Page 210 - DMTH503_TOPOLOGY

Topology

Notes  covers X. For, given x let m be the smallest integer such that x B . m Then x belongs to a
element of  , by definition.
m

Note Some of the properties of a paracompact space are similar to those of a compact
space. For instance, a subspace of a paracompact space is not necessarily paracompact; but
a closed subspace is paracompact. Also, a paracompact Hausdorff space is normal. In other
ways, a paracompact space is not similar to a compact space; in particular, the product of
two paracompact spaces need not be paracompact.
Theorem 1: Every paracompact Hausdorff space X is normal.
Proof: The proof is somewhat similar to the proof that a compact Hausdorff space is normal.
First one proves regularity. Let a be a point of X and let B be a closed set of X disjoint from a. The
Hausdorff condition enables is to choose for each b in B, an open set about b whose closure is
b
disjoint from a. Cover X by the open sets , along with the open set X – B; take a locally finite
b
open refinement  that covers X. Form the subcollection D of  consisting of every element of 
that intersects B. The  covers B. Furthermore, if D , then D is disjoint from a. For D intersect
B, so it lies in some set , whose closure is disjoint from a. Let
b
V =  D;
D 
then V is an open set in X containing B. Because  is locally finite,

V =  D,
D 

so that V is disjoint from a. Thus regularity is proved.
To prove normality, one merely repeats the same argument, replacing a by the closed set A
throughout and replacing the Hausdorff condition by regularity.
Theorem 2: Every closed subspace of a paracompact space is paracompact.
Proof: Let Y be a closed subspace of the paracompact space X; let  be a covering of Y by sets
open in Y.
For each A , choose an open set A of X such that AY = A. Cover X by the open sets A,
along with the open set X – Y.

Let  be a locally finite open refinement of this covering that covers X.
The collection  = {B Y : B }
is the required locally finite open refinement of .

Example 3: A paracompact subspace of a Hausdorff space X need not be closed in X.
Solution: Indeed, the open interval (0, 1) is pracompact, being homeomorphic to , but it is not
closed in .
Lemma 3: Let X be regular. Then the following conditions on X are equivalent:
Every open covering of X has a refinement that is:
1. An open covering of X and countably locally finite.

2. A covering of X and locally finite.

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