# 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.

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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.

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

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 = 

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:

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

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.

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.

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.

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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