Page 132 - DMTH503_TOPOLOGY

Topology

Notes contains Y, and it is disjoint from the open set
U = U ... U
y1 yn
formed by taking the intersection of the corresponding neighborhoods of x . For if z is a point
0
of V, then z V for some i, hence z U  yi and so z U.  Then U is a neighbourhood of x , disjoint
yi 0
from Y, as desired.
Theorem 3: The image of a compact space under a continuous map is compact.

Proof: Let f : X Y be continuous; let X be compact. Let  be a covering of the set f(X) by sets
open in Y. The collection
–1
{f (A) | A }
is a collection of sets covering X; these sets are open in X because f is continuous. Hence finitely
many of them. Say
–1
–1
f (A ), ...., f (A ), cover X, then the sets A ...., A cover f(X)
1 n 1 n

Note Use of the proceeding theorem is as a tool for verifying that a map is a
homeomorphism

Theorem 4: Let f : X Y be a bijective function. If X is compact and Y is Hausdorff, then f is a
homeomorphism.
Proof: We shall prove that images of closed sets of X under f are closed in Y; this will prove
–1
continuity of the map f . If A is closed in X, then A is compact by theorem (1). Therefore by the
theorem just proved f(A) is compact. Since Y is Hausdorff, f(A) is closed in Y by theorem (2)

Example 3: Show by means of an example that a compact subset of a topological space
need not be closed.
Solution: Suppose (X, I) is an indiscrete topological space such that X contains more than one
element. Let A be a proper subset of X and let (A, I ) be a subspace of (X, I). Here, we have
1
I = {, A}. For I = {, X}. Hence, the only I – open cover of A is {A} which is finite. Hence A is
1 1
compact. But A is not I-closed. For the only I-closed sets are , X. Thus A is compact but not
closed.
Theorem 5: A closed subset of a countably compact space is countably compact.
Proof: Let Y be a closed subset of a countably compact space (X, T).

Let {G : n  N} be a countable T-open cover of Y, then
n
Y   G .
n
n
But X = Y Y
Hence X = Y{G : n  N}
n
This shows that the family consisting of open sets Y, G , G , G ,.... forms an open countable cover
1 2 3
of X which is known to be countably compact. Hence this cover must be reducible to a finite
subcover, say
 n 
Y, G , G , ..., G so that X = Y   G i 
1 2 n   i 1 

n
 Y   G i

i 1

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