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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. Recall that a space is Hausdorff if every pair of distinct points x1 x2 can be separated by disjoint open neighborhoods: x1 x2 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 A1 U1, A2 U2 and U1 U2 = f U2 U1 A1 A2 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 xX - A1. We want to separate A1 and x by disjoint open sets: U A1 V X If A1 = f then the proof is easy ( U = f), so we’ll look at the case where A1 f. If a A1, then a x, so we can separate these points since X is Hausdorff: a Ua and x Va A1 Ua Va a X Ua Va =f We now do this for every a A1: we get a collection C = {Ua : a A1} of open sets whose union contains A1. For each Ua, there is a corresponding open set Va containing x such that Ua 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 A1(Uª1 Uª2 Uªn) Let U = Uª1, ... Uªn be the desired open set containing A1. The corresponding open sets around x yield the desired neighborhood: xV = (Vª1Vª2.... Vªn). Then U V = f since for each i, Uª i Vª i = f. Thus we’ve shown that A1 and x can be separated, so that Xt is regular. U A1 V X 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 A2 X - A1. By regularity, we can separate x from A1 : xUX , A1VX . VX A1 A2 X UX Then C 2 = {Ux : x A2} is an open cover of A2, which is compact since it’s closed, so there is a finite subcover: A2 ( Ux1 Ux2 ..... Uxm) = U1 where U1 is disjoint from U2 = (Vx1 Vx2 .... Vxm). We’ve separated A1 from A2. U1 A1 U2 A2