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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 $\operatorname{ri}(L)\cap F\neq\varnothing$, then $L\subseteq F$.
Here, $\operatorname{ri}(L)$ 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.

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 $a=(0,0)$, $b=\{(0,y)\mid y\in[0,1]\}$, and $c=(1,1)$. We have $a\leq b$. 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 $\{p\}\in\operatorname{Part}(C)\operatorname{Part}(F)$. 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 $F(C)$, 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.
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