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 (ii):
Let \(F\) be a face in \(P\) be defined by the valid inequality \(cx \leq c_0\). Then

$$ \begin{align*} F = P \cap \{ x \in \mathbb{R}^d \mid cx = c_0 \}. \tag{1} \end{align*} $$

Let \(G\) be another face in \(P\) defined by the valid inequality \(bx \leq b_0\). Then

$$ \begin{align*} G = P \cap \{ x \in \mathbb{R}^d \mid bx = b_0 \}. \tag{2} \end{align*} $$

Take any point \(x \in P\), then \(x\) must satisfy both inequalities. Therefore, their sum

$$ \begin{align*} x(b + c) \leq c_0 + b_0. \end{align*} $$

is also a valid inequality for \(P\). Let \(H\) be the face defined by this inequality. So

$$ \begin{align*} H = P \cap \{ x \in \mathbb{R}^d \mid x(b+c) = b_0+c_0 \}. \end{align*} $$

We claim that \(H = F \cap G\). Let \(x \in F \cap G\). Then \(x\) satisfies both (1) and (2). Hence, it also satisfies their sum \((b+c)x = b_0+c_0\). Moreover, since \(x \in P\), then \(x \in H\). Conversely, let \(x \in H\), then \(x\) satisfies

$$ \begin{align*} bx + cx = b_0 + c_0. \tag{3} \end{align*} $$

Moreover, since \(x \in P\), then \(bx \leq b_0\) and \(cx \leq c_0\). Hence the only way to satisfy \((3)\) is for \(bx = b_0\) and \(cx = c_0\). But this implies that \(x \in F\) and \(x \in G\). Therefore, \(x \in F \cap G\). So \(F \cap G\) is face in \(P\). \(\blacksquare\)

References