Proposition 2.3
Let \(P \subseteq \mathbb{R}^d\) be a polytope, and \( \begin{align*} V := \operatorname{vert}(P) \end{align*} \). Let \(F\) be a face of \(P\).
  1. The face \(F\) is a polytope, with \(\operatorname{vert}(F)=F\cap V\).
  2. Every intersection of faces of \(P\) is a face of \(P\).
  3. The faces of \(F\) are exactly the faces of \(P\) that are contained in \(F\).
  4. \(F=P\cap\operatorname{aff}(F).\)

Proof (i):
Let \(F\) be defined by the valid inequality \(cx \leq c_0\). Since \(P\) is a polytope, it an intersection of finitely many halfspaces and it is also bounded. By defintion \(F = P \cap H\) where \(H = \{x \in \mathbb{R}^d \mid cx = c_0 \}\). Hence \(F\) is also an intersection of finitely many halfspaces. Next we show that \(\operatorname{vert}(F) = F \cap V\) as follows:

\(F \cap V \subseteq \operatorname{vert}(F)\): Let \(v \in F \cap V\). Suppose for the sake of contradiction that \(v \notin \operatorname{vert}(F)\). Since \(F\) is a polytope, Proposition 2.2(i) implies that \(F = \operatorname{conv}(\operatorname{vert}(F)).\) Since \(v \in F\), it can be written as a convex combination of points in \(\operatorname{vert}(F)\). Since we assumed that \(v \notin \operatorname{vert}(F)\), all of these vertices are distinct from \(v\). Since \(\operatorname{vert}(F) \subseteq F \subseteq P\), it follows that \(v\) is a convex combination of points in \(P \setminus \{v\}\). But \(v \in V = \operatorname{vert}(P)\), and Proposition 2.2 shows that a vertex of \(P\) cannot be written as a convex combination of points in \(P \setminus \{v\}\). This is a contradiction. Therefore \(v \in \operatorname{vert}(F)\).

\(\operatorname{vert}(F) \subseteq F \cap V\): Let \(x \in \operatorname{vert}(F)\). We know that \(x \in F\) so we just need to show that \(x \in V\). Since \(x \in F \subseteq P\). Hence we can write \(x\) as a convex combination of the \(V\) such that

$$ \begin{align*} x = Vt, \quad t \geq 0, \quad \mathbf{1}t = 1 \end{align*} $$

Let \(F\) be defined by the valid inequality \(cx \leq c_0\). Since \(x \in F\), \(cx=c_0\). Substituting \(x=Vt\) gives

$$ \begin{align*} c_0 = cx = c(Vt) = (cV)t \tag{1} \end{align*} $$

But since \(cx \leq c_0\) is a valid inequality, then every point of \(P\) must satisfy it. Hence every vertex \(v_i\) satisfies \(cv_i \leq c_0\). Hence \(cV \leq c_0 \mathbf{1}\). Since \(t \geq 0\), we can multiply each side by \(t\) to get

$$ \begin{align*} (cV)t &\leq (c_0 \mathbf{1})t \end{align*} $$

Hence by (1) we get that

$$ \begin{align*} c_0 = (cV)t &\leq (c_0 \mathbf{1})t = c_0 \end{align*} $$

Therefore, equality must hold so we get

$$ \begin{align*} (cV)t=(c_0\mathbf 1)t \end{align*} $$

But since \(\mathbf 1 t = 1\) and \(t_i \geq 0\) for all \(i\), the left-hand side is a weighted average of the numbers \(cv_i\). Since each \(cv_i \leq c_0\) and the weighted average equals \(c_0\), then every \(v_i\) with \(t_i>0\) must satisfy

$$ \begin{align*} cv_i=c_0 \end{align*} $$

But this implies that every \(v_i\) with \(t_i>0\) lies in \(F\). Thus \(x\) is a convex combination of vertices in \(V \cap F\). Since \(x \in \operatorname{vert}(F)\), Proposition 2.2 implies that \(x\) cannot be written as a convex combination of other points in \(F\). Therefore every vertex \(v_i\) with \(t_i>0\) must satisfy \(v_i=x\). Since \(v_i \in V\), then \(x \in V\) and since we already know that \(x \in F\), then \(x \in F \cap V\). \(\blacksquare\).

References