Number of faces in a simplex

Question

Let $D \subset \mathbb R^d$ be a $d$-simplex with vertices $V=\{ v_1,v_2, \ldots, v_{d+1} \}$

Prove that for every $W \subset V$ $\operatorname{conv}(W)$ is a face of $D$

My definition of a face of a polytope is the following:

Let $H$ be a supporting hyperplane of the polytope $P$, the faces of $P$ are the sets $P \bigcap H $

A simplex is a $d$-polytope with $d+1$ faces of dimension zero (vertices).

How can I use this to proof the statement above ? I am just learning about polytopes and have no clue how to proceed

Answer 1

Let's assume without loss of generality that we have the standard simplex. Why can we do this? Because any realization of the simplex, by definition, has the same face lattice. Note you can prove something far stronger here - any two simplices of the same dimension are in fact affinely isomorphic.

So we have that $D = \operatorname{conv}(e_1, e_2, ..., e_{d+1}) $. Now let $W \subset \{e_1, ..., e_{d+1} \} = V$. Let $c = \sum_{e_i \in V \setminus W} e_i$. For example, if $W = \{e_1, e_2, ..., e_{d-1} \}$, then $c = e_d + e_{d+1}$.

Consider the hyperplane $H = \{ x \in \mathbb{R}^{d+1} | \langle c,x \rangle = 0 \}$. Notice that $\langle c, e_i \rangle$ is $0$ if $e_i$ is in $W$, and $1$ if $e_i$ is not in $W$.

It follows quickly that that $D \cap H = \operatorname{conv}(W)$. Can you show this? Hint: Take $x \in D$, write $x = \sum_{i = 1}^{d+1} \lambda_i e_i$ s.t. $\sum \lambda_i = 1$, $\lambda_i \geq 0$, when is $\langle c, x \rangle = 0$ ?

The geometric intuition behind the last step is that all in $V \setminus W$ lie on the positive side of the hyperplane, while the vectors in $W$ lie inside the hyperplane. So in order to stay inside the hyperplane, you can only take the directions offered to you by the vectors in $W$.