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 xX - 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:
xV = (Vª1Vª2.... 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 : xUX , A1VX .
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