Basic Analysis – Gently Done Topological Vector Spaces

4: Separation 33
We claim that U and V are disjoint. Indeed, if x ∈ U ∩V, then x ∈ U a ∩V b for
some a ∈ A and b ∈ B. This means that d(x,a) < 1
2 ε a
d(a,b) ≤ d(x,a)+d(x,b)
< 1
2 (ε a +ε b )
≤ max{ε a ,ε b }
and d(x,b) < 1
2 ε b
which is a contradiction. We conclude that U ∩V = ∅, as required.
. Hence
We can give an alternative proof using the continuity of the distance between
a point and a set in a metric space. We recall that if x is any point in a metric
space (X,d) and A is any non-empty subset of X, the distance between x and A
is given by
dist(x,A) = inf{d(x,a) : a ∈ A}.
One shows that x ↦→ dist(x,A) is a continuous mapping from X into R and that
x ∈ A if and only if dist(x,A) = 0. Now suppose that A and B are disjoint
closed non-empty sets in X. Put U = {x ∈ X : dist(x,B)−dist(x,A) > 0} and
V = {x ∈ X : dist(x,B)−dist(x,A) < 0}. Then U is the inverse image of the
open set {t ∈ R : t > 0} under the continuous map x ↦→ dist(x,B) − dist(x,A)
and so is open in X. Similarly, V is open, being the inverse image of the open set
{t ∈ R : t < 0} under the same continuous map. It is clear that U ∩V = ∅. Now,
if a ∈ A, then dist(a,A) = 0. If dist(a,B) were 0, we would conclude that a ∈ B.
But B is closed so that B = B and we know that a /∈ B. Hence dist(a,B) > 0
and so a ∈ U. Thus A ⊆ U. Similarly, B ⊆ V and the result follows.
Theorem 4.4 Every compact Hausdorff space is normal.
Proof Suppose that (X,T) is a compact Hausdorff topological space. We shall
first show that if z ∈ X and F is a closed set not containing z, then z and F have
disjoint neighbourhoods.
For any x ∈ F, there are disjoint open sets U x and V x such that z ∈ U x and
x ∈ V x (by the Hausdorff property). The sets {V x : x ∈ F} form an open cover of
F. Now, F is a closed set in a compact space and so is compact. Hence there is
a finite set J in F such that F ⊆ �
x∈J V x
. Put U = �
x∈J U x
and V = �
x∈J V x .
Then U and V are both open, z ∈ U, F ⊆ V and for any x ∈ J, U ⊆ U x so that
U ∩V x = ∅. Hence U ∩V = ∅.
Now let A and B be any pair of disjoint non-empty closed sets in X. For each
a ∈ A there are disjoint open sets U a and V a such that a ∈ U a and B ⊆ V a , by the
above argument. The sets {U a : a ∈ A} form an open cover of the compact set A
and so there is a finite set S in A such that A ⊆ �
a∈S U a
V = �
a∈S V a
. Put U = �
a∈S U a and
. Then U and V are both open and A ⊆ U and B ⊆ V. Furthermore,
U a ∩V ⊆ U a ∩V a = ∅ and so U and V are disjoint and the proof is complete.