Page 212 - DMTH503_TOPOLOGY

Topology

Notes This step involve a new trick. The previous trick, used several times, consisted of ordering the
sets in some way and forming a new set by subtracting off all the previous ones. That trick
shrinks the sets; to expand them we need something different. We shall introduce an auxiliary
locally finite closed covering  of X and use it to expand the element of .
For each point x of X, there is a neighbourhood of x that intersects only finitely many elements
of . The collection of all open sets that intersect only finitely many element of  is thus an open
covering of X. Using (3) again, let  be a closed refinement of this covering that covers X and is
locally finite. Each element of  intersect only finitely many elements of .
For each element B of , let

(B) = {C : C ad C X – B}

Then define E(B)X = X  C
C (B)
Because  is locally finite collection of closed sets, the union of the elements of any subcollection
of  is closed by lemma, therefore the set E(B) is an open set. Furthermore, E(B) B by definition.
Now we may have expanded each B too much; the collection {E(B)} may not be a refinemet of .
This is easily remedied. For each B , choose an element F(B) of  containing B. Then define
 = {E(B) F (B)| B }.
The collection  is a refinement of A. Because B (E(B) F(B)) and  covers X, the collection 
also covers X.
We have finally to prove that  is locally finite. Given a point x of X, choose a neighbourhood
W of x that intersects only finitely may elements of , say C , ..., C . We show that W intersects
1 k
only finitely many elements of . Because  covers X, the set W is covered by C ,...C . thus, it
1 K
suffices to show that each element C of . Now if C intersects the set E (B) F(B), then it intersects
E(B), so by definition of E(B) it is not contained in X–B; hence C must intersect B. Since C
intersects, only finitely many elements of , it can intersect at most the same number of elements
of the collection .

Theorem 3: Every metrizable space is paracompact.
Proof: Let X be a metrizable space. We already know from Lemma 2 that, given an open covering
 of X, it has an open refinement that covers X and is countably locally finite. The preceding
lemma then implies that  has an open refinement that covers X and is locally finite.

Example 4: The product of two paracompact spaces need not be paracompact. The space
is paracompact, for it is regular and Lindelöf. However, × is not paracompact, for it is
Hausdorff but not normal.

Self Assessment

3. Show that Paracompactness is a topological property.
4. If every open subset of a paracompact space is paracompact, then every subset is
paracompact. Prove it.

24.3 Summary

 Let X be a topological space. A collection  of subsets of X is said to be locally finite in X
if every point of X has a neighbourhood that intersects only finitely many elements of .

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