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Published byHoward Elwood Modified about 1 year ago

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The proof is quite similar to that of a previous result: a compact subspace of a Hausdorff is closed. Theorem: If topological X space is compact and Hausdorff, then X is normal.

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Recall that a space is Hausdorff if every pair of distinct points x 1 x 2 can be separated by disjoint open neighborhoods:

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Recall that a space is normal if every pair of disjoint closed subsets A 1 and A 2 can be separated by disjoint open sets: there exist open sets U 1 and U 2 with A 1 U 1, A 2 U 2 and U 1 U 2 =

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We’ll prove the theorem by first showing that a compact Hausdorff space X is regular. Let A be a -closed subset of X with x X - A 1. We want to separate A 1 and x by disjoint open sets:

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If A 1 = then the proof is easy ( U = ), so we’ll look at the case where A 1 . If a A 1, then a x, so we can separate these points since X is Hausdorff: a U a and x V a

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We now do this for every a A 1 : we get a collection C = {U a : a A 1 } of open sets whose union contains A 1. For each U a, there is a corresponding open set V a containing x such that U a V a = . Since A 1 is compact, (it’s a closed subset of a compact space) there exists a finite subcover {U ª 1, U ª 2,...,U ª n } such that A 1 (U ª 1 U ª 2 U ª n ) Let U = U ª 1, ... U ª n be the desired open set containing A 1.

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The corresponding open sets around x yield the desired neighborhood: x V = (V ª 1 V ª 2 .... V ª n ). Then U V = since for each i, U ª i V ª i = . Thus we’ve shown that A 1 and x can be separated, so that X is regular.

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Now we use this regularity to prove normality: if A 1 and A 2 are disjoint, nonempty, closed subsets of X -- the empty case is trivial -- let x A 2 X - A 1. By regularity, we can separate x from A 1 : x U X, A 1 V X.

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Then C 2 = {U x : x A 2 } is an open cover of A 2, which is compact since it’s closed, so there is a finite subcover: A 2 ( U x 1 U x 2 ..... U x m ) = U 1 where U 1 is disjoint from U 2 = (V x 1 V x 2 .... V x m ). We’ve separated A 1 from A 2.

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