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The attitudes of patients' mothers toward mental illness at Veterans Administration Neuropsychiatric Hospital, Marion, Indiana, 1968

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A study of the development of the Catholic colored clinic, Atlanta, Georgia, October 1944-January 1947, 1947

Burke, Mary E. H. (Author)

Tested differences and correlations of scholastic achievement in primary grades 1-3 between a continuing teacher and rotating teachers grades 1-3 for the three years of primary school

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Burns, Barbara (Author)

A CLASS OF SPACES IN WHICH COMPACT SETS
ARE FINITE
A THESIS
SUBMITTED TO THE FACULTY OF ATLANTA UNIVERSITY
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS
FOR THE MASTER OF SCIENCE DEGREE
BY
BARBARA BURNS
DEPARTMENT OF MATHEMATICS
ATLANTA, GEORGIA
JULY 1970
TABLE OF CONTENTS
Page
INTRODUCTION 1
Chapter1.TOPOLOGICAL PRELIMINARIES 2II.SCME PROPERTIES OF SPACES WITH NO ISOLATED POINTS .... 7III.MI-SPACES 12
BIBLIOGRAPHY 18
ii
INTRODUCTION
A mathenatical note entitled "A Class Of Spaces In Which Compact
Sets Are Finite” by Murry R. Kirch appeared in the January 1969 issue of
the American Mathematical Monthly. The term of space has been intro¬
duced by Norman Levine to describe a topological space in which each
compact set is finite. The term pseudo-finite has also been used to de¬
scribe such spaces and will be used in this paper. Levine has shown
that first countable T, spaces which are pseudo-finite must be discrete.
Kirch shows that there exists a wide variety of nondiscrete pseudo-
finite spaces. The purpose of this paper is to explain and provide
proofs for all theorems and lemmas found in Kirch's paper. Examples
are also provided.
Preliminary information on topological spaces is found in Chapter
I* Pseudo-finite spaces have no isolated points. Some results of
spaces with no isolated points are presented in Chapter II. The main
results of this paper are found in Chapter III.
1
CHAPTER I
TOPOLOGICAL PRELIMINARIES
A topological background is needed to introduce the main results of
this paper. The necessary definitions, lemmas, and theorems are pre¬
sented in this chapter.
Definition 1. A topological space (X,T) is a set X of points and a
family T of subsets of X which satisfy the following axioms:
1) The empty set 0 and the whole space X are members of T.
2) The union of any number of members of T is a member of T.
3) The intersection of any finite ninnber of members of T is a
member of T.
The family T is called a topology for X, and the members of T are said
to be open sets of the topological space (X,T).
Definition 2. If X^ is a subset of a topological space (X,T), the in¬
duced or relative topology for X^ is the collection T^ of all sets
which are intersections of X^ with members of T. (X^,T^) will be
called a subspace of (X,T) if and only if T is the relative topology.
Definition 3. Let X be any set. The family of all subsets of X yields
a topology for X which is called the discrete topology for X, and X
together with its discrete topology is called a discrete topological
space or simply a discrete space.
Definition 4. A subset A of a discrete topological space is discrete
2
3
if and only 1£ £x} is open relative to A for each x ^ A.
Definition 5. Let X be any set. The family consisting of 0 and X
yields a topology for X called the indiscrete (or trivial) topology, and
X together with its indiscrete topology is called an indiscrete topolog¬
ical space or simply an indiscrete space.
Definition 6. A point x 6: A is an interior point of A if and only if x
belongs to some open set which is contained in A; i.e., x ^ S
A where sx is open.
Definition 7. A set A of a topological space X is open if and only if
each of its points is an interior point.
Definition 8. Given a topological space (X,T), a point x ^ X is a
limit point of a subset A of X if and only if every open set G contain¬
ing X contains a point of A different from x; i.e., x ^ G ^ T, then
(A/^G) {xj ^ 0. The set of all limit points of a set A is
called the derived set of A and is denoted by A'.
Lemma 1.1. If a set A is a subset of a set B, then every limit point of
A is also a limit point of B; i.e., A^B^ A'^ B*.
Proof. By definition x € A' if and only if (G ^ )/l A 9^ 0
for every open set G containing x. But B ^ A; therefore G
B ^ G "V fxj A ^ 0. Thus x€A'B'; i.e.. A'CZ B'.
Definition 9. A subset A of a topological space X is a closed set if
and only if its complement, A^ , is an open set.
4
Definition 10. The closure of a set A is the intersection of all closed
sets containing A. In other words, if ij is the class of
all closed subsets of a topological space X which contain A, then A ■*
Proposition 1.2. Let A be the closure of a set A. Then the following
hold: 1) A is a closed set since A is the intersection of closed sets,
2) if C is a closed set containing A, then AO A O C, and 3) A is
closed if and only if A “ A.^
Lemma 1.3. Let A be a subset of a topological space X. Then the
closure of A is the union of A and its set of limit points; i.e., A =
A O A'.^
Lemma 1.4. If a set A is a subset of a set B, then the closure of A is
a subset of the closure of B; i.e., if A O B then A O B.
Proof. If A0 B, then by Lemma 1.1 A* CZ- B*. Hence A Kj A* CZ B O B'
or by Lanma 1.3 A O B.
Lemma 1.5. Given a topological space X. A set A has no limit points
if and only if it is closed and discrete.
Proof. Suppose A has no limit points; i.e.. A' “ 0. A A' by
Lemma 1.3. SoA“AWa'=aO0'='A. Therefore A is closed. Let
X £ A. X is not in A' since A' is empty. Therefore there exists an
_
H. L. Royden, Real Analysis. New York: Macmillan Company, 1968.
2
Ibid.
5
open set G containing x and not containing any points of A other than x.
Hence x C G, G is open and G (A - {xj ) = 0. G /I A = f X J which
is open relative to A by definition. Therefore fxj is open in A.
Thus A has the discrete topology.
Now let A be closed and discrete. Suppose jS. Let x ^ A'. Since
A is closed, A = A = AU A'. Therefore x € A. A is discrete, so f x3
is open in A. There exists an open set G such that A
Since x ^ A*, G must contain a point of A other than x. This is a con¬
tradiction of the fact that A G = fx} . Hence the assumption A* 0
is false. Therefore A' = 0.
Definition 11. A subset A of a topological space X is said to be dense
in B C X if B is contained in the closure of A; i.e., B C A. In par¬
ticular, A is dense in X or is a dense subset of X if and only if A = X.
Lemma 1.6. If A is a subset of B and A is dense, then B is also dense.
Proof. Let A C B and let A be dense. By Lemma 1.4 Ad B A d B.
A = X by definition of a dense set. X = A d B d X. Thus "b*= X.
Definition 12. A cover of a set A is a collection of sets such
that A 1,Jjy G<<,. If all the sets of a cover are open, we say we have
an open cover.
Definition 13. A topological space X is said to be compact if and only
if every open cover has a finite subcover. A set K in a topological
space X is said to be compact if and only if, with the relative topology,
the subspace K is compact; equivalently, a set K in X is compact if and
6
only if every open cover of K by open sets in X has a finite subcover.
Lemma 1.7. Every closed subset of a compact space is ccmpact.
Proof. Let A be a closed set of a compact space X. Let G be a cover
of A by open sets in X. Since A is closed, A*^ is open. Hence the
members of G together with the open set A^ is an open cover of X.
Since X is compact, this open cover of X contains a finite subcover of
X. In other words, there are a finite number of open sets 0, ,...,0^
in G such that 0, ...LJo^ LJ A^ = X. Thus ^0, ,...,C^J covers
A and A is cbmpact.
CHAPTER II
SOME PROPERTIES OF SPACES WITH NO ISOLATED POINTS
Lemma 2.1. Let X be an infinite set. Then there exist two sets M and
N such that the following hold:
1) M U N = X
2) M n N = 0 and
3) M and N are each infinite
Proof. Let fx^ , x^ , ...] be a countably infinite subset of X.
Let M = f x^, x^y , x^ , ...J and let N = X'v.M. Since N is the com¬
plement of M, we clearly have MN = 0 and M U N * X. Clearly M is
infinite, and N is also infinite since it contains the infinite set
Definition 14. A point x in a topological space is said to be an iso¬
lated point if and only if {^x J is open.
Lemma 2.2. Let X be a topological space having no isolated points. If
S is an infinite subset of X and x ^ S, then there exists a set M such
that the following hold:
1) M is infinite
2) M ^ S
3) X 6 M
4)
Proof. Let S be an infinite subset of X and x 6 S. Since S is infinite,
7
8
S - £x} is also infinite. By Lemma 2.1 there exists Mj , such that
M, n a, * 0, M| = S - [x] , and Mj and are each infinite.
Case I. All open sets containing x intersect . If U is open and
X ^ U, then U n M, 0. Choose M = LJ f x J . Since is infinite,
M is infinite. - {xj ^ S and x6S, soM = M^ LJ x J
d S. Hence M d S. Clearly x6 M. Now suppose 0 is open and x € 0.
By Case I hypothesis 0 intersects M | . Let y £■ 0/1 M, . Then y 6 M,
and M| S - {x} so y x. Also since y ^ M, and M| CX = 0,
y ^ . Thus y (: ^ ^ J ® M. But y £ 0, so 0 M.
Case II. There exists an open set containing x which does not intersect
M| . Let U be such an open set containing x. Now choose M = Mj
£ X J . Since M| is infinite, M is infinite. M|d S - {x ] ^ S
and X g S, so M = Mj ^ J ~ S* Clearly x £■ M. Now suppose 0 is
open and x^O* x^O/Hu and 0/^ U is open, since it is the inter¬
section of two open sets. Since there are no isolated points, {xJ is
not open. Hence fx^ 0 ^ U. But {x] d 0 U, so 0/1 U con¬
tains a point y different from x. Now y £ U, but U does not intersect
M| by case hypothesis, so M, . Also since y x, we have Mj
= M. But y £• 0, so 0 So we have the desired result in all
cases.
Definition 15. If S is a subset of the topological space X, then the
interior of S (written Int S) is defined to be the union of all open
sets which are subsets of S.
Theorem 1. Let X be an infinite topological space having no isolated
9
points. Then there exists an infinite subset S of X such that Int S
= 0.
Proof.
A. Choose Xj X; this is possible since X is infinite and hence non¬
empty. Since X is infinite, by Lemma 2.2 there exists a set M, such
that
1) M| is infinite
2) M, ^ X
3) X| €■ M,
4) if Xj 6 0 and 0 is open then 0 ^ M
B. Choose 6 Mj " { Xj J ; this is possible since M is infinite
and hence Mj - fxj] is non-empty. Since M| is infinite, M| - f x, }
is infinite and thus by Lemma 2.2 there exists a set such that
1) is infinite
2) M^ ^ M, - fx,}
3) x^ e M^
4) if x^ 6 0 and 0 is open then 0
C. Now suppose X| , Mj , ... , x^ , have been chosen where
1) M,,..., M^ are each infinite
2) xj £■ Mj where i ■ 1, ... , n
3) M; ^ M;_, - fx..,} where i = 2, ... , n
4) if X,' (r 0 and 0 is open then 0 ^ M* where i = 1, ... , n
D. Choose £ M^ - Cx^} ; this possible since M^ is infinite
and hence M^ - fx^} ^ 0. Also since M^ is infinite, M^ - £x^} is
10
infinite. So by Lemma 2.2 there exists a set such that
1) is infinite
2) - Cx}
’'n+i ^ ^n+i
4) if ^ 0 and 0 is open then 0 ^
By mathematical induction we have defined a sequence X| , M, , x
... such that
1) M,' is infinite where i = 1,2,...
2) M/ ^z. where i = 2,3,...
3) x; €• M;
4) if xj £ 0 and 0 is open then 0 ^ M; where i = 1,2,...
Let S = fx, , x^, , Xj , •••3 • Suppose i'^j where i, j are positive
integers. Mj ^ - fx^,} ^ Mj., C ..
^ ^ M; - fxj] ^ Mj . Thus Mj ^ M j . Also xj^ ^j — M
— fxj j . So x; ^ Xj . Thus i j x - # x^* , so S is infinite. Let 0
be open and 0 S. Suppose 0 # 0. Then let xJ be the element of 0
having smallest subscript. Hence x^ € Mj and since x* ^ M,* ^ Mj
for all i^ j we see that all other elements of 0 will also be in M j .
Thus 0 S Mj . From the induction 0 is open and Xj 6 0, so 0 ^ Mj .
This is a contradiction. Hence it must be that 0=0. Thus the only
open sets which lie in S are the empty set. By definition 15 we thus
have that Int S =* 0.
Lemma 2.3. A Tj space with no isolated points is either empty or infinite.
Proof. Let X be a T, space with no isolated points. Since X is T^ ,
11
each point is closed and hence each finite subset is closed being the
union of a finite number of closed sets. If X # 0 and X is finite, then
let X € X. X - fx] is finite and hence closed. Therefore fx3 is
open so X is an isolated point. This is a contradiction. So either X
is empty or finite.
CHAPTER III
MI-SPACES
The main results of this paper are embodied in this chapter. I
will show that a group of spaces called MI-Spaces has the property that
each compact set is finite. The term pseudo-finite will be used to desig¬
nate this property.
Definition 16. A topological space X is said to be an Ml-space if it
possesses no isolated points and each dense set is open.
Lemma 3.1. An Ml-space is a T, space.
Proof. Let X be an Ml-space and let x£ X. fxi is not open since X
is an Ml-space. Thus every open set containing x must contain a point
other than x. Hence every open set containing x contains a point of
f xl^ . Therefore X ^ (x J** . Since fxj^ CZ. (xj*’
we have X = {x J 1 x ] ^ {x . Thus fx ? ^ is dense.
By definition of an Ml-space each dense set is open, so fx } ^ is open.
Therefore {x J is clos-ed. Therefore X is a T, space.
Lemma 3.2. The empty set is an Ml-space.
Proof. The empty set has no points; hence no isolated points. The only
subset of 0 is 0 itself. 0 is open, hence each dense set is open. Thus
0 is an Ml-space by Definition 16.
Lemma 3.3. In an Ml-space each set with empty interior has a dense
12
13
complement.
Proof. Let X be an MI-space and let X such that Int A = 0. Let y
^ X. Let 0 be an open set such that y ^ 0; 0 cannot lie in A since 0
0 and Int A = 0. Therefore 0 ^ A. Hence 0 O A^ 0. Since 0 is
arbitrary, every open set containing y contains a point of A*'. There¬
fore y is a point of closure of A^; i.e., y t A*" . Hence X ^ A^ .
Clearly A®" X, so A = X. By definition A is dense.
Definition 17. A set A of a topological space X is totally isolated if
and only if it has no limit points whatsoever.
Theorem 2. In an Ml-space each set with empty interior is totally iso¬
lated.
Proof. Let X be an MI-space. Let A^ X such that Int A = 0. Suppose
A has a limit point x.
Case I. X € A. I will show A ixj , is a non-open dense subset of
X. First I will show A^' U tx} is not open. Since x is a limit
point of A, every open set containing x contains a point of A different
from X. But no open set lying in A^ L/ f xj contains a point of A
different from x, so no open set containing x lies in A^ f x } .
Hence A ' U {x3 is not open. A is dense by Lemma 3.3. Since
A^ A^U {xj ,A*' U £x5 is dense by Lemma 1.6. Thus it has
been shown that KJ (xj is a non-open dense subset of X. This
contradicts the definition of X being and MI-space.
. C.
Case II. xffA;i.e., x6A . I will show that A itself is a non-
14
open dense subset of X. Since x is a limit point of A, every open set
containing x contains a point of A different from x. In particular,
every open set containing x contains a point of A; i.e., a point not is
A . Hence no open set containing x can lie in A . Therefore A is
not open. A^is dense by Lemma 3.3. So again we have a contradiction.
So our original assumption that A has a limit point is false. Therefore
A* =0. So by Definition 17 we have A is totally isolated.
Theorem 3. In an Ml-space each totally isolated set has empty interior.
Proof. Let X be an Ml-space and let A be a totally isolated subset of X
such that X € Int A. There is an open set 0 in X such that x 6 0 ^ X.
Since A is totally isolated, each point y ^ A is such that /yj is open
relative to A. x £ A, so (xj is open relative to A. There is an open
set U in X such that U A = (xJ . Since O^A, U/H 0 ^U0A =
f X J . Therefore U no- {x} . Hence fxj is open. This gives
us a contradiction since X contains no isolated points. That is, no
point such that fx] is open.
Theorem 4. A subspace of an Ml-space is also an Ml-space provided it
has no Isolated points.
Proof. Let X be an Ml-space and let K be a subspace of X with no iso¬
lated points. Let A ^ K such that A is dense in K. Since A ^ A Kj
then A O is dense in X by Lemma 1.6. By definition of an Ml-space
each dense set is open. Thus A U is open in X. (A \J H K “
A since A^.R. Hence A is an open set in K. Thus each dense set in K
is open in K. By hypothesis K has no isolated points so K is an Ml-space.
15
Theorem 5. If X is an MI-space and X # Cl, then X is infinite.
Proof. Let A ^ X. Then A is finite and A=iJ x6aJ,
a union of a finite set of closed sets. So A is closed. Hence each
subset of X is closed and thus each subset of X is open, so X has the
discrete topology. Since X 0, let x € X. Then £x} is open, so x is
an isolated point; a contradiction, since X is an MI-space. Hence there
are no finite non-empty MI-spaces.
Theorem 6. Each non-empty Ml-space is non compact.
Proof. Let X be an MI-space which is not empty. By Theorem 5 X is in¬
finite. Let A be an infinite subset of X with Int A ® 0. The existence
of such a set follows from Theorem 1. In an Ml-space each set with
empty interior is totally isolated by Theorem 2. Hence A is an infinite
totally isolated subset of X. Let x € A, then {x} is open relative to
A, since A is totally isolated. A® LJ i x^aJ - The
class {£x}j x6aJ is a cover of A and is an open cover of
A since all subsets of a discrete space are open. A is closed and discrete
by Lemma 1.5. No proper subclass of CL is a cover of A, for if we take
out one {xj we would no longer have a cover of A. Hence CL is in¬
finite since A is infinite. Therefore the open cover CLot A contains
no finite subcover. Hence A is not compact. Since A is closed, X is
not compact by Theoran 1.7.
Theorem 7. Each Ml-space is pseudo-finite.
Proof. Let K be a compact subset of the Ml-space X. Int K K and
Int K is open in K since it is the intersection of an open set, Int K,
16
with K. So K - Int K is closed in K since its complement, Int K, is
open in K. Hence K - Int K is a closed subset of the compact set K. By
Lemma 1.7 K - Int K is compact.
Suppose X 6 Int (K - Int K) . Then there exists an open set 0 such that
X 6 0 Cl K - Int K. Clearly 0 ^ K, so x £ Int K and x ^ K - Int K.
But X € Int (K - Int K) C- K - Int K. This is a contradiction, so Int
(K - Int K) is empty. So K - Int K is totally isolated by Theorem 2.
By Lemma 1.5 K - Int K has the discrete topology. Let C * f Cxj /
X £ (K - Int K) J . This is an open cover of K - Int K since the rel¬
ative topology is discrete. The only subcover of C is C itself. If
K - Int K is infinite, then there is no finite subcover of K - Int K, so
it would not be compact. Since it is compact, then K - Int K is finite.
Let I be the set of isolated points ofK; i.e., 1= fx|x^K and
£ xj open relative to kJ . Let x6 I. Suppose x € Int K. Since
X £ I, there exists an open set 0 such that 0 O K = fx } . So 0 O
Int K = £x 3 • Hence fx} is open in X. This is a contradiction
since X is an Ml-space and thus has no isolated points. Therefore x ^
Int K. But I !^-K and x € I so x ^ (K - Int K) . Thus I ^ (k - Int K) .
I is finite since it is a subset of a finite set. I = u f c x3 /
X 6 IJ , a union of open sets in K. Hence I is open in K. K - I is
closed in K so again by Lemma 1.7 K - I is compact. Hence K - I is a
compact subset of X with no isolated points. Hence K - I is an Ml-space
by Theorem 3. By Theorem 4 K - I is empty since each non-empty Ml-space
is non-compact. Since K - I = 0, then K I. Clearly as before I ^ K.
Therefore K = I. Hence K is finite since I is finite. Thus each compact
17
subset K is finite. Hence the whole space X is pseudo-finite.
BIBLIOGRAPHY
Hu, Sze-Tsen, Elements of General Topology. San Francisco, California:
Holden-Day, Inc., 1965.
Kelley, John L., General Topology. New York: Van Nostrand, Inc., 1955.
Pervin, William J., Foundations of General Topology. New York: Ac¬
ademic Press, 1964.
Royden, H. L., Real Analysis. New York: Macmillan Company, 1968.
18


A class of spaces in which compact sets are finite

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