face of a convex set
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 $\mathrm{ri}(L)\cap F\ne \mathrm{\varnothing}$, then $L\subseteq F$.
Here, $\mathrm{ri}(L)$ denotes the relative interior of $L$ (open segment of $L$).
A zerodimensional 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.

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The intersection^{} of two faces of $C$ is a face of $C$.

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

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

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

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If $C$ is compact^{}, then $C$ is the convex hull^{} of its extreme points.

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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^{}.

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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 $(a\vee b)\wedge c=b\wedge \mathrm{\varnothing}=\mathrm{\varnothing}$.

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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).

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Additionally, in $F(C)$, zerodimensional 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.
References
 1 R.T. Rockafellar, Convex Analysis, Princeton University Press, 1996.
Title  face of a convex set 

Canonical name  FaceOfAConvexSet 
Date of creation  20130322 16:23:08 
Last modified on  20130322 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 