16.2.2 More Separation Theorems
There are other separation theorems which can be proved in a similar way. The
next theorem considers the separation of an open convex set from a convex
Theorem 16.2.14 Let A and B be disjoint, convex and nonempty sets with B open.
Then there exists f ∈ X′ such that
for all a ∈ A and b ∈ B.
Proof: Let b0 ∈ B,a0 ∈ A. Then the set
is open, convex, contains 0, and does not contain a0 −b0. By Lemma 16.2.3 there exists
f ∈ X′ such that
for all a ∈ A and b ∈ B. Therefore, for all a ∈ A,b ∈ B,
Before giving another separation theorem, here is a lemma.
Lemma 16.2.15 If B is convex, then int
≡ union of all open sets contained
in B is convex. Also, if int
≠∅, then B ⊆int
Proof: Suppose x,y ∈ int
. Then there exists
0 and a finite set A ⊆
Then V is open, V ⊆ B, and if λ ∈
is convex as claimed.
Now let y ∈ B and x ∈ int
and let xλ ≡
. Define the open cone,
Thus C is represented in the following picture.
I claim C ⊆ B as suggested in the picture. To see this, let
and so C ⊆ B as claimed. Now this shows xλ ∈ int
and this proves the lemma.
Note this also shows that B = int
Corollary 16.2.16 Let A,B be convex, nonempty sets. Suppose int
A ∩ int
∅. Then there exists f ∈ X′,f≠
0, such that for all a ∈ A and
b ∈ B,
Proof: By Theorem 16.2.14, there exists f ∈ X′ such that for all b ∈ int
a ∈ A
Thus, in particular, f≠0. By Lemma 16.2.15, if b ∈ B and a ∈ A,
This proves the theorem.
Lemma 16.2.17 If X is a topological Hausdorff space then compact implies
Proof: Let K be compact and suppose KC is not open. Then there exists p ∈ KC
for all open sets V p containing p. Let
Then C is an open cover of K because if q ∈ K, there exist disjoint open sets V p and V q
containing p and q respectively. Thus q ∈
. This is an example of an open cover of
which has no finite subcover, contradicting the assumption that K
is compact. This
proves the lemma.
Lemma 16.2.18 If X is a locally convex topological vector space, and if every point is
a closed set, then the seminorms and X′ separate the points. This means if x≠y, then for
some ρ ∈ Ψ,
and for some f ∈ X′,
In this case, X is a Hausdorff space.
Proof: Let x≠y. Then by Theorem 16.2.5, there exists f ∈ X′ such that f
separates the points. Since f ∈ X′
, Theorem 16.1.6
for some A a finite subset of Ψ. Thus
and so ρ
0 for some ρ ∈ A ⊆
Ψ. Now to show X
is Hausdorff, let
Then the two disjoint open sets containing x and y respectively are
This proves the lemma.