face of a convex set

Let $C$ be a convex set in ${ℝ}^n$ (or any topological vectorspace). A face of $C$ is a subset $F$ of $C$ such that

1. $F$ is convex, and

2. given any line segment $L\subseteq C$, if$\mathrm{ri}(L)\cap F\ne \mathrm{\varnothing }$, then $L\subseteq F$.

Here, $\mathrm{ri}(L)$ denotes the relative interior of $L$(open segment of $L$).

A zero-dimensional face of a convex set $C$ is called an extremepoint of $C$.

This definition formalizes the notion of a face of a convex polygon ora convex polytope and generalizes it to an arbitrary convex set. Forexample, any point on the boundary of a closed unit disk in${ℝ}^2$ is its face (and an extreme point).

Observe that the empty set and $C$ itself are faces of $C$. Thesefaces are sometimes called improper faces, while other facesare called proper faces.

Remarks.Let $C$ be a convex set.

• The intersection of two faces of $C$ is a face of $C$.

• A face of a face of $C$ is a face of $C$.

• Any proper face of $C$ lies on its relative boundary,$\mathrm{rbd}(C)$.

• The set $\mathrm{Part}(C)$ of all relative interiors of the faces of $C$ partitions $C$.

• If $C$ is compact, then $C$ is the convex hull of its extreme points.

• The set $F(C)$ of faces of a convex set $C$ forms a lattice, where the meet is the intersection: $F_1\wedge F_2:=F_1\cap F_2$; the join of $F_1,F_2$ is the smallest face $F\in F(C)$ containing both $F_1$ and $F_2$. This lattice is bounded lattice (by $\mathrm{\varnothing }$ and $C$). And it is not hard to see that $F(C)$ is a complete lattice.

• However, in general, $F(C)$ is not a modular lattice. As a counterexample, consider the unit square $[0,1]\times [0,1]$ and faces $a=(0,0)$, $b=\{(0,y)\mid y\in [0,1]\}$, and $c=(1,1)$. We have $a\le b$. However, $a\vee (b\wedge c)=(0,0)\vee \mathrm{\varnothing }=(0,0)$, whereas $(a\vee b)\wedge c=b\wedge \mathrm{\varnothing }=\mathrm{\varnothing }$.

• Nevertheless, $F(C)$ is a complemented lattice. Pick any face $F\in F(C)$. If $F=C$, then $\mathrm{\varnothing }$ is a complement of $F$. Otherwise, form $\mathrm{Part}(C)$ and $\mathrm{Part}(F)$, the partitions of $C$ and $F$ into disjoint unions of the relative interiors of their corresponding faces. Clearly $\mathrm{Part}(F)\subset \mathrm{Part}(C)$ strictly. Now, it is possible to find an extreme point $p$ such that $\{p\}\in \mathrm{Part}(C)-\mathrm{Part}(F)$. Otherwise, all extreme points lie in $\mathrm{Part}(F)$, which leads to

  $$\mathrm{Part}(F)=\mathrm{Part}(\text{convex hull of extreme points of }C)=\mathrm{Part}(C),$$

  a contradiction. Finally, let $G$ be the convex hull of extreme points of $C$ not contained in $\mathrm{Part}(F)$. We assert that $G$ is a complement of $F$. If $x\in G\cap F$, then $G\cap F$ is a proper face of $G$ and of $F$, hence its extreme points are also extreme points of $G$, and of $F$, which is impossible by the construction of $G$. Therefore $F\cap G=\mathrm{\varnothing }$. Next, note that the union of extreme points of $G$ and of $F$ is the collection of all extreme points of $C$, this is again the result of the construction of $G$, so any $y\in C$ is in the join of all its extreme points, which is equal to the join of $F$ and $G$ (since join is universally associative).

• Additionally, in $F(C)$, zero-dimensional faces are compact elements, and compact elements are faces with finitely many extreme points. The unit disk $D$ is not compact in $F(D)$. Since every face is the convex hull (join) of all extreme points it contains, $F(C)$ is an algebraic lattice.