Let \begin{equation} S := \left\{x\in\mathbb{R}^{m\times n}: \begin{aligned} & 0\leq x_{ij}\leq 1,\ i=1,2,\ldots,m,\ j=1,2,\ldots,n \\ & \sum_{i=1}^n x_{ij} = 1,\ j=1,2,\ldots,n \\ & a_k\leq\sum_{i=1}^k\sum_{j=1}^n x_{ij} \leq b_k,\ k=1,2,\ldots,m,\ a_k,b_k\in\mathbb{N} \end{aligned} \right\}\neq\emptyset, \end{equation} and $E_S:=\{x\in S: x \text{ is an extreme point of } S\}$. Show that $E_S = S\cap\{0,1\}^{m\times n}$.
I can prove that $E_S \supset S\cap\{0,1\}^{m\times n}$. For any $x\in S\cap\{0,1\}^{m\times n}$, suppose there exist $y,z\in S$ and $\alpha\in (0,1)$ such that $x=\alpha y + (1-\alpha)z$, then $y=x=z$ due to $0\leq y,z\leq 1$. Thus, $x$ is an extreme point of $S$, i.e., $x\in E_S$.
However, I cannot prove that $E_S \subset S\cap\{0,1\}^{m\times n}$. Could someone help me?
I have an idea for $E_S \subset S\cap\{0,1\}^{m\times n}$. However, I cannot finish the whole proof since discussing a decimal loop in matrix $x$ seems inevitable. Could someone help me? Thanks.
Assume that there exists an $x\in E_S$ but $x\notin S\cap\{0,1\}^{m\times n}$. Denote the indices of the active constraints in $a_k\leq\sum_{i=1}^k\sum_{j=1}^n x_{ij} \leq b_k$ by $\mathcal{I}(x)$. For convenience, assume $x_{11}\in(0,1)$. Then there exists $l\neq 1$ such that $x_{l1}\in(0,1)$.
If $1\notin\mathcal{I}(x)$, there exists $y,z\in S$ \begin{equation} \begin{cases} y_{11}=x_{11}+\epsilon,\ y_{l1}=x_{l1}-\epsilon \\ z_{11}=x_{11}-\epsilon,\ z_{l1}=x_{l1}+\epsilon \\ y_{ij} = x_{ij}=z_{ij},\ \text{for other } i,j \end{cases}\text{ (Incorrect!)} \end{equation} such that $x=\frac{1}{2}y+\frac{1}{2}z$, which contradicts the assumption that $x$ is an extreme point of $S$.
Otherwise, $1\in\mathcal{I}(x)$, there exists $p\neq 1$ such that $x_{1p}\in(0,1)$, and $q\neq 1$ such that $x_{qp}\in(0,1)$.
If $l\notin\mathcal{I}(x)$, there exists $y,z\in S$ \begin{equation} \begin{cases} y_{11}=x_{11}+\epsilon,\ y_{l1}=x_{l1}-\epsilon,\ y_{1p}=x_{1p}-\epsilon,\ y_{qp}=x_{qp}+\epsilon \\ z_{11}=x_{11}-\epsilon,\ z_{l1}=x_{l1}+\epsilon,\ z_{1p}=x_{1p}+\epsilon,\ z_{qp}=x_{qp}-\epsilon \\ y_{ij} = x_{ij}=z_{ij},\ \text{for other } i,j \end{cases}\text{ (Incorrect!)} \end{equation} such that $x=\frac{1}{2}y+\frac{1}{2}z$, which contradicts the assumption that $x$ is an extreme point of $S$.
Otherwise, $l\in\mathcal{I}(x)$, there exists $r\neq 1$ such that $x_{lr}\in(0,1)$. Then there exists $y,z\in S$ \begin{equation} \begin{cases} y_{11}=x_{11}+\epsilon,\ y_{l1}=x_{l1}-\epsilon,\ y_{1p}=x_{1p}-\epsilon,\ y_{lr}=x_{lr}+\epsilon \\ z_{11}=x_{11}-\epsilon,\ z_{l1}=x_{l1}+\epsilon,\ z_{1p}=x_{1p}+\epsilon,\ z_{lr}=x_{lr}-\epsilon \\ y_{ij} = x_{ij}=z_{ij},\ \text{for other } i,j \end{cases} \end{equation} such that $x=\frac{1}{2}y+\frac{1}{2}z$, which contradicts the assumption that $x$ is an extreme point of $S$.