PlanetMath: face of a convex set

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face of a convex set

(Definition)

Let be a convex set in $\mathbb{R}^n$ (or any topological vector space). A face of is a subset of such that

is convex, and
given any line segment $L\subseteq C$ , if $\operatorname{ri}(L)\cap F\ne \varnothing$ , then $L\subseteq F$ .

Here, $\operatorname{ri}(L)$ denotes the relative interior of

(open segment of

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

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 itself are faces of . These faces are sometimes called improper faces, while other faces are called proper faces.

Remarks. Let be a convex set.

The intersection of two faces of is a face of .
A face of a face of is a face of .
Any proper face of lies on its relative boundary, $\operatorname{rbd}(C)$ .
The set $\operatorname{Part}(C)$ of all relative interiors of the faces of partitions .
If is compact, then is the convex hull of its extreme points.
The set of faces of a convex set forms a lattice, where the meet is the intersection: $F_1 \wedge F_2 := F_1\cap F_2$ ; the join of is the smallest face $F\in F(C)$ containing both and . This lattice is bounded lattice (by $\varnothing$ and ). And it is not hard to see that is a complete lattice.
However, in general, is not a modular lattice. As a counterexample, consider the unit square $[0,1]\times [0,1]$ and faces , $b=\lbrace (0,y)\mid y\in [0,1]\rbrace$ , and . 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, is a complemented lattice. Pick any face $F\in F(C)$ . If , then $\varnothing$ is a complement of . Otherwise, form $\operatorname{Part}(C)$ and $\operatorname{Part}(F)$ , the partitions of and 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 such that $\lbrace p\rbrace\in \operatorname{Part}(C)-\operatorname{Part}(F)$ . Otherwise, all extreme points lie in $\operatorname{Part}(F)$ , which leads to
$\displaystyle \operatorname{Part}(F) = \operatorname{Part}($ convex hull of extreme points of $\displaystyle C)=\operatorname{Part}(C),$
a contradiction. Finally, let be the convex hull of extreme points of not contained in $\operatorname{Part}(F)$ . We assert that is a complement of . If $x\in G\cap F$ , then $G\cap F$ is a proper face of and of , hence its extreme points are also extreme points of , and of , which is impossible by the construction of . Therefore $F\cap G=\varnothing$ . Next, note that the union of extreme points of and of is the collection of all extreme points of , this is again the result of the construction of , so any $y\in C$ is in the join of all its extreme points, which is equal to the join of and (since join is universally associative).
Additionally, in , zero-dimensional faces are compact elements, and compact elements are faces with finitely many extreme points. The unit disk is not compact in . Since every face is the convex hull (join) of all extreme points it contains, is an algebraic lattice.