Convex hull of a 0/1 set in $ \mathbb{R}^d$ Reading in my textbook, I found the following example:
$$ $$Let S $ \subseteq $ {0,1}$^d$ be an arbitrary 0/1 set in $ \mathbb{R}^d$ and the Polyhedron Q = conv(S). It can be shown easily that the set of the vertices of Q is equal to S.
I don't find this easy to show, can anyone help me out with a proof?
 A: Let $V$ be the set of vertices of $Q$. From the definition of
the convex hull follows that $V \subseteq S$.
To prove that $V = S$ it is sufficient to show that every point
of $S$ is a vertex of $Q$. To show that we will use the following basic fact which states that a point of a polyhedron is a vertex if and only if it is an extreme point of the polyhedron.

A point $x$ of a polyhedron $P$ is a vertex of $P$ if and only
  no points $z,w \in P,$ both different from $x$, exist such that $x = \lambda z + (1-\lambda)w,\ \lambda \in (0,1)$.

Consider now the point $s=(s_1,\ldots, s_n) \in S$ and two points $z,w \in \text{conv}(S)$. Assume that we can write $s$ as a convex combination of $z,w$, $s = \lambda z + (1-\lambda) w$.
If there exist an $i$ such that $s_i = 0,\ z_i,w_i >0$, then
\begin{align}
s_i &= \lambda z_i + (1-\lambda) w_i\\
0 &= \lambda z_i + (1-\lambda) w_i > 0
\end{align}
since $\lambda z_i, (1-\lambda)w_i >0$.
If no such index $i$ exists then there exists an index $j$ such that
$s_i = 1, z_i,w_i < 1$. Then \begin{align}
1 &= \lambda z_i + (1-\lambda) w_i\\
\lambda &= \frac{1-w_i}{z_i -w_i} > 1
\end{align}
where wlog we assumed $z_i > w_i$.
Thus each point of $S$ is a vertex of $Q$.
