Theorem: If topological X t space is compact and Hausdorff, then X t is normal.

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

2. Recall that a space is Hausdorff if every pair of distinct points x1 x2 can be separated by disjoint open neighborhoods:

3. Recall that a space is normal if every pair of disjoint closed subsets A1 and A2 can be separated by disjoint open sets: there exist open sets U1 and U2 with A1Ì U1, A2Ì U2and U1Ç U2 =f

4. We’ll prove the theorem by first showing that a compact Hausdorff space Xt is regular. Let A be a t-closed subset of X with xÎX - A1. We want to separate A1 and x by disjoint open sets:

5. If A1 = f then the proof is easy ( U = f), so we’ll look at the case where A1¹ f. If a Î A1, then a ¹ x, so we can separate these points since X is Hausdorff: a Î Ua and x Î Va

6. We now do this for every a Î A1: we get a collection C= {Ua : a Î A1} of open sets whose union contains A1. For each Ua, there is a corresponding open set Va containing x such that UaÇ Va = f. Since A1 is compact, (it’s a closed subset of a compact space) there exists a finite subcover {Uª1, Uª2, ... ,Uªn} such that A1Ì (Uª1È Uª2È...È Uªn) Let U = Uª1, È ... È Uªn be the desired open set containing A1.

7. The corresponding open sets around x yield the desired neighborhood: xÎV = (Vª1ÇVª2Ç....Ç Vªn). Then U Ç V = f since for each i, Uª iÇ Vª i = f. Thus we’ve shown that A1 and x can be separated, so that Xt is regular.

8. Now we use this regularity to prove normality: if A1 and A2 are disjoint, nonempty, closed subsets of Xt -- the empty case is trivial -- let x Î A2Ì X - A1. By regularity, we can separate x from A1 : xÎUX , A1ÌVX .

9. Then C2 = {Ux: x Î A2} is an open cover of A2, which is compact since it’s closed, so there is a finite subcover: A2Ì ( Ux1È Ux2È ..... È Uxm) = U1 where U1 is disjoint from U2 = (Vx1 Ç Vx2Ç .... Ç Vxm). We’ve separated A1 from A2.