The following definition of a face of a convex set in a real vector space is sometimes useful.

Let $C$ be a convex subset of $\mathbb{R}^n$ Before we define faces, we introduce oriented hyperplanes and supporting hyperplanes.

Given any vectors $n$ and $p$ in $\mathbb{R}^n$ define the hyperplane $H(n,p)$ by $$ H(n,p) = \{ x \in \mathbb{R}^n \colon n \cdot (x - p) = 0 \}; $$ note that this is the degenerate hyperplane $\mathbb{R}^n$ if $n=0$ As long as $H(n,p)$ is nondegenerate, its removal disconnects $\mathbb{R}^n$ The upper halfspace of $\mathbb{R}^n$ determined by $H(n,p)$ is $$ H(n,p)^+ = \{ x \in \mathbb{R}^n \colon n \cdot (x - p) \ge 0 \}. $$ A hyperplane $H(n,p)$ is a supporting hyperplane for $C$ if its upper halfspace contains $C$ that is, if $C\subset H(n.p)^+$

Using this terminology, we can define a face of a convex set $C$ to be the intersection of $C$ with a supporting hyperplane of $C$ Notice that we still get the empty set and $C$ as improper faces of $C$

Remarks. Let $C$ be a convex set.

• If $F_1 = C\cap H(n_1,p_1)$ and $F_2 = C\cap H(n_2,p_2)$ are faces of $C$ intersecting in a point $p$ then $H(n_1+n_2,p)$ is a supporting hyperplane of $C$ and $F_1\cap F_2 = C\cap H(n_1+n_2,p)$ This shows that the faces of $C$ form a meet-semilattice.
• Since each proper face lies on the base of the upper halfspace of some supporting hyperplane, each such face must lie on the relative boundary of $C$