Page 131 - DMTH503_TOPOLOGY

Unit 13: Compact Spaces and Compact Subspace of Real Line

Notes
Example 1: The real line is not compact, for the covering of by open intervals

 = {(n, n + 2)/ n  }
contains no finite sub-collection that covers .

Example 2: The following subspace of is compact
X = {0} {1/n  }.
+
Given an open covering  of X, there is an element U of  containing O. The set U contains all
but finitely many of the point 1/n; choose, for each point of X not in U, an element of 
containing it. The collection consisting of these elements of , along with the element U, is a
finite sub-collection of  that covers X.

Lemma (i): Let Y be a subspace of X. Then Y is compact if and only if every covering of Y by sets
open in X contains a finite sub-collection covering Y.
Proof: Suppose that Y is compact and  = {A } is a covering of Y by sets open in X. Then the
x  T
collection
{A  Y J}
x
is a covering of Y by sets open in Y; hence a finite sub-collection
{A  Y, ...., A  Y}
 1  n
covers Y. Then {A , ..., A } is a sub-collection of  that covers Y.
 1  n
Conversely, suppose the given condition holds; we wish to prove Y compact. Let  = {A } be a

covering of Y by sets open in Y. For each , choose a set A open in X such that

A = A Y
 
The collection  = {A } is a covering of Y by sets open in X. By hypothesis, some finite

sub-collection {A ,...., A } covers Y. Then {A , ..., A } is a sub-collection of  that covers Y.
 1  n  1  n
Theorem 1: Every closed subspace of a compact space is compact.
Proof: Let Y be a closed subspace of the compact space X. Given a covering  of Y by sets open
in X, let us form an open covering  of X by joining to  the single open set X – Y that is
B =   {X – Y}
Some finite sub-collection of  covers X. If this sub-collection contains the set X – Y, discard X – Y;
otherwise, leave the sub-collection alone. The resulting collection is a finite sub-collection of 
that cover Y.

Theorem 2: Every compact subspace of a Hausdorff space is closed.
Proof: Let Y be a compact subspace of the Hausdorff space X. We shall prove that X – Y is open.
So that Y is closed. Let x be a point of X – Y. We show there is a neighborhood of x that is disjoint
0 0
from Y. For each point y of Y, let us choose disjoint neighborhoods U and V of the points x and
y y 0
y, respectively (using the Hausdorff condition). The collection {V / y Y} is a covering of Y by
y
sets in X; therefore, finitely many of them V , ...., V cover Y. The open set
y1 yn
V = V .... V
y1 yn

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