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§5 Compact topological spaces
The notion of a compact topological space plays an important role
in topology and its applications.
The prototype of this notion is the following property of a
segment(Lebesgue, 1903): any open covering of a segment [a, b]
of the real line contains a finite subcovering.
Borel later established that any closed bounded subset of a finite
dimensional Euclidean space posses the same property.
any open covering of
a segment [a, b] of
the real line contains
a finite subcovering
The concept of a compact space was introduced by Vietoris in
1921.
Def. A collection A of subsets of a space X is said to cover X, or
to be a covering of X, if the union of the elements of A is equal
to X. It is called an open covering of X if its elements are open
subsets of X.
Def. A space X is said to be compact if every open covering A of
X contains a finite sub-collection that also covers X.
Def. A collection C of subsets of X is said to have finite
intersection property if for every finite sub-collection A of C,
the intersection  A is non-empty. The collection C is also
called a centered system.
Prop. 2. A space X is compact iff the intersection of any centered
system of closed subsets of X is non-empty.
Proof. →Suppose X is compact, i.e., any collection of open subsets
that covers X has a finite collection that also covers X . Further,
suppose {Fi: i  I} is a centered system of closed subsets. We
claim that {Fi: i I } is non-empty.
Suppose not, i.e., {Fi: i  I } = , then
X = X \ ({Fi: i I}) = {X \ Fi: i I}.
Since each Fi is closed, the collection {X \ Fi: i  I} is an open
cover for X . By compactness, there is a finite subset J  I such
that X = {X \ Fi: i J}, and then X = X \ ({Fi: i J}) , so
{Fi: i J} =  , which contradicts with the fact that {Fi: i  I}
is a centered system.
←The proof in the other direction is analogous. Suppose the
intersection of any centered system of closed subsets of X is nonempty. To prove that X is compact, let {Fi: i  I} be a collection of
open sets in X that cover X . We claim that this collection contains
a finite subcollection that also covers X .
The proof is by contradiction. Suppose that X  {Fi: i  J} holds
for all finite set J  I . Then, we have {X \ Fi: i  J} = X \ {Fi: i
 J}  , and thus the collection of closed subsets {X \ Fi: i  I} is
a centered system.
It follows that   {X \ Fi: i  I} = X \ {Fi: i  I}. This
contradicts the assumption that {Fi: i  I} is a cover for X .
A very closely related property to the concept of compactness is
the countable compactness. It was introduced by Frechét in 1906.
Definition. 3. A space X is called countably compact iff every
countable open cover of X has a finite subcover.
Prop. 4. Every compact is countably compact.
§5.2. Fundamental properties of compact spaces
Thm. 2. If f : X  Y is continuous surjection and X is compact ,
then Y is compact.
Proof. Let {U :   A} be an open cover of Y, then { f 1 (U  ) :   A}
is an open cover of X. It has a finite subcover
{ f 1 (U1 ), f 1 (U 2 ), , f 1 (U n )} , and U i  f ( f 1 (U i ))(i  1,2, , n),
n
n
n
Therefore Y  f ( X )  f (i 1 f 1 (U i ))  i 1 f ( f 1 (U i ))  i 1U i , and
{U1 ,U 2 ,,U n } is a finite open subcover of {U :   A} .
Corollary: Any continuous real valued function on a compact
space is bounded and reaches its maximum and minimum
values. (from Thm. 1 and Thm. 2)
Remark. Compactness is not hereditary property, [0,1] is
compact, but its open subspace (0,1) is not.
Theorem 3 . A closed subspace of compact space is compact.
Proof. Let X be a compact space, Y  X is a closed subspace,
then for any open cover {V :   A} of Y, we have V  U  Y
where U  is an open subset of X.
Y
X
Y
X
{U :   A}  { X \ Y } is an open cover of X, hence has a finite
sub-cover, say {U 0 ,U1 ,,U n } {X \ Y } , then {V0 ,V1,,Vn } is a finite
sub-cover of {V :   A} . Thus, Y is a compact subspace.
Theorem 4 . Every compact subspace Y of a Hausdorff space
X is closed in X.
Proof . Let x  X \ Y . for any y  Y , there exists a pair of
disjoint open subsets U x , y and V y such that x U x , y , y V .
Then {Vy : y  Y } is an open cover of Y, hence has a finite
sub-cover {V0 ,V1,,Vn } , denote the corresponding open sets
n
U

U
,
U
,

,
U
containing x by 0 1
i0U i , then U is an open nbd.
n . Let
n
of x disjoint from i 0 Vi  Y , this shows that Y is closed.
Note. This is converse of Thm 3 in the class of Hausdorff space.
Another words, in the class of Hausdorff spaces, compactness is an
absolute closeness.
Ex 3. X is infinite set with finite complement topology.
X is compact, since for any open cover, we fix one element
of it, the complement is finite, so can be covered by finite
many elements. And any subspace of X is compact by same
reason. But any infinite subset(other than X) is not closed .
This space is not a T2 space, since any 2 nonempty open subsets
intersect.
Thm. 2. If f : X  Y is continuous and X is compact, Y is Hausdorff,
then f is a closed map.
Proof. Let A be a closed subset of X, then A is compact. Since f is
continuous, f (A) is compact subspace of Y by Thm. 2. Now, f (A)
is closed in Y, since Y is Hausdorff space.
Corollary. A condensation(1-1, onto and continuous map )
from compact space to a Hausdorff space is a homeomorphism.
Above two results can not extend to a continuous map to a T1 space.
Ex 4. I = [0,1] with usual topology. X = [0,1] with finite
complement topology. Then X is T1 space but not a Hausdorff space.
id: I  X is a condensation, I is compact, but ix is not a
homemorphism. [0,0.5] is closed in I, but i[0,0.5]=[0,0.5] is not
closed in X. In fact, any closed set in X is finite.
§5.4. Pseudo-compact space
Definition 4: A completely regular T1 space X on which each
continuous real valued function is bounded is said to be pseudocompact.
Theorem 11: A metrizable space is compact iff it is pseudocompact.
§5.5. Compactness and separation axioms
In compact spaces, T2, T3, T4 are the same.
Ex.3 of§5.3 shows that compact T1 space need not be a
Hausdorff space.
Thm. 13. Every compact Hausdorff space X is normal.
Proof. Step1. Every compact Hausdorff space is regular.
Let x  A where A is closed in X. Since a closed subspace of
compact space is compact, x and A can be separated by disjoint
open nbds of x and A(refer to the proof of Thm.4), therefore X is
regular.
2. Let A, B be disjoint closed subsets. For any x  A , x  B, they
can be separated by disjoint open nbds Vx ,U x where x Vx , B  U x .
{Vx : x  A} is an open cover of A.
A
B
A is a compact subspace of X, because compactness is closed
hereditary property. Therefore, it has a finite sub-cover
denote by {V0 ,V1 ,,Vn }, the corresponding open nbds of B
n
by{U0 , U1 , , Un }. Let U  i 0 U i , V  in0 Vi , then U, V are
disjoint open subsets, and B  U , A  V .
Def. A topological space X is said to be completely regular if, for
each set A and any point x  X\A which is far from A, there is a
continuous function f: X R on X such that f(x) is far from the
set f(A).
Prop. A topological space X is completely regular if, for
each point x  X and for each closed set A which does not
contain x, there is a real-valued continuous function f on X such
that f(x) = 0 and f(y) = 1 for all y  A.
Prop. Every completely regular is regular.
Prop. Every subspace of a compact Hausdorff space is completely
regular.
§5.6 Compactification of topological spaces
Def. 5. Compact space cX is called a compactification of X if, X is
a subspace of cX and X  cX .
One point compactification aX  X  { } , the open sets to be the
entire space and the sets which is open in X.
The space aX does not satisfy the T1 separation axiom, since one
point subset in X are not closed in cX.
Remark. not every T2 or T3 space has such a compactification .
Thm. 14. Every completely regular T1 space has a Hausdorff
compactification.
The proof of this theorem uses Tikhonov’s theorem.
Thm. 15. Every normal T1 space X has a Hausdorff
compactification  X s.t. any two disjoint closed sets of X has
disjoint closures in  X .
This is called a Stone-Cech compactification.
Theorem 15 is used to prove the following result.
Thm 16. If f : X  R is continuous and bounded, X is T4 , then f
can be extended to  X .
The following generalization of Theorem 16 is useful.
Thm 17. If f : X  Y is continuous, X is completely regular T1 , Y
is compact T2, then f can be extended to  X .
§5.7 Locally compact spaces
Def. 6. X is locally compact if x  X has an open nbd. O s.t.
O is compact.
Prop. 7. Locally compact Hausdorff space is completely regular.
Proof. 1. Locally compact Hausdorff space is regular.
Let x  U , U be an open nbd. of x. By locally compactness,
there is a compact nbd. D, i.e., x  D ,U  D is an open nbd. of x in
compact subspace D.
Since compact Hausdorff space is regular, there is an open nbd.
V of x in D s.t. clDV  U  D. V is also a nbd. of x in X . And

clDV  V since D is closed in X. Thus, x V  V  U  D  U .
So, X is regular.
2. Let x  B , B is closed in X , then U = X \ B is an open nbd of
x such that x  D   D  X \ B , D is compact Hausdorff subspace
of X, hence completely regular, therefore in D there is a function
g : D  [0,1] such that g ( x)  0, g ( y )  1 for y  D \ D .
Define h : X \ D  [0,1] s.t. h  1, Then h is also continuous.
By passing lemma,
 g ( x), z  D
f ( z)  

h
(
z
),
z

X
\
D


is continuous and f ( x)  0, f ( y )  1 for y  B  X \ D .
Prop. 8. X is locally compact Hausdorff iff it is an open subspace
of a compact Hausdorff space.
1. Let X be a space, prove that X is T1 space iff for each x  X,
the intersection of all open neighborhoods of x is {x}.
2. Let T be the usual topology of R, T 1 = {G – E: G  T , E 
Q}. Prove that (R, T 1) is a Hausdorff space, but not regular.
3. Let X be a regular space, A a compact subset of X, Y  X. If A
 Y  cl(A) , then Y is a compact subset of X.