face of a convex set

Let $C$ be a convex set in ${ℝ}^n$ (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 $\mathrm{ri}(L)\cap F\ne \mathrm{\varnothing }$, 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 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 ${ℝ}^2$ 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, $\mathrm{rbd}(C)$.

• 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.

Title	face of a convex set
Canonical name	FaceOfAConvexSet
Date of creation	2013-03-22 16:23:08
Last modified on	2013-03-22 16:23:08
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	12
Author	CWoo (3771)
Entry type	Definition
Classification	msc 52A99
Related topic	ExtremePoint
Defines	face
Defines	proper face
Defines	extreme point
Defines	improper face