Let $C$ be a convex set in $\mathbb{R}^n$ (or any topological vector space). 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 $\operatorname{ri}(L)\cap F\ne \varnothing$ , then $L\subseteq F$ .

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

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

Observe that the empty set and $C$ itself are faces of $C$ . These faces are sometimes called improper faces, while other faces are 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, $\operatorname{rbd}(C)$ .
• The set $\operatorname{Part}(C)$ 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 $\varnothing$ 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 $a=(0,0)$ , $b=\lbrace (0,y)\mid y\in [0,1]\rbrace$ , and $c=(1,1)$ . We have $a\le b$ . However, $a\vee (b\wedge c)=(0,0)\vee \varnothing=(0,0)$ , whereas $(a\vee b)\wedge c = b\wedge \varnothing=\varnothing$ .
• Nevertheless, $F(C)$ is a complemented lattice. Pick any face $F\in F(C)$ . If $F=C$ , then $\varnothing$ is a complement of $F$ . Otherwise, form $\operatorname{Part}(C)$ and $\operatorname{Part}(F)$ , the partitions of $C$ and $F$ into disjoint unions of the relative interiors of their corresponding faces. Clearly $\operatorname{Part}(F)\subset \operatorname{Part}(C)$ strictly. Now, it is possible to find an extreme point $p$ such that $\lbrace p\rbrace\in \operatorname{Part}(C)-\operatorname{Part}(F)$ . Otherwise, all extreme points lie in $\operatorname{Part}(F)$ , which leads to $$\operatorname{Part}(F) = \operatorname{Part}(\mbox{convex hull of extreme points of }C)=\operatorname{Part}(C),$$ a contradiction. Finally, let $G$ be the convex hull of extreme points of $C$ not contained in $\operatorname{Part}(F)$ . 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 $F\cap G=\varnothing$ . 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.

Bibliography

1

R.T. Rockafellar, Convex Analysis, Princeton University Press, 1996.