TITLE: Product of two polytopes is a polytope
QUESTION [5 upvotes]: Please have a look at my attempt for this problem.

Let $x = \begin{pmatrix} x_1\\ x_2 \\ \end{pmatrix}, x_1 \in P_1, x_2 \in P_2$. 
I want to show that $x \in conv\{P_1 \times P_2\}$, i.e. $x$ can be represented as the convex combination of some points of $P_1 \times P_2$.
Without loss of generality, suppose that $d_1 \geq d_2$. By Caratheodory's theorem, $x_1$ can be represented by at most $d_1 + 1$ points of $P_1$, i.e. there exists $\{v_1,...,v_{d_1+1}\} \subset P_1$, and $\alpha_1,...,\alpha_{d_1+1}$; $ \alpha_i \geq 0$, $ \sum \alpha_i=1$, such that: 
$$x_1 = \alpha_1v_1+...+\alpha_{d_1+1}v_{d_1+1}$$
$x_2$ can be represented similarly, with points $\{w_1,...,w_{d_2+1},...,w_{d_1+1}\} \subset P_2$, and coefficients $\beta_i$'s, such that $\beta_{d_2+2},...,\beta_{d_1+1}$(if exist) are all $0$'s:
$$x_2 = \beta_1w_1+...+\beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1}$$
So now we have $x$ as:
$$x = \begin{pmatrix} x_1\\ x_2 \\ \end{pmatrix} = \begin{pmatrix} \alpha_1v_1+...+\alpha_{d_2+1}v_{d_2+1}+\alpha_{d_2+2}v_{d_2+2}+...+\alpha_{d_1+1}v_{d_1+1}\\ \beta_1w_1+...+\beta_{d_2+1}w_{d_2+1} + 0.w_{d_2+2}+...+ 0.w_{d_1+1} \\ \end{pmatrix}$$
It is here that I got stuck. The points $\begin{pmatrix} v_i\\ w_i \\ \end{pmatrix}$ above are certainly in $P_1 \times P_2$, but are the coefficients right? Shouldn't the coefficients be in $\mathbb{R}$, instead of $\mathbb{R}^2$?
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Edit: the polytopes here are all convex polytopes.
Edit 2: Caratheodory's theorem: https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)

REPLY [2 votes]: For every $i$ in $\{1,2\}$, let $\mathcal{E}_i=\{v^i_1,\ldots, v^i_{m_i}\}$ be the set of extremal points of $P_i$. We can regard $P_i$ as a subset of $\mathbb{R}^{n_i}$ for some natural number $n_i$ and the dimension of $P_i$ is simply the dimension of the linear spawn $V_i$ of the set $G_i=\{v^i_1-v^i_1, v^i_2-v^i_1,\ldots, v^i_{m_i}-v^i_1\}$ in $\mathbb{R}^{n_i}$.
We know that a point $x_i$ in $\mathbb{R}^{n_i}$ belongs to $P_i$ if and only if there exists non-negative real numbers $t_1^i,\ldots, t_{m_i}^i$ such that $\sum_{j=1}^{m_i}t^i_j = 1$ and $\sum_{j=1}^{m_i}t^i_jv^i_j = x_i$.
Now, consider the cartesian product $P_1\times P_2$, which is a subset of $\mathbb{R}^{n_1}\times \mathbb{R}^{n_2} \cong \mathbb{R}^{n_1+n_2}$. For every $k_1$ in $\{1,\ldots,m_1\}$ and $k_2$ in $\{1,\ldots,m_2\}$ let us call $v_{k_1k_2}=(v^1_{k_1},v^2_{k_2})$. We will prove that $P_1\times P_2$ is the convex hull $\mathcal{C}$ of the set $\mathcal{E}_1\times\mathcal{E}_2 = \{v_{k_1k_2}\mid 1\leq k_1 \leq m_1,\enspace 1\leq k_2 \leq m_2\}$ in $\mathbb{R}^{n_1+n_2}$.
Let $x=(x_1,x_2)$ be a point in $P_1\times P_2$ then we can write $x = (\sum_{k_1=1}^{m_1}t^1_{k_1}v^1_{k_1},\enspace \sum_{k_2=1}^{m_2}t^2_{k_2}v^2_{k_2})$, in the way it was detailed above. So
$$
x = \bigg{(}\sum_{k_1=1}^{m_1}t^1_{k_1}v^1_{k_1},\enspace \sum_{k_2=1}^{m_2}t^2_{k_2}v^2_{k_2}\bigg{)} =\sum_{k_1=1}^{m_1}\sum_{k_2=1}^{m_2}t^1_{k_1}t^2_{k_2}(v^1_{k_1}, v^2_{k_2}) = \sum_{k_1=1}^{m_1}\sum_{k_2=1}^{m_2}t_{k_1k_2}v_{k_1k_2},
$$
where $t_{k_1k_2}=t^1_{k_1}t^2_{k_2}$ for every $k_1$ and $k_2$. Note that, for every $k_1$ and $k_2$, $t_{k_1k_2}$ is a non-negative real number and also
$$
\sum_{k_1=1}^{m_1}\sum_{k_2=1}^{m_2}t_{k_1k_2} = \sum_{k_1=1}^{m_1}\sum_{k_2=1}^{m_2}t^1_{k_1}t^2_{k_2} = \bigg{(}\sum_{k_1=1}^{m_1}t^1_{k_1}\bigg{)}\bigg{(}\sum_{k_2=1}^{m_2}t^2_{k_2}\bigg{)} =(1)(1)=1.
$$
Thus $x=(x_1,x_2)$ belongs to the convex hull $\mathcal{C}$ of the set $\mathcal{E}_1\times\mathcal{E}_2$. So $P_1\times P_2 \subseteq \mathcal{C}$.
On the other hand, $\mathcal{E}_1\times\mathcal{E}_2 \subseteq P_1\times P_2$ and $P_1\times P_2$ is convex,  so $\mathcal{C}\subseteq P_1\times P_2$. Thus $\mathcal{C} = P_1\times P_2$.
Finally, the set $G=\{v_{k_1k_2}-v_{11}\mid 1\leq k_1 \leq m_1,\enspace 1\leq k_2 \leq m_2\}$ is equal to the cartesian product $G_1\times G_2$, of the sets defined in the first paragraph. So the linear spawn $V$ of $G$ in $\mathbb{R}^{n_1+n_2}$ is simply the cartesian product $V_1\times V_2$ of the linear spawns of $G_1$ and $G_2$. Thus the dimension of $P_1\times P_2$ is equal to the sum of the dimension of $P_1$ and $P_2$.
Note that we do not need the set $\mathcal{E}_i$ to be extremal, only its convex hull is $P_i$. But it is interesting to note that $\mathcal{E}_1\times \mathcal{E}_2$ is exactly the set of extremal points of $P_1\times P_2$.