Gruber P. Convex and Discrete Geometry

Proof. The first step is to show the following:
(3) Let F ∈ F(P). Then F ♦ ∈ F(P ∗ ).
Let F = conv{x1,...,xk}. Then
14 Preliminaries and the Face Lattice 255
F ♦ = � y ∈ P ∗ : (λ1x1 +···+λkxk) · y = 1forλi ≥ 0, λ1 +···+λk = 1 �
= � y ∈ P ∗ : xi · y = 1fori = 1,...,k � =
= P ∗ ∩
k�
{y : xi · y = 1}.
i=1
k�
{y ∈ P ∗ : xi · y = 1}
(If F =∅, this shows that F ♦ = P ∗ if, as usual, the intersection of an empty family
of subsets of E d is defined to be E d .) Each of the hyperplanes {y : xi ·y = 1} supports
P ∗ (or does not meet P ∗ ) by the definition of P ∗ . Thus F ♦ is the intersection of k
faces of P ∗ and hence is a face of P ∗ by Theorem 14.6, applied to P ∗ .
In the second step we prove that
(4)
♦ = ♦P is one-to-one and onto.
For this, it is sufficient to show that
(5)
i=1
♦ P ∗ ♦P = identity and, dually, ♦P ♦ P ∗ = identity.
Let F ∈ F(P). We have to prove the equality F = F ♦♦ . To show the inclusion
F ⊆ F ♦♦ ,letx ∈ F. The definition of F ♦ then implies that x · y = 1 for all y ∈ F ♦ .
Hence x ∈ F ♦♦ by the definition of F ♦♦ . To show the reverse inclusion F ⊇ F ♦♦ ,
let x ∈ P = P ∗∗ , x �∈ F. Consider a support hyperplane H ={z : z · y = 1} of
P such that F = H ∩ P. Then z · y ≤ 1 for all z ∈ P. Hence y ∈ P ∗ . Further,
z · y = 1 precisely for those z ∈ P for which z ∈ F and thus, in particular, x · y < 1.
By the definition of F ♦ we then have y ∈ F ♦ . This, together with x ∈ P ∗∗ and
x · y < 1, finally yields that x �∈ F ♦♦ by the definition of F ♦♦ . The proof of the
equality F = F ♦♦ and thus of (5) is complete. (5) implies (4).
The definition of ♦P and proposition (4) readily imply the next statement.
(6) Let F, G ∈ F(P), F � G. Then F ♦ � G ♦ .
In the third step of the proof the following will be shown.
(7) Let F ∈ F(P). Then dim F ♦ = d − 1 − dim F.
If F =∅or P, then F ♦ = P ∗ , resp. ∅ and (7) holds trivially. Assume now that
F �= ∅, P. An argument similar to the one in the proof of Theorem 14.6 shows that
F is a (proper or non-proper) face of a facet of P. Using this, a simple proof by
induction implies that there is a sequence of faces of P, say∅=F−1, F0,...,F =
Fk,...,Fd−1, Fd = P ∈ F(P), such that
(8) ∅=F−1 � F0 � ···� F = Fk � ···� Fd−1 � Fd = P,
where dim Fi = i.