INVERSE LIMIT 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

Stephen Boyd Williams, B. A.

Denton, Texas

December, 1974

N,,

Williams, Stephen B., Inverse Limit Spaces. Master

of Science (Mathematics), December, 1974, 107 pp., biblio-

graphy, 6 titles.

Inverse systems, inverse limit spaces, and bonding

maps are defined. An investigation of the properties that

an inverse limit space inherits, depending on the condi-

tions placed on the factor spaces and bonding maps is made.

Conditions necessary to ensure that the inverse limit space

is compact, connected, locally connected, and semi-locally

connected are examined.

A mapping from one inverse system to another is defined

and the nature of the function between the respective inverse

limits, induced by this mapping, is investigated. Certain

restrictions guarantee that the induced function is contin-

uous, onto, monotone, periodic, or open. It is also shown

that any compact metric space is the continuous image of

the cantor set.

Finally, any compact Hausdorff space is characterized

as the inverse limit of an inverse system of polyhedra.

PREFACE

The purpose of this thesis is to examine the inverse

limit spaces of inverse systems and to use inverse systems

to characterize some well known topological spaces.

In Chapter I, inverse systems, inverse limit spaces

and bonding maps are defined. An investigation of some

of the properties that the inverse limit of an inverse

system can be forced to inherit from the factor spaces

of the system is made.

A mapping from one inverse system to another is de-

fined in Chapter II and the relationship between the

inverse limits of these systems is examined. The second

chapter will culminate with the use of inverse limit

systems to prove that any compact metric space is the

continuous image of the cantor set.

In Chapter III, simplices, complexes, and polyhedra

are defined. Some preliminary theorems are proven, then

it is shown that any compact Hausdorff space is homeo-

morphic to the inverse limit of an inverse system of

polyhedra.

Many definitions and theorems of general topology

such as those found in Willard's General Topology

are used without reference. Most of the theorems in

this paper can be found in Capel's "Inverse Limit Spaces"

iii

Gentry's "Some Properties of the Induced Map" and Nagata's

Modern General Topology. However, all proofs are original.

iv

TABLE OF CONTENTS

PagePREFACE . ............. ....11l

Chapter

I. INTRODUCTION............ . 1

II. MAPPINGS BETWEEN INVERSE LIMITS,...... 41

III. AN INVERSE LIMIT CHARACTERIZATIONOF COMPACT HAUSDORFF SPACES*. ...... 79

BIBLIOGRAPHY.. . .. . . . ....... 107

V

CHAPTER I

INTRODUCTION

The intention of this chapter is to examine the proper-

ties inherited by 1lm(X ,f ,D) from the factor spaces of

HX . To accomplish this, some preliminary definitions and

theorems are needed.

Definition 1.1; Let {X }aD be a collection of topo-

logical spaces, indexed by the directed set D. For a : Sin

D, let f :X +a X be a continuous function such that

f :X -* Xa is the identity function, and for ac5y 5f in D,

f = f a0fy. Then (Xa3 f ,D) is an inverse system and

f is called a bonding map.

Let H Xa be the topological product of the X .s, here-acD

after denoted by fXca when no confusion may result. Let

P a:HX + X be the a-th projection function. For x llXa'

let P (x) =x

Definition 1.2: Let X = {x eX a|for ca:5 in D,

f (x ) = x a } and give X, the relative product topology.

Then X., is called the inverse limit space of the inverse

system (Xaf ,D). For a c D, let 7r = POaIXXOO + X .

Then Tr is continuous. For f 6 D, let

S = {x 6Xalif as 3 in D, then f (x X l) = .

Then X = sFS.D

1

2

For notational purposes, let X = lim(Xa , f D).

Whenever D is the positive integers, hereafter denoted by

Z+, then (Xt If5z +) will be called an inverse sequence.

Unless otherwise specified, the hypotheses of the theorems

in Chapter I will assume that (X3f ,D) is an inverse

system and X= lim(X ifD).

Theorem 1.1: If ao5 in D, then f o r = 7.

Proof: Let x e X . Then f 0 o TF(jx) = f ( 7(X)

Since (x) = x ,fa TrX) f 6).x But x c X*. There-

fore, f (x ) = x . Since T[O (x) = x , f o7T(x) x .

Hence, f,,07 = 7Ta

Theorem 1.2: A base for X, is

{7ra) 1 (U)a D and U is open in Xa

Proof: Let x c X. Then, for a E D, xU . X . There

exists an open set Ua of X such that xa 6 U. Therefore,

x 6 Tj(U). Hence, X = U 7j~1(U). LetacD

x 6 7T~(UT) Y 1f ~(Us)

where U a and U are open in X and XS respectively. Since

a and 6 are in D, there is 6 e D such that a:5X and 5X.

Let U = f ~(Ua) fx ~1(Us). Then U is open in X and

x U . Hence, x 6 rx1 (Ux). If y e A- 1(U), yX E Ux.

But fx(yx) = ya. Therefore, Y. 6 U. Hence, y c '7a 1(U ).

Similarly, y c Nr~1(U ). Consequently,

7TX_1(U X)C:7T ~1(U 06 n7T (U ).

3

Therefore, {T'(UQla)ea c D and U is open in Xa} is a base

for some topology on X. Let V be a basic open set of HX

Then, V = n P "(UcU) where F is a finite subset of D andaeF O

U is open in X. Let H be a basic open set of X,. Then,

H = X OrW where W is a basic open set of HX,. Then

w = nP a 1(U ). There exists f3 E D such that a:5 for allaF-E

* e E. Let U = fn f t(Ua). Then U is open in X.Let

x c H. Then x e X. W. Since x e W, xa 6 U for all a E.

Since x e X, x f ~'(Ua) for all a E. Therefore,

x U . Hence, x c TrK-(U ). Therefore, HSr Erf(U3

Let z 71(U ). Then z e XO and z6 U. Hence,

z a flfnf (Ua). Therefore, f (zn) e U for all a E.

Since z e XC,, f (z ) = z . Therefore, za C U for all

a 6 E. Hence, z s enP . (U) . Consequently, z F W nX,.

Therefore, T 1 (U ) = H. Now,

{TrO 1 (U)1ao c D and U is open in X }

is a base for X

Theorem 1.3: The collection {S }aFD is descending in

the sense that if a and are in D, then there exists \ in D

such that ScS qS,.

Proof: If a and 0 are in D, there exists A D such

that as A and 5X, Let x c S . If a F D and a A, then

f CxA) = x . Let 6 !3 in D. Then 6 A. Therefore,

14

= f X6(x )=f 0o f X (x )=f (X~)

Hence, x e S . imilarly, x 6 S . Therefore, S

Theorem 1.4: If each X is Hausdorff, then each S

is closed in Xa. Hence, X, is closed in HX.

Proof: Let x E IHX \S Then there exists "A e D such

that A <a and fa(x ) x . Since X is Hausdorff, there

exists open disjoint sets U and V in X, such that x X U

and fX (xa) E V. Now, f ~j'(u) nf ~'(V) = 4 and each is

open in X , Since f (x ) e V, xaE : fx 1 (V). Hence,

X E Pj'( faj'(V)). But xs6 U. Therefore, x e P X(U).

Let H = Px_(U)Q)P'1(f fl(V)). Then x c H and H is open

in HX . If y c S. H, then P (y) c f ~1(V). Hence,

Y = f (ya ) e V. But y e PX~ 1 (U). Therefore, y c U.

Consequently, Y c UqnV. This is a contradiction to

UqV = $. Hence, S a H = p. Therefore, IX \ S is open.

Thus, S is closed. Since X = q ns, X, is closed in RX

aF-D

Lemma 1.1: If 4 is a collection of compact, nonempty,

connected sets in a Hausdorff space, X, such that for each

A and B in 14 there is s ome C in 1 s uch that CcAqB, then

rni is compact, nonempty, and connected.

Proof: Since each C c ?( is compact and X is Hausdorff,

each C c 2 is closed. Hence nu is closed. Let C 6 2.

Then nu CC. Hence, n2 is a closed subset of a compact

set. Therefore, rn u is compact.

Suppose that n is empty. Let V 6 2. Then, for

x 6 V, there exists U x 6 such that x U . Since {x} and

5

U are compact and disjoint, there exists open disjoint sets

H and K in X such that x 6 H and U K . Letx x x x~ xH = {H |x F-V}.

Then H covers V. Since V is compact, there exists M, a

finite subset of H, covering V. Let M ='{H ,H ,...,H }1 2For each H. s M, consider the corresponding Ki and the

m m m mU cK . Then qU q K. But ( nK.)((UHi) =

= i=l i= i=l

m mHence, since Vo .UHi, V n( nK.) = . Therefore,

i=l i=l

mv ( q U) =

i=l

This violates the hypothesis that the finite intersection

of elements of 14 contains an element of -. Hence, ni4 7 .Suppose that Q2 is not connected. Then there exists

A and B, each closed in rw, such that A mB = p, A / $ / B,

and Q2= A(UB. Let C c . Then f .sC. Therefore, AE C

and B cC. Since A and B are closed in fl4 and n( is

compact, A and B are compact. Now AnB = 4 and X is Haus-

dorff. Therefore, there exist open disjoint sets W and V,

of C, such that A oW and B,=V. If x c C and x [ , then

there exists C V such that Cx _cC and x ( Cx. Hence,

x E C\C X which is open in C. Therefore,

M =*{C\Cx such that x c C\qn{}uf{(Wv)} covers C.

Since C is compact, there exists a finite subcollection N

of M covering C. Now, (W UV) must be an element of N.

6

Let the other elements of N be denoted by C\C ,C\C ,...,C\C1 2 m

mThen there exists C t c? such that C'c qn0c. If x C C',

i=1

x C\Ci for any i. Hence, x 6 W UV. Therefore, C'-.;W UV.

If C' cW, then n ?..( SW. But n.W = A UB. Hence, A UB.L=W.

However, B CV and wIv = $. Therefore, B = $. This is a

contradiction to B # (. Hence, C' .W. Similarly, C'4 V.

Now C, W (p C' rV. Therefore, since (C W) U(ca v)= C',

(C ,qw) n((Cqnv) = (p and each of C' TW and C'QV are open in

C', C' is not connected. This contradicts the hypothesis

that C1 is connected. Therefore, fv is connected.

The following example illustrates the necessity of,,

compactness.

Example 1.1: For each n c Z+, let Xn be the box in R2

with corners at (0,0), (1,0),(0,1) and (1,) such thatn n{(x,0)|0<x<l} and {(x,1 )|0<x<l} have been deleted. Direct

{X }n= by set inclusion. Then {X }= = satisfies then n=l n nlhypothesis of the lemma, except for compactness. However,

ne = {(0,0),(1,0)}. Hence, )C is not connected.

The preceding theorems now bear fruit by assisting in

the proof of Theorem 1.5.

Theorem 1.5: If each X is compact and Hausdorff, then

X is compact. If, in addition, each X (p, then X (C .

In addition, if each Xa is connected, then X is connected.

Proof: If XU is compact and Hausdorff for each ca c D,then HiXa is compact and Hausdorff. Hence, by Theorem 1.4,

7

X0 is closed in HXa. Consequently, since a closed subset

of a compact set is compact, X, is compact.

Suppose that for each a e D, X a p. Let a c D. Then

there is xa 6 Xa, For each c D such that f a,

f (xa) 6 X Let x HX such that for each f :a in D,

X f (x ) and, for y e D such that y$a, x is some ele-

ment of X . Then for 6 <a in D, f (x ) = x . Therefore

x c S . Hence, S O $ for each a. Thus, by Lemma 1.1,

XO / $.

Suppose that X is connected for each a E D. Let

a 6 D, Suppose that S is not connected. Then Sa = A UB

where A and B are closed in S . Further, A 1B = $ and

A ; $ 7 B. Since A and B are closed in Sa, they are closed

in HX . If x e A and y 6 B then, x F HX \ B, which is open.

There exists a basic neighborhood V, of x, such that V is a

subset of HX \B. Let

V = E P (V ),3sF

where V is open in X and F is a finite subset of D. If

P 6 (Y) Cv6 for all 6 e F, then y c P6 1 (V6) for all 6 EF.

Hence, y 6 n PTV ). This contradicts the suppositionSEF

that y c B. Therefore, there is some 6 c F such that

PY(x) 1 PY(y). Since X is Hausdorff, there exist disjoint

open sets H and K of X, such that P (x) E H andY Y Y Y Y

P (y) F KY.Y Y

8

Let ,Xca for all ( e F. Then fx j1 (HY) qf ~' (K ) =

and P Cx) P (71.,H ence P CA)nP (B) = $. Let s c X .

Let t be an element in IIX such that P (t) = f (s) for each

<a in D. Then t c S . Hence, P (S ) is onto X . If

h ca in D, then S _oS. Since A>a, SX.S for each a c F.

Therefore, S_ A UB. Hence,

Xx = P(S x)cPx(A UB) cPx(A) UPx(B).

Since P is continuous and A and B are compact, Px(A) and

P (B) are compact. Hence, Px(A) and Px(B) are closed in

Xx. Therefore, X is not connected. This is a contradiction

to the supposition. Hence, S is connected. Therefore,

by Lemma 1.1, X., is connected.

It is noteworthy to point out that as a result of

Theorem 1.5, if each factor space is a continuum, then X0

is a continuum.

Theorems 1.6 through 1.11 will be necessary to estab-

lish the background for proving Theorems 1.12 through 1.16.

Theorem 1.6: If X0 is compact and Hausdorff for each

a in D, then

7Ta(X) = q{f (XS)ja:5 in D}.

Hence, if each f is onto, then 7 is onto.

Proof: Let z c r (X). Then, there is z F X such

that Tr(z) = z . Since z E X , if a in D, f ~z )._x

Hence, z f(X ) . Therefore, z a E: {f (X )jac5 3 in D}.

Let x e {ff (X )|a 53 in D}. Suppose that x a y (X ),

Then P'-(xa)qnx.=0p. Therefore, if x C Pa~1(x), thereaa

9

exists 2. s D such that x { Sx. Hence, x 6 HX\S2 . Letx a x

K ='{A|x P 1(x )}.

Then,

M O= {IX \ A }x x

covers P-(x ). Since PJ1(x ) is closed in HX, it is

compact. Therefore, there exists a finite subcollection

7/ of?7 covering P a'(xa). Let

7/ =OIXS\ S such that 2 X6M},

where M is a finite subset of D. There exists a E D such

that A a for each A e and a5a. Since a , a,

xa Cfa(X ). There exists xa c X such that x = f (x ).Let y c IX such that y is f (x ) if A!5a and some point

in X if Xt;a. Then y F S and y 6 P ~1(x ). Since

y C PaI(xa), y HX1\S c ?/. But Aa. Hence, S _CS .Therefore, since y 6 S ay 6 SX. This is a contradiction

to y EHX \S . Therefore, PJ~1(x )q X 0 0 $. Thus,

Sa :a ( -Hence, q{f, (Xe)|a J z in D}cira(Xc,) .There-

fore,

Ta(X )= O {fo (X )|a 'f3 in D}.

Since each f is onto, f (X ) = X for each ca5 in

D. Therefore, 7%a (X ,) = Xa.

To settle an obvious question, it will now be shown

that if each f is onto, then f (X ) must be onto X for

each a j5. If not, there exists f 6 D such that f (X ) is

10

not onto X . Hence, there exists x c X such that

f (X ) rx} =

But each 7r is onto. Therefore, there exists y 6 X00 such

that 7T (y). =x, However, by Theorem 1.1,

foaa C7 (Y) = 70cjY-Hence,

f ( OT(y)) =X.

This contradicts the supposition that

f (X) Qn{x} =

Therefore, f (X ) is onto Xa.

Theorem 1.6 is important because, as the following

example shows, there are nonempty inverse limit spaces such

that the 7 function does not map onto each factor space.

Example 1.2: For each n E Z+, let Xn be the closed

unit interval. Define f X X by f (x) = andn+ln: n+l n n+l n 2f nn :Xn Xn to be the identity map. For m>n, let

mn Xm+Xn be fn+l n0f n+2 n+l 0' 0' f m m-1. Since each

mn is continuous, (XnfmnZ+) is an inverse system. For

each n, fn+l n(0) = 0. Hence, v = (0,,0,...) 6 X. There-

fore, XC O p. If n' l and x 6 [0,1], fn1 (x) = 2n-lx.

Let y = (p,2p, ...,2 n-1 p),...) be an element of X, such

that y # v. Then p # 0. Hence p>0. There exists me Z

such that 2ml(P)>1 . Therefore,

,(y 1= (x)>l.

This is impossible since 7Tm(Y)cE[,]. Therefore, y = v.

11

Hence, X, has only one element, namely v., Now, since

V CQ,0,.,.1, T.nCX1 ={O} for each n, Hence, TJn(XO) is

not onto for any n.

Lemma 1.2 Let 0 be a collection of compact subsets of

X, a Hausdorff space, indexed by the directed set D, such

that for a and f in D, there exists A D such that

X S (),q . Then, if U is open in X and o2cU, there is

C E a such that CSU.

Proof: Suppose that for each C 6 3, C U. Then for

each a 6 D, there exists xa ECa such that x U. Hence,

for each a, X\U nC, 4 . Since U is open, X\U is closed.

Therefore(X\U)qca is closed. But since C. is compact

(X\U)OCa is compact. For a c D, let H =(x\U)nac. Let

H = {Haja F D}.

If H. and H are elements of H, there exists 6 c D such

that C&. Ca oC ,"Let x 6 H6. Then, x s(X\U)(OC 6 . There-

fore, x c C6 and x 4 U. But since x c C6, x c C. 'Since x i U, X 6 (X\U)OC and x c (x\u)qc Therefore,

x e H .nH3 6-Hence, -c;Ha H .V By Lemma 1.1, H $.

Let y c H. Then y U. But if y c H, y e Ca for each

a. Therefore, y c n . Now qacU. Hence y s U. This

contradicts y { U. Therefore, there is C c C such that

C CU:

The following theorem is now made easy by Lemma 1.2

Theorem 1,7: If X. is compact and Hausdorff for all

a c D and K is open in i X such that X OcK, then there

12

exists a ED such that

XO cS c K.

Proof:, Since each X is compact and Hausdorff, llX

is compact and Hausdorff. Since each S. is closed in IX

each S is compact. By Lemma 1.2, the theorem is proven.

For the remaining theorems of Chapter 1 it will be

assumed that if (X 3f ,D) is an inverse system, then each

X is Hausdorff.

Theorem 1.8; Let A be a compact subset of X,

Au = 7a (A), B = HA, and g = f |A .Then the following

are true:

(a) (A , g ,D) is an inverse system and eact gq is

onto.

(b) If A = lim(A ,g D), then A = Bf'X=A.+o fca O 0

(c) X\A = U{ r 1(X \A )|ac D}.c a a

Proof:

(a) Since f is continuous, |A is continuous.

Therefore, each g is continuous. Let x c A . Then there

exists y e A such that y = x. Therefore, if Sa in D,

then

g (x) = f (y ) = y

Since y e A, y F A Hence g A + A If x c A ,

gc(x) = f (x) = x.

Therefore g Is the identity map on A. Suppose that

a 5A 5 in D, Let x e A V Then there exists x E A such

13

that iT Cxl = x . Since x c A, g9(x) = f (x ). But

x 6 X Therefore,

fV (x ) = f o f (x ) = f (x ) = x.

Since x c A1 ,

f Cx g = g (xX

Hence,

Since g (x) = x and xX E A,

Therefore,

g (x ) = g 9 gX(x

Hence, (A , g ,D) is an inverse system.

Let a<1 in D and x 6 A . Then there exists x in A

such that = Tr(x). Then,

Tr (x) = x F _ A

Since x X, f (x ) = x. But

g9j(x f) = if(x x =a

Hence,f is onto.

(b) Let i be the a-th projection function of HA

restricted to A , Let x 6 A. Then for each a and 1 in D

such that cc 1, g1 jx ) = x(a. Hence, f (x =x . There-

fore x e X. Since x c A0, x HIA . Hence, x c X nB.

Therefore, A -c XO B,

Let x 6 B nX, . Suppose that x A. Then for each

y 6 A, there exists A E D such that x y . Therefore,y y y

14

there exist disjoint sets U and V, , open in X. , suchy y y

that x EU and y Vx . Now, each of 7T (U )y y y y y y

and f ~(V 1 is open in A and their intersection isx . x 00

y y

empty. Let

{ I (V )jy 6 A}.

Then W covers A. There exists a finite subset V of *

covering A, Let

7 = { J (V ))IA }

where M is a finite subset of D. There exists a D such

that acya for all a e M. For each A M,

g IuX)ng (Vx) = e.

Recall that U is the open set of X that is disjoint from

V and contains x . If z e Aa, there is z e A such that

S(z) = z. Since z A, z ' (V) 7. Hence, zX VA*

Since z 6 X, z a E g aj,(VX). Therefore {g x 1 (Vx)IA M

covers A , But x )g (Vx) for any A M. Therefore,

x a Aa, But if xg G Aa then x { HA . This is a contra-

diction to x E B. Hence x F A. Therefore A X_ qB cA.

Let x e A, Then x FXs and for each aE : D, x eA .

Therefore, x E Xq B, Hence AcX OB.

Let x F BnX. Since x c B, xc A for each a.

Since x 6 XCO,, for a: , f5x) =fx . But x 6 A . Hence,

g a(x = f a(x )=x a .

15

Therefore, x F A, Thus AX qBGA,. Hence,

A0 = X O B = A.

(c) Let x E X,0 \ A. Suppose that for each a sD,

x A . Then x c TA .nX. By part b, x 6 A. This con-

tradicts x e XQO\A. Therefore, there exists A E D such that

x A.-Hence, xX E XX\AX. Therefore, x C T[A 1 (Xx\ AX).

Therefore, x 6 U{ '(Xa\Aa)ja D}. Hence,

Xw , A E U (6 1(Xa\Aa)a E D).

Let x 6 U{rj-(Xa\Aa)Ia e D}. Then there exists

A c D such that x 6 r 1 (XX\AX). Hence, x AX. There-

fore, 7Xj1 (xX) OA = 4. Since x 6 Tr-j(x,), x A. Hence,

x F X \A. Thus,

X,\ A = LTr a'(Xa\Aa)I|ac D} .

Theorem 1.9: Let A and B be compact subsets of X,,

C = AqnB, C = a(C), and h = C Then

C = lim(Ca~,h ,D) = C..

Proof: By Theorem 1.8, part a, (ca, h ,aD) is an in-

verse system. Let 7a (A) = Aa and F a(B) = B . for each

a F D. For a:" in D, let g, = f and j = f W B

Then, by Theorem 1.8, part b, A0 = lim(A, ,g,D) = A and

50= lim(Ba VD) = B. Let x S Co.. Then for each a,

x C. Therefore, xa F Acx , Baand for a:5 in D,

hO(x ) = f C(x$) = fWA BCx) = Xa

Therefore, x c A00 and x Boo. Consequently ~c B,. Let

x c AqB,. Then for @ in D,

gPCx)( = Cxx) =x .

But

g(x ) = f A (x ) = Xa

and

j (x ) = ff3 B(X ) = x

Hence,

X f O A qB C x f3 tICjxf)

Therefore, x 6 C . Now, A OB = C . Since A = A and

Bo = B, C0 = AnB = C. Hence,

C = lim(C ,h ,D) = C0 .4- a O 0

Theorem 1.10: If E is cofinal in D, then (X ,f ,E)

is an inverse system and X = lim(X ,f ,D) is homeomorphic00 + Qa t3

to Y = lim(X ,f ,E).

Proof: For each a t in E, f 3 is continuous and f

is the identity map on X If a56 5 in E, then a56 t3

in D. Hence, f = f o f1 . Therefore, (X ,f ,E) is

an inverse system. Let 'ir be the a-th projection function

of II X restricted to X, and r be the a-th projectionacD

function of II X restricted to Y . Define a relationactE C0

g;X + Y by g(x) = t where for all A E, f (t) = (x).

It is apparent that g:X +÷Y,, for if x X and a:5 in E,

then f (t ) f Cx ) = x = t . Hence, t c Y . If

g(x) = w and g(x) = t, then for all a cE, i (w) =7r (x)= (t).T o=nin

Therefore, w = z. Hence, g is a function.

16

17

Suppose that x y in X . Then there exists X s D

such that xX y . Since E is cofinal in D, there exists

6 E such that A 56. Hence xX yX. Therefore,

Hence, g(x} ) g(yL. Therefore g is one to one.

Let y c Y If a 6 D, there exists a'c E such that

a5 at. Let x 6 II X such that for each a sCDcD a

71 a x) = f aO(7r I(Y))

If a' and a" are elements of E such that ac5 o' and c 5oa",

then there exists ac" in E such that at <ct"' and a" l<a"'".

Then,

But

fct 0 f y = f ? (i n(y))Ot Ot a Ot a a a

and

a a O 1 t,, =' f tf YOtI ?,(Y))

Therefore,

fc 7T aGi t(~Y)) f= tOt 'o Ct (Y

Let a f in D. Then,

HF(X)n=f(Y)

Hence,

f ((x))OTf (f (To (

f f O'=t y f ,(-f ,(y)= T ( x)

18

Hence, x 6 X. and

g(x) = y.

Therefore, g is onto.

Let x e X0 and g(x) = y in Y. Let U be a basic open

neighborhood of y in Y. Then U = Ta '(U ) where U is open

in Xa and a c E. Now a (y) = Tra (x) . Therefore, U is open

in X and a (x) e U. Hence, x 6 '7r -(U ) which is open

in X, Let Tra '(U ) = Ut . If s E U1, (g(s)) a U . There-

fore, g(s) T rij(CU 0 ). Hence g(U t ).U. Consequently, g

is continuous.

Let y e Y . Then g'(y) = x 6 X. Let U = r ~1(U )

be a basic neighborhood of x in X00. Then, 'a (x) c Ua. But,

by the way in which an inverse element was found in proving

g onto, Tr (x) =f a( ,(y)) where caz5a t and a' E. There-

fore, 7a (y)F- f 1 a(U ). Let

f ~1(U ) = U '.

Hence, y ' U ). Since a' c E, a , 1(U 0) is open

in YO. Lets 6 7 '(U,) . Then g~(s) c X0 and

I (s a) C U ,. Hence,

f TECs(0)) = I(g-'(s)) E US a lota t a

Therefore, g-'(s) 7 ~t(U ). Now, g~(F ,71 (U ,)) c 1(U )

Hence, g~1 is continuous. Therefore, g is a homeoporphism.

Hence, Y. is homeomorphic to X,.

Definition 1.3; A continuous function. f, from Y onto

Z, is monotone if and only if for each p E Z, f~ 1 (p) is

connected.

19

Lemma l,3; A continuous function f from Y onto Z,

where Y is compact and Hausdorff and Z is Hausdorff, is

monotone if and only if for each connected set C in Z,

f (C) is connected in Y.

Proof: Suppose that f is monotone. Let C be a set

connected in Z. Suppose that f'1(C) is not connected in Y.

Then, f~'(C) = AUB where A( B = p, A $ / B, and A and B

are closed in f '(C)., Let p c C. Suppose that

f -,(p) n A $/f~-,(p) n B.

Then,

f~(p) = (f(p)nA) U(f-1 (p) nB).

But,since A and B are closed in f1'(C), A = H f 1(C) and

B = Knf-'(C) where H and K are closed in Y. Therefore,

f 1 (p)qnA = f 1 (p)qnH,

fc(p)qnB = f~1(p)flK

and each is closed in f~1(p). Hence,

f'(p) = (f~l(p)nH) u (f-1 (p)qnK).

But (f~'(p)qH) and(f' 1(p)q K) are disjoint and nonempty.

Hence, f '(p) is not connected. This is a contradiction to

the supposition that f is monotone. Therefore, f~1(p)c A

or f 1 (p)cB.

Let

M = {p E CIf~I(p)c A}

and

N = {p CJf 1 Cp)_cB}.

20

Then MN =, C = MU N and M # # N. Suppose that M is

not closed in C, Then, there exists p E C\M such that,

if U is open in C and p c U then UN M X$ Since Z is

Hausdorff, for each y c Z such that y p, there exist

disjoint sets Uy and Vy, open in Z such that p 6 Uy and

y F V . Since f is continuous, f'(Vy) is open for each y.

Therefore,

{ ={ff'(Vy)jy # p}

covers Y\f_'(p). Since p M, f~'(p)c B. Suppose that

y c f~ 1 (pffiH. Then, y E f'1(C)q nH. Hence, y E A. But

y t f~(p) c B, Therefore, A nB 4. This is a contra-

diction to AnB = $, Hence f~'(p)qH = 4. Therefore,

HE Y\f'(p}. Hence, ( covers H. Since H is closed, it is

compact. Therefore, there exists a finite subcollection

of V covering H. Let

= {f(V Y)Ii in}

nbe this finite cover. Furthermore, let U = n U

i=l Yi

Then, p 6 U and U is open in Z. Hence, Un C is open in C

and contains p. Let W = US C. If W #M X$, then there is

q c Wq M. But q c M. Therefore, f~'(q) _A. Since q C W,

nq E r U , it follows that f~ 1(q).Gf" 1(U ) for each i.

I=1 i

Hence,

nf W7I(q)gn f-il(U )

21

But,n

( q f 1(U )) A = d.1=1

This contradicts the fact that f-1 (q)._A. Therefore,

W m = p. Hence, M is closed in C. Similarly, N is closed

in C. Therefore, C is not connected. This contradicts

the supposition that C is connected. Hence, f'1(C) is

connected in Y.

Suppose that f' CC) is connected in Y for each C

connected in Z. Since {p} is connected in Z for each p,

f '(p) is connected in Y. Hence, f is monotone.

Lemma 1.4; Let X= lim(X ,f ,D) where each X is

compact and Hausdorff. If A B = ( and each of A and B

are closed in X, then there exists 6 e D such that

76(A) nTr 6(B) = (.

Proof: Let x e A. Then, for each y c B there exists

A e D such that x # y . Since X is Hausdorff, there exist

disjoint sets U and V , open in XV, such that x e U

and y EV . Let

?'= {7T 1(V )|y cB}.

Then, ?r covers B. Since B is closed in X , B is compact.

Therefore, there exists a finite subcollection ofrcovering

B. Let

r{T[- 1(V ) 1 F}

where F is a finite subset of D, be this finite cover.

22

There exists a E D such that at y for each y E F. Let

H = U{fj' ()|X F}

and

K = U{fcxj' (UJ)|X c F}.

Then each of H and K is open in X and H nK =a a a a aHowever, x a K and 7,r (B) H . Hence, for each x c A,

there is 3 c D such that there exists M and N disjointx

and open in X, such that x F M and 7 (B)cN . LetSx 13 -x

T{1(M )|x E A}.

Then Z covers A. Therefore, there exists a finite sub-

collection 9, of 4 covering A. Then, there exists 6 e D

such that, if a c D and 1(m ) e 9, then 6 !a. Leta x

U = q{f ~1(1 )|r ~1(M ) e #u{ , ( x a x '

Then U is open in X6 and 76(B). U. Let

V = U{f6 (N )|r '(7 ) E}.

Then, V is open in XI and Tr6(A) cV. If s E U, then

f 6(s) c NY for each N such that ia ~1 (Mx ) c . There-

fore, f6y(S) 4 m for each M such that 1( G.

Hence,

f 6U1(f (s))f '(M ) =6f oSa a x

for each NM such that 'r '(M ) 9. Therefore, s 4; V.

23

Similarly, if e V, U. Hence, Un V =. But,

7T (B} U and 71 (A) 1&V . Therefore,

T6(A)rn7T 6 (B) = $.

Theorem 1.11: Let (X ,f ,D) be an inverse system of

compact Hausdorff spaces such that each f is monotone.

Then, each i-r is monotone.

Proof: Suppose that there exists a s D such that ff

is not monotone. Then, there exists x c X such that

7 aI(x ) = A UB where each of A and B is closed in 7T (-'(x )and A(B = 4 and A 7 4 7 B., Since A and B are closed in

7 aI(x ), they are closed in X, and IIX . By Lemma 1.4,

there exists 6 6 D such that 74(A) Rr)7 6 (B) = $. Since X6is Hausdorff, there exists disjoint sets H6 and K, open

in X6., such that 6 (A) cH6 and ff6 (B) .K6. There exists

a F D such that 6 ca and ac a. Let

H=f 1(H6

and

K = f ~(K

Then, HnK = and each of H and K is open in X . If s c A,

sa a ( f 6 (H6 ).If s c B,

s E f ~1(s )cf 1(Ka 6 6 a (K6

If pa 6 fa U1(x ) then there is p e X such that 7a (p) =p .

Now,

p = f (p ) = f (x ) = xS Ga a aot, a

24

Therefore, p E x ). It then follows that either p 6 A

or p E B. Hence, p H or p c K. But, this means that

f !(,x ) is separated by the nonempty, disjoint, open

sets H and K. This contradicts the supposition that f., is

monotone. Hence, 71 must be monotone,

Definition 1,4: A space Y is locally connected if for

each y c Y and each open set U containing y, there exists

an open connected set V such that y e VZU.

Definition 1.5: A space Y is semi-locally connected

if for each y E Y and each open set U containing y, there

exists an open set V such that y e V.GU and Y\V has only

finitely many components.

For the remaining theorems of Chapter 1, it is assumed

that each f is monotone.

The following theorem returns to the examination of

properties passed to X by the factor spaces.

Theorem 1.12: Let (X ,f ,D) be an inverse system of

compact Hausdorff spac es . If for all and ain D, f is

mono t one and X6 is locally c onne ted (semi-loally c connected ,

then Xead is locally connected (semi-locally connected).

Proof: Suppose that for all f and a in D such that

a, f is monotone and X is locally connected. Let

x e6 X and U be open in X00 such that x c U. Then, there

exists a basic neighborhood of x, 7Tr 1 (U ), such that

Tr (U ) _CU. Since X is locally connected and x E UX,

there exists V , connected and open in X such that

25

xx 6 VXU , Since Tis monotone, 1Tr1(V,) is connected.

Butx 7T ('V )_qj ~(Ux)SU. Hence, X is locally

connected.,

Suppose that for all ( and a in D such that f3 a,

f is monotone and X is semi-locally connected. Let

y c U where U is open in X. Then, there exists a basic

neighborhood, ra 1(U ), of y such that Ra '(U.).9u. There-

fore, ya c U and Ua is open in Xa. Since Xa is semi-

locally connected, there exists V open in X such thata a

ya a CUa and Xa\U has only finitely many components.

Let C2,,C2 , ...,Cn denote these components. Now,

y E 7T '(V ) CU.

Suppose that there exists more than finitely many components

of X0\ 7TaI(V a'). Then, there exists C, a component of

X,\ 'ij(Va,), such that C X TFa '(Ci) for any i. Since

Tr a (Ci) is connected for each i, if CC'rr~'(C.), for some

j, then

C = T '(C.).

But C f (C ) for any i. It then follows that CP 'c.)(for any i. However, 7Tr (C) is connected in X and

Ta (C) SX \V . Hence, 7T (C) EC for some j. But,

a, '~, ,a, a JCC gT OT (ra(C))-SIa,'(C *

Therefore , C C1f (C.). Consequently,

C = a ~'(C.).

This contradicts C a '(C.). Hence, X is semi-locally

connected,

26

The following example illustrates that Theorem 1.12

will not be true if each f is not monotone.

Example 1.3: For each n F Z+ let Xn be the unit

circle in the complex plain. Give Xn the relative topology.

An element of Xn e 1, will be denoted, in degree, by 0.

Let fn+1 n; Xn + Xn be defined by

fn+l n(e) = 20

and

f (0) = 0,

Then each f n+1 nis continuous. For m >n, let f :Xm + Xn

be defined by

fmn e) = fn+l n 0 fn+2 n+l o fm m-1(8).

Then (Xn fmn',Z+) is an inverse sequence. Hence, an inverse

system, Let

X = lim(X ,f ,z ).+ n mn +

Suppose that X, is locally connected. Let U be an

open, connected, nonempty subset of Xn not containing 00.

Then,

U = {a EX n < t < S}

where 00-<5 and 6:53600. Let

H = {y n+

and

K=(YcXr+ 1 if- + 1800 < y < + 1800}.

Then H and K are disjoint regions and 01 H UK. If 0 U,

then < e<a .,But fn 1(e) = + 1800}.n+1 n 2 2'

27

Hence, < '< -and + 18O< + 18o0<p6 + 180' . Therefore,2 7 f2 2

fn+1 n (e) L;HU K.

Consequently, fn+ n '(U).SHUK, Furthermore, fn~l nn+l n

meets both H and K. If a E H U K, then a E H or a K.

But if r F_ H, then (< $.< Therefore, 3< 2 a< 6. Hence,

<f n+i nCa<6, Consequently, fn+l n(a) c U. Also, if

a P K, then + 1800< a< + 180, It then follows that

2( + 1800)< 2a< 2( + 1800). Hence, < 2a < 6. Therefore,

fn+l n(a) e U. It has now been shown that

fn+l n(U) = H UK.

By induction, f (n 1-(U) is the disjoint union of 2n-l

mutually disjoint regions. Further, if X c U, fn 1 (x)

meets each region. Let z 6 X . If p c Tn (7~1(z1)), then

7n (P)r r1 I(z1 ) A q. Therefore, there exists s 6 X such

that s c Tn -(P)nq 1 '(z 1 ). Now sn = p and s = z .

Hence, fn 1 (P) = z1 . Therefore, p c fn 1 '(z1 ). It then

follows that 7 n r 17(z 1 )) Cfn 1 ~1(z1 ). Let p c fn 1 ~1 (z1 ).

Let s eX0 such that sn=np. Since s = fn n (p) = z ,

s F Tr 1 (z1). Consequently, in n(s) r ( ~i 1(z )). But

sn = p. Hence, p 6 n- (.I7(z1 )). Therefore,

nl 1 1(z1)) = n 1

Now, U can be chosen such that 1800 U. Then,

7F " (1800).s7 1 1(U).

Since X, is locally connected for each z c 711(1 8 0 0 ), there

28

exists V an open connected subset of X such that

z 6 V z-Tj(U, Then,

?r= {VzIz & 1-1 (1800)}

is an open cover of 7 (1801). Since 7 1(1800) is closed

in X., it is compact. Therefore, there is a finite sub-

collection 2,of V/, covering Tr1 (18 00). Let

'{V ,V2 ''''Vm .

There exists n e Z+ such that 2n-1>m. Now, fn ~1(U)

is the union of 2n-1 disjoint regions and fn 1 180)

meets each region. For each i, 7T (V.) is connected.

Therefore, n n(V.) is a subset of one of the regions.

Since there are more than m regions, there exists at least

one region R, such that n(Vn ) OR = for each i. But

m7-1(1800) c U V.

i=l

Therefore,

n T1 (18oo)) OR = q.

However,

Sn OT (1800)) = fn 1 (18 00 )

and f n 11(1800) meets each region. Consequently, it meets

R. Therefore, there exists some j such that n (V.) OR .

This contradicts the fact that n (V1) OR = for each i.

Hence, X is not locally connected.

Theorem 1.13: Let (X ,f ,D) be an inverse system

of compact Hausdorff spaces. Let C be a closed subset of

X, and a and b be distinct points of Xc,\C. For a c D, let

29

C =. (C). If, for a E D, X \C = A UB where A and Ba c. a a a a aare disjoint regions containing a and b respectively,

a athen there exists disjoint regions A and B in X containing

a and b respectively, such that X\ C = A UB.

Proof: By Theorem 1.8,

X\C = U{7r ~'(X\C )Ia s D}.Sa a a

But Xa\C = A UB . Hence,a a a a

X\C = U{7rj (A UB )fa FD}.

Since

7T 1'(A UBa) = Tr ~ (A )UT ~-(B ),a a a a aX C = U{Tr~'(A ) U 71(B )1a 6sD

= { UT ~1(A )}U{ U 7 <~1(B )}.

acD a a asD a

Let

M, = U 7r 1 (A )

and

N = U ~(B ).a.:D

Then, a c M and b e N.

Let x E m'N. Then there exists A and 6 in D such

that x e XA ) and x e f6~ 1(B ) There is a c D such

that A 5a and 6:5a. It then follows that x P Aand

x6 c B 6 . Suppose that zX6 A and f~'1(z X). A. Then,

there exists z c X such that z f A and f (z ) = z .

Hence, z a B or z a C . If z r C , there exists

z e C such that P (z) = z . Since z e C, z c X . There-

fore,

f(z ) = z = 7A (z).

30

Hence, z EC , But z EC and z A cA Consequently,

z e AnC . However, AA/C = . Therefore, zo C .

It must follow that z E B . Since each f is monotone,

f ~(z ) is connected, Thus, f '(z )CB But, since

A is connected and z e AA, f ~'(A A).9B , However,

a O f '1(A ) Hence, a c B .. This contradicts the

hypothesis. Therefore, f 1(Ax)cA . Similarly,

f 1(B )_cB Since xx AA and x6 B6, x cEf x )_cAand x e f 6

1 (B6 )-B., Hence, A flB = . This contra-

dicts the hypothesis, Therefore, MqnN =

Suppose that M = HUK where HOK = 4, H X X # K and

each of H and K are open inM. Then, ae H or a cK.

Suppose that a 6 H, Suppose also that for some a c D,

y C 7r '(A ), Now, ra -(A )cH UK and a -'(A ) is con-

nected. If y K, i- 1(A )_cK. However, a 6 'rr '(A )and a K, Therefore, for each a D, r -'(A ) OH.

Hence, K = $. This contradicts the supposition that H and

K disconnect M. Consequently, M is connected. If a K

then, by a similar argument, H = $. The supposition that

M is not connected is again denied. Hence, M is connected.

Clearly, by substituting N for M, N is also connected.

Now, X\C = MUN where M and N are disjoint regions con-

taining a and b, respectively.

It is noteworthy to point out that if D is countable

and each X is a separable metric space, then X is a

separable metric space,

31

Definition 1.6; Any space homeomorphic to [01] is

an arc.

Definition 1,7: A connected space is irreducibly

connected between two points if no proper connected subset

contains both points.

The following two theorems are beyond the scope of

this paper and will be assumed. Theorem A has been proven

by Wilder (5) and Theorem B by Whyburn (4).

Theorem A: A space Y is irreducibly connected

between a and b if and only if for y c Y\{ab}, Y\y = A UB

where A and B are disjoint regions such that a e A, b s B.

Theorem B: A separable metric continuum Y is an arc

if and only if Y is irreducibly connected between some two

points.

The following two lemmas and Theorem 1.15 are proven

in preparation for showing that if Xa is an arc for each

az, then X, is an arc.

Lemma 1.5: If a continuum X is irreducibly connected

between a and b and X is irreducibly connected between

c and d, then

{a,b} = {c,d}.

Proof; Suppose that a X y # b. Suppose that X is

irreducibly connected between a and y. Then X\b = H UK

where a E H, y K, i(K = $ and each of H and K is a

region. But, since X is irreducibly connected between

a and b, X\y = MUN, where M and N are disjoint regions

32

containing a and b, respectively. There exists r K

such that r y, If not, {y} is both open and closed,

contradicting the supposition that X is connected. Since

r A y, r E M or r s N, If r s M, then r c mOK. Hence,

MUK is open and connected. Therefore,

X = (MUK) UN,

But (MUK) ON A $. If not, X is not connected. Since

M nN = 4, K N A $. Hence, K UN is connected and open.

But, (K UN) UH = X. Therefore, (K UN) OH A $. Otherwise,

X is not connected. However, H nK = $. Consequently,

N H A $. It follows that NUH is connected and open.

Since y E N and y 6 H, N UH is a proper subset of X. But,

a e N and b c H, This contradicts X being irreducibly

connected between a and b. Therefore, X is not irreducibly

connected between a and y. A similar argument shows that

X is not irreducibly connected between b and y. Hence, if

{a,b}fl{c,d} A $,

then,

{a,b} = {c,d}.

Suppose that {a,b} O{cd} = $. Then a A c A b and

a A d A b. By the previous argument, X is not irreducibly

connected between a and c or between a and d. By Definition

1.7, there must be a proper connected subset H containing

a and c, There must also be a proper connected subset K

containing a and d, Now, if X 7 H UK then H UK is a proper

connected subset of X. But C F H UK and d E H UK. There-

33

fore, X is not irreducibly connected between c and d. This

contradicts the hypothesis. Hence, X = H UK. However,

b e X. Consequently, b 6 H or b c K. But, a c HqK. It

then follows that X is not irreducibly connected between

a and b. This also contradicts the hypothesis. As a

result,

{ab} = {c,d}.

Therefore, Lemma 1.5 guarantees that if a continuum

X is irreducibly connected between two points they are

unique.

Lemma 1.6: If f is a monotone function from a con-

tinuum X onto a continuum Y and X is irreducibly connected

between a and b, then if Y is irreducibly connected between

c and d,

{f(a),f(b)} ='{c,d}.

Proof: Suppose that f(a) = f(b). Then, f~'(f(a))

is connected and contains both a and b. Hence, f~'(f(a)) =X.

Therefore, Y = f(a). Consequently, c = d and Y is not

irreducibly connected between c and d. Consequently,

f(a) # f(b), Suppose that Y is not irreducibly connected

between f(a) and f(b). Then there exists a proper connected

subset U of Y such that f(a) U and f(b) c U. Since f is

monotone, f-'(U) is connected in X. But a and b are both

elements of f'CU). Hence, f1'(U) = X. However, U is a

proper subset of Y. Therefore, there exists x E Y such

that x ( U. Consequently, f~1(x) qf~'(U) = p. But

f '(U) = X. Hence, f '(x) = 4. This contradicts f being

34

onto. Therefore, Y is irreducibly connected between f(a)

and f(bY.. By Lemma l,5, {f(a),f(b)} = {c,d}.

Theorem 1.14; If, for each a e D, Xa is a continuum

irreducibly connected between two points, then X, is a

continuum irreducibly connected between two points.

Proof: By Theorem 1.5, X, is a continuum. Let a e D.

Let

E = 'sO Da0}.

Then E is cofinal in D. Hence, by Theorem 1.10,

YCO= lim(Xa ff E)

is homeomorphic to X. For A E, let TX be the X-th

projection function of II X restricted to Y. LetY,,beAXE

irreducibly connected between a. and b.. By Lemma 1.6,

for each C E, there exists as and b such that X is

irreducibly connected between as and b where f (a$ ) = a

and f (b ) = b . If A 5 in E, let X be irreducibly

connected between a and b,. Then

f,,(a) = a .

If not, f (a) = b (aX and b are unique). Therefore,

fa(a ) = f o f (a,) = f(bX) = b .

But,

f (aa) =aa

This contradicts a b . Similarly,

f (b ) = b .

For 6 E, let H = 1T (a ) and K% = 6 (b6). Each of

35

H and K6 is closed, compact, connected, non-empty and

Hausdorff. Let

V = {H616 e E}

and V = {K6 |6 cE}.

Let H and H be elements of t(. There exists a E such

that <cy and , o. If s c HV, then s= a and

s = f (a ) = a .

Hence, s c H . Similarly, s H . Therefore, H ._cH nH

By Lemma 1.1.,1Q/ cp. Similarly, rn? 4. Let a c nwand b e r7. For each 6 e , (a) = a6 and T6(b) = b6

Let t c Y,\{a,b}. Then, there exists r E such that

a t b . Let

F = {6 c EfnT6}.

Then F is cofinal in E. Then Z = lim(X ,f ,F) is homeo-00 + a B

morphic to Y . Let f:X+-* Y, and g;Y÷+ Z, denote these

homeomorphisms. For each A F,

X\(g(t)) = MUN

where M and N are disjoint regions containing (g(a))

and (g(b))X respectively. Recall that from Theorem 1.10,

(g(a)) = a . By Theorem 1.13, Z\ g(t) = MUN where M and

N are disjoint regions containing g(a) and g(b), respec-

tively. Therefore,

Y 0\ t = g 1 (M) U g'(N)Hence, Y. is irreducible between a and b. Consequently,

X, is irreducible between some two points.

36

Theorem 1.15: If D is countable and for each a s D,

X is an arc, then X is an arc.

Proof: By Theorem B, for each a E D, X is irreducibly

connected between some two points. By Theorem 1.14, X, is

a continuum irreducibly connected between some two points.

As was previously noted, X, is a separable metric space.

Therefore, by Theorem B, X, is an arc.

Definition 1.8: Any space homeomorphic to S', the

unit circle in the plane, is a simple closed curve.

Definition 1.9: A space Y has property Q if it is

connected, non-degenerate and for every pair of distinct

points x and y, Y\{x,y} = A UB where A and B are nonempty,

disjoint regions.

This paper will assume the following theorem by

Whyburn (4).

Theorem C: A separable metric continuum Y is a simple

closed curve if and only if Y has property Q.

Theorem 1.16: If, for each acED, X is a continuum

having property Q, then X, is a continuum having property Q.

Proof: By Theorem 1.5, X is a continuum. Let

x ; y e X. Then there exists a c D such that x y . Let

E = {X c Dja <x}.

Since X has property Q for all c D, X \ {x ,y a} = A UBwhere A and B are disjoint nonempty regions of X . Let

6 e E. Then x6 . Hence, X\{x' 6Y6} = H UK where H

and K are disjoint nonempty regions of X6. Let t c f 1(A ).6 Ba'S

37

If t = x, then f (t) = x . Hence, x e A . But x A

Therefore, t x, Similarly, t y . Hence,

f a(A) c HUK.

In a like manner,

f ~(B )gH UK.

Suppose that f6a~'(Au ).H and f,,(A ) 1K. Then

f~ '(A ) meets both H and K. But, f6 is assumed to be

monotone. Hence f 1(A ) is connected. Therefore,

f '(A ) cannot meet both H and K. Consequently,

f 1(A ) cH

or

f '(A) _cK.

Similarly,

f 6-(B )cH

or

f ~1B )cK.

For 6 c E, denote the region containing f 1(A ) as A

and the region containing f6 1(B ) as B .

Now, E is cofinal in D. Hence.

Y = lim(X ,f ,E)C* + tc

is homeomorphic to X . Let 7a be the a-th projection

function of II X restricted to Y . Let g be the homeo-

morphism from X onto Y, as described in Theorem 1.10.

Then Cg(x)) = x and (g(y)) = y for all f c E.

Let t Y0\ {g(x),g(y)}. Then, there exists A E

such that (g(x)) = x t y = ( Therefore,

38

t A or tX B .Hence, t s X'(A ) or t e i~1(BX).

Consequently,

t 6C U Tr (A ))U( U T'(B ))cE FE

Let

A =U '(A

and

B = U EI1(B ).

Then

Y\00 {g(x),g(y)} cA UB.

Let t AUB, If t j Y .\{g(x),g(y)}, then t = g(x) or

t = g(y). Suppose that t = g(x), then t = x for all

6 E. Hence, t A and t B for any 6 E. There-

fore, t 4 AUB. Similarly, if t = g(y), t 4 AUB. This

contradicts t c A UB. Hence,

t F Y \{g(x),g(y)}.

Consequently,

A UB = Y \ {g(x),g(y)}.

Since A is nonempty, there exists v a A . Now each f

is monotone. Therefore each f is onto. Hence, i is

onto for each a c E. Therefore, it follows that there

exists v e Y such that i (v) = v . Since v A A ,

f '(v )CA for each E. Hence,

v O W ~'(A ).aE:E a 06

Therefore, A is connected. Similarly, B is connected.

Since A and B are open in X for each a c E, T ~'(A ) and

39

7T (3 are open in Y' Q Consequently, A and B are open

in Y0. Suppose that w E AQnB, Then w E F (A6 ) and

w 6 7 (B ) for some 6 and in E, Hence, w6 . A and

w e B . There exists A E such that a SA and cy X5.

Therefore, wA A and w B .B This contradicts the

fact that A ( B = $ for all E .E, Hence, AnB = p.

Therefore,

Y0\{g(x),g(y)} = AUB,

where A and B are disjoint regions of YC. Hence, Y.0 has

property Q. As a result, XC, has property Q.

An obvious result of Theorem 1.16 is that if D is

countable and if for each a sD, X is a simple closed

curve, then X0, is a simple closed curve.

CHAPTER BIBLIOGRAPHY

1. Capel, C. E., "Inverse Limit Spaces," Duke MathematicalJournal, 21 (1954), 233-245.

2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1968.

3. Whyburn, Gordon T., "Analytic Topology," AmericanMathematical Society Colloquium Publications,2 8-(1942)0.

4. Wilder, Raymond L., "Topology of Manifolds," AmericanMathematical Society Colloquium Publications,32 (1949).

40

CHAPTER II

MAPPINGS BETWEEN INVERSE LIMITS

Before attempting to characterize compact metric spaces,

it will be helpful to investigate the following situations.

Let (X ,f ,D) and (Y ,g ,D) be inverse systems

where for each a in D, f and g are continuous. By

a mapping Q:(.X ,f ,D) + (Ya,g ,D) is meant a collection

of functions {Qa} such that Q :X -+ Y and for a:5

in D,

Q o f = g o Q .Qa 0 fa = a 0

Define a relation Q :X + Y by

Q,({xaaFD9 = {Q Q(x aOD*

For H X , let -T be the ot-th projection function restrictedUED

to X., and for H Y., let TF be the a-th projection functionaE:D

restricted to Y00. Let x 6 X0 such that x = {x }D Then,

Q00(x) = {QU (xa)}acD

Let a 5in D. Now,

T(jQ0 0(x)) =QU~(xU).

But

g(QSox$)) = that(x

So that

9 CQ ( ))= Q,(x9,.

41

42

Therefore,

7F( Q" 'x)) = a(--" Qc(x)))

Hence, Q,,x) c Y . It then follows that Q (X ) gy.

Let x = y in X . Then, for each a s D, x = y

Since Q is a function, Q (x ) = Q (y ). Therefore,

{Q Cx )} = {Q~ (y)}D.{Q a IacD {Qa a aE:D

Consequently,

Q (x) = Q(y).

Now Q is a function,

To say that Q has a certain property will mean that

Q has that property for each a c D. By placing the proper

restrictions on Q, Q. can be forced to possess certain

properties, as is shown by the following theorems.

Theorem 2.1: Let each of (X3,fD) and (Y,g ,D)

be inverse systems and Q be a mapping from (X ,f ,D)

to (Y ,g D). Then, if Q is continuous (one to one and

onto) (a homeomorphism), then Q, is continuous (one to one

and onto) (a homeomorphism) .

Proof: Suppose that Q is continuous. Let s E X C

and Qo(s) = w. Let U be open in Y, such that w c U. Then,

there exists a basic neighborhood, if (U9, of w such

that w F V 1 (U 6 ) CU. Now, w6 e U and

Q(s) ='{Q (s )} D= {wa}D.

Since Q6 is continuous and Q(s = w6 E U., there exists

V, open in X, such that Q(V)cU6 and s6 6'V Also,

s e 'r'(V I which is open in X . Let p (.V

43

Then, p, C V6. Hence, Q(P9 16. But Q0 (p) c Y and

a (Q a (Pa acDSince Q6(p1 s Up, Qp) e j1(U )~U. Consequently,

Q00 (7T (V6)) cU. As a result, Q. is continuous.

Suppose that Q is one to one and onto. Also, suppose

that Q xl = Q (yl. Then,

{Qa Cx )}ac = {Qa(ya)}aDa a aD a a asD'Hence, Q (x ) = Q (y ) for all a. Since each Q is one

to one, x =y for all a. Therefore, x = y. It then

follows that Q is one to one. Let t c Y . Then for each

a, ta c Y a. Each Q is one to one and onto. Consequently,

for each a, there is a unique p in X such that Q (p) =t

Let p be the element in TIX such that the a-th projection

of p is pa for each a. Let as5 in D and suppose that

f (p) p.' Since % is one to one, Q (f (p )) Q(p ).

Hence, Q (f (p t . However,

QtC a (pag~a= 0 Q (p ) 9g t )=taThis contradicts the previous statement. Therefore,

f (p) =p

Hence, p c X, But

QO(P) = Q (p )}aeD = t.

Consequently, Q is onto.

Suppose that Q is a homeomorphism. Then, Q is con-

tinuous, one to one and onto. By the first two arguments,

Q. is continuous, one to one and onto. Let y c Y, and

QI 'Cy) = x. Let U be open in X00 such that x csU. Then,

44

there exists a in D such that U is open in X and

x C 7 T (U since Q is a homeomorphism, Q (U ) = V

is open in Y . Therefore, 7~-(V ) is open in Y.. Since

x 6 ff _1U ), x a U . Consequently, Q (x ) c V However,Q x cx T -1(ccx 1

Qaa x ) a Hence, y ' (V ). Let w c T ~a(V ).

Then w 6 V , As a result,

Q -'(w ) Q ~(V ) = U

It then follows that Q ~'(w) E: '(U ). From this it

is observed that Q ~ is continuous. Hence, Q is a homeo-

morphism.

Definition 2.1: If f is a function from a space X to

X, x e X and n c Z+, then fn(x) means f composed with it-

self n times acting on x.

Definition 2.2: A function f from X to X is periodic

if there exists some positive integer n such that, for each

x C X, fn(x) = x. The least such n is the period of f.

If (X, d) is a metric space then f is called almost periodic

if for each c > o, there is a positive integer n such that

for all x c X, d(x,fn(x))< c.

Theorem 2.2: Let ((Xnd ),f ,Z ) be an inverse se-nn mn +

quence of metric spaces and

Q:((Xnjd ),f Z+)÷+ ((xn,dn3m, Z+

be continuous. Then, if Q is periodic, Q is almost per-

iodic.

Proof: It will be helpful to recall that, if for

each n c Z+, CXndn) is a metric space, then a metric d

45

00

for R (X d ) ia given byn=l n, n

d (x ,y )d(x,y} = 2 nn 1

n=1 2[1 + dn n nCxy)]

where

00

and

t- 00

y ={yn In n=1For each i c Z+, let Q have periodicity n1 and let

c > o. Then, there exists k Z+ such that Z<C.p=k+l 2P

Let m be the product of the n Is for 1:55k.1

and let

Let z c X

d (Qm (z ),z ) = sp pp p p

Thenco

000d(Q (z),z) =

p=l

k

p=l

sp

2[ + s]

po

p=k+1

s-.p

2 [1 + SP]

sp

L 2 + s ]p

For p:k, let

... -* nk= rp.

Then Qmp

n rP P. Hence, Qm(z ) = z. Therefore, forp p p p

p ! k, sp = 0. Consequently,

dCQm (Czi,zI =00

p=k+l 2P[1 + s ]p

00

p-k+l 2p

As a result, Q, is almost periodic .

n + 2 +..-n ~ - np+1

The following example illustrates that requiring each

Qn to be periodic will not force Q to be periodic.

Example 2.1: For each n c Z+, let Xn be the unit

circle in the plane. Give each Xn the relative metric

topology, dn. An element, e o, of Xn will be denoted by

0, expressed in degrees. Define Q :X + X byn n n

Q (e) =+ 3600n n

Clearly, each Qn is continuous and of period n. Define

fnn :Xn Xn to be the identity map and fn+l n:X n+l - Xn

by

f =6 n+l()n+l n n

Obviously, each fn+l n is continuous and onto. For m> n in

Z+, let f :X X be f a f 0mn m n n+ln n+2 n+l0..'fm m-l'

Then fmn is continuous and ((Xn3,dn). fmnZ+) is an inverse

sequence of metric spaces and Qn is onto for each n. Also,

fmn 0 Q M(e) = f ( + 360

n+l n+2 m + 360n n+ lm-l n

m 3600=-0 + .n n

However,

Q f (e) = Q (n+l n+2 mn mn n n n1 n-1

= Qn(m( m) = + 360.T=n n n

Therefore,

mn om = n ofmn

Now, suppose that Q, is of period s. Then, Qs(x) = x

for each x E X, and Q(x) ='{Q5 (x )}neZ+ If P> s, then00n~n nZ.I ,te

there exists y c X such that Q (y) / y. Let x e 7r 1 (y).p p Pp

Then.,

QSc)={Q s(x ) =X.

Qx) ={nxn neZ+

Therefore, Q (y) = y. This is a contradiction to Q (y) # y.p , p

Hence, Q. is not of period s.

Theorem 2.3; Let ((Xndn ) f Z+) be an inverse se-

quence and Q:((Xn,dp, fm,Z+) + ( (Xndn),f ,Z+) be con-

tinuous. If Qn is almost periodic and Xn is compact for

each n, then Q, is almost periodic.

Proof: Let a >o. Then, there exists b such that

o <b minimum {l,a}. There exists k E:Z+ such that k> b

00

and p= +1 < . Let 0< 6< Since each Xn is compact,p=k+l 2P -b

each f is uniformly continuous. Therefore, for each i

such that i <k, there exists 6. such that if U is set of

Xk of diameter less than 6i, the diameter of fki(U) is

less than 6. There exists c such that

0< c< minimum {6, 1 1<1 i k} .

But Qk is almost periodic. Consequently, there exists m

such that for x Xk, dk(Q m(.x),x)< 2-. If x E Xk, then

there is an open set UX of diameter less than c such that

{x,Q (x)}}SU. But fkiCUx) has diameter less than 6.

48

Ience, d (fki x),fki (Q Cx)) <6. Therefore, dk(Qi(f W),

fki(x)) <6. Let y 6cX and d (Qm(y ),y ) = s . Then,PkPpp p

dCQm(y),y s=P5 -00 5p(m3+

p=1 2P[ 1 + s I p=k+l 2 [1 + s ]p p

But for p5k, s 6< kb Hence,

b

s < _ kbk(l-l)k

Consequently, s [l + ] , so that s < + sp(b). Now,

s < [l + s ]. Thus, 1+sP < b. Therefore,pkp l+ s P ip

k s k k1Di() I 51- _=1

p=l 2P[1 + s p=1 2p 2 k 2 k 2pp

As a result,

b 00 5 00

d(Qm b++p+ -l<b +b<a2 p=k+l 2P[1 + s ] p=k+l 2P 2 2

p

It is then evident from this that Q is almost periodic.

Theorem 2.4; Let (X , f,D) and (y , g ,D) be inverse

systems and Q:C(X a ,fD) + (Y , g ,D) be continuous. If

Q is monotone and X is compact and Hausdorff for each a,

then Q00 is monotone.

Proof: Let y 6 YO, such that Q00 1(y) X p. For each

a, let Q '(y ) = A . Since Q is monotone, A is a con-

tinuum. Hence, .II A is a continuum. For ac 35 in D,acD

let J,, = f$aIAK Then j a:A + A . For if x c A,

49

Q (xI = y 5. Hence,

g o0Q (W = g(y)=y.

However,

0oQa(x) = Q of (x)=ya.

Therefore,

f (x) 6 Q -1(y ) = A.

Now, (A ,j ,D) is an inverse system and A = rlim(A, j ,D)

is a continuum.

Lets 6 A. Then, for as in D, j a(s ) = s.

Therefore, f (s = a. Consequently, s c X . Since

A SHA , A iHA aX0. Let s e HA F)X. Then, s F X. As

a result, for a:50 in D, f (s ) = s . But s HA . Hence,

s c A So that j (s ) = s . Therefore, s c AM. Now,

A = X FUTA.00 0o a

Let s c A0. Then, s c X. and for all a e D, s e A

Consequently,

Q (s ) = y .

As a result, Q,(s) = y. It then follows that A SQ ~1 (y).

Let s E QW~ 1(y). Then s c X and, for o c D, s a Q ~1(y ).

Hence, s A . Therefore, s c HA , so that s E HA qX = A

Hence, A00 = Q00 1(y). Since A0 is a continuum, Q0 ~1(y) is

connected. It then follows that Q is monotone.

Theorem 2.5: Let

(aXndn), fmn, z+and

(CYn ,can nZ

50

be inverse sequences and

Q;C(Xn, nJfmn,3Z+) * n n), gmn,3Z+)

be continuous and open. Suppose that if gn+i n n+i1 n

and Q (x y, then

nn+1 n+1 n+1 xn

Then Q, is open.

Proof: Let 7 n be the n-th projection function of

R Xn restricted to X. and Tn be the n-th projectionn nZ+

function of R Yn restricted to Y,. Let U = n ~ 1(U )nZn+ n

where Un is. open in Xn. If x e U, then xn 6 Un. Hence,

n n) n(U)n Since Q,(x) = {Qn(x)}n=l'

Q(x) E 7n (Qn(Un))

Therefore,

Q, (U)_nIn(Un)).

Let x c 7n'(Qn(U) )). Then, xn n)Q.(U Hence, there

exists kn E -U such that Qn(kn) x n. But x c Y,. Conse-

quently, xm m for each m and g (Xm ) = Xn for all n 5m.

By the hypothesis, since Q (k ) = x and g n(xn+l) = Xnin n n n1nn1 n

Qn+ii'(xn+)mfn+i n(kn) *34p.

Let kn+1 be an element of this intersection. By induction,

for each m c Z such that m > n, there exists k c X such

that Qm(km=xm andf(m m-1(k ) = kM-1. Let k cllXn such

that for n!E i in Z+, 7Ti(k) = k and for n< i,

7(k) = f .(k ).7i ni n

51

Then k E X and Q Ckl ='{Q (k )} = x. Since k c Un n n=l n nk E 7 n (U)n, Therefore,

Q(k) = x c Q(U),

Hance, ' 7T CQn (U ))ZQ(U), It follows thatn n n"

Q (U) = ~7 Q (U )).n n n

Consequently, QcjiU) is open in Y . Now, Q, is open.

The following example proves that requiring each Qn

to be open is not sufficient to guarantee that Q will be

open.

Example 2.2: Consider Example 1.3. Let H be

{fnl~'({o0 ,18O})|n scZ+ and l< n}.

For each n c Z+, let Xn and y n be the unit circle with H

deleted. Give Xn and Yn the relative topology from the

plane. An element of Xn or Yn will be represented the

same as in Example 1.3. Let f+ nXn++ X be defined byn+l n' n+l n

fn+li n( e = 26. For m >n, let fmn;Xm - Xn be defined as

fn+ n n+2 n+0 fm M-1 Then (Xn ,f Z+) is an

inverse sequence. Define gn+1 nYn+l + Yn by gn+ is

4e if 00< 0 < 900, 3600 - 2e if 900< 0< 2700 and 20 if

270' < e<3601 . Then gn+1 n is continuous. For m> n let

mn m n begn+n1 n n+2 n+l 0 m M-1. Then

gm is continuous. Now (Yn +) is an inverse sequence.

Define Qn.;Xn n nY by Qn() is 3600 -aif 0 0< < 1800 and

0 if 18G 0 < 0 <3600. Then each Qn is continuous and open

for each n.

52

If Q < 0 < 900 then 0 < 20 < 1800 and 2700 < 3600 - 0< 3600.

Therefore,

S(n+ e() =Qn(20) = 360Q 26

and

gn+1 n 0n+i(0) = n+1 n(36 00 0)

2(3600 - e) = 3600- 20.

If 900 < 0< 1800, then 1800 < 26 < 3600 and 1800< 3600 - < 2700.

The refore,

n fn+i n(0) = Qn(20) = 20

and

n+1 n 0n+j 0( = ) n+1 n(3600 )

- 360o - 2(360 0a) = 20.

If '1800 < 0 <2700, then 0 < 20 < 1800. Hence,

Qn nofn+ n(0) = Qn(20) = 360o - 20

and

gn+1 n *0 n+i 0 ) = gn+1 n(8) = 3600 - 20.

If 270 < 0 <3600, then 18o <26< 3600. Therefore,

Qn *fn+i n(0) =n (20) = 20

and

gn+1 n 0 Qn+1 0 = n+ 1 n(e) = 20.

Now Q is a mapping.

Let 7r be the i-th projection of fXn restricted to

X and ~ii be the i-th projection of HYn restricted to Y,.

Let V = Tr iCU) where U is open in X1 and 14Qo c U. Then,

x = (14o0 ,700,350,...) c V.

Q0(x) = (2200,2900,3250,-...) F Q (V).

53

Let QCx) = y. For each open set H of Y such that y c H,

there existsHi open in Y such that y 6 F '(H ) SH.

Since y C ' 1i 1(Hi)-, yi c Hi. But y1 is such that

180, < y. < 3600. Therefore, there is eO6 Y +1 such that

0 < 6 <900 and 4e = yp. Hence, gi+l i(e) = y . But

0G < 6< 9Q0 . Hence, 001<-< 90O. Therefore,

g+2 1+6

By induction,

z = (2200,2900,...,yi,4,+,...) y

But z, = y . Hence, z c fi (Hi) CH. Now z Q00 (v).

Therefore, Hq(Y \ Q (V)) 4p for each H. Hence Q00(V) is not

open in Y , Therefore, Q is not open.

If each of (X ,f ,D) and (Y,g ,D) is an inverse

system of compact Hausdorff spaces and Q is a continuous

mapping from (X fD) to (Y ,g, then Q will be

closed. Example 2.3 illustrates that requiring Q to be

closed will not guarantee that Q is closed.

Example 2.3: For each n c Z+, let Xn = Z+. Give

X the discrete topology. Define f :Xn+ X to ben n+l n' n+l nthe identity function. For m>n, let f :X + X be

mn m n

fn+l n 4 n+2 n+l0 '0' f m m-1. Then (Xnf mn Z+) is

an inverse sequence. For each n s Z+, let

n p{ l1pin}.

Give Y the discrete topology also. Define gn+ nn+l

1 1 1 1 1 1 1by gn+1 n(-) if f and if -<, For m> n, let

54

gmn:Xm + Xn beg n+1 n n+2 n+1 a 0" mo m-1. Then

gm is continuous and CY n mn,Z+) is an inverse sequence.

For each n, let Q ;X -* Y be defined by Q (z) = ifn n n n zz n and 1 if z >n, Then each Q is continuous and closed.nn

Let t X+ such that t < n+l. Then, t <n and 1 1andn+l scn -t

1 1n+l <:. Hence,

Qn f (t) = Q (t) 1n n+1 n n t

and

n+1 n Q0n+Qit) = gn n =

If t X n+1 and t = n+lthen,

Q f (t) Q (t)

n n+1 n n n'

Now,

gn+ n Q1t)1n+1 n+1 nt '

If t C Xn+1 and t >n+l, then t > n. Therefore,

Q o f (t) = Q (t) =1n n+1 n n n

But

n+1 n n+l n+l n n+ 1 '

Now Q is a mapping.

An element N of X is described by N =(n,n,n,..),

1 1 11 1and Q(-N) = (1,1,,..1, ,,,,...). Now

1 1 1y = 17,]P...* ) 6 Y .

A basic open set U of y is T ~'(U ) where c U . Butn n n n

N = (n,n,n,. .) X,, and Q (n,n,n,...)= (1 1 ., 1 n

where - is the n--th coordinate. Therefore Q (N) c 'n 1(U 'nny Xs n n

Qonaequently, Q,.XO) is not closed. HenceQ. is not closed.

55

Definition 2.3: A space X is totally disconnected

provided it has no nondegenerate components.

Definition 2.4; A function f from a space X to a

space Y is called light if for each y 6 Y, f~'(y) is to-

tally disconnected.

Theorem 2,6: Let (X ,f ,,D) and (Y ,g ,D) be inverse

systems and Q be a continuous mapping from (X ,f ,D) to

(Ya', D). If Q is light then Q, is light.

Proof: Let ff and TF be the respective projection

functions of X, and Y. Suppose that there exists t 6cY,

such that QW_1(t) is not totally disconnected. Then there

exists H, a nondegenerate connected subset of Q '~(t). For

each u, 7T(H) is connected. If x 6 H, then x c f (H).

But, Q (x ) = ta. Consequently, xa Qa~1(t ). Since Q

is light, Q a1(t ) contains no nondegenerate connected

subsets. Hence, 7a (H) must be a single point. Let

7T(H) = k .

If x and y are elements of H such that x # y, then, for

some a, xa y a. As a result, 7a (H) is not a single point.

This is a contradiction to 7F (H) = ka, a single point.

It then follows that Q is light.

Although Theorem 2.7 breaks the tradition of investi-

gating Q 0,, it is interesting and will be useful later.

Theorem 2.7: Let (X ,f ,D) be an inverse system. If

X is totally disconnected for each a, then X00 is totally

disconnected.

56

Proof; Suppose that there exists H, a nondegenerate

connected subset of X,,,. Then, TVCH) is connected for each

a. Hence, (TH} must be a one point set. But, H is non-

degenerate., Consequently, there exists x and y in H such

that x A y. Therefore, there exists c D such that

x ; y. As a result, {xVy }.r (H). This is a contra-

diction to 7 (H) being a single point. It then follows

that X is totally disconnected.

Definition 2.5: A function f from a space X to a

space Y is compact if for each compact subset K of Y,

f- (K) is compact.

Theorem 2.8: Let (X ,f ,D) and (Y ,g ,D) be inverse

systems of Hausdorff spaces. If Q:(X ,f ,D) + (Y ,g ,D)

is continuous and compact, then Q, is compact.

Proof: Let K be a compact subset of Y. Let T

and TV be the respective projection functions of X and

Y . Since TV is continuous for each a., v (K), denoted by

K , is compact. However, Q is compact for each a. There-

fore, each Q a1(K ) is a compact subset of X . Hence,

R (Q1(K)) is a compact subset of IX . Let t c Q0 ~1(K).

Then, there exists w 6 K such that Q0 0(t) = w. For each a,

Qa(t ) = w aEK .

Consequently, t 6 Q '(K ). As a result, t H(Q ~1(K )).S a a a a

It then follows that Q"'(K).S (Q '(K )). Since K iso K a a

compact and Y is Hausdorff, K is closed in Y . Therefore,

57

Q '(K) is closed., But, Q~(K) SIICQ '(K )) which isca a

compact. Hence, Q ~'(K) is compact. It is then evident

that Q, is compact.

Now that it is understood how certain properties can

be built into Q, it will be helpful, in characterizing

compact metric spaces, to examine the following seemingly

unrelated material.

Definition 2.6: If x and p are elements of a space

X, then x and p are separate din X if there exist open

disjoint sets H and K such that x c H, y c K and X = H UK.

Definition 2.7: For an element p in a space X,

Cp = {x X such that x and p cannot be separated in X}.

is called the quasi-component of p.

Theorem 2.9: If p c X, then Cp = flX where

Y = {Up e U and U is both open and closed in X}.

Proof: Let x E C . If U c ( and x U, then U isp

open, X\U is open, p c U and x c X\U. Since X = U U(X\U),

U and X\U separate x and p. This contradicts x c CP.Hence, x c U. Therefore, Cp .g nK.

Let x c n('. If x Cp, then there exist disjoint

sets W and V, open and closed in X, such that p c W, x e V

and X = W UV, Now, p e W and W is open and closed in X.

Hence, W F . Since x V, x W. Therefore, x n

This contradicts the supposition that x F . Conse-

quently, x c CP . It then follows that fl(CP, . As a result,

p pp

58

It is now revealed in Theorem 2.10 that in a compact

Hausdorff space componenets and quasi-components are iden-

tical.

Theorem 2.10: If p is an element of X, a compact

Hausdorff space, then Cp is a component.

Proof: Let z 6 X\Cp. Then, there exists disjoint

sets U and V, open in X, such that X = U UV, z c V and

p e U. Suppose that w 6 Vnc . Then p F U and w c V.

Hence, w CP . Therefore, V(n Cp = . Now, z V._X CPConsequently, X\Cp is open. As a result, Cp is closed in

X. It then follows that Cp is compact.

Suppose that CP is not connected. Then, there exists

H and K, disjoint, closed, nonempty subsets of Cp_, such

that Cp = H UK. Since H and K are closed in Cp, H and K

are both compact, Therefore, there exist disjoint sets

U and V, open in X, such that HcU and KLcV. But X\(UUV)

is closed and compact. If x E X\(U UV), then x C CP.

Hence, there exist disjoint sets U and V , open and closedx x

in X, such that

X = UxUV ,

p 6 Vx and x e U . Also, Cc CV . Now, ( = {UxIx s X\(UUV)}

covers X\(UUV). Consequently, there is a finite subcol-

lection of 'U covering X\(UUv). Let

W* ={Ul,U2 , ...Un

be this finite cover, M = V Then, C c M. Sincei=l

59

each Vi is open and closed in X, M is open and closed in X.

nLet N = U U It then follows that X\(U UV) cN and N is

1=1

open and closed in X. Clearly, mON = . If z c M,

z 4 N. Consequently, z X\(U UV). As a result, z s (U UV).

Therefore, Mc (U UV). Since C cM, C c(MOU) U(MNOV).

Each of M and U is open in X. Consequently, M OU is open

in X. Suppose that MNU is not closed in X. Then, there

exists z E X\(MOU) such that if z S H and H is open in X,

then H O(r ON) $, But V is open and U OV = . Hence,

V O(MOU) = $. Therefore, z V. If z U, then z U Uv.

As a result, z 6 X\(UUV). It then follows that z c N.

However, N is open and NONm = $. Consequently, NO(mnOU) =.

This contradicts the statement that if H is open and z c H

then HO(mNOU) # . Therefore, z N. It then follows

that z e U.

If z E M, then z c MOU. However, z E X\(mOU).

Hence, z 4 M. Hence, z c X\M which is open in X. But

(x\M) ON = q. Therefore, (X\m) O(mOU) = 4. This is

also a contradiction to the previous statements about z.

It now follows that mnOU is closed in X. Now

x = (mOU)U (x\ (mnOU)).

Hence, CP;mnOU or Cpgcx\(mu). If CpCmnOU, then

Cp SU. Since uOv = 4, Cp Ov = $-. However, Cp = H UK

and K CV. Consequently, K = 4. This contradicts the

supposition that K $. Therefore, C cX\(m nu). Hence,

60

C n (rm r =ui =. But Cp SM. Thus, C U = $. Since HCU,P P p

H -. This. contradicts H / $p It now follows that C isp

connected..

If C is a connected subset of X and CP is a proper

subset of C, then there exists x c C such that x Cp.Since x Cp, there exists disjoint sets UX and VX, open

and closed in X, such that x c U , p c V and

X = UK ,UVx x

But Ux and Vx separate C. Therefore, C is not connected.

This contradicts the supposition that C is connected.

Consequently, Cp is a component.

Theorem 2.11: A compact Hausdorff space X is totally

disconnected if and only if for distinct points x and y

in X, there is a set U, open and closed in X, such that

x e U and y 4 U.

Proof: Let x and y be distinct points of a totally

disconnected, compact Hausdorff space X. Since X is com-

pact and Hausdorff, CX is a component. But, since X is

totally disconnected, C = {x}. Now Cx = n (where

Y(= {U cXIx c U and U is open and closed in X}.

Since y C , there exists U c ( such that y U. Hence,

x F U, y U and U is open and closed in X.

Let X be a compact Hausdorff space such that for

distinct points x and y in X, there exists a set U, open

and closed in X, such that x c U and y U. Suppose that

X is not totally disconnected. Then, there exists H, a

61

nondegenerate component of X. Let x and y be distinct

elements of H, By supposition, there exists an open and

closed subset U of X such that x E U and y U. Hence,

y E X\U and X\U is open in X. However, U H is open in H

and (X\U) (nH is open in H. Consequently,

H = (UrnH) U ((X\ U) oH),

x c UqH, y e (X\U)nH and (U H) n((x\u)nH) = $. There-

fore, H is not connected. This contradicts the statement

that H is a component. Hence, X is totally disconnected.

Theorem 2.12: If (X,d) is a compact metric space

and V is finite open cover of X, then there exists c> o

such that if V is an open set of X having diameter less

than o, then V is a subset of some element ofV.

Proof: Let V ( = {Ul,U 2 , ... ,Un } be a finite open cover

of X. Suppose that for each c > o, there exists Vc, an

open set of diameter less than c, such that V0c-U1 for

1l5in. For each m E Z+, let xm Vm where V is an open

set of diameter less than 1. Then the sequence {x }M n n=l

has a cluster point x. Since x c X, x 6 U for some i.

There exists S(x,6), a spherical open neighborhood of x

of diamter less than 6, such that S(x,6)._U . However,

there is m c Z such that < . Since {x } clusters+ m T n n=l

at x, there exists p e Z such that p > m and xF S (x,).

Let y c Vp . Then d y,x):5d(y,xp) + d(xp,x). But,

6d( y,x )< and d(xx) < Conse quently,p p p 6 6 6

dCy, x) < + < + = <zCS

62

As a result, y E SCx,6), Hence, V S(x Thisp 1)CU1 Ti

contradicts the supposition that VC ui for any i. There-

fore, the theorem is proven.

Definition 2.8: Let 74 and V be covers of X, then 2

refines V, written as ?(< V, means that for each A E

there exists B F V such that AcB.

The following theorem establishes a connection between

inverse systems and theorems 2.9 through 2.12.

Theorem 2J13: If (X,d) is a totally disconnected,

compact, metric space, then X is homeomorphic to the in-

verse limit of an inverse sequence of finite spaces.

Proof: Let U be open in X and x 6 U. Since X\U

is closed in X, it is compact. For each y 6 X\U, there

exists Vy open and closed in X, such that y e Vy and

x V. Then ? ='{V |y E X\U} covers X\U. Hence, there

exists a finite subcollection of ?r covering X\U. Let

W={V V,23*''V}n

be this finite cover. Now, UW is open and closed in X

and does not contain x. Let z E X\ Uk. Then, z U2.

Therefore, z X\U. Consequently, z F U. Hence, X\U2 CU.

As a result, x 6 X\ U9c U. Now, X has a base of open and

closed sets.

Let n e Z+ Then, for each x c X, there exists Hxsuch that x 6 Hx and H is open and closed in X and has

a diameter less than 11. Therefore, W( = {HxIx Xcovers X.2n

63

Since X is compact, there exists a finite subcollection

$ of K covering X. Let

= {H1 ,H2 ,..j,Hm).

Let U = H3,n-l

U2 = H2\ H,..,Un = Hn\ U H

Continue this process as long as the U1 s remain nonempty.

Let

S= {UIlim}

be this collection of nonempty U is. For each i,

i-l i-1H\ UH = (X\ UH )rnH .

p=lp p=lp

Consequently, U is both open and closed in X. Let x c X.

Then, x E Hi for some i. Let j be the least such i. Now

j-1x 6 H.\ UH = U..

i p=j p J

Therefore, tW covers X. Suppose that 1-5j m and 1 5 m

and j Z. Either j <k or k < j. Suppose that j< k.

Then,k-1

U = H\ U Hp=l P

So, if x E U., x 42 H, Hence,

j-1

x4H.\ UH =U..J p=lp J

j --l -1If x e U , since j < , xE UH H. As a result, x 6 H\ UH

p=l p=lP

Hence, x 42 U." Therefore u% RnU, = $. Similarly, if k <

64

U qU = $. Then t4 is a cover of mutually disjoint, closed

1and open seta each having diameter less than - By2 n

Theorem 2.12, there exists 6 > 0 such that if V is a subset

of X of diameter less than 6, then V CU 1 ?{. Let A be

less than the minimum of {6 , .- By the previous con-2n+l

struction, there exists a finite open cover of X by mutually

disjoint, open and closed sets of diameter less than A.

Let ?r be this cover. Then, 7f< V. By induction, for each

n c Z+, there exists n, a cover of X by mutually disjoint,

open and closed aets of diameter less than such that2 n

4n+l< nn Let {? iil be a sequence of such covers. For

each n, let Yn be {'nI with the discrete topology. Define

n+l n n+l n by

fn+l n (U) = V

where V is the unique element of Yn such that USV. Then,

each f n+1 nis a continuous function. Let f :Xn - Xn be

the identity map and for m >n,

f f of o .fmn n+ln 0 n+2 n+l 0 '' m m-l'

Then, (Yn3fm, Z+) is an inverse sequence. Let

Y = lim(Y ,f Z)S n + mn,+and y c Y0 where y = '{U} such that U s ?cI. Define

F:YO X by000

000

F(y) = FC{U } ) = ui=i s

Since f n+ n(CU n+1 =U n,3U n+1 U n. But X is compact and

65

each U1 is closed. Therefore, U1 is compact. By Lemma 1.1,

00

q u1=1

If x and y are elements of n U1 such that x # y, theni=1

there is n cEZ such that -<d(xy). If x 6 Un, y Un*

00 00 00

Hence, if x c q U,, y n U1 . Therefore, q U1 is aI=1 1=1 1=1

single point. Consequently, if F(y) = p and F(y) = q,

then p = q. As a result, F is a function.

Suppose that F(X) = F(y) where x ='{U 1 }O and

y = {V }0 0 . If U1 V. for some I, then U i V = Hence,

00 00

1=1 1=1

This contradicts FCx) = F(y). Consequently, U1 = V. for

each i. Therefore, x = y. It then follows that F is one

to one.

Let x 6 X. Then, for each n, there exists Un n

such that x c U andUn is iniquein Y . Let y iYn n n n

such that Pn(y) = Un for each n. Suppose that for some m,

f m+ m(U)m+1 )7Um. Then

mm+ m(Um+ )lUm

But x E Um+l. Therefore, x fm+1 fm(U m+1). Since x 6 UM3

m+1 m(Um+) num X '

This is a contradiction to fm+1 M(Um+ 1 )Um = . Conse-

quently,

m+l m(Um+ ) =um0

66

As a result, y , YO.. But00

E~y) = 0 U. = x

1=1Hence, F is onto.

Let t ='{w } e Y and Z be an open set of X such

that F(t) e Z, Then, there exists a basic open neighborhood

U = S(F(t),6), the spherical neighborhood about F(t) of

radius 6, such that F(t) c U.SZ. Let

H = {y = {U } 1 e Y.1there exists n ycZ+ such that U nCU}.

00

Since t = {w } and F(t) = { fl 1W}LU, by Lemma 1.2,i=1

there exists m e Z+ such that wmLU. Therefore, t c H.

Hence, H / p. Let y = {U H. Then, for some j,00

U U. But F(y) = q U. U.. As a result, F(y) c U.i=i'

Consequently, F(H)5U. Let

z = {V } O H.

Then for some n, V CU. For each i, {V.} is open in Y..n T

Hence, 7Tr '({Vn}) is open in Y and contains z. If

n n { u={s}00 - I(r{V

ii-l n n1then {s n{V}, Therefore, s = V. But V U. Sincen . 1sn n n :U icsn =Vn' nU, It then follows that s c H. Now, H is

open, t E H and F(H)cUcZ. Consequently, F is continuous.

Now, F is a continuous function from a compact space onto

a Hausdorff space. Hence, F is a homeomorphism. Therefore,

X is homeomorphic to Y00 the inverse limit of an inverse

sequence of finite spaces.

67

Definition 2.9: A space X is perfect if each element

of X is a limit point of X.

Theorem 2,14; If U is a nonempty open subset of X,

a compact, Hausdorff, totally disconnected, perfect space

and if n is a positive integer, then U is the union of

exactly n open, mutually disjoint, nonempty sets.

Proof; Since U is nonempty, there exists x E X such

that x c U. Since X is perfect, x is a limit point of X.

Therefore, there exists y 6 U such that y X x. Since X is

compact, Hausdorff and totally disconnected by Theorem 2.11,

there exists a set V., open and closed in X, such that x c V

and y V. Now, x e VqCU which is open in X and

y 6 (X\V) qU which is also open in X. Let V nU = U1 and

((x\v) qU) = U2 . Then U = U1 UU 2 , U1 qU 2 = , and

U 1 X U2, By induction, for any n c Z+, U is the union

of n disjoint, open, nonempty sets.

Theorem 2.15: Any two compact, perfect, totally dis-

connected, metric spaces are homeomorphic.

Proof: Let X and Y satisfy the above hypothesis. For

a set H, let tHl denote the cardinality of H. For each

n E Z+, let "9 and V be - finite covers of X and Y, re-+n n

spectively, of mutually disjoint, closed and open sets such

that each element of "h and each element of 7/ has diametern

less than and %+l and n+1 n, By Theorem 2.14

can be made the same as IVnI for each n. For each

68

n c Z let X be'{0n} with the discrete topology and Y+ 1n nbe {V} with the discrete topology. Let

XO = lim(Xn fZ+)

and

Y = lim(Y ,g ,Z+ n mn +

be the same as in Theorem 2.13. Then, X is homeomorphic

to X, and Y is homeomorphic to Y,. Define Q :X - Y by

Qn(U ) = Vi,

for each n. Then, each Qn is a homeomorphism. Now,

n+1n Qn+l(Un+l) = + n(Vn+l) = Vnand

n n+ln( n+l n(Un n'

Therefore, by Theorem 2.1, Q :X-+ Y is a homeomorphism.

Hence, X is homeomorphic to Y.

Theorem_2.16: Every compact metric space is the con-

tinuous image of the cantor set C.

Proof: Let (X,d) be a compact metric space. Let

' = {A,1 1 A2 1 ,.*,An 11

be a cover of X by closed sets of diameter less than 2.

For each i such that 1515n1 , let Ki be a finite cover

1of A 11by closed sets of diameter less than. Since

A is closed in X, it can be assumed that each elementn

of Ki is a closed subset of Ai l:Let ?2 = U K be denotedi=l

by

2 2 ,A2 2, ,An22

69

Then 2 . and, if x Ai, there exists A 2 : 2 such

that x e A. j2A1 . By continuing this process a sequence

{2U } of finite covers of X by closed sets can be ob-

1tained, An element of 14 has diameter less than -. Eachn 2

Vn has the properties that n+ <4 n and if x 6 A iCVn

then there exists A. n+ n+1such that x E Aj nlA in'

For each m, let

2m = {Am,AA2m,...An m Km

For each i such that 1:i5 n , let

B(1,) = {(x,i,l)fx E A }

Let K be open in B(1,1) if and only if {xj (xi,l) E K} is

open in Au. Let n

H=1nb p B(i nly.i~l

Define a set V to be open in H1 if and only if Vn B(j,l)

is open in B(j,l) for each j.

For each Aj2 such that A .CS:A , letA. j2~ p1

B(j,p,2) = {(x,jp,2)Ix c Aj2

Letn1

H2 = U (UB(jip,2)|Aj2CAp ).p=l

Define open sets in the same manner as for B(il) and H1 .

For each AP 3 such that A, 3 -cA.j2 cA 1p, let

BC,j,p,3) = {(x,Qj,p,3)Ix E A }.

Letn

H3= U CUB , jp,3)jAP 3 -!A j2SAp1)'p=l

70

Define open sets in the same manner as for B(j,2) and H2.

By continuing this process, a sequence of sets {H i} i is

obtained. Now,n

H ,,CUBi,n)jA EA. . ..Hn n, U*CUBjj n -J JH n-1 .nl

i=1 n n-1

The method used to define open sets of Hn guarantees

that each B(inin-13,...,iin) is both open and closed in

Hn* It is also apparant that the B's are disjoint. Since

Hn is the union of the Bts and the B's are open, Hn is the

union of the open sets defined on Hn. Let V { ='{U }acDbe

a collection of open sets of Hn indexed by D. Let

K = B(jinin-1 ,'''j 1 ,n).

Since each U is open in Hn'

M = {x|(x,j ,jn-,'''i 1 ,n) s K(OU }

is open in A. . Therefore UM is open in A. . ButEnn asD Jnn

U Mc a ={x|(xjj n- 1 '.,1,n) 6 KnO(U L4)}.

Hence Uk is open in K. Therefore UV is open in H .

Let U and U 2 be open in Hn. Then for i c {1,2},

Mi = {x|(x'jn jn 1''', 1 ,n) P U K}

is open in A. . Therefore M 1Nm2 is open in A .But

MN m2 = {XI ('xj n l,n) 6 (U qU 2 )rK}.

Hence, U 1U 2 is open in K. Therefore, UJn U2 is open in

Hn, By induction, this can be extended to the intersection

of any finite number of open sets. Now the open sets of

Hn form a topology for Hn

71

For each n, let Q H + X be defined bynfn

QnC(-xnn-lj''l,*,jln)). = x

Clearly Qn is a function from Hn onto X. Let

y = (x,j n' 1n-l''j'1,n)

be an element of Hn and V be an open set of X containing

x. Since V is open in X, K = VFnA. is open in A.Jnn jnn

Let

U {(',jn'jn-l''j 1 ,n)|w e K}

and

Then, U c T.

open in T.

open subset

then V K.

quently, Qn

Define

T = B(jn'$n-l''' ,n).

Since {y(yj ,jn-l''063 1 ,n) c UI = K, U is

Since the B's are disjoint in Hn and U is an

of T, U is open in Hn. Clearly, y c U. If

h = (v,jnjn-l,,'',j'1 ,n) c U

Hence, v 6 V. Therefore Qn(U).V. Conse-

is continuous.

f *H n+ H byn+l n n+l n

S(+ , n n+1 .n'''ln+l))

Let f be the identity function. Clearly, f is ann n+l n

function. Let

t = (xjn n-l ',,j 1 ,n) Hn'

Then x e A. . There exists A. c n such that3n Jn+ln+l n+l

x c A. cA. . Thenin+1 n ann

S= (x jn+1lUn'''',j 1 ,n+1) e Hn+1

72

and fn+ n(s) = t. Hence f n+1 nis onto. Let

n ln1n+l1jn +'1,n+l) c Hn+1

and V be an open set of Hn containing fn+1 n(y). Let

N = B( jn'0n-l n'., n).

Then fn+i n y) e N, Since V is open in Hn3

M = {z (z,j n'in-l .j 1 ,n) E VON}

is open in A.n Since A. CA. , K = A.l mJ

nan+ln+1 J an+ln+l

is open in A j n+lAlso, x c K. Let

U =f{(w,jn+1,jn '...,jln+l)Iw e K}.

Now UL N and U is open in N. Hence U is open in Hn+i*'If

S = (t~jn+in ,'.'.,j 1,n+l) s U,

then t E K. Hence, t c A.in+n+1NM. Therefore, t c V.

Hence, fn+l n (s )c V. Now, fn+i n(U)L V. Since y s U

and fn+n) = x V, f n+1 n is continuous. For m> n,

let f :H + H be defined bymn m n

fmn ( f)1=0fn+ n +2 n+1 ' ' mm-

Then f is continuous and onto. Therefore, (Hn ,f Z+)

is an inverse sequence.

For each n in Z+, let Xn = X and in+1 be the identity

function from Xn+l onto Xn. For m> n, i:X + Xn be

in+1 n 0 1n+2 n+1 ..*.'0im m-l. Then (Xn mn ', Z+) is an

inverse sequence. Further, n;QHn X is continuous andn n n

onto. If

n+ 1 -jn'.''* 1,n+l) Hn+1

73

then

n0f n+inCY) QnC vjjn'in-,'' ,n)) = v.

But

in+l + in+l(v) = V.

Therefore, by Theorem 2.1, Q00:H0 -+ X is continuous.

Let w = (x,x,,..) c X . Then x A1 1for some i.

Hence (x,11 ,l) cH1 . There exists A1 22 F2 such that

x F A1 22 A*. Hence (x,12,11,2) c H2 and

f2 1 x,12 'I1 ,2) = (x,11 ,1).

By induction, there is an element s of H such that the00

n-th projection of s is (xin,1in-l.'. I 1 ,n). Hence,

Q (s) = w.

Therefore, Q is onto. Since X is homeomorphic to X,

X is the continuous image of H.

To complete the theorem it will be sufficient to show

that H is the continuous image of H X C and that H X C

is homeomorphic to C. Let

s = (x,jn ,jn-l '' '''d i,n)

and

t = (y'i ,i n-l-''''-' 1,n)

be elements of Hn such that s # t. If there is p such that

1 p n and j / iP, then s and t are in different B's.

Hence, there exists disjoint open sets

B(inI3n-l ''''',n) = T

74

and

S= B Q n 'j n.,- 3 '' ,.n)

containing t and s, respectively. If jP = iP for all pthen x # y and a and t are in T. Therefore, x and y are

elements of A ,Since x # y, there exist open disjoint

sets M and N of A. such that x e M and y e N. Let

J = {Cv n'nn-l'..,j,n)|v M}

and

I ={(w,j njn- ',.',i,n)|w N}.

Then c J and t6 I andJr)I=p. By the method of de-

fining I and J, each of I and J is open in T. Further,

J cT and I CT. Hence, I and J are open in H . Therefore,n

Hn is Hausdorff.

Let 7W = {VU},D be an open cover of Hn. Let

R = B(jnin-l''..,j,n)

be one of the disjoint collections of Hn' For each a c D,

V qR must be open in R. Therefore,

M Hn{(xlj '...,J ,n) E V mR}

is open in A . If y A , then s = (y,j ,...,jln) c R.

Therefore, there exists 6 6 D such that s 6 V Hence,

s 6 MN6R. Now, {MaRca D} covers A. . Since A. is

closed in X, A is compact, Hence, there exists a finitejfn . hreeitnafnt

subset F-of D, such that {M a e F} covers A. . Let

nz= (vj' in e R.

75

Then v e A 4. Therefore cM- for some a c F. Hence,

z e V . Therefore {V }aEF covers R. Hence, Hn is compact.

Now, H is compact and Hausdorff for each n. Therefore,

11H, is compact and Hausdorff. Hence, HO, is compact.

Let w A y be elements of X0,. Then there exists an n

in Z+ such that 7Tnr(w) A 7rn(y). Let

in(w) = Wn =(xn'j nn-l''...,jln)

and

=Tnn = (vn in -lW''' i1,n).

If for some p such that 1 p!5n, ip 7 j , then

Wn , BCjn .u n-l ''',n) = W

and

yn cB(in in-,'..,ii,n) = T

and W T. Therefore, WnT = . Hence, r (W)1n

closed in X., contains w and does not contain y.

for all p, then x A v. There exists m 6 Z+ such

1 d(.x V)1< d2x ) Therefore, xc Am e t( and x c A22mm n

where i . Further, A. nA.m lm1m jmm

is open and

If j =i1p p

that

. evj mmm

= $. Then

,ff.m(w) = (x, jm 1 ,.0.j 1,m)

and

=rm( Cvim im..-1' i 1,m).

Since j im, by the above argument there exists an openmm

and closed set of X. containing w and not y. Therefore by

Theorem 2.10, X, is totally disconnected.

76

Since X0 is compact and Hausdorff, it is regular. If

X is second countable then by Urysohnts metrization theorem,

X, is metrizable, Now, if H has a countable base for each00 n

n, then TtHn will be second countable. Hence, X will be

second countable and metrizable, Since X is a compact

metric space, X is second countable. Let 7 = {V}i be

a countable base for X. For each BKjnVn'..'''iln), let

M n' =l''xj n'dn-l''' ,il,n)|x s VqA }.

Then,

MiQn.n-l''...,jln)

is an open subset of

BQjn'$n1-1 ' ' ' 1 ,n)

and for each B there are only countably many such sets. Let

B(jnn-''..',jin) = R

and U be open in R. Then

.{Xj (x~,n'n-l''n,.1,j,n) c U nR} = P

is open in Ajn, But 7/ covers X. Hence, there is a sub-

collection C of 7 such that (U3)nA. = P. LetJnn

= {Vp1,VP

2

Then, Mp Q nn .,j 1 ,n) c U. If

t = (w,jn' n-l''.,j 1 ,n) E U,

then w c P. Therefore, there is V P -6 such that w c V .

jn

77

{M (nj'V1n) }0 is a base for 3(j njn-n1 '''.jl,n).{PiQn Jn-I's ~nl

As a result, Hn is second countable. Therefore, H, is

metrizable.

Let K = HOO X C. An element of K can be written as

(h,c) where h s H0 and c c C. Since the projection function

from K to H00 is continuous and onto, H00 is the continuous

image of K.,

Since each of H00 and C is a compact metric space,

K is a compact metric space. If (h1 ,c1 ) # (h2 ,c2 ) in K,

then h1 $ h2 or c1 c2 . If h1 X h2 , then since H. is

totally disconnected, there exists U open and closed in H00

such that h 1 U and h2 U. Consequently, (h1 ,c1 ) 6 U X C

and (h2 ,c2 ) is not. A similar argument holds for the

assumption that c1 c2 . As a result, K is totally dis-

connected. Let (hic) K and U be a basic open set of K

containing (h,c). Then U = U1 X U 2 where U1 is open in

H and U2 is open in C. Since C is perfect, there exists

c E U2 such that c c1 . Therefore, (h,c1 ) e U and

(h,c) (h,c1 ). It then follows that K is perfect. Now,

by Theorem 2.15, K is homeomorphic to C. Since H00 is the

continuous image of K and X is the continuous image of H0,,

X is the continuous image of C.

CHAPTER BIBLIOGRAPHY

1. Gentry, K. R., "Some Properties of the Induced Map,Fundamenta Mathematicae, LXIV(1969), 55-59.

2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.

78

CHAPTER III

AN INVERSE LIMIT CHARACTERIZATION OF

COMPACT HAUSDORFF SPACES

To characterize compact Hausdorff spaces, it will be

necessary to establish the proper background.

Definition 3.,1: Let En be euclidean n-space. Then,

,n 1 2 na point ai in E is expressed by ai (a ,a ,j.., ).Let

{aO,al,...,ar} be r + 1 linearly independent points in

En. Define [acial,...,ar I to be

1 2 n i{(x ,x ,...,x )Ix

r r= 2a 1.a,5ii5n, and 2 D .

j=0 j=1 0

= 1 and for each j, O .l1}.

Then, [a0 ,a1 ,.. .,ar] is called an r-simplex and each a.

r .is called a vertex. If x is such that x = x.a. where

j=0 J

r+l for each j, then x is called the barycentre of

[a 0 ,a,...,ar-]

To simplify notation, whenever x = (x 1 x2 ,...,xn) is

r .expressed by x = X.a., for 1 i<n, the superscript i

j=o c su

will be deleted if no confusion results.

79

8o

Definition 3,2: If a1 ,ai ,...,a are s + 1 pointso 1 s

of {aoal...,ar}, then [ai ,ai ,...,a ] is called ano 1 5

s-face of [a0 ,a1 ,.., arI.

Definition 3.3: Let K be a finite collection of

simplices in En auch that every face of a simplex of K is

in K and the intersection of any two simplices of K is

either empty or a face of each. Then K is called a complex

and UK is a polyhedron.

It is important to notice that each simplex of K is

closed and compact. As a consequence, UK is compact.

Theorem 3.1: Let Xn be an n-simplex in K and

A(Xn) be the barycentre of Xn. Let n0 > n1 > ... >ni be

a finite descending sequence of positive integers and

n0 n n. n.X ,X ,...,X be such that each X 0 is an n.-simplexUO a 1 i a1i

nk n n0 nwhere X is a face of X kl. Then, [A(X ),...,A(X )]

ak ak-1 a 0 Ot

is a simplex.

Proof: For notational purposes, let

n. 0 1 n.X X =[a ,a .. ,a

n a aa jaj

andn.

ACX 3) = Ba. n.

3 J

Suppose that'{BnnBn ,n1..,n } is not linearly independent.

81

Since X is a face of X , the following arrangement3 ~j-1l

can be made, Let

B be the barycentre of [a0 ,a1 , ... ,an]

Bn be the barycentre of [a0,a1 ,...,a]

Bn be the barycenre of [a 0 ,a1 ,... ,an .ni

Since the barycentres are not linearly independent, there

exists B such thatnp

B =cB +c +...+C Bn On ln 1nl n-p 0 1 n - n -

+ c B +.. + Bnp+l np+1 i

where all the c 's are not all 0. Therefore,13n c0 n0 c n

-n 1 a = s + ... + n- asn+ln0+1nn1 + +12Ss=s=0

+ p+l np+lC n

p+l -On 1+1 s

As a result,

c n0c n n0 =Da + .*. + pa1

n0,+10an +1 s in+ ias= p -l 5=0 p s=0

n n+ p+ L p+l c i i

+ j aF+,,.. + IFa.np+l+1s=0 n1 s=0 S

82

c.For o 5j <i and j p, let K. = Let K

S n.+l p n+13 P

Then, n n

Q= K Za + .,. + K a + .. + K Das=0 Es=0 S=0

Hence,

i i0= E K )a0 + ( EK )a +

m=0 m m=0 ml11-1

+ ( E K )a + ( mK )am=0 m n1 m=0 m n1+l

+ ( K K n)an+2 ,..+ E Km)am=0 m g2m=0 m ni1y

0an1+1+ K0an+2 O+n + K0an0

However, {a 0 ,a1 ,..n.,anI } is linearly independent. Therefore

jeach coefficient must be 0. mLKet c. E K . Now, c = K .S m=0m

If Kp = KO, then K ~ 0. This is a contradiction~p 0, 0 n +1p

to each coefficient being 0. As a result, Kp # KO'

Therefore, K0 = 0. Hence c0 = 0. But,

0 = c1 = K + K = 0 + K .

If n =n , K = 1 -0. This contradicts each coeffi-p 1 1 n 1+1X

cient being 0. Therefore, n / n1 ,Hence, K = 0. Con-

sequently, .= , Thus, c1 = 0. By induction, n nn +1 p

83

for any i such that 0 5i5n This contradicts the assump-

tion that {Bn 13 n ,'. n.} was not linearly independent.0 1 in

Hence {B n ,'.'Bn } is linearly independent. Now,

n n[ACX)0 ,...,ACX )] is a simplex.

0i

Theorem 3.2: Let K be a complex and Xn be an n-simplex

of K., Let n 0>n1 > ... > ni be a finite descending sequence

n n nof positive integers and X ,X ,...,X be such that each

a0

n. nk nklX is an n.-simplex where X is a face of X . Let

a k ak-l

nB be the barycentre of X k. Then [B ,B ,...,B ] is

a simplex. If K is the collection of all such simplices

formed from X n for each Xn in K, then K is a complex.

Proof: Since K is a complex, it contains only

finitely many simplices. Each simplex of K gives rise

to only finitely many simplices of K .6Consequently,

K has, only finitely many simplices.

Let Xni = [B0 ,B,**.,Bn ] be a simplex in K where

B is the barycentre of [bo,b,...,bn]0 n01

B1 is the barycentre of (b0 ,b1 ,...,bbn ]3

Bn is the barycentre of [b,b ,.. .,b ].

84

Let [B ,B ,.. ,, ]be a face ofX . ThenUa alajn

Baois the barycentre of [b 0,b1 ,...,ba]

is the barycentre of [b 0 ,b1 ,.. .,b al

B is

However, since

the barycentre of [b 0 , b,...,b ].

for each i, [b0,b ,..,b3] is a face of

[b.b ,*w .,bno], it is a simplex of K. Therefore,

[B a 0 3Ba 1'04.,B a ]jI c K101

The following lemma will help to show that if the

intersection of elements of K is nonempty, then the in-

tersection is a face of each.

Lemma 3.l: Let X = [Bo,B1 , ...,B ] be an element of

K where

Let S =

B0 is the barycentre of [b 0,b, .. ,bm 0

B is the barycentre of [b 0 , b1 ,.3. ,bIm

Bp is the barycentre of [b 0 ,b,...,bm ]p

[a0,ai,...,an] be a face of [b0,b1 ,...,bm ] such

that XOS 7k. Then XqS is a face of X.

85

Suppose that there does not exist i such that o 5i:p

and B e S. Then for each i such that o 5i 5p, there

exists b such that o j< m and b {aOa , .. ,an}.

Let x E X n~S, Then,

x X + 1 B 1 + ++ . . pBp

=0a0 + 61a 1+...+na n'

Since each Bi is the barycentre of [b 0,b, ...,bm],

A m0

o+s=0 pm +s=0

0a0 + ... + nnaX.

For each j such that o:j!p, let K. -=. Then,

pX Z= K.)b0 +. +( K.)bj=0 JO j0 p

p--1 p-i+ ( ZK)b +1 + . (EK.)bm

j=0 p j=0 Jmp-1

+ .,.,, + K 0bm 1 + . . . + K 0bm 0

6a0 + . .+. + 6a . +K0 0 n n

However, there is b. such that b. a for any i such that

o i :n. Hence, the coefficient of b . is 0. Since the

coefficient of b. is K.,

i j=0 0

+ Kp =0.

86

Therefore,

0 1 p

This contradicts the statement that if x c X then the

sum of the coefficients must be 1. Consequently, there

exists i such that oSi p and B S.

Let {B ,B ,.,.,B } be the collection of B 'sa0 al at

such that o:5i.<p and Bi c S. Let x s Xq S. Since x c S,

x ao + 61a 1 + ... + 6a nn*

Since x X,

x = A0 B0 + 1B1 + ... +AB

For each i such that o -<i !p, let K =m.iTheni m.+P Thn

p px = ( K)b0 + ... + ( K)b

m=O m=O p

p-1 p-1+(rK )b + + K)+ ... bm=m m +1 m mM=O p M=0 p-

+.,. + K0 b n1 +1+ + K0bn0

Suppose that o 5 i:5p and Bi i' B for any j such that

o < j !5t and 0. Let s be the largest such i. Then

there exists br {b 0 ,...,bm } such that br {aO,...,an}9

Now, for each j s, ar c {b 0 ,.*..bm} . The coefficients

sof b is DK. = 0, Hence, U = A1 = . = A 0r j=0s

87

By assumption, 7 0. Therefore, x [EBa,.0..,Ba].

Clearly, if x E [E3B,*a .' B at], x X s, Therefore,

0 1t0Ba ,Ba

which is a face of X,

Now to complete the proof of Theorem 3.2, let X and

T be simplices of K such that XnT # . Let

Xs = [BQ,B...,Bp]

and

Xt = [HOH .. .Ht

be the smallest faces of X and T respectively such that

X qTcXs and xrnTg;Xt. Let

B0 be the barycentre of [b 0 ,b1 ,.. .,bn0=P

B be the barycentre ofD[b,3 , b.n.1.,bn

Bp be the barycentre of [b 0 , b1 ,.. .,bn ] and

p

H0 be the barycentre of [h0,h,. . .,hm0 =L

H1 be the barycentre of [h 0,h,.. .hm]1

Ht be the barycentre of [h 0 ,h1 ,...,hm t

Now, let

[beb , .. ] (Minh 0 ,h1 , ...,hm ] = [a0 ,a1 ,...a ] = R.0 0p

88

The XOTZCR and R is, a face of P and L. By Lemma 3.1,

X QR is a face of X S But XQ T.GXs DR, hence, xs Q R = X sAlso, Xt R is a face of X and XQTCX R. Hence,t -tXt nR = Xtq. It can be assumed that R is the smallest

face of P and L containing X nT, If a1 c{asa ,. .a },

then a. c {b,b ,...bn00

} n{hh ,3...,5h }. For each j0

such that o j p, there exists x e Xs such that

x = AB$B + X11 + .. + AB. + ... + A B

and A. 0. For each j, let K. .n0 n1 n+

x = K(L b ) + K (Lb ) + ... +

S=0 s=0

Then,

n

K ( L bp S=0 s

p= ZL K )b

s=0 s

p+.Z+ K s)b ns=0 np

p-1+ ( Ky)b n+1

s=0 np

p-1+ .,.+s )b n

s0 sn0+ .,. +K~b.

+ ... +Kb0 1

Since A. 0, the coefficients of b ,b ,.. .,bn.b0' bj3*3bn.3

zero. But, x=60a0 + ... Therefore,

{bobl,..,b } c {a .,ap,...ap

Consequently,

As a result,

{b 0 ,b ,..,bn }_{a0 ,a1 ,...,a }.

{ b Q b j ,.) *. b n } = ' { a ,a . . . ,a0p }

are non

89

By a similar argument,

{hohihh,.,hm } = {aoa ,,,,a{h hm0 0 ' ' 'p

Therefore, B0 = H0 .

If Bi A H for some j such that b <j t, then either;

Case 1; there exists b c {b 0 ,b1 ,. ..,nb } such that

b {h ,h. .. ,hI}

or

Case 2: there exists h e {h 0,h1 ,...,hM } such that

h {bo,b ,..,,b }.

Suppose that B1 A H1 . Let x E [B,B ,...,Bp] where

x = 0 B0 + . B and A1 7 0. If x e [HoHi,...,Ht],

x 0 n0n = - b sn0 S=0 s

+

60 sM0=m----- h +

Let K = and L. =i n +1 1 m.1+1A

x np.. +-- Z b

np+1s S=Os

6t mt.. + TE h

mt+1s=0s

Therefore,

px = ( D K )b0 +...

m=0

+ .,, + K0hn+1

tCZLm h0 +,,m=0

+ ., + Lhm+1

p+ ( Zi K )b

m=0 m np

+ *.. + K0bn0

t+ E 2 L )hm

m=0 mint

+ , O

p-1+ ( Z K )b

m=0 m np+1

t.-1+ C Z Lm)hm

m=0 t

then,

90

IIf Case 1 is true, then the coefficient of b is Z) K

m=0 m

for 1. However, since b {{h0 ,h1 ,...,hm }, the co-

efficient of b must be Lo. Therefore,

LO = K0 + K1 + ,,, + K1 ,

PIf Lo - Ko, then K =E Lm for some p 21, Hence,

m=0

Lo= L0 + L + ... + Lp + K +... + K.

As a result, K1 = 0. It must follow that A = 0. This

is contrary to the supposition that A 1 j 0. Therefore,

Ko = L, But, Lo = K0 + K + ... + K. Hence, K = 0.

It then follows that A = 0. This contradicts A 0.1 1Hence, Case 1 is not true. Therefore, Case 2 must be true.

iNow, the coefficient of h is Lm for some i'l: and it is

m=0

also Kg . If Ko / Lo, then Lo = Ko + K + ... + K. for

some j e l. But, Ko = Lo + L + ... + Li for some i1.

Consequently,

Ko= K0 + K + . +..+K. + L + + L .

Therefore, K1 = 0. Hence, A = 0. This contradicts

A1 l 0. As a result, Ko = Lo. It then follows that since

Ko = LO + L1+ .o + L , L1 = 0. Now, 61 = 0.

Suppose that if B # Hr for 1l5 r5Z, then Ko =LO and

L = L2 =..' = L =0.

Suppose that B1 H +1, Then, if Case 1 is true, the

91

I Icoefficient of b is Ek , for 1, and it is also Z Ls=0 s=0

for some j such that 0< j< k + 1. Therefore, K1 = 0.

Hence, A1 = Q. But, X13 0. Consequently, Case 2 must

be true. As a result, the coefficient of h is K0 and also

iFL for some i Z! + 1, Then,s=0

K = L 0+ L1+... + L +1 + ... + L .

Since K0 = L0 , L+l = 0. Consequently, 6%+l = 0. Hence,

by induction, if B1 Hp, then 6p = 0. However, on the

tcontrary, F 61 = 1. Therefore, 60 = 1. Since 6 0 =

i=0 0

A = 0. This contradicts A1 $0. It then follows that

x P [Ho, .. ,Ht]. Consequently, B1 is not a necessary

vertex of [B0QB 1 ,. ..,Bp]. But, by the supposition, B1 is

necessary. Therefore, B = H for some j such that O<j t.

By a similar argument, H = B for some i such that 0< i 5p.

Suppose that B = H. and j >1. Then

{b0,b ,1...,b n 1} $ {h0,h ,...,)h }

which is a proper subset of {hh1,...,hm }. But

{h0 ,h1 ,..., hm } = {b0 ,b1 ,...,bn }

Therefore, {b 0 ,b1 ,...,b } is a proper subset of {b 0,b1,...,b }.

However, 1 1. This is a contradiction. Hence, H = B

There exists, b & {b0 , ., .,b n} such that b {b 0 ,...,bn .

92

Since B=H1, b E{h0,''.,hm} and b {h0 ,...,h '

Therefore, if x X Xt andx t

x =XA B + . .+ X B =6 H + .0.0 + 6tH'0. Q +PApB 0 0 t t3

then A0 =60*1

Suppose that for each i< j, B = Hi and if x 6 X nxtsuch that

x=AB+ . AB 6H + ... + 6H0p p 0 O t t3

then A = Suppose that B7 H. Let x csX

such that x = A B + .. + AB and A. 0. Then,0 0 p p

px = ( ZKm)b0 + ... + ( ZK)b

m=0 m=0 p

p-1+(K)b + +Kb

m=0 m n+1 0 n +1M- p 1

+,.+ K 0bn 0a

If x E Xt, then x = 60H0 + ... + 6t H. Hence,

t tx = CL )h

L= m)0 m= mZ m t

t"-1+ (2Lm)h + ... + L0h

M0m mt+I m+1

+ 0.w. + L 0h m 0

If Case 1

some i1 4j

1is true, then the coefficient of b is nD K for

and Z Lm for some s <4. Hence,m=0

0+1K+ ... + K + ... + K = L +

93

Since s <j, K = 0. Therefore, A. = 0. But A. # 0.

Consequently, Case 2 must be true. Hence, the coefficient

i 3of b is t-Lm for some i j and Z Km for some s< j.

m=m m=O

Therefore,

KO + . +Ks =L0 +... + L + ... +L

Since s <j, L = 0. Hence, 6 = 0. By a similar argument,

if B. / H. , then 6 = 0.3 j +l' +1Suppose that if B. H. for 0 s< i then 6 = 0.

3 3+s j+5Suppose that B. :Hj+. If Case 1 is true, then the co-

kefficient of b is Z K for some k >j and r L for some

p=OP p= 0 P

k <j +i. Hence, K. = 0. Therefore, A. = 0. But A. 0.

Then, Case 2 must be true. Consequently the coefficient of

kb must be L for some k tj + i and 2 K for Z< j + i.

s=0 s=0

Therefore,

L0 + L1+... + L. + + LS 1j+i kK +K 1 + ... + K

Hence, L = 0. As a result, 6 = 0. Therefore, by

induction, for all sk j, 6 = 0. Hence, 6 + . + 6 = 1.

Therefore, A0 + A1 + ... + = 1. Hence A. =0. This

is a contradiction to A # 0. Therefore, x [Ho, H1 ,...,Ht].

Consequently, B is not a necessary vertex of [Bo,B1 ,...,B ].

This is a contradiction. Therefore, B. = Hk for some kQ.

By a similar argument, H. = B for some k. The same3 k

94

argument that proved H = B1 will hold for H. BJ Je

Therefore,

[BOB 'EEp] = [HO,Hi, ...,Ht].

It now follows that K is a complex.

Now, K1 is called the barycentric subdivision of K.

It is noteworthy to point out that if K is the barycentric

subdivision of K, then UK = UK.

Theorem 3.3: If T = [AO,A**..,Ap] is a simplex in

En where d is the usual metric on En, then the diameter

of T is the maximum {d(A.,A.)I0 i p,o j :5p}.

Proof: Let

D = maximum {d(Ai,A )o5isp and Oj p}.

Let x and y be in T. Let A be a vertex farthest from y

and d(y,Ai) = 6. Then A. 6 S(y,6), the spherical neighbor-

hood around y of radius 6, for each j. Consequently,

TcS(y,6). If x T, d(x,y) d(y,A1 ). Let A. be vertex

farthest from A and X = d(Ai,A .). Then Tc S(AiA).i -3s

Therefore, d(y,A) 5d(A1 ,A3) :D. Consequently,

d (,y) :5 d (y, A 1) 5 d(A ,A )< D.

Hence, the theorem is proven.

Theorem 3.4: If K = [A0,A , *..,Am] is a simplex of

diameter D in En, K1 is the barycentric subdivision of K

and D is the diameter of K then D !5 D.1' lm+l

Proof: Let d be the usual metric on En. By Theorem

3.3, D1 is. the maximum of the distances between the vertices

95

of K 1, Let D= d(B.,B.) where i sj, E1 is the barycentre

of [AQ,A1 ,...,A1 ] and B. is the barycentre of [A05,A ,...,A].

Then, B A and B = A for 1 <r <n. Therefore,

n i

d(B.,B ) = ( (k ZAr i :Ax) 2)2r=1 i+lp=Q P 3 +1p=p

n1C 2 (( -) Ar 2 A2)r=1 + + p=0 P + 1 p=I+lP

n'_r__ 1 r 2 2

r=1 (+1)(j+1) p jp=+

n1= 2 7' ( (U) A r_ 1 Ar)2 2

n1. p jr=1 j+1 1+=p=0 p=I+1

r=l j+1 1+1p= Ap=+In= i11=:(Li('E Ar Ar 2

r= +1 +1 OP p ilp

ar1 +p=0 p=+1n 1 1

However, (+)=1, () (j-1) and AIs a vertex of K

for each 1 sj. Consequently, x = (x 2 ,..,n) where-

xI =l2 Ar, and y = (y 2yy )Ewherey =k Ar

p=r = 3 p=I+1

are elements of K, Hence,

n 1(C , Ar 1 Ar 2 ) 2 2D.

r=1 +1_ 0 ~ =1+1

96

Therefore, dCB ,B ) .5j-'D. But, j-i 5j and j 5m. As a

result, d lBB}<-D <m+1,D

Dieflhition 3.4. If K is a complex in En, then the

mesh of K is the maximum of the diameters of simplices of K.

Theorem 3.5: If{K }00 is a sequence of complexesi 1=1

such that for each i, K is the barycentric subdivisioni+lof Ki and M is the mesh of Ki, then{Mi}+0 as i+oo.

00Proof: Suppose that {K i}j 1is a sequence of complexes

satisfying the hypothesis and that the mesh of K. is M .1 i.

Let k = [AOA1 ,. .,Am] be a simplex of K of diameter M1

such that k has a maximum number of vertices. Then, if

k2 is a simplex of K2 , by Theorem 3.4, the diameter of

km M Hence, M2M ml M Furthermore, k has at

most m vertices. If k3 is a simplex of K3 , then the dia-

meter of k is less than or equal to m M Hence,3 m+l 2 H

M3 mE M2 By induction, M m:l(_ n-lM Since N is a

constant and ( 0n+s as n+oo, {M1}I 1-*0 as i+o.

This paper will assume without proof that if N is a

finite indexing set, then I= N I where I is theaoN 6?a

closed unit interval for each a, is a polyhedron.

Theorem 3.6; If N is a finite indexing set, F is a

closed subset of IN and U is an open subset of IN such that

FE Uc IN, then there exists a polyhedron P such that

Fc POc P c U.

97

NProof: Since EU and Iis a compact metric, there

exists an open subset V of IN such that F cVcVLU. Now,

since IN ia assumed to be a polyhedron, there exists a

complex H such that IN =UH. Let {H}L1 be a sequence

of complexes such that H = H and H.+ is the barycentric

subdivision of H . For each i, let Mi denote the mesh of

NNH, Since V SU, VU(I \U) = $ and each of 7 and IN\U is

closed in IN. Let (IN\U) = W. There exists 6> 0 such

that if x e V and y c W, d(x,y) 6, where d is the usual

Nmetric on IN. There exists j e Z+ such that M4. <6. Let

R = {S|S is a simplex of H. and S n7 }

Then R covers V. Let

K = {sjs is a face of some S c R}.

Now, K is complex, UK is a polyhedron and V. UK. Suppose

that x e UK and x U. Then x c W and there exists s e K

such that x e s. Since s c K, s is a face of some S c R.

Therefore, SqV v 4 and x E S. Let y F VM S. Then, since

y E 7 and x E W, d(x,y) 6. However, x E S, y c S and

S E R. Since S 6 R, S is a simplex of H.. Therefore, the

diameter of S is less than 6. Hence, d(x,y)< 6. This

contradicts the statement that d(x,y) 6. Consequently,

x e U. As a result, UK.9U. Now, F1V. V U K SU. It

now follows that F c UKOC U K.SU.

Theorem 3.7; If Y is a compact Hausdorff space, then

Y is homeomorphic to the inverse limit of an inverse system

of polyhedra.

98

Proof; Let Y be a compact Hausdorff space. Then Y

is homeomorphic to a subspace of H I , denoted by IAcA cA

where I is the closed unit interval and A is some indexing

set. Denote the homeomorphic image of Y in IA as X. Then,

AX is a closed compact subset of IA. Let

13 ={B|B is a nonempty finite subset of A}.

Now, for B and D in/s, B 3 D if and only if IB -riD

For each B e '1, let fAB1 IA÷B be defined by

f A({x a} :) = {xa } pBfAB caoacA a cwB'

Then, fAB is continuous and onto for each B cS and fAB

Bis a closed subset of I By Theorem 3.6, there exists P,

a polyhedron, such that fAB(X) CPO cP CB. Let

9 = {P IP is a polyhedron such that for some B c 1,

Bf AB (X) CP CPCI }

Let &be an indexing set for . If A e.6, and B and B

are elements of /3 and

fAB(-Q. ciB

and

fAB(X) _P _oP _IB

then B = B . If not, IB 1 B p. Therefore, for each

A & ,, there is a unique B c 8 such that fABXCPOC CIB

Let

E C= {(AB)(A,B) x

and fAB(X) O LIB

99

Then, E is an indexing set for P. Let (XB) c E, be

denoted by ,X 2

If C-B-CA, where C and R are elements of /3, then

define C +f Iby

fBC U a aE:B = a aeC'

Then each fBC is continuous. If x = {x a} A 1A, then

fAC(X) = {xaaC

However, fAB(ax) = aEB and f(BCx a IaB = xa aC. Hence,

fAC =fBC 0AB

Direct E in the following manner. Let rB and 6D be

elements of E. Then, B and D are unique in /3 such that

FAB * B WB cIB

and

f (X) gP0 O p LAD 6 D DcID

Now, define B 6 D if and only if B cD and fDB(6D B

Suppose that AD e E. Then DOCD and

D D D

Therefore, XD XD

Suppose that X V 5% and B 56 D. Then,

fAF(X) CPF FcIF

f AB (X) g P -P 01fABB SB

and

AD 6D 6D

100

Now, since FOCB and BSCD, F!D. AlsPo, D )D and

fBF CPO P. Consequently, fDF(P ) cP . It thenB y D ~ F4

follow-s that

The remaining property, necessary to show that "t<"

is a direction for E, is that of a common follower for two

elements of E. The following lemma will make this easy

and be helpful later.

Lemma 3.2: If 0B is in E and D is in / such that

B _cD, then fAD(X fDB B(P )

Proof: Suppose that x c fAD(X) and x f DB' B

Now, fDB (x) EfDB(fAD(X)). Therefore, fDBx) fAB(X).

But, x f DB B). Hence, fDB(x).P B Consequently,

fDB (x) fAB(X). This is a contradiction to fDB(x ) fAB .As a result, x fD B)( . It then follows that

fAD fDB( B

Now, to complete the proof that " " is a direction

for E, let XF and B be elements of E. There exists D c 1

such that B9 D and FcZD. By Lemma 3.2,

DB DE

which is open in ID. By Theorem 3.6, there exists a poly-

hedron P such that

f D(X)BP P BfI(PO )qf ( O CDAD DB B DF XF

101

Now, P e 9 and must be indexed by some 6 in E. SinceDP I , M =D, ThereforeD E.B ,fB6D E BtD B(BP

f (P)c P0 Fc D and BED. Therefore, XF 53 D and

B :6D. Consequently, " " is a direction for E.

Let PB<D in E. Then BcD. If B = D, let

f B PD -*PB be the inclusion map. If B is a proper

subset of D, let

f P + P .bef DBI .D B D B Dp

Now, (PsB f 6 ,E) is an inverse system.

Let x = {xa} A E X. For P DP

f (X) PO c cFAD D f cf D ADD D 6Dt D

AD~ ~ a c DD

Let x be the element in II P such that for each 6D D6E,FSE F

the 6D-th

If B 6D

P6D x

projection of x , P (x), is fAD ({X aA)

in E, then (xB )=XfAB{xa} A) and

fAD ix} . Consequently,

DB % Cx )) = (f (x} )AD a DaDB D caA

Since fAD( a{a} A D

and

102

f (f'({1x} ))6 D B' AD- a ac A

= f DB (fAD a XTatA)

AB QA6A)x )-B

Therefore, x s X = lim(P6 f ,E).00 +66IE3 )

Let g be a relation from X to X00 defined by g(x) X

and =D DX6Suppose that x = y in X where x = {x }

and y = {y }a Then, g(x) = x is such that

Tr1 (x ) =f ({x } )7 B AB aaEA

and g(y) = y such that

Y

BAB)=

a aA

for each 1B in E. Since x = y, x = y for each a cA.

Hence,

fAB({Xa}EA) AB aaAfor each B F-8. Therefore, 7TB B(X) = (y ) for each B'

t IConsequently, x = y . As a result, g is a function.

Suppose that x A y in X. Then there exists a c A

such that x aXy. Let B c 1 such that a e B. Then,

fAAB B) There exists XB c E such thatff (X) CAOThereCB

fABE#X) B- AB

B BBConsequently., . A 3 x ) X 7 X(y ).Therefore,, g(x) 74.g(y).

Hence, g is one to one.

Let x e X0 and 1B e E. Suppose that 7rT13 (x) fAB(X).

Then there exist disjoint open sets U and V of IB such that

103

fA(X)cU and 7rBCx) e V. Hence, fB(xX) unpo which isAB ABB

open in IB. Consequently, there is S B c E such that

fAB( X) c ~ycP , cUnp ,B B B

No w, TrB) B.However, BoB and fBB(PVB B P PB * B B B3? B

Therefore, B:. B. Since x X. and B:5 B'

Tr (x) = f I (Tr'B(x)).

Since B = B, f BOB is the inclusion map. Consequently,

f ' (7T'(x)) = Tr ' (x).

f OB B B W BTherefore,

Tr (x) = r (x) P .

Hence, )rX e P cU rP B.Now, T B(x) c U andB B B

7T rB(x) 6 V. This contradicts the statement that Uq- = p.

Therefore, if x E X, Tr B (X) ,fAB(X) for each B F E.

Suppose that there does not exist y c X such that g(y) = x.

Then for each y c X, there exists r3B E such that

fAB (y) B (x). Therefore, there exists Uy SB and Vy 0BBopen and disjoint in I such that f AB(y) Uy and

Tr (x) E V .Hence,y B

( =.{fAB 1(Uy)|y c X} covers X.

104

There exists a finite subcollection H of ?( covering X.

Let

H ='{fAD-(uD)1 6D D

where 9 is a finite subset of E. There exists M _1 such

that B cM for each B s ' By Lemma 3.2, fAN(X)cf ~1(PO )y ema 32, fAM (X-. fMB BB

for each .B . Let R = n fMBXP0 ) which is open in

I Then, fAMCX).CR. There exists am E such that

fAM(X) cP 0 c:P CR.AM m- m'

Consequently, fMB am B and B cM for each B e

Hence, B5m for each B c7. As a result,

7T r (x) = f GB(7Tra M(x))770B MOB MN

for each B c7. Now, since 7rM (x)cfAM(X), there exists

z c X such that fAM(z) = 7TrM(x). But z fAD(U6 D)for

some 6D 69 .- Therefore, fAD(z) U6 . However,

(6D) = 6D M

fM6 D fAM(z)) = fMD 0 fAM(z)

fAD(z).

Consequently, UD nv6D . This is a contradiction to

U 8 V = . As a result, there is y c X such that g(y) =x.HD D

Henc e, g is onto .

105

Suppose that y 6 X and that U is open in X., such that

g(iy) c U. Then there exists a basic open neighborhood K

such that g(y) K.U., Let K = 7r 4 (U6 ) NowD D

Tr, (g(y)) 6 U6 and fAD is continuous. Therefore, there

exists V open in X such that y e V and fAD (V) ;;U 6 ID f

z E V, fAD(z) E U6 .*Therefore, g(z) c K. Consequently,

g is continuous. Since X is compact Hausdorff, XO is

Hausdorff and g is a one to one continuous function from

X onto X,, g is a homeomorphism. Now, Y is homeomorphic

to X,, the inverse limit of an inverse system of polyhedra.

CHAPTER BIBLIOGRAPHY

1. Nagata, Jun-iti, Modern General Topology, Amsterdam,North Holland Publishing Company, 1968.

2. Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.

106

BIBLIOGRAPHY

Capel, C. E., "Inverse Limit Spaces," Duke MathematicalJournal, 21 (1954), 233-245.

Gentry, K. R., "Some Properties of the Induced Map,"Fundamenta Mathematicae, LXIV (1969), 55-59.

Nagata, Jun-iti, Modern General Topology, Amsterdam,North Holland Publishing Company, 1968.

Willard, Stephen, General Topology, Reading, Mass.,Addison-Wesley Publishing Company, 1970.

Whyburn, Gordon T., "Analytic Topology," American Mathe-matical Society Colloquium Publications, 23 (1942).

Wilder, Raymond L., "Topology of Manifolds," AmericanMathematical Society Colloquium Publications,32 (19' 9) .

107