18.2. SEPARATION THEOREMS 491

ThenρA (z− xλ )< (1−λ )r

and so

ρA

(z

1−λ− x− λy

1−λ

)< r.

Therefore,z

1−λ− λy

1−λ∈ BA (x,r)⊆ B.

It follows

(1−λ )

(z

1−λ− λy

1−λ

)+λy = z ∈ B

and so C⊆ B as claimed. Now this shows xλ ∈ int(B) and limλ→1 xλ = y. Thus, y ∈ int(B)and this proves the lemma.

Note this also shows that B = int(B).

Corollary 18.2.16 Let A,B be convex, nonempty sets. Suppose int(B) ̸= /0 and A∩ int(B)=/0. Then there exists f ∈ X ′, f ̸= 0, such that for all a ∈ A and b ∈ B,

Re f (b)≥ Re f (a).

Proof: By Theorem 18.2.14, there exists f ∈ X ′ such that for all b ∈ int(B), and a ∈ A,

Re f (b)> Re f (a) .

Thus, in particular, f ̸= 0. By Lemma 18.2.15, if b ∈ B and a ∈ A,

Re f (b)≥ Re f (a) .

This proves the theorem.

Lemma 18.2.17 If X is a topological Hausdorff space then compact implies closed.

Proof: Let K be compact and suppose KC is not open. Then there exists p ∈ KC suchthat

Vp∩K ̸= /0

for all open sets Vp containing p. Let

C ={(V p)C : Vp is an open set containing p}.

Then C is an open cover of K because if q ∈ K, there exist disjoint open sets Vp and Vq

containing p and q respectively. Thus q ∈(V p)C. This is an example of an open cover of K

which has no finite subcover, contradicting the assumption that K is compact. This provesthe lemma.