COMPACT TOPOLOGICAL SPACES

APPROVED:

3 or Proftssor

EEHSLORLSLIPESSHI^OFWLEWEFLM

- I cr^JLtry^g, Dmui of the graduate School

COMPACT TOPOLOGICAL SPACES

THESIS

Presented to the Graduate Council of the

North Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF SCIENCE

By

Thomas M. Conway, B.S.

Denton, Texas

June, 1964

TABLE OF CONTENTS

Chapter Page

I. INTRODUCTION . . I

II. COMPACT SPACES AND THE SEPARATION PROPERTIES. 13

III, OTHER IDEAS OF COMPACTNESS 31

BIBLIOGRAPHY 40

111

CHAPTER I

INTRODUCTION

The purpose of this paper Is to Investigate some prop-

erties of compact topological spaces and to relate these

concepts to the separation properties. The first chapter

provides necessary background. The second chapter deals

with properties of compact spaces, conditions equivalent to

compactness, and the relationship of these ideas with the

separation properties. The third chapter includes several

different notions of compactness.

The fundamental concepts of logic will be assumed

throughout, and theorems in Chapter I will be offered with-

out proof*

Definition 1-1, If each of A and B is a set, then the

complement of B with respect to A, A - B, is the set of all

points of A that are not points of B. The empty set will

be noted by 0.

Theorem 1-1. Let G be a collection of subsets of a

aet X. Then (a) X - ( ^ ) - - g)

and (b) X - ( Q p ) = y g ( x - g).

This is the well-known theorem of DeMorgan.

Theorem 1-2. If A is a countable set and B Is a finite

set, then Al JB Is a countable set,

theorem 1-3. If a set A has n elements, then there are

2 n subsets of A.

Definition 1-2. I*et X be a nonempty set, and let x be

a collection of subsets of X. The collection x is said to

be a topology of X if x satisfies each of the following

three conditions:

(a) X e x and $ e x ;

(b) if Q O , then ^ g * T ; and

{C) if A € T and Bet, then a O B € x ,

If t is a topology for X, the elements of x are called x~open

sets of the topology, and the pair (X, x) is called a topo-

logical space. If It is clear that ACT, then A will be re-

ferred to as an open set,

Definition 1-3. If (X,T) is a topological space and

ACX, then the statement that A Is closed means X - A is

open.

Definition 1-4. Let (X,x) be a topological space, and

let p € X. A subset N of X Is called a x-neighborhood of £

if there is a x-open set A such that pe A and ACN.

Theorem 1-4, If (X,x) is a topological space and G

is a finite collection of x-open sets, then is an open

set,

Theorem 1-5. If (X,t) is a topological space and ACx,

then the set A is T~open If and only if A contains a t-neigh-

borhood of each of its points.

Definition 1-5. If (X,x) Is a topological space,

A(JC, and peX, then the statement that j> is an accumulation

point of A means that if U is a neighborhood of p, then U

contains a point of A different from p. A' will denote the

set of all accumulation points of A.

Definition 1-6, If (X,t) is a topological space and

p € X, then the statement that p is an (^-accumulation point

of a subset A of X means every neighborhood of p contains

infinitely many points of A,

Definition 1-7. If (X,t ) is a topological space and

ACX» than the t -closure A of A is the intersection of all

closed sets which contain A.

Theorem 1-6, If (X,t) is a topological space and ACx,

then A Is t - c l o s e d if and only If A ' C a .

Theorem 1-7. If (X,t) is a topological space and

ACX, then aLJa* is closed and 5 <• A^A' »

Definition 1-8. If (X,t) is a topological space and

ACX, then the statement that £ is an interior point of A

means that A is a t-neighborhood of p. The interior A° of

A is the set of all Interior points of A.

Definition 1-9. If (X,t) is a topological space and

A O , then the statement that U is a neighborhood of A means

AQj°. A closed neighborhood of A Is a neighborhood B of A

such that B Is closed.

Definition 1-10, If (X,T) IS a topological space and

p € X, then the neighborhood system of p is the collection

of all neighborhoods of p. The neighborhood system of X Is

the collection of all neighborhood systems of each point

in X.

Definition 1*11. Let (X,T) be a topological space» A

family P of neighborhoods of X is a base for the neighborhood

system of X means that if p c X and A is a neighborhood of p»

then there Is an element B of F such that B is a neighborhood

of p and BCA.

Definition 1-12. If (X,T) Is a topological space, then

a base for T is a collection a of sets such that X ** &WA

and for each point p In X, In U and V are elements of a con-

taining p, then there is an element W of a such that p € W

and WC(uPlV).

Definition 1-13. If (X,T) is a topological space and

YCX, then the statement that (Ytn ? is a sub space of (X,T )

means that TJ « {ujfor some T in t, U = TOY}. T\ is said to

be the topology for Y relative to r. The statement that Y

is a closed subspace of X means that Y is closed and (Y,rj)

is a subspace of (X,t).

Theorem 1-8, Let (X,T), (Y,I ), and (Z,pi) be topologi-

cal spaces. If (Y ) is a subspace of (X,t) and ) is

a subspace of (Y,rj), then Z is closed as a subspace of Y if

and only if % Is closed as a subspace of X,

Definition 1-14. A relation is a set of ordered pairs.

The domain D(R) of a relation R is the set of all first

terms of the ordered pairs in the relation* The range Rn(R)

of the relation R is the set of all second terms of the or-

dered pairs in R.

Definition 1-15. A function T is a relation such that

no two ordered pair® of T have the same first element. If

no two ordered paira in T have the same second element,

then T is said to be reversible.

Definition 1-16. If T is a function and the ordered

pair (x,y) cT, then y is denoted by T(x). The element y is

called the image of x under the function.

Definition 1-17« & function T is said to be a mapping

of X into Y if the domain of T is X and the range of T is

a subset of Y, A function T is said to be a mapping of X in i i f >niiu i m T t r t u t f m r n i n m n -mm*

onto Y if the domain of T is X and the range of T is Y.

Definition l~l8. A function f of X into Y is said to

be a one-to-one mapping of X into Y if f is a mapping of X

into Y and f is reversible.

Definition 1-19. If T is a function, then T"1 »

{U,y) |(y#x)« t} ,

Definition 1-2Q. Let T be a function from X into Y.

6

(a) If ACX, then T[A] - {y€ Y|y - T(x) for some x in A) •

{b) If A O , then T"1 [A] « £x CX|T(X) C A).

Definition 1*21. Let (X,*r) and (5f,r| ) be topological

spaces, and let f be a mapping from X into Y, The mapping

f is saW to be continuous if f"x[G3 la T-open whenever 0

is Tj -open# and the function f is said to be a T-fi continuous

mapping of X into Y. mi nut jiu rWftiiijftMWi • iflwimiminWfr mwm- m Mjiimw waui ii>m>»ipr «#*«*<

Definition 1-22* Let (X,f) and (Y,rj) be topological

spaces. A t-fj continuous mapping f of X into Y is said to

be bicontinuous If f [0] is an i -open set whenever G is a

T-open set (l, p. 39).

Definition 1-23. Let (X,T) and (Y,T] ) be topological

spaces, and let f be a mapping of X into Y, The function f

is said to be a t-o homeomorphism of X into Y if f Is a jni.il ii«i«m—.i*|ni i "ir'' inMniimi'niy tn>m • «»»»

one-to-one bicontinuous mapping of X onto Y (l, p. 40),

Definition 1-24. If R and S are relations, then the

composition R*S of R and S Is the set of all pairs (x,z)

such that for some y, (x,y)c S and {y,z)f R,

If X is a non-void set, then there Is a function f on

the set of all subsets of X such that if BCX, then f [B] eB.

The function f is said to be a choice function. This Is

known as the Axiom of Choice.

Definition 1-25. A collection G of sets is said to be

a covering of a set X if xC^jg. If 1® a covering of X

and S s is a covering of X such that G aCGii then G e is called

a subcovering of Gx.

Definition 1-26. If (X,T) is a topological space, then

a covering G of X is said to be a T-open covering of X if

every element of G is a t-open set. A covering 8 of a set

X is finite if G has only a finite number of elements.

Definition 1-27. A topological space (X»t) is said to

be compact if every T-open covering of X has a finite sub-

covering .

Definition 1-28. If (X,-r) is a topological space and

YQC, then the statement that Y is compact means the topology

for Y relative to T is compact.

Definition 1-29. The statement that a family of set©

has the finite intersection property means the intersection

of the members of each finite subcollection of the family

is non-void.

Definition 1-30. A topological space is said to be

locally compact if and only if each point has a compact neigh-

borhood.

Definition 1-31. A binary relation * directs a set D

if D is non-void and the following conditions are true;

(a) if m*n and p are elements of D such that m*n

and n*p, then m*pj

(b) if m€0, then m&nj and

8

(c) if m and n are elements of D, then there Is a p

In D such that nrfp and n^p.

Definition 1-32. A directed get is an ordered pair

(D,s=) such that ^ directs D,

Definition 1-33. A net is an ordered triple (T#D,^)

such that T is a function whose domain contains D and (D,<£)

is a directed set,

Definition 1-3^. The statement that the net (T,Dj^) is

in a set A means that for each element n in D, T(n)€ A, A

net N » ( T i s eventually in a set A means there is an

element m in D such that if n € D and m^n, then T(n) e A. The

net N is frequently in A means that for each m in D there

is an n in D such that m^n and T(n)e A.

Definition 1-35. If (X,i) is a topological space and

p e X, then the statement that p is a cluster point of a net

N means N is frequently in every neighborhood of p.

Definition 1-36. If (X,i) is a topological space, p € X,

and H is a net in X, then the statement that N converges to

£ means that if U is a neighborhood of p, then N is even-

tually in U.

Definition 1-37. A net (S,K, £) is a subnet of the net

(T,D,£) means that there is a function N on K into D such

that the following conditions are true:

(a) S » T«N and

(b) for each m in D, there exists a k in K such that

if k£z, then ra^(z)*

Theorem 1-9• If (X,t) is a topological space, H is a

net in X» and p € X, then p is a cluster point of N if and

only if some subnet of N converges to p.

For brevity, the set of all positive integers will be

denoted by I+.

Definition 1-38. & sequence S is a net such

that the relation «*= it the usual ordering -for the positive

integers.

Theorem 1-10, If (X,t) la a topological space, S is a

sequence of points in X, p€ X, and Sf is a subsequence of S

such that S* converges to p, then p is a cluster point of S.

Theorem 1-11» Every sequence in a topological space

has a cluster point if and only if every infinite set has

an -accumulation point*

Definition 1-39. Let (X,i) be a topological space,

A is a topological space such that if

p t X, q e X, and p jl q, then there is an open set

which contains one point and not the other,

^ A *?«.~gPaoe is a topological space such that if

pel, q« X, and p q, then there is an open

set which contains p and not q.

^ A ^,a~8tPace is a topological space such that if

10

Pi € X, ps c X, and Pi £ pzt then there exist dis-

joint t-neighborhoods kx of px and A2 of p2, A

T^-space Is also said to be a Hausdorff space.

(4) A regular space Is a topological space such that

if p € X and P is a closed subset of X such that

p/F» then there are x-open sets A and B such

that p € A, FCB, and APlB « <t>. A T«-apao» is a

regular Tx-space,

(5) A n o r ^ i space 1© a topological space such that

if Fx and Fs are disjoint t-closed subsets of X,

there exist disjoint t-open sets Gx and G2 such

that F xC0i and F 2C0 2. A TA-gpace is a normal

Tx-space.

Theorem 1-12, A space (X,t) la regular if and only if

for each point p e X and every t-neighborhood A of p there

is a t-closed neighborhood V of p such that VGA.

Definition 1-40. Let R be the set of all real numbers.

A subset G of B is said to be tl-open if G « or if 8 ^

and for each p e G there is an open interval J such that p e J

and JCG.

Theorem 1-13. If # consists of the collection of all

subsets of R that are $~open» then (R,t!) is a topological

space (l, pp.15-16),

Definition 1-41. The topological space (R#0) will be

referred to as the usual topology for the real numbers.

11

Definition 1-42, If each of A and B Is a set, then

the cross product A x B of A and B Is {(x,y)|xcA and y€B}-.

Definition 1-43. Let (X,t ) and (Y,t)) be topological

spaces. The product topology for X and Y is the ordered

pair (Z,n.) such that Z « X x Y and \L « {U x v |u e t and ?e rj>.

Theorem 1-14, Let be the usual topology for the

real numbers. If ACR» then A is compact if and only If A

is closed and bounded.

Theorem 1-15. Let Ai#A2,«#.,An and B be subsets of

some set X. Then (A^Ob) « AjJPlB.

CHAPTER BIBLIOGRAPHY

Mansfield, Maynard J I n t r o d u c t i o n to Topology, Prlncse-ton, Mew Jerifey* J5. Van Sosflr&n^1"CoT, ' i n c 1 9 6 3 •

12

CHAPTER II

COMPACT SPACES AMD THS SEPARATION PRQPERTIBS

The theorems of this chapter eoncern the general Ideas

of compactness and relate them to the separation properties.

Theorem 2-1. If (X±t) is a topological space such

that X has only a finite number of points* then (X,.t) is

compact»

4.

Proof, There is an n i l such that X has n distinct

points, By Theorem l-3> there are at most 2n open sets in

t. Since every covering G of open sets is a subset of t,

G could contain at most 2n open sets. Hence, (X,T) is com-

pact.

Theorem 2-2, If (x,r) is a topological space and KCX#

then K is compact if and only if every cover of K by t-open

sets contains a finite subcover.

Proof, Suppose every cover of K by t-open sets con-

tains a finite subcover, Let "n be the topology for K rela-

tive to t. Now let G be an rj-open cover of K. For each

A in G, there is a T in T such that A » TP|K. Denote one

such T by Ta» Then ]^QtA Z)^^A » K, Thus there exist a

finite number of set® A1#Ae<a..#An in G such that T^»

13

14

T« ,..., covers K. By Theorem 1-15#

s n

A * i - ^ ' T*i n K) - ( A t * I »° k - k-

Thus, K is compact.

Suppose that TJ la the topology for K relative to t.

Let G be a collection of t-open seta covering K, Then H »

{AJfor some T in t, A • tOk} IS an tj-open cover of K. Thus

there exist a finite number of sets HlfH2,,,,,Hn in H which

cover K. For each i e I+» % » TPlK for some T in 0. De-,n. .n.

note one such T by Tlt Then (fJTi)D(^Hi) « K. Thus, G

contains a finite subcover of K.

Theorem 2-3* If (X,t) is a compact topological space

and Y is a t -closed subset of X, then the sub space (Y,r|) is

compact•

Proof. Let G be a t-open cover of Y. Since Y is t-

closed, X - Y is t-open. Then G* « GU(X - Y) is a t-open

cover of X. Since (X,t) is compact, there is a finite sub-

cover H of 6* which covers X. Thus H is a t-open covering

of Y, and since (X - Y)OY - <£, H - (X - Y) is a t-open

covering of Y, But [H - (X - Y) ] d Q - Therefore, by Theorem

2-2, (Y,T| ) is compact.

Theorem 2-4, The intersection of the members of an

arbitrary family of closed and compact subsets is closed

and compact.

15

Proof, Suppose \X,t) Is a topological space and a Is

a family of closed and compact subsets of X, Since jj a(X - A)

is open and |(X - A) • X - ( j ^ A ) , then A is closed.

Now choose Ffa, and denote Y » Am A' *1 k® the topology

for Y relative to t. Too# let \l be the topology for F rela-

tive to t. By Theorem 1-8, (Y,tO is a closed subspace of

(F,ti). Since (P,ii.) is compact, by Theorem 2-3, (Y,rj) is

compact.

The intersection of two compact subsets of a topologi-

cal space may fall to be compact.

Example 1, Let (R,0) be the usual topology for the

real numbers* Let (A,r\) be such that A * {0 ,1} and i\ • {A,

Then (A,^) is a topological space* Let (Z,|i) be the product

space of (R,U) and (A,tj), Then Z is the union of the line

y « 1 and the line y * 0. Define *» { ( x , l ) |x € R and xc

UQ » { (x,Q)| x e R and x c (V*, 1] >, Vi ® { ( x , l ) | x e R

and x € {x/a, l]>, and V0 * { (x,0)J x c R and x e [O,3/* )} ,

y«l

0

u,

(0,1) (%h)

(V|>o) Up (1^0)

y«l

y=0

o t r n 4-

(1,1)

(O>O)VQ ( 7 y i )

How let U » Ui^Uo and V « v xU^o • Then, using Theorem 1-14-,

it can be shown that U and V are compact subsets of Z. Then

UOV « { (x,y)| x,y€R, X e (V4 ,3/4 ) and y » 0 or y « 1}.

16

y*l (y«.i) (y«,D

oOv

(y.i) (a,l,o)

Let % « {(x,y)|x e i) tod y • 0 or y • 1)}

Gs - {(x»y)U« (x/4,ts) and y » 0 or y - 1};

t " Then

Gn 88 C (x#y) I* < (xA#tn) ana y » o or y « 1}.

D6flH6 t i m tig m % J > t • J t ** (t j«,x % ) "S* 2

G w {0x#0a, ... •On,...} is a collection of tJ-open sets cover-

ing UOV, but no finite aubcollection of G covers UO"^- Thus,

upjV is not compact.

Theorem 2-5. If (X,t) is a compact topological space

and K is an infinite subset of X, then K has at least one

accumulation point in X.

Proof. Suppose K has no accumulation point in X. Then

for each point p of X, there is a f-open neighborhood Qp of

p that contains no point of K other than possibly p. The

collection G - {Gp|pe X> is an open covering of X. Since

(X,t) is compact, there are n points pi*p2, «*»#Pn i n X such

that G* * {GpjJ 1 € I + and I^N} is a covering of X. For each

i In I+, Gpi contains at most one point of K. Thus K has at

IT

most n points. This contradicts the hypothesis that K is

an Infinite set.

The converse of this theorem is not true.

Example 2. Let R be the set of all real numbers* and

let t consist of all open Intervals which contain the num-

ber one. It can be shown that (R,t) is a topological space.

Now suppose A is an infinite subset of X. There are two

points p and q in A such that either p< 1 and q <1 or p> 1

and q > 1. Suppose then that p and q are both greater than

1, and let p > q. Every open set which contains p also con-

tains q. Hence, p is an accumulation point of A in X, A

similar argument for q> p and also for both p and q less

than one would prove that A has an accumulation point. Now

let Q m { {—2 > 2) j ( -3*3)* *»*#( "*11 l)#***}* Then G 1b a

covering of R, but no finite subcover of G exists.

Theorem 2 - 6 . Let (Y#n) be a subspace of a topological

space (X,t). If Z is a non-empty subset of Y, then (Z,ix)

is compact as a subspace of (Y,tj} if and only if it is com-

pact as a sub3pace of (X, t ) .

Proof. Let (Z#|x) be a compact subspace of (X, t ) and G

be a p.-open cover of Z. There exists a finite subcollectlon

GlfGa, ...,Gn of G which covers Z. There is a T in t such

that Gj » t O z for each i € I + and jten. Denote one such T n , _ v

as Ti, Then ( I J t^^ z. For each Ti, (TiflY) € rj. Denote it

18

by Ki» Thus, for each G^ of G, there is an element Kj, of

rj such that GjL «• KjOz» Thus, (Z,n) is compact as a sub*

space of (Y,t} ).

Suppose (Z,\i) is a compact sufespace of (¥,tj) and let

G be a n,-open cover of Z, There exists a finite subcollec-

tion GA,Gg, ...,Gn of G which covers Z. Let 1e I+. There

IS an N of •q such that Gi » nO z. Denote one such N by K^»

There is a T in t such that * tOy. Denote one such T

as T±. Thus, Gi - KjHZ « (TiOYjOZ - TiHCYPlZ) « T^OZ.

Hence, (Z,^) is a compact subspace of (X,t).

Theorem 2-7* A topological space (X,t) is compact if

and only if each family of closed sets which has the finite

intersection property has a non-void intersection*

Proof. Suppose each family of closed sets which has

the finite intersection property has a non-void intersection,

Suppose that (X,t) is not eorapact. Then there is a collec-

tion a of T~op@n set© such that no finite subset of a covers

X, For some n in I+, let {An} denote a subcollection of a

containing exactly n elements. For each 1 in 1+ with i^n,

let « X * Aj[» Define { % } to be the set of all such Bj. »s.

Suppose now that ^ Bi « 0. If p € X, then p/l ( 1QB i).

There is a j in I + such that p/Bj. Then p € Aj. Thus iVla1 n

must cover X, a contradiction. Therefore, f) g J <j># j isss x X

B {y|for some A in a, y « X - A}. Then B is a family of

19

closed sets having the finite Intersection property. Thus,

bcgb j* 0. I«et If A c a, then ptX - A, This con-

tradicts the hypothesis that a covers X. Hence, (X,T) is

compact»

Now let (X,T) fee a compact space, and, let P be a family

of closed sets having the finite intersection property.

Suppose m 0, Let a « {A|for some B of p, A « X - B}.

Choose peX. Since p/^e^k, p^b for any b in p. Thus,

there is an A of a such that p c A. Hence, X « AVOA. Since

(X,T) is compact, there Is a finite subcollection AiaA2,

.. Ayj of A which covers X. Define {%} such that Bj •

X - Ai for each i in I+ and i n. Then $ « X - X * X ~ n n

which is the contradiction sought. There-

fore • 4

Theorem 2-8. A topological space is compact if and

only if every class of closed sets with empty intersection

has a finite subclass with empty Intersection.

Proof. Suppose every class of closed sets with empty

intersection contain® a finite subclass with empty intersec-

tion. Too, suppose (X,T) is not compact. Then there is a

collection G of open sets covering X such that no finite sub-

collection of G covers X. Then F * {X - A|A C G} is a collec-

tion of closed subsets of X. The family P has the finite

Intersection property, for otherwise G would contain a finite

20

subcover of X. Then » a Q ( X - A) - I - (aU0A) - X - X

» <p. Thus there is a finite subclass F* of F such that

Q ^ f « 0. Denote F* - lFj.,Fa, . ..*Fn} and Fx * X - Alt

F 2 = X - k Z ! ... ,Fn = X - An where Ai € G for each 1 e I+.

n n n n

Then $ m ^ F i . /J.(X - Ai) « X - iVxAi. Thus, X - ^ A »

which Is the desired contradiction.

Now let (X,t) be compact. Suppose there is a class F

of closed sets with the finite intersection property and

such that fQpf * <P. But this contradicts Theorem 2-7. Thus,

f c V *

Theorem 2~9. If (X , t) is a topological space, then

(X,T) is compact if and only if every cover of X by base

elements has a finite subcover.

Proof. Let 0 be an open cover of X, and let a be a

base for t. Define E * ( A | * e a and for some B in G, AC®.

Then H(^a and |jg||h * X. Thus, there is a finite subcover

H x ,H 2 , , . . ,Hn of H. If i e I + and i^n, then % C a for some n n

A in a. Denote one such A by Ai, Then « X.

Thus, (X,T) is compact*

If the space is compact, then every cover of X by base

elements contains a finite subcover.

Theorem 2-10, Let (X,t) and (Y,^) be topological spaces,

and let f be a continuous mapping of X onto Y, If (X,t)

is compact, then {Y,tj) is compact.

21

Proof. Let G be ami-open covering of Y* Define F «

{x for some g €G, x « F"1[g]}» Then I is a t-open covering

of X, There exist n points p1# p2, ...»Pn in X such that

F* • {FPl| Fpj^CP* it and l = n} is a covering of X, Now

let G* » {Gp± | GpjL » fCFPl3 for i € I+ and i^n}. Clearly,

G # C @ and covers Y, Hence, (Y,ri ) is compact.

Theorem 2-11, If (Y,r\) is a compact subspaee of a

Hausdorff space (X,t), then Y is t-closed.

Proof. Suppose x e (X - Y). Then for each element y in

Y, there are open sets Uy and Vy such that x c Uy, y € Vy, and

VyOlJy - 0. The collection G * {A for some y in Y, A »

VyOY} is an T\ -open covering of Y. There exist n points yA,

y2, ... ,yn in Y such that {Vy^p^Yj i e 1+ and i^ n) Is a

covering of Y. Therefore, Y^^U^Vy^)« Let G « ^Cl^i*

Then G is a t-neighborhood of x and * 4>t other-

wise, there exists a k in I + such that Uy^OVy^ f There-

fore, GOY * <$>. Hence, GCX. -Y. Thus, X -Y contains a t-

neighborhood for each of its points. By Theorem 1-5# X -Y

is t-open, Hence Y is f-closed.

Theorem 2-12. If (X,t ) is a compact space, (Y,tj) is

a Hausdorff space, and f is a one-to-one t-tj continuous map-

ping of X onto Y, then f is a homeomorphism.

Proof. Let G be a f-open set. Then X - G is t-closed.

Let tj be the topology for X -G relative to t, By Theorem

22

2-3t (X - 0,t\ ) 1® a compact space. Thus# by Theorem 2-10,

f[X-G} is compact. By Theorem 2-11, f [X - G) is rj-closed.

Let y c f [X - 0], Then there is a point x in X - G such that

y * f(x). Since f is one-to-one, x is the only element of

X such that y « f(x). Thus, y / f [GJ, Therefore, y c Y - f [G J.

Now let z e Y ~ f [G)» Then there is at least one x in X - 0

such that z m f(x). Hence, z e f (X - G), Since f[X - G] is

t}-closed and fix-©3 * Y - f fGj, it follows that f[Gj is r\~

open. Therefore, f is a homeomorphism.

Theorem 2-13. If A is a compact subset of a Hausdorff

space (X,T) and p is a point of (X- A) , then there are dis-

joint neighborhoods of p and of A.

Proof. Let tj be the topology for A relative to t. By

Theorem 2-11, A is t-closed. Thus p is not an accumulation

point of A, For each q of A, there exist disjoint open sets

Vq and Uq such that pe Uqand q e Vq. Let G » £y| for some

q € A, y » VqOA}, G is an r -opep cover of A. Hence, there

exist a finite number of points px, ps, ..,,pn of A such .a n

that (®p^'^pg* * * * c o v e r s A, A(_ * and p® iQupi-.n, t

Choose x c *P1- For some i in I+, x c Vp1. Then x i Up^.

Thus, x i . Hence, (j§ivPjL) O(j^Upi) • 0. Therefore,

there are disjoint neighborhoods of p and of A,

Theorem 2-14, If A and B are disjoint compact subsets

of a Hausdorff- space (X,t), then there are disjoint neigh-

borhoods of A and B.

23

Proof. Choose p « B. By Theorem 2-13, there exist dis-

joint open sets Vp and Up such that ACVp and pcUp. Define

G » {QpJ for some p in B, Gp » UpOB}. Let M- he the topology

for B relative to t. Since (B,p.) is compact, there exist n

points PiiPe* «.*#Pn of B such that {Qp1#Op2, ...,Qpn} ,n, v n ,n.

covers B. Then CCj^Upj:) and A C C ^ ^ p i ) • Choose **£^jUpi«

Then xe Up.for some 1 in I+, Thus, x iVPl. Hence, x i /». Vp,. n, n . J~1

Therefore, ^Q^Pi) " Thus, there exist dis-

joint neighborhoods of A and B.

Corollary 2-1. Each compact Hausdorff space is normal.

Proof, This fact followi as an immediate consequence

of Theorems 2-3 and £-14,

The converse of Corollary 2-1 is not true. For an ex-

ample of a T4~space that is not a compact Hausdorff space

one may refer to Example 6-3.7 on page '73 of Maynard Mans-

field's Introduction to Topology.

Theorem 2-15, If (X,T) is a regular topological space,

A a compact subset of X, and U a neighborhood of A, then

there Is a closed neighborhood V of A such that vCu.

Proof, Choose peA, Then U is a neighborhood of p.

There Is an open set V such that peV and VCu. Since V

is open, X - ¥ is closed* Since p /X - V and (X,t) is regu-

lar, there exist disjoint open sets Gi and G 2 suoh that

(X - V)(3i p e G2. Then (X - Q x)Cu and X - 0X la closed.

24

SaCUj otherwiae, there is a point q of G2' such that q/u.

Then q e Gj,. Since Gj. is open, Gj, contains an open neighbor-

hood of q, which contradicts the fact that every open set

containing q contains a point of G2 distinct from q. Thus,

for each point p in A, there is an open set such that

D € ¥p and 7pC"U. Let G • {VpjpeA}. Since A is compact,

there exist n points px,ps, ...,pn of A such that {Vp1,Vp2,

«..,Ypn) covers A, Thus, (1^1Vpi)Cu and is closed.

Corollary 2-2, Each compact regular space is normal.

Proof. Let (X,t) he a compact regular space. Let A

and B be two disjoint closed subsets of X. Thus, ACX - B

and X - B is open. By Theorem 2-15, there is a closed neigh-

borhood V of A such that VC(X - B). Then ACv°. Too,

B Q X - V). Since X - V is open and V°f}(X - V) « <*> , then

the space is normal.

Theorem 2-16, If Y is a compact subset of a regular

space (X,t), then Y is compact.

Proof. Let p 6 X - Y. For each point q in Y, either

(1) there is an open set containing q which does not contain

p or (2) every open set containing q also contains p. If

(2) is true, then p is an accumulation point of Yj otherwise,

there is a neighborhood U of p such that no point of Y is

in U. Since (X,t) is regular, there is a closed neighbor-

hood K of p such that K(Zu, Then X - K is a neighborhood

25

of q which does not contain p, which is the contradiction

sought. Define W m {p| p e X - Y and for some point q in I,

p is in every neighborhood of q), Clearly, (W^jY )Cf, and

every coyer of Y by t-open sets also covers W^JY. Then let

p e X - (Wv JY). For each q in Y, there is a neighborhood %

of q such that pjffZ. Since the space is regular, there is

a closed neighborhood A of q such that A(_Z. Then X - A

is an open neighborhood of p. Therefore, there exist two

disjoint open sets Vq and Uq such that p € "Vq and q^Uq.

Then Y C(>''UQ). Since Y is compact, there exist a finite <m * n

number of points qi#qa, .. ,,qn of Y such thatOUq. covers n n

Y (and thus, covers V{JY). Then (j[JxVqi)0 (iyiUq1) » <t>.

Thus, (^^Vq^ )Pi (W' JY) « Hence, X - (W(JY) contains an

open neighborhood for each of its points. By Theorem 1-5#

X - (W(Jy) is open, Therefore, Wl JY is closed, Hence,

^C(WL/Y) • Therefore, Y * (WijY) and is compact.

The space in Theorem 2-16 must be regular as the follow-

ing example will show.

Example 3. Let{R,T) be the topology as defined in

Example 2. Define K » {!]. K is compact by Theorem 2-1.

But K = R, for if pe R and p 1, then every open set con-

taining p also contains 1. Therefore, p is an accumulation

point. Clearly, no cover of X by finite open intervals

could be finite. Therefore K is not compact.

26

Prom the definition, it is clear that every compact

space is locally compact. The converse Is not true.

Example 4, Let (R,tt) be the usual topology for the

real numbers. Since R is not bounded, by Theorem 1-14,

(R,$) IS not compact. But the space is locally compact,

since each real number is an element of some closed in-

terval ,

Theorem 2-17, If (Z,ia) is a closed subspace of a lo-

cally compact space (X,t), then (Z,m.) is locally compact,

Proof, Choose pe Z. There is a compact neighborhood

U of p. Let r\ be the topology for U relative to -t. Define

¥ * z O u . Now let G be an t -open cover of V, For each

point qi of U - V, there is anrj-open set W with qiW such

that W contains no point of Vj otherwise, q is an accumula-

tion point of Z and of V. Since Z is closed, qe Z* But

q cU - V - U - (ZPlU) = UH[X - (ZOU)]. Thus q € Z, which

is the contradiction sought. Denote one such W by Wq* Be-

fine A « {Wq q € (U - V)}. Then A-vJV Is an rj-open cover of

U, Thus, there is a finite subcover H of U» Define 0* «

{b |B€ G and Be H}, If qi?, then qe B for some B In 0*.

By Theorem 2-2, V is compact. Therefore, (Z,m-) is locally

compact,

Theorem 2-l8+ If (X, t) la a locally compact space,

(Y,"q) is a topological space, and f is a bicontinuous mapping

27

of X Into Y, then (Y,tj) is locally compact.

Proof. Choose p € Y. Then p » f(a) for some a in X.

Since (X, t) is locally compact, there is a compact neigh-

borhood ¥ of a. Let G toe an n-opan cover of f [v]. Since

f is a continuous mapping, the inverse image of each open

set is open. For each •n-open set A, let B& denote its in-

verse image. Then B m is a -r-open cover of V. Since

V is compact, by Theorem 2-2, there exist a finite number . n

of sets AX,A2, .».,An in 0 such thatAJB* covers V. Then n laEX 1

U Ai covers f[V]; for if q € f[v], then there is a point x Xs®! in V such that f (x) * q. Then xe for some J in I+,

J

Thus, f(x) m q and qeAj. Therefore, p has a compact neigh-

borhood. Hence, the space (Y,n) is locally compact.

The mapping f in Theorem 2~l8 must be both open and

continuous.

Example 5. Let X be the set of all real numbers. If

ACX, then A c t . Then (X, t ) i s a topological space. Let

« {b|bCR and H - B is countable}. It can be shown that

(R,t|) is also a topological space. Let f be the identity

function from X onto Y. Then f is continuous, for the in-

verse image of every open set in R is an open set in X; how-

ever, f is not open, for the image of a finite set in X is

not open in Y. Too, (X,t) is locally compact, for every

discrete space is locally compact (l, p.146). Let U be an

28

•?}-neighborhood of the point 0 in Y. Let V « (pj^pg,

Pjj,...) be an Infinite set of distinct points of U, Define

Gx - (X - V)U{Pi)i a* - (X - V)U{P2)i ...|Gn « (X - V)U

(Pn) #.... Then G » {G^Gg, ...,Gn,...} is an n-open cover

of U, tout no finite subcover exists. Thus, (Y,T\) IS not

locally compact.

Theorem 2-19. If (X,T) is a locally compact Hausdorff

space, then (X,T) is regular.

Proof. Let p € X and U be a closed set not containing

o. Since (X,T) is locally compact, there la a compact

neighborhood V of p. By Theorem 2-11, ¥ is closed. If

VOU • <t>$ then X - ¥ is an open set containing U, pe V°,

and V°n(X - V) * 4>* Therefore, suppose u O v Since

uPlV is a closed subset of V, uPlV is compact. For each

point q in uf)V, there exist disjoint *t-open sets ¥q and

Uq such that p € Vq, qeUq, and VqC v» There exist a finite

number of points qi#q2j in uOv such that ^^Uqj n *

covers Uf IV. Then A « /_,\.vqi i s a n open neighborhood of p

such that lOu » If not, then there is a point k such

that k € UOV and kel, Then k c Uq,, for some ie I+. Then

Uq^ contains a point of "Vq , a contradiction. Since X ~ A

contains U and peA°, then A° Pl(X - 1) * 4>* Therefore, the

space is regular.

Theorem 2-20. If (X,T) is a locally compact regular

29

space, then the collection F of all closed compact neighbor-

hoods of each point In X Is a base for the neighborhood

system for X.

Proof. Choose p in X, and let U be a neighborhood of

p. Let H € P such that H is a neighborhood of p, and let r\

be the topology for H relative to t. Define W « uOh. By-

Theorem 1-12, there is a t - c l o s e d neighborhood V of p such

that VCW. Let p. be the topology for 7 relative to H. By-

Theorem 2-3, i® compact as a subspace of (H,tj), By-

Theorem 2-6, (V,p.) is compact as a subspace of (X,t). Thus

V.€ P. Therefore, F is a base for the neighborhood system

for X

Theorem 2-21, If D is a neighborhood of a closed com-

pact subset A of a regular locally compact topological space

(X,t), then there is a closed compact neighborhood V of A

such that A C v C U .

Proof. Choose p€A. By Theorem 2-20, there is a

closed compact neighborhood V* of p such that V*Cu« De-

note one such neighborhood by Vp. Define G * {V$j p € A ) .

Then A C U s . Since A is compact, there exist a finite

number of points Pl,p2, ..,sPn of A such that U ^ D a . n Is®1 *

Then a closed compact neighborhood of A, and

<HVCu-

CHAPTER BIBLIOGRAPHY

1, Kelley, John L., General Topology, Princeton, New Jer-&ej, D. fan Noatrand 'do., Inc., 1955.

30

CHAPTER III

OTHER IDEAS OF COMPACTNESS

There are other interesting notions of compactness.

A few of these concepts will now be investigated.

Definition 3-1. A topological space is said to be

countably compact if and only if every countable open

cover of the space has a finite subcover.

Definition 3-2. A topological space is said to be

sequentially compact if and only if every sequence has a

convergent subsequence•

Definition 3-3. A topological space is said to be

bloompaet if and only if every infinite set has an accumu-

lation point.

Theorem 3-1* A topological space (X,t) is countably

compact if and only if every sequence in the space has a

cluster point.

Proof. Let each sequence in X have a cluster point.

Suppose ( x , i ) is not countably compact. Then there is a

countable T-open cover G « (G^Gs, .. .#Gn» ...) such that

no finite subcover of G covers X. Define S(l) cX - Gx j

S(2) e X - (GiU02)j .. .;S(n) € (X - .U G, b Let S de-

note the sequence (Sn). Clearly, S(n) exists for each n.

31

32

Let p be a cluster point of S. There Is a j in I + such that

p € Oj, There Is a kcl + with k > j such that S(k) € Gj, But

S(k) € (X - - iWiGi). And 0*0 (X - jfcUt) » <*>. k

Hence, GjmvX - iWj® ) «* 0, a contradiction. Thus, (X,t )

is countably compact.

Now let (X,T ) be count ably compact, and let S be a se-

quence in X, Suppose S «= {SN3 has no cluster point. Let

Ai m {S(m)| m € I + and m>i}. Suppose * <p. Then X ®

X "* " ifI+^X ~ For each i in I+, let Gi •

X - Then j^j4^i is a countable cover of X, Hence,

there exist a finite number of positive integers kx,k2,

...,kn such that (Gki»Gk2> «»-#

Gkn} covers X. For each 1

in I + and i n, there is an Njt in I+ such that if m € I+

and m>Nk^# then S(m)^ (X - %k±) • Let N m maximum of

(Nkx,Nk2> ...#Nkn3. If m> N, then S(m) iX - for each

i— n. This contradicts the definition of a cover. Thus,

(ji j+ i) Therefore, there is a point p so that p e ffi

for each i c I+. If p is not a cluster point of S, then

there is a neighborhood U of p and a positive integer j

such that if i> J, then S(i) /U, Therefore, uPlA-j «• 0.

Consequently, p/Xj, a contradiction. Thus, p is a cluster

point of S.

Theorem 3-2, A topological space (X,T) is compact if

and only if each net in X has a cluster point.

33

Proof, Let each net In X have a cluster point. Let a

be a family of closed subsets of X having the finite inter-

section property. Define 0 » (A(for some finite subset a«

of a, A m |^,&}. Then {3 has the finite Intersection prop-

erty. Let ($,—) be such that if Bx€ $ and Ba e p, then

Bi— B2 if and only if BaCBi» Then {$,—) is a directed set,

For each B in £, let T(B) e B, This ,can be don© using the

Axiom of Qboice, Now let T m {(A,B)| A e £ and B m T(A)}.

Then r\ « (T,£,£) is a net in X, and by hypothesis, -n has a

cluster point, say p. Suppose B€£, If A € p such that

B—A, then T(A) € AGB. Therefore, q is eventually in the

closed set B, Thus, p €B, Hence, Since ^B 4

#, then g^A 4 0 . By Theorem 2-7, (X,t) is compact.

Now let rj - (T,N,—} be a net in the compact space

(X,T ), For each n € I+, denote AN «• {T(m)|m € 1+ and n^rn}.

The collection of all sets AN has the finite intersection

property. Thus, the family of all closures I n also has the

finite intersection property. Since X is compact, there is r\ _

a point p in X such that p c ^ A n . Suppose p is not a clus-

ter point of t|, There is a neighborhood U of p such that

r\ is not frequently in U. Therefore, for some n in N, if

n£ m, then T(m)/U, Thus, Up|An *» <f>, Consequently, pft~Rnt

a contradiction. Therefore, p is a cluster point of r,,

Corollary 3-1. A topological space (X,t) is compact

34

If and only If each net In X has a subnet which converges

to some point in X,

Proof. Apply Theorems 1-9 and 3-2.

Theorem 3-3. A TA-space is countably compact if and

only if the space is bicompact,

Proof. Let (X,t) be a bicompact Tj.-space, and let A

be an infinite ®ub»et of X. By Theorem 2-5, A has an accurau«

lotion point, say p. Let U be a neighborhood of p. There

is a point % of A ®uch that p $ % and qx € U. There is an

open set Vx such that p e Vx, Vx, and There is a

point q2 of A such that qs € Vj. and p q2. Then there is

an . open set containing p such that q 2^V 2 and ¥sC^"j,»

By using mathematical induction, there exists an infinite

set {qx,q2, .. .,qn,..,} such that q^ e U for each i € I+,

Since every neighborhood of p contains infinitely many

points of A, p is an ^-accumulation point of A. By Theorem

1-11, every sequence in X has a cluster point. Thus by

Theorem 3-1* (X,t) is countably compact,

Now let (X,t) be a countably compact Tj.-space* Let A

be an infinite subset of X, Define S(l)€ A{S(2)e A - S(l)j ,3*

...jS(n)c (A - U S(l) I.... Then S - {S1#S2, ...,Sn,...}

is a sequence with the property that if i and «J are positive

integers and i J, then S(i) + S(j). By Theorem 3-1, S has

35

a cluster point, say p, Let U be a neighborhood of p.

Since S is frequently In U, there Is an 1 € I* such that

S(i) eU. If S(l) « p, there is a j in I + with j >1 such

that S(j) € U. In either case there 1b a point q of A such

that q e U and q / p, Thus p is an accumulation point of

A, Hence, (X,t) is bicompact.

Theorem 3-^» If (x,f) is a countably compact Tx-space,

then (X,t) is countably compact If and only if every in-

finite open cover of X has a proper subcover.

Proof, Suppose the space Is countably compact. Sup-

pose further, that there Is an infinite open cover G of X

such that no proper subset of 0 covers X. If A c G, then A

contains a point q such that If Bi 6 and B ^ A, then q/^B,

Define K = {x x € A for some A in G and if B e Q such that

A / B, then x ft B] * Thus, K is an infinite set in a countably

compact Tx-space. By Theorem 3-3* there is an accumulation

point of K, say p. There is an element A of 8 such that

p c A. Then there is a point q of K such that qc A and p / q«

There is an open set V such that pcV, q / V, and V"Ca, Thus,

there is a point s of K in V such that s ft p and s € A. But

A contains q and s and q s, a contradiction. Therefore,

every infinite open cover has a proper subcover.

Now suppose every infinite open cover has a proper sub-

cover. Suppose further, that A is an Infinite set with no

36

accumulation point. For each point p of A, there Is an open

set V with p e V such that V contains no other point of A.

Denote one such V by Vp, and let U »{lfp p eA). Since A is

closed, X ~ A is open, Thus, UU(X - A) is an open cover

of X, No proper subcover exists, which contradicts the hy-

pothesis. Therefore, A has an accumulation point, By

Theorem 3-3 (X,t) is countably compact.

Definition 3-^. The statement that the topological

space (X,t) satisfies the first axiom of countablllty means

that if p € X and a is the neighborhood system of p, then

there exists a countable subset £ of a such that every ele-

ment of a contains an element of p.

Definition 3*5« The statement that the topological

space (X,T) satisfies the second axiom of countablllty mean®

that there exists a countable base for t,

The following theorem will not be proven.

Theorem 3-5* If (X,T) satisfies the first axiom of

countablllty and p e X, then there exists a sequence

of neighborhoods of p such that for each nil4",

Un+j^CUn, and If U is a neighborhood of p, there is a Jc c I+

such that UicCu (l, p.50).

Theorem 3-6. If (X,T) satisfies the first axiom of

countablllty, then (X,T) is countably compact if and only

if it is sequentially compact.

37

Proof. Suppose the space Is sequentially compact, By

Theorem 1-10, every sequence has a cluster point. Then by

Theorem 3-1# the space is countably compact.

Now let the space be countably compact, and let S be a

sequence In X, By Theorem 3-1* $ has a cluster point, say

p. By Theorem 3-5# there exists a sequence Ux,Ua,

of neighborhoods of p such that for each n e I*, U n+xCu n#

and if U is a neighborhood of p, there is a k in I + such that

UJCOJ. There is a k in I4" such that S(k)eUi, Denote one

such k by N(1). There is an m e I + with m > k such that

S(m)€U2, Denote one such m by N(2). Continue this process,

and let N - N(l),N(2), ...,N(k), Define 3« - S»N. If

O is a neighborhood of p, then there is an i e I"*" such that

U-jO. Then S'(Nj.) c Ui and if N(k)>N(i), then S* (%} € % ( ! % .

Thus S« converges to p. Therefore, (X,t) is sequentially

compact,

Theorem 3-7. If (X,T) satisfies the second axiom of

countability, then (X,t) is countably compact if and only

if it is compact•

Proof. If (X,T) is compact, then the space is countably

compact, Therefore, let the space be countably oompact. Let

G be an open cover of X, Since the space satisfies the sec-

ond axiom of countability, there exists a countable open

base for t, say a. Let £ « {K for some A in G, KCA and

38

&f a), 6 is a cover of X and p O « There exists a finite n

ssubset {BX,B2» ..«*%} of p such that covers X. For

each if I4", B^Ch for some H in 0, Denote one such H by n

Gl» Then U Qi is a cover of X, and if i e I4", then % € G « jtssl

Therefore, (X,t ) is compact.

CHAPTER BIBLIOGRAPHY

1. Kelley, John L., General Topology, Princeton, Hew Jer-s@y» *>• Van Nostrand Co., Inc., 1955.

39

BIBLIOGRAPHY

JCelley, John L. , General Topology, Princeton, N, J., D. Van Nostrand Co., Inc., 1955.

Mansfield, Maynard J., Introduction to Topology, Princeton, N.J., B, Van Nostrand Co., Inc., 19fc>3.

40