Prove that the polyhedron $$P =\{(x_1, \ldots , x_m, x_{m+1}): 0\leq x_m \leq 1, 0 \leq x_i \leq x_{m+1} \text{ for } i = 1, \ldots , m \}$$ has integer extreme points.
My Attempt:
I have a similar proof in my notes to prove that the polyhedron $$P =\{(x_1, \ldots , x_m, y) \in \mathbb R_{+}^{m+1}: y \leq 1,x_i \leq y \text{ for } i = 1, \ldots , m \}$$ has integer vertices. This is done by taking the form: $$\mathcal P := \bigg\{ \begin{bmatrix} \mathrm x\\ y\end{bmatrix} \in \mathbb R^{n+1} : \begin{bmatrix} \mathrm x\\ y\end{bmatrix} \geq A \begin{bmatrix} \mathrm x\\ y\end{bmatrix} \leq \begin{bmatrix} 0_n\\ 1\end{bmatrix} \bigg\}$$ Then proving that the matrix $A$ is totally unimodular. However in this case the matrix a contains a column with all but one elements as $-1$. i.e: \begin{bmatrix} 1 & 0 & \dots & 0 & -1 \\ 0 &1 &0 & \dots & -1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 &0 & \dots & 1 \end{bmatrix}
My Question:
How would you adapt this proof to prove for an integer extreme point?
Consider $\mathbf a=(a_1,\ldots,a_{m+1})\in P$.
If some $a_{k}\notin \Bbb Z$, let $$u_i=\begin{cases}1&\text{if }a_i=a_{k}\\0&\text{otherwise.}\end{cases}$$ Then clearly $\mathbf a+h\mathbf u\in P$ for $|h|< \min\{|1-a_{k}|,a_{k}\}$, thus showing this is not en extreme point.
We conclude that any extreme point must have all-integer coordinates.