Gruber P. Convex and Discrete Geometry

14 Preliminaries and the Face Lattice 253
intersects F along G.Ifδ>0 is sufficiently small, then all vertices of P not in G are
in int H − G . Hence HG is a support hyperplane of P with G = HG ∩ P. This shows
that G is a face of P.
(ii) If G =∅, F, we are done. If not, choose support hyperplanes HF, HG of P
such that F = HF ∩ P, G = HG ∩ P. This, together with G�F, implies that HF �=
HG and HF ∩ HG is a hyperplane in HF. Since HG supports P and F ⊆ P, HF
and F �⊆ HG, the hyperplane HF ∩ HG in HF supports F. Since G = HG ∩ P,
G ⊆ F ⊆ P, and F ⊆ HF, wehaveG = HG ∩ P = HG ∩ F = HG ∩ HF ∩ F.
Hence G is a face of F. ⊓⊔
The next result shows that F(P) is a lattice.
Theorem 14.6. Let P ∈ P and define binary operations ∧ (intersection) and ∨
(join) on F(P) as follows:
(1) F ∧ G = F ∩ G,
F ∨ G = � � H ∈ F(P) : F, G ⊆ H � for F, G ∈ F(P).
Then � F(P), ∧, ∨ � is a lattice with zero ∅ and unit element P, the face lattice of P.
Proof. A finite family of sets, which is closed with respect to intersection and contains
the empty set ∅ and the union of all sets, is a lattice with respect to the operations
∧, ∨ as defined in (1). The zero element of this lattice is ∅ and the unit element is
the union of all sets. For the proof of the theorem it is thus sufficient to show the
following.
(2) Let F, G ∈ F(P). Then F ∩ G ∈ F(P).
If F ∩ G =∅or F = G or F = P or G = P, we are done. Otherwise we may
assume that o ∈ F ∩ G. Choose support hyperplanes
of P, such that
HF ={x : u · x = 0}, HG ={x : v · x = 0}
F = HF ∩ P, G = HG ∩ P.
Since F �= G, wehaveHF �= HG. Thus HF ∩ HG is a support plane (of dimension
d − 2) of P with
F ∩ G = (HF ∩ HG) ∩ P.
The hyperplane
is a support hyperplane and
H ={x : (u + v) · x = 0}
H ∩ P = (HF ∩ HG) ∩ P = F ∩ G,
concluding the proof of (2) and thus of the theorem. ⊓⊔