ O]}Show that

(1) (R, τs) is a Hausdorff topological space (notation: Rs).(2) {[a, b) : a, b ∈ R} is a basis of τs and dimRs = 0.(3) top(d) τs, where top(d) is the usual topology on R.(4) Rs is separable.(5) Rs /∈ B2 (i.e., there is no countable topological basis).(6) Rs is not metrizable.

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Exercise 2.20.

(1) Characterize all intervals of R up to the homeomorphisms.(2) For which intervals X, Y of R there exists a continuous onto function

f : X → Y ?

Exercise 2.21.

(1) Characterize all digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in sans serif font up to the home-omorphisms.

(2) Characterize all capital letters of the English alphabet in sans serif font up tothe homeomorphisms.

Remark: sans serif font is a font without additional ”calligraphic tails”. Seehttp: // bueler. github. io/ M404F16/ letters. pdf

Exercise 2.22.

(1) Show that every nonempty open subset O of Rn is locally connected (but notnecessarily connected).

(2) * Show that there exists a subspace of R2 which is (pathwise) connected butnot locally connected.

Remark: X is locally connected means that for every point a ∈ X and every neigh-borhood U ∈ N(a) there exists a neighborhood V ∈ N(a) such that V is connected.That is, if there exists a local basis at every a ∈ X which contains only connectedmembers.

3. Topological products

Exercise 3.1. Let X be a topological space. Prove that X is Hausdorff if and only ifthe diagonal is closed in X ×X.

Exercise 3.2. Let X =∏

i∈I Xi be a topological product. Prove that:

(1) X ∈ T1 if and only if Xi ∈ T1, ∀i ∈ I.(2) X ∈ T2 if and only if Xi ∈ T2, ∀i ∈ I.(3) If Ai is closed in Xi then

∏i∈I Ai is closed in X.

Exercise 3.3. * Let f : X → Y be a continuous function. Prove:

(1) The graph of this function Gr(f) is homeomorphic to X.(2) If, in addition, Y is Hausdorff then Gr(f) is closed in X × Y .

Exercise 3.4. * Show that the topological product Rs × Rs (called Sorgenfrey plane)is a separable space and contains a discrete subspace of cardinality |R|. Conclude thatthe separability is not a heredity property in Hausdorff spaces.

Exercise 3.5. Let α be a subbase of a topological space (Y, τ) and f : X → Y be amap. Prove that f is continuous if and only if f−1(O) is open in X for every O ∈ α.

Exercise 3.6. Let fi : Y → Xi be continuous for every i ∈ I. Prove that the naturallydefined ”diagonal map”

f : Y →∏i∈I

Xi, f(y) = (fi(y))i∈I

is continuous.Hint use 3.5.

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Exercise 3.7. Let fi : Yi → Xi be continuous for every i ∈ I. Prove that the naturallydefined ”product map”

f :∏i∈I

Yi →∏i∈I

Xi, f((yi)i∈I) = (fi(xi)i∈I)

is continuous. If every fi is homeomorphism then f is homeomorphism.

Exercise 3.8.

(1) Give an example of a topological product X1 × X2 such that the projectionp1 : X1 ×X2 → X1 is not a closed map.

(2) Let X =∏

i∈I Xi be a topological product. Prove that every projection pi :X → Xi is an open map.

Exercise 3.9. Let X = X1 ×X2 be a topological product and a = (a1, a2) be a givenpoint in X. Prove that

i1 : X1 → X, x 7→ (x, a2)

andi2 : X1 → X, y 7→ (a1, y)

are topological embeddings.Remark: The similar result is true for infinite products.

Exercise 3.10. Prove that X = X1 ×X2 is connected if and only if X1, X2 both areconnected.

Hint: One may use 3.9 .

Exercise 3.11. Prove that:

(1) R \ {0} ' {1, 2} × R.(2) R2 \ {0} ' S1 × R.

4. Compactness

Exercise 4.1. Prove:

(1) Every finite topological space is compact.(2) a discrete space X is compact if and only if X is finite.(3) (cofinite topology) Every set with the cofinite topology is compact.(4) (cocountable topology1) In R with the cocountable topology compact subsets are

exactly finite subsets.(5) * There exists a T1-space X which is not T2 and every compact subset of X is

closed in X.(6) Z with the p-adic topology is not compact.(7) Cantor set is compact.(8) Cantor cube (see 1.10) is compact.(9) The set On := {A ∈ GLn(R) : A−1 = At} of all orthogonal n×n real matrices

is a compact subset in the metric space Matn×n(R) of all n× n real matrices.

Exercise 4.2. Let C be the Cantor set (see 1.16 and 1.10). Show that:

(1) C ' {0, 1}N (where {0, 1}N is the topological product).

1A subset is said to be cocountable if its complement is countable

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(2) C ' Cn ' CN.(3) * C is a homogeneous topological space.

Exercise 4.3.

(1) Prove that Rn is locally compact for every n ∈ N.(2) Prove that RN is not locally compact.

Remark: X is locally compact means that for every point a ∈ X and every neighbor-hood U ∈ N(a) there exists a neighborhood V ∈ N(a) such that V is compact.

Exercise 4.4. Prove:

(1) In the Banach space l∞ the ball B[v, r] is closed bounded but not compact forany vector v ∈ l∞ and any r > 0.

(2) * l∞ is not locally compact.

Exercise 4.5. (1-point compactification)Let (X, τ) be a locally compact Hausdorff space which is not compact. Let p /∈ X

and X∗ := X ∪ {p}. Define

τ ∗ := τ ∪ {X∗ \K : K is compact in X}.(1) Prove that (X∗, τ ∗) is a compact Hausdorff topological space and the natural

injection i : X → X∗ is a dense topological embedding.(2) Show that (up to homeomorphisms):

a) X∗ = {0} ∪ { 1n

: n ∈ N} for X = N (or, every countable infinite X);b) X∗ = S1 for X = R;c) X∗ = Sn for X = Rn.

(3) Conclude that every discrete space (X, τdiscr) is a subspace of a compact Haus-dorff space.

Exercise 4.6. * Prove that in every compact metric space there are only countablymany clopen subsets.

Exercise 4.7. Prove that every locally compact Hausdorff space is completely regular(T3 1

2).

Exercise 4.8. Let K be a compact subset of a metric space (X, d). Prove that thereexist k1, k2 ∈ K such that diam(K) = d(k1, k2).

Exercise 4.9. Let X be a compact space. Prove that the following conditions areequivalent:

(1) There exists a topological embedding i : X → [0, 1]n into the n-dimensionalcube [0, 1]n for some n ∈ N.

(2) There exists a family of continuous functions fi : X → [0, 1], i ∈ {1, · · · , n}which separates the points(meaning that for every x 6= y in X there exists i such that fi(x) 6= fi(y)).

Exercise 4.10. Prove that the following conditions are equivalent:

(1) X ∈ T3.5.(2) X is embedded topologically into a Tychonoff cube [0, 1]S (for some set S).(3) X admits a compactification.

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Exercise 4.11.

(1) Let X be compact and metrizable. Show:X ∈ B2, X ∈ Sep, card(X) ≤ card(R).

(2) Give an example of a topological space X which is compact Hausdorff but notmetrizable.

5. Quotients

Exercise 5.1. Let f : X → Y be a continuous onto function.

(1) Give an example of f which is not a quotient.(2) If f is open then f is a quotient.(3) If f is closed then f is a quotient.(4) Define the induced equivalence relation on X: a ∼ b if and only if f(a) = f(b).

Then we have the following commutative (f = f ◦ α) diagram:

X

α ��@@@

@@@@

@f // Y

X

f

OO

where α(x) = [x] (the equivalence class of x) and f([x]) = f(x) is the induced

map. Define on X the factor-topology. Now prove that f is a continuous 1-1

onto map. Moreover f is quotient if and only if f is a homeomorphism.

Exercise 5.2. Let f : X → Y be an onto continuous map, Y is Hausdorff and X iscompact. Prove that f is a quotient.

Exercise 5.3. (”Gluing points”)

(1) On the interval X = [0, 1] consider the equivalence relation induced by the

identification: 0 ∼ 1. In the quotient set X define the quotient topology. Show

that X is homeomorphic to the circle S1.(2) On the square X = [0, 1] × [0, 1] consider the equivalence relation induced by

the identifications:

(0, t) ∼ (1, t) ∀ 0 ≤ t ≤ 1.

Show that the quotient space X is homeomorphic to the cylinder S1 × [0, 1].(3) * On the square X = [0, 1]× [0, 1] consider the equivalence relation induced by

the identifications:

(0, t) ∼ (1, t), (t, 0) ∼ (t, 1) ∀ 0 ≤ t ≤ 1.

Show that the quotient space X is homeomorphic to the torus S1 × S1.

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Exercise 5.4. Prove that if the composition f2 ◦ f1 : X → Z of continuous mapsf1 : X → Y and f2 : Y → Z is a quotient map, then f2 : Y → Z is a quotient map.

X

f1 @@@

@@@@

@// Z

Y

f2

OO

As a corollary conclude: let f : Y → Z be a continuous onto map. If X is asubspace of Y and the restriction fX : X → Z is an onto quotient map then f is aquotient.

Exercise 5.5.

(1) Let Y be a subset of a topological space X and f : X → Y be a continuousonto retraction (that is, f(y) = y for every y ∈ Y ). Then f : X → Y is aquotient.

(2) (”complex projective line”) In the unit circle T := {z ∈ C| ||z|| = 1} definethe equivalence relation as follows

v ∼ −v ∀v ∈ T

Show that the corresponding quotient space T/ ∼ is homeomorphic to thecircle.

Exercise 5.6. Let X := {(x, y) ∈ R2| 0 ≤ x or y = 0} show that the projectionp1 : X → R, (x, y) 7→ x is a quotient map which is not open and not closed.

6. Hints and solutions

Subsection 1

1.1 Consider X = [1, 2] ∪ {3} (as a metric subspace of R).Define a1 = 1, r1 = 2.5, a2 = 2, r2 = 2.5. Then

B(a1, r1) = [1, 2] ( B(a2, r2) = [1, 2] ∪ {3} but r2 < r1.

1.2 Let x ∈ B(b, r). This means that d(b, x) < r. Now by the axiom (m3) we get

d(a, x) ≤ d(a, b) + d(b, x) < R− r + r = R.

Therefore, x ∈ B(a,R).

1.3B(a, r) =

{{a} for r ≤ 1

X for r > 1B[a, r] =

{{a} for r < 1

X for r ≥ 1S(a, r) =

{∅ for r 6= 1

X \ {a} for r = 1

1.4 Possible answers:

(1) X = N \ {4}, Y = N(2) X = [1,∞), Y = (1,∞)

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1.5 Observe that Tz(B(θ, r) = B(z, r).

1.6 See Lecture Notes.

1.7 (1) Bmax[θ, 3]={all f ∈ C[0, 1] such that the graph belongs to the rectanglebounded by the lines x = 0, x = 1, y = 3, y = −3}B1[θ, 3]={all f ∈ C[0, 1] such that its graph (together with the lines x = 0, x = 1)

bounds an area ≤ 3}.(2) Observe that ||f ||1 ≤ ||f ||max for every f ∈ C[0, 1]. Hence, Bmax[h, r] ⊂ B1[h, r]

for every h ∈ C[0, 1].

1.8 Straightforward. For (5) observe that for every pseudometric d we have

∩{B(a, r)| r > 0} = {x ∈ X : d(a, x) = 0}.

1.9(3) dp(0, p

n) = 1pn→ 0.

(4) We have to show that B(0, r) is a subgroup of Z. It is equivalent to show thatx− y ∈ B(0, r) for every x, y ∈ B(0, r). Observe that

d(0, x− y) = d(0 + y, x− y + y) = d(y, x) ≤ max{d(y, 0), d(0, x)} < r.

Open ball Br(a) is also closed. One proof follows from the equality Br(a) = Bpm [a]for every 1

pm< r ≤ 1

pm−1 .

Another proof (for every ultrametric space): for every given r > 0 the relationx ∼ y if and only if d(x, y) < r is an equivalence relation. Each equivalence class[a] of a is the open ball Br(a). In particular, any union of such subsets are open.Now observe that the complement Br(a)c is open (being a union of other equivalenceclasses).

(5) Let b ∈ B(a, r) (so, d(a, b) < r). We have to show that B(a, r) = B(b, r). Firstwe show that B(b, r) ⊆ B(a, r). Indeed, for every x ∈ B(b, r) we have d(b, x) < r.Then we get

d(a, x) ≤ max{d(a, b), d(b, x)} < r.

This means that x ∈ B(a, r). Hence, B(b, r) ⊆ B(a, r). Similarly you can checkB(a, r) ⊆ B(b, r).

Remark: This proof remains true for every ultrametric space.

1.10 Some arguments are similar to the case of 1.9.

1.11(1) Consider for example, the following sequence e1, e2, · · · in l2. Then this sequence

is coordinate-wise converging to the zero element θ := (0, 0, · · · ) of l2 but {en}n∈N isnot converging with respect to the metric not being even Cauchy. Indeed, ||ei−ej|| =√

2, ∀i 6= j.(2) (N, d∆) ↪→ l2, n 7→ 1√

2en}.

(3) Consider (for example) the following linear shift

f : l2 → l2, (x1, x2, · · · ) 7→ (0, x1, x2, · · · )

14

(4) On X := {A,B,C,D} define

d(A,B) = d(B,A) = d(B,C) = d(C,B) = d(C,D) = d(D,C) = d(A,D) = d(D,A) = 1

d(A,C) = d(C,A) = d(B,D) = d(D,B) = 2

Then (X, d) is a metric space. It cannot be embedded isometrically into any Eu-clidean space Rn. Indeed, otherwise B should be the midpoint of the interval AC.Similarly, D should be the midpoint of the same interval AC. This implies that Band D are the same points, a contradiction.

In fact, (X, d) cannot be embedded isometrically into l2. Indeed, otherwise it canbe embedded into a 4-dimensional linear subspace of l2. On the other hand, everysuch subspace of l2 is isometric to the Eucledean space R4.

1.12 A = f−1(0), where f : X → R, x 7→ d(x,A). Clearly, {0} = ∩n∈N(−∞, 1n).

Now observe that

A = ∩nOn, On := f−1(−∞, 1

n)

Equivalent proof:

A = ∩nOn, On := B(A,1

n) = {x ∈ X : d(A, x) <

1

n}

1.13For example, the sequence an := 1+p+p2 + · · ·+pn is a Cauchy sequence in (Z, dp)

which is not converging. This sequence is Cauchy because

d(an, an+i) ≤1

pn→ 0

for every i ∈ N.This sequence is not converging in (Z, dp). Indeed, assuming the contrary let x ∈ Z

such that x = limn an (with respect to the p-adic metric). Then for every given k ∈ Nthere exists sufficiently large n ∈ N such that

d(x, am) <1

pk∀m ≥ n.

Since d(x, am) < 1pk

we have am − x ≡ 0 mod pk. Therefore,

am − x = 1 + p+ p2 + · · ·+ pk−1 + (pk + · · ·+ pm)− x ≡ 0.

Clearly, pk|(pk + · · ·+ pm). So,

(1 + p+ p2 + · · · pk−1 − x) ≡ 0 mod pk

This means that pk|(1 + p+ p2 + · · · pk−1 − x).On the other hand, since x ∈ Z is a given constant we can suppose that for

sufficiently big k ∈ N we have

0 < 1 + p+ p2 + · · · pk−1 − x < pk

Then pk cannot divide 1 + p+ p2 + · · · pk−1 − x.We get a contradiction.

15

About Cantor cube (from Exercise 1.10). Later we show that it is compact (being atopological copy of the Cantor set). So, being a compact metric space it is complete.

1.14 See Lecture Notes.

1.15 id : (C[−1, 1], || · ||1)→ (C[−1, 1], || · ||max).

1.18(1) Let {yn}n∈N be a Cauchy sequence in (Y, dY ). We have to show that {yn}n∈N

converges in (Y, dY ). Then {yn}n∈N is a Cauchy sequence in (X, d). Since (X, d) iscomplete there exists a limit lim

n→∞yn = x ∈ X. Since Y is closed then it is sequentially

closed. So, x ∈ Y . This implies that {yn}n∈N converges in (Y, dY ).(2) By a characterization of closed subsets in metric spaces it is equivalent to show

that Y is sequentially closed in X. Let {yn}n∈N be a converging sequence in X. Thenthere exists lim

n→∞yn = x ∈ X. This sequence is Cauchy in (X, d). Since dY is a

restriction of d we obtain that {yn}n∈N is Cauchy in (Y, dY ), too. Since it is completewe obtain that there exists lim

n→∞yn = y ∈ Y in Y . Then necessarily x = y.

Section 2

2.1 Check for example, the axiom (t2). Let O1, O2 ∈ top(d). For every x ∈ O1 ∩O2

(if it exists) one may choose ε1, ε2 such that B(x, ε1) ⊆ O1 and B(x, ε2) ⊆ O2. ThenB(x, ε) ⊆ O1 ∩O2 for ε := min{ε1, ε2}. Hence, O1 ∩O2 ∈ top(d).

2.2 Counterexample for unions:

X := {a, b, c}, τ1 := {∅, {a}, {a, b, c}}, τ2 := {∅, {b}, {a, b, c}}

Then τ1, τ2 are topologies on X but not τ1 ∪ τ2.

2.3(1)⇒ (2). Let us show that {a} is closed. By the axiom T1 for every x 6= a one may

choose an open nbd Ox ∈ N(x) such that a /∈ Ox. Then ∪{Ox : x 6= a} = X \ {a} isopen (by t3, the union of open subsets). Therefore, its complement, {a} is closed.

(2) ⇒ (3). Every finite set is an union of finitely many singletons.(3)⇒ (4). Every finite set F is closed. Hence, its complement X \F is open. This

implies that all co-finite sets are open in (X, τ). Therefore, τcofin ⊂ τ .(4)⇒ (1). Let a 6= b. Since the cofinite subsets are open. In particular, X \{b}, X \{b} ∈ τ . Define U := X \ {b}, V := X \ {b}. Then b /∈ U ∈ N(a) and a /∈ V ∈ N(b).

2.4(1) It is straightforward to check the axioms (t1), (t2), (t3). So, (Z, τ≤) ∈ TOP.Since Z is linearly ordered, for every distinct m,n ∈ Z we have m < n or n < m. Let

m < n (the second case is similar). Then m ∈ Om and n /∈ Om. Hence, (Z, τ≤) ∈ T0.

16

For every pair U, V of nonempty open subsets U∩V contains one of the Ok for somek ∈ Z. So, U ∩ V is nonempty. This implies that there is no topological partition of(Z, τ≤). This means that (Z, τ≤) ∈ Conn.

(2) Indeed, if yes, then there exists a (pseudo)metric d on Z such that top(d) = τ≤.We have two cases (a) d(0, 1) = 0; or b) d(0, 1) > 0. If (a) then every open setcontaining 0 contains also 1 and vice versa. Contradiction to T0 property establishedin (1).

If (b) then B(0, r)∩B(1, r) = ∅ for every r < 12d(0, 1). On the other hand by defi-

nition of τ≤ every open set containing 1 contains also 0. So we get a contradiction.

(3) Assuming the contrary let f : (Z, τ≤) → R be a continuous function such thatf(m) 6= f(n) for some m < n. Take disjoint open neighborhoods U, V of f(m) andf(n) in R, respectively. Then f−1(U) and f−1(V ) are disjoint open neighborhoods of

m and n in (Z, τ≤). On the other hand, every open set containing n contains also m.

This contradiction finishes the proof.

2.5 τcofin := {∅} ∪ {F c : F is a finite subset of X}.Straightforward to prove that (X, τcofin) ∈ TOP. Every singleton {a} is closed

because {a}c is open being cofinite.(X, τcofin) ∈ Conn for every infinite set X. Indeed, let U, V be open subsets such

that U 6= X, V 6= X,U 6= ∅, V 6= ∅. Then U and also U c both are cofinite. ThenU c, U both are finite but this implies that X is finite, a contradiction !

As before, intersection of two cofinite subsets is not empty (otherwise, X is finite).This implies that (X, τ≤) is not Hausdorff for infinite X.

Finally, (Z, τcofin) is not (pseudo)metrizable. Indeed, since the intersection of twocofinite subsets is not empty we obtain that (Z, τcofin) is not Hausdorff. Therefore,(Z, τcofin) is not metrizable. Assuming that it is pseudometrizable with some d wenecessarily have d(a, b) = 0 for some distinct a, b ∈ Z. Then {a} is not closed (becauseits closure contains also b). On the other hand, (X, τcofin) ∈ T1 (every singleton isclosed). This contradiction finishes the proof.

2.6 Check the axioms for τf,Y :(t1) Observe that f−1(X) = Y , f−1(∅) = ∅. Hence Y, ∅ ∈ τf,Y .(t2) Use f−1(O1) ∩ f−1(O2) = f−1(O1 ∩O2) (and the fact that τ is a topology)(t3) Use ∪i∈If−1(Oi) = f−1(∪i∈IOi).

2.7 See Lecture Notes.

2.8 See the homeworks.

2.10 Take (X, τ) ∈ TOP such that (X, τ) /∈ T0. Now consider id : (X, τdiscr) →(X, τ). This map is continuous (2.8) and (X, τdiscr) ∈ T0. Similarly, for other cases.

2.11 Which topological properties from the following list are hereditary:

(1) connectednessNOT

(2) Ti, i ∈ {0, 1, 2, 3}

17

YES(3) compactness

NOT(4) B2

YES(5) B1

YES(6) Metrizability.

YES

2.12 For (1), (2), (3) see Lecture Notes.(4) Let A,B are disjoint closed subsets. If one of them, say A, contains p then

p /∈ B. Therefore, B necessarily is a subset of R. Moreover, B is countable becausep ∈ Bc ∈ τ and Bc is cocountable. Now observe that U := X \ B ∈ N(A) andV := B ∈ N(B) are disjoint (open) neighborhoods of A and B.

If A ⊂ R, B ⊂ R then choose U := A, V := B.

2.15 (1) Let’s check (t2). If x ∈ U1 ∩ U2. Then S(a1, x) ⊆ U1, S(a2, x) ⊆ U . ThenS(a1a2, x) ⊆ U1 ∩ U2.

(2) Every S(a, b) (with a 6= 0) is infinite.(3) The basis sets S(a, b) are both open and closed. They are open by definition,

and we can write S(a, b) as the complement of an open set as follows:

S(a, b) = Z \a−1⋃j=1

S(a, b+ j).

(4) Assume in contrary that the number of primes is finite. Then the finite union∪{S(p, 0) : p is prime} of closed sets (using (3)) is closed. Now using Z \ {−1, 1} =∪{S(p, 0) : p is prime}. we conclude that Z\{−1, 1} is closed. Then {−1, 1} is open,a contradiction (by (2)).

2.16(1) YES. The closure clX(A) of A in X contains B (because, clX(A)) ⊇ clB(A) =

B). Clearly, in general clX(clX(A)) = clX(A). On the other hand, B is dense in X,so clX(B) = X. Summing up we get clX(A) ⊇ clX(B) = X.

(2) NOT. The disjoint subsets A = Q and B = Qc are both dense in R.(3) YES. Let U, V are open dense subsets in X. Our aim is to show that cl(U∩V ) =

X. It is equivalent to prove that O ∩ (U ∩ V ) 6= ∅ for every open nonempty subset Oin X. Since cl(U) = X we have O ∩ U 6= ∅. The intersection O ∩ U is a nonemptyopen set. Since cl(V ) = X we have (O ∩ U) ∩ V 6= ∅.

(4) NOT. Consider the metric space Q. For every element q ∈ Q its complementOq := Q\{q} inQ is open inQ. However, the (countable !) intersection ∩{Oq : q ∈ Q}is empty.

2.17

(∂(A))c = (A \ int(A))c = (A ∩ (X \ int(A)))c = (A)c ∪ int(A) = int(Ac) ∪ int(A)

18

2.18 It is convenient to use the following formula from 2.17

(∂(A))c == int(Ac) ∪ int(A)

which implies that we have the following ”disjoint union”

X = int(Ac) ∪ ∂(A) ∪ int(A).

Let x ∈ int(A). Then ξA(x) = 1. {1} is an open nbd of 1 in the two element spaceY := {0, 1}. The preimage ξ−1(1) = A is a neighborhood for every x ∈ int(A) (notethat always, A ∈ N(x) ∀x ∈ int(A)). This proves (by local definition of continuitypoints) the continuity of ξA : X → {0, 1} at points x ∈ int(A).

Completely similar proof shows the continuity of ξA : X → {0, 1} at points x ∈int(Ac).

Assume now that x ∈ ∂(A). We are going to show that ξA is discontinuous at x.We have two cases: ξA(x) = 1 or ξA(x) = 0. In the first case: ξA(x) = 1, observethat for every nbd V ∈ N(x) its image is not a subset of U := {1} ∈ N(1). Thatis, ξA(V ) * {1}. Indeed, as we know ∂(A) = A ∩ Ac. So, V ∩ Ac is not empty. Fory ∈ V ∩ Ac we have ξA(y) = 0. Therefore, ξA(V ) * {1}. The second case ξA(x) = 0is similar.

2.19 See Lecture Notes.

2.201. There are three classes up to homeomorphisms(enough to use homeomorphisms coming from elementary functions ...)R ' (a, b) ' (a,∞) ' (−∞, b)[a, b) ' (c, d] ' [a,∞) ' (−∞, b][a, b]2. So by (1) the question on continuous images it is enough compare only three

representatives: R, [a, b), [a, b].R→ (a, b]→ [a, b] the corresponding continuous onto functions are quite easy.(a, b]→ R this case is not so easy. Take for example f : (0, 1]→ R, f(x) = 1

xsin( 1

x)

2.21 (1) There are four equivalence classes up the homeomorphisms:{1,2,3,5,7} are homeomorphic to the closed unit segment [0, 1](warning: take into account that 1 has no ”bottom tail” in sans serif font){4,6,9}{8}{0}

(2) There are five classes up the homeomorphisms:C I J L M N S U V W ZD OE F G T YH KA R

Hint (one of the typical cases): T and K are not homeomorphic. K has a point k0

such that K \ {k0} has four Conn-components while T has no such a point. Another

19

approach: K has four end-points (= not separating points) while T has three end-points.

Remark: For detailed explanations see the following file of Rafael Lopez:https://arxiv.org/pdf/1410.3364.pdf

2.22 (1) Hint: recall that Rn is homeomorphic to any open ball B(v, ε), v ∈ Rn.With more details: Let O be a nonempty open subset of Rn and v ∈ O. Every open

neighbourhood of v in the space O looks like U ∩O where U is an open neighborhoodof v in Rn. Hence, U ∩ O is open in Rn (and in O). Now choose ε > 0 small enoughsuch that B(v, ε) ⊂ U ∩O. Clearly, B(v, ε) is connected being homeomorphic to Rn.So, O contains arbitrarily small connected neighborhoods, as desired.

Note that O is not necessarily connected. Indeed, take for example, O = (0, 1) ∪(3, 5) in R.

(2) Define X (as a subspace of R2) union of countably many closed intervals asfollows:

X = I0 ∪ I1 ∪ · · · In ∪ · · · , n ∈ Nwhere each In is the standard linear interval in R2 between the vectors (1, 0) and (0, 1

n)

and I0 is the interval between (1, 0) and (0, 0). These (connected, of course) intervalshave a unique common point (1, 0). Clearly, X is connected (and even pathwiseconnected). However, X is not locally connected at (0, 0). Take a neighborhoodU of (0, 0) in X such that (1, 0) /∈ U . Then every neighborhood V of (0, 0) inX such that V ⊆ U is not connected. Indeed, observe that V meets almost allintervals In (use that the sequence (0, 1

n) converges to (0, 0)). So, there exists n0 ∈ N

such that V ∩ In 6= ∅, ∀n ≥ n0. However In is not contained in V for every n(because (1, 0) /∈ V ). It follows that V ∩ In0 is a clopen subset in the space V . Since∅ 6= V ∩ In0 6= V we conclude that V is not connected.

2.14 Assuming the contrary let f1(z) 6= f2(z) for some z ∈ X. Since Y is Hausdorffwe may choose disjoint open neighborhoods U ∈ N(f1(z)) and V ∈ N(f2(z)). Thecontinuity of f guarantees that O1 := f−1

1 (U) ∈ N(z), O2 := f−12 (V ) ∈ N(z) are

also open neighborhoods. The subset A is dense in X. Therefore, there exists a ∈A∩O1 ∩O2. Then f1(a) 6= f2(a) (because f1(a) ∈ U , f2(a) ∈ V ). This contradictioncompletes the proof.

Section 33.1 See the homeworks.

3.2 (1) is a particular case of (3) (because T1 is equivalent to the closedness of thesingletons). No problem to prove this also directly similar to (2).

(2) For every distinct pair of points u, v ∈ X there exists i ∈ I such that ui 6= vi.Since Xi is Hausdorff one may choose ui ∈ Ui ∈ τi and vi ∈ Vi ∈ τi such thatUi ∩ Vi 6= ∅. Then p−1

i (Ui) and p−1i (Vi) are disjoint neighborhoods of u and v in X.

(3) Homeworks 9.6. In fact, a stronger result is true:

cl(∏i∈I

Ai) =∏i∈I

cl(Ai)

20

for every family Ai ⊆ Xi.

3.3 (1) It is enough to show that

h : Gr(f)→ X, h(x, f(x)) := x = p1(x, f(x))

(the restriction of the projection p1 : X × Y → X on Gr(f)) is a homeomorphism.This function is 1-1. Indeed, (x1, f(x1)) = (x2, f(x2))⇔ x1 = x2.

This function is onto. Indeed, clearly, p1(x, f(x)) = x for every x ∈ X.This function is continuous (as a restriction of the continuous projection).We need only to show that the inverse function

h−1 : X → Gr(f), x 7→ (x, f(x))

is continuous. It is equivalent to show that

h−1 : X → X × Y, x 7→ (x, f(x))

is continuous. Let’s use the local criterion of continuity. Namely, we have to showthat h−1 is continuous at every given x0 ∈ X. Consider any neighborhood O of(x0, f(x0)) ∈ X × Y in the product X × Y . We need to prove that there existsU0 ∈ N(x0) in X such that h−1(U0) ⊂ O. Since the ”open rectangles” is a base of theproduct topology we can suppose (without restriction of generality) that O = V ×W(where V ∈ N(x0) and W ∈ N(f(x0)). By the continuity of f there exists U ∈ N(x0)in X such that f(U) ⊂ W . Define U0 := U ∩ V . Then U0 ∈ N(x0) and f(U0) ⊆f(U) ⊆ W . Hence, h−1(U0) ⊂ V ×W .

(2) Assume, in addition, that Y is Hausdorff. We have to show that Gr(f) is closedin X × Y . It is equivalent to check that the complement is open. Take any point inthe complement (x, y) /∈ Gr(f). Then y 6= f(x). Since Y is Hausdorff we may choosedisjoint open neighborhoods

U ∈ N(y), V ∈ N(f(x)), U ∩ V = ∅.

By the continuity of f there exists an open neighborhood O ∈ N(x) in X such thatf(O) ⊂ V . Observe now that O×U is an open neighborhood of (x, y) in X ×Y suchthat (O × U) ∩Gr(f) = ∅. This proves that the complement of Gr(f) is open.

3.4 Q×Q is dense in the ”Sorgenfrey plane” Rs×Rs. Hence, Rs×Rs is separable.The topological subspace

X := {(x,−x) : x ∈ Rs} ⊂ Rs × RsIs discrete, as a topological subspace. Indeed, for every x0 ∈ Rs consider O :=[x0, x0 + 1)× [−x0,−x0 + 1]. Then O is an open neighborhood of x0,−x0) in Rs×Rs.On the other hand, the intersection O ∩X = {x0} is just the singleton {x0}. Hence,x0 is an isolated point in X0.

Clearly, X, being discrete and uncountable, is not separable (because discrete spaceis separable if and only if it is countable).

3.5 The preimage f−1 preserves intersections and unions ...

3.6 The ”elementary open boxes” is the standard subbase of the product topology.By 3.5 it is enough to show that the preimage f−1(p−1

i (Oi)) is open in Y for every

21

Oi ∈ τi. Now observe that f−1(p−1i (Oi)) = f−1

i (Oi) (because pi ◦ f = fi) and applythe continuity of fi : Y → Xi.

3.7 As in 3.6 use 3.5.

3.8 (1) p1 : R×R→ R is not a closed map. Indeed, A := {(x, 1x) : x > 0} is closed

in R× R but p1(A) = {x ∈ R : x > 0} is not closed in R.(2) Let X =

∏i∈I Xi be a topological product.

We have to prove that every projection pi0 : X → Xi0 is an open map. Since everymap preserves the union (and the union of open sets is open) it is enough to showthat the image of every basic open box is open. Every basic open box O in X lookslike

O =∏i∈I

Oi

where there exists a finite J ⊆ I such that Oi = Xi for every i /∈ J . Now it is clearthat pi0(O) = pi0(

∏i∈I Oi) = Oi0 .

3.9 Straightforward using the definitions (product and subspace ...)

3.10 Let u = (u1, u2), v = (v1, v2) ∈ X1 × X2). It is enough to show that theybelong to the same component. Equivalently, that there exists a connected subsetA ⊂ X1 ×X2) such that u, v ∈ A. Now define

A := (X1 × {u2}) ∪ ({v1} ×X2).

Now observe that each of these two subsets are connected by 3.9 and they have anontrivial intersection. Namely the point (v1, u2).

3.11 (1) R \ {0} ' {1, 2} × R.Indeed, the function

h : R \ {0} → {1, 2} × R, h(x) =

{(1, ln(x)) for x > 0

(2, ln(−x) for x < 0,

is a desired homeomorphism.(2) R2 \ {0} ' S1 × R.Hint:

C∗ → S1 × (0,∞), rcis(α) 7→ (cis(α), r)

is a homeomorphism, where C∗ is the space of nonzero complex numbers.

Section 4

4.1 (2) Hint: the covering {{x} : x ∈ X} of X is open if and only if X is discrete.(3) Hint: every cofinite subset already covers almost all elements.(4) If F ⊆ X is finite then F is always compact for any topology on X.

22

If A ⊆ X is not finite then it contains a sequence B := {bn : n ∈ N} with distinctelements. Consider the following family of subsets in X:

α := {On : n ∈ N}, On := {b1, b2, · · · , bn} ∪ (A \B).

Since X carries the cocountable topology and every Ocn = {bn+1, bn+2, · · · } is count-

able, then α is an open family in X which covers A. No finite subfamily of α cancover A. This implies that A is not a compact subset in the subspace topology.

(5) Take X := (R, τcocount) the reals in the cocountable topology. Then X isT1 because every singleton is closed (in fact, every countable subset is closed). Atthe same time X is not T2 (because any two cocountable subsets necessarily meet).Finally, observe that the compact subsets of X are exactly all finite subsets as itfollows from (4).

(6) Z with the p-adic metric topology is not compact because (Z, dp) is not complete.(7) Cantor set is a bounded closed subset of [0, 1] by the construction.(8) It is enough to show that the Cantor cube (see 1.10) is homeomorphic to the

Cantor set. Observe that Cantor set C is exactly the set of points in [0, 1], where inthe ternary representation we allow to use only 0 or 2 (but not 1). That is,

C = {∞∑k=1

ak3k, where ak ∈ {0, 2}}.

The function

f : {0, 2}N → C, (a1, a2, · · · ) 7→∞∑k=1

ak3k

is continuous onto 1-1. By Tychonoff theorem the topological product {0, 2}N iscompact. So, f is a homeomorphism. Now observe that {0, 2}N is naturally homeo-morphic to the product {0, 1}N which in turn is homeomorphic to the Cantor cubefrom 1.10.

(9) Hint: Observe that the metric space of all n×n real matrices under the natural

matric is isometric to the Euclidean space Rn2and use Heine-Borel theorem.

4.2 (3) * Hint: look at {0, 1}N as a group under the addition of binary sequencesmodulo 2. Now show that every translation Ta : {0, 1}N → {0, 1}N is a homeomor-phism.

4.3 (1) Hint: By Heine-Borel theorem every ”n-dimensional cube” [a, b]n is compact.Every point of Rn has a neighborhood which is homeomorphic to [a, b]n.

(2) In order to show that RN is not locally compact note that every neighborhoodU of any point in RN contains a basic box

∏k∈NOk, where Ok = R for almost all k.

So, some projection of U is not compact. This means that U itself is not compact.

4.4 (1) Hint: Consider D := {en : n ∈ N} ⊂ l∞. Then D is discrete as a subspace.Since D is infinite we get that D is not compact (4.1.2). On the other hand, D is abounded closed subset of l∞. Moreover, D is a closed subset of B[θ, 1] (where θ is thezero element). This implies that B[θ, 1] is not compact.

(2) Hint: Similar to (1) show first that B[θ, r] for every r > 0 and observe thatevery neighborhood of θ contains B[θ, ε] for some ε > 0.

23

4.5 (1) Homeworks.(2) (a) X∗ ' {0} ∪ { 1

n: n ∈ N} for X = N;

Hint: consider the function

f : N∗ → {0} ∪ { 1

n: n ∈ N}, f(p) = 0, f(n) =

1

nObserve that f is continuous, onto and 1-1. In fact, a homeomorphism because it isa function from a compact space into a Hausdorff...

(b) and (c) Use the stereographic projection homeomorphism Sn \ {p} ' Rn. Ob-serve that the 1-point compactification is unique (up to the homeomorphisms).

At least, for the case n = 1 one may give an alternative proof: X∗ ' S1 for X = R.

Since S1 is homeomorphic to T it is enough to show X∗ ' T. Consider the function

f : R∗ → T, f(x) = cis(2πx

1 + |x|) for x ∈ R, f(p) = (1, 0).

The rest as in (a).

4.7 Hint: by 4.5, X topologically is embedded into the one-point compactificationX∗ which is a compact Hausdorff space. Therefore, X∗ is T4. Hence, also T3.5. Nowcheck that T3.5 is a hereditary property. So, X also is T3.5.

4.8 Hint: By Tychonoff theorem X ×X is compact. Now show that the distancefunction d : X×X → [0,∞) is continuous and apply generalized Weierstrass theorem.

4.9(1) ⇒ (2) Take fi = pi, i ∈ {1, · · · , n} the projections. Observe that the family of

all projections always separate the points for any product.

(2)⇒ (1) Suppose that fi : X → [0, 1], i ∈ {1, · · · , n} are continuous and separatethe points of X. Consider the following ”diagonal function”:

f : X → [0, 1]n, f(x) = (fi(x))i∈I .

This function is continuous (see 3.6). Also, f is 1-1. Indeed, if x 6= y in X thenfi(x) 6= fi(y) for some i. Therefore, the vectors f(x) and f(y) are distinct in the cube.Since X is compact and [0, 1]n is Hausdorff we know that f is a closed map. Since fis also 1-1 we obtain (by theorem about topological embeddings) that f : X → [0, 1]n

is a topological embedding.

4.10 (1) ⇒ (2): X ∈ T3.5. So, by definition there exists a family of functionsS = {fs : X → [0, 1] : s ∈ S} which separates points and closed subsets of X (forexample, S = C(X, [0, 1])). Consider the diagonal function

f : X → [0, 1]S, f(x) = (fs(x))s∈S.

Then f is a topological embedding.(2) ⇒ (3): Let ν : X → [0, 1]S be a topological embedding. Denote by Y the

closure of the image. That is, Y := f(X). Then Y ∈ Cop ∩ T2 and the induced mapν∗ : X → Y is dense. Therefore, ν∗ : X → Y is a compactification of X.

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(3) ⇒ (1): Let f : X → Y be a compactification. Then X is homeomorphic tof(X). So, it is equivalent to show that f(X) ∈ T3.5. This follows from the hereditaryproperty of T3.5 taking into account that f(X) ⊂ Y and Y ∈ Comp∩T2) ⊂ T4 ⊂ T3.5.

4.11 See Lecture notes.

Section 5

5.1 (1) Consider, for example, any 1-1 onto continuous map which is not homeo-morphism. Say, f : (R, τdiscr)→ (R, τ(d)).

(2) Let f : X → Y be an onto continuous open map. We have to show that f isa quotient. Equivalently, we need to check that f−1(A) is open in X implies that Ais open in Y . Now, if f−1(A) is open then f(f−1(A)) is also open because f is open.On the other hand, f(f−1(A)) = A (because f is onto).

(3) Let f : X → Y be an onto continuous closed map. We have to show that f isa quotient. Equivalently, we need to check that f−1(A) is open in X implies that Ais open in Y . Turning to the complement, it is equivalent to show: f−1(A) is closedin X implies that A is closed in Y . Now, if f−1(A) is closed then f(f−1(A)) is alsoclosed because f is closed. On the other hand, f(f−1(A)) = A.

(4) If f : X → Y is a homeomorphism then it is a quotient map. Now, f : X → Y

is a quotient as a composition f ◦ α of two quotient maps.

Conversely, let f be a quotient. We have to show that f : X → Y is a homeomor-phism. It is continuous 1-1 and onto. Therefore, it is enough to show that this map

is open. Let O be open in X. We have to prove that f(O) is open in Y .

Since f is 1-1 we have O = f−1(f(O)) = f−1(A), where A := f(O). Then

α−1(f−1(A)) = f−1(A) is open in X (because α is continuous). Since f is a quo-

tient we can conclude that A is open. On the other hand, A = f(O).

5.2 f : X → Y is an onto continuous map, Y is Hausdorff and X is compact.Therefore, f is a closed map. Hence, f is a quotient by 5.1.3.

5.3 (1) Consider the function

f : X := [0, 1]→ Y := S1, f(x) = cis(2πx) = (cos2πx, sin2πx)

This function is a quotient by 5.2. On the other hand, the equivalence relation ∼induced by f on [0, 1] is just 0 ∼ 1. So by 5.1.4 we can conclude that X ∼ S1.

(2) Consider the function

f : X := [0, 1]× [0, 1]→ Y := S1 × [0, 1], f(x, y) = (cis(2πx), y)

The rest as in (1).(3) Consider the function

f : X := [0, 1]× [0, 1]→ Y := S1 × S1, f(x, y) = (cis(2πx), cis(2πy))

The rest as in (1).

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5.4 We have to show that f2 : Y → Z is a quotient. Let f−12 (A) is open in Y . By

the continuity of f1 we obtain that f−11 (f−1

2 (A)) is open in X. Since f = f2 ◦ f1 is aquotient and f−1(A) = f−1

1 (f−12 (A)) we can conclude that A is open.

5.5 (a) Observe that the restriction of f on Y is the identity function. So, f is thequotient by 5.4.

(b) The function f : T → T, f(t) = t2 is continuous onto from a compact space, Tis Hausdorff. So, this map is closed and onto. We obtain that f is a quotient.

5.6 The map p1 : X → R, (x, y) 7→ x is continuous (as a restriction of the projec-tion) and onto. Its restriction to the x-axis {0} × R is a homeomorphism. So, ourmap is a quotient.

It is not open: the subset [0, 1) × (2, 3) is open in X but its image [0, 1) is notopen in R.

It is not closed: the subset {(x, 1x) : x > 0} is closed in X but its image (0,∞)

is not closed in R.

References

1. D. Leibowich, Set Topology (in Hebrew) 1993.2. S. Lipshutz, General Topology, 1965.3. E.G. Milewski, The topology problem solver, 1994.4. J.R. Munkres, Topology, 2000.5. O. Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, V. M. Kharlamov, Elementary Topology Problem

Textbook, 2008.

Department of Mathematics, Bar-Ilan University, 52900 Ramat-Gan, IsraelURL: http://www.math.biu.ac.il/∼megereli