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Tuesday, December 30, 2014

Unions of Paracompact Spaces (6.41.7)

James Munkres Topology, chapter 6.41, exercise 7:

MathJax TeX Test Page Let X be a regular space. Show that (a) if X is a finite union of closed paracompact subspaces, or (b) if X is covered by the interiors of countably many closed paracompact subspaces, then X is paracompact.

Proof: (a) Suppose X=K1K2 where K1,K2X are closed and paracompact. By induction the general case will follow from this one. Suppose {Uα} is an open cover of X. Then {UαK1} and {UαK2} are open covers of K1 and K2 respectively, so let B1={Bβ} and B2={Bγ} be locally finite refinements of the two. We claim C=B1B2 is a locally finite refinement of {Uα} covering X. The refinement and covering conditions are evident, so we proceed to demonstrate local finiteness; assume xK1. If U is a neighborhood of x such that K2U intersects only finitely many members of B2, then U(XK1) is a neighborhood of x intersecting among the same finite family from B2, and is disjoint from the members of B1. So we may assume xK1, and similarly xK2. As such let U and V be neighborhoods of x such that K1U and K2V intersect only finitely many members of B1 and B2 respectively. If UV intersects an element of B1, then since that element is within K1 so too does UVK1UK1, so that only finitely many elements of B1 are intersected. Similarly too for B2, and now UV is a neighborhood of X intersecting only finitely many elements of B.

(b) Let X=int Ki where each Ki is paracompact. Again let {Uα} be an open cover of X. For each i, let Bi be a locally finite refinement of {UαKi} covering Ki, and further let Ai={Bint Ki | BBi}. Then Ai is a locally finite open cover of int Ki for each i, so that A=Ai is a countably locally finite open refinement of {Uα} covering X, so that X is paracompact by Lemma 41.3. 

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