Let $$X = \{ x \in \mathbb R^n: 1/2 \leq x_1 \leq 1 \text{ and } x_{i-1} \leq x_i \leq 2x_{i-1}\forall i=2,\ldots,n \}.$$ Prove that $X$ is a polyhedron and determine all of its extreme points.
My attempt:
First, we'll get $X$ in the form $X = P(A, b) = \{ x \in \mathbb R^n \mid Ax \leq b\}$. In order to do so, we can write the constraints as $$ -x_1 \leq -1/2,\,\, x_1 \leq 1, \\ x_1 - x_2 \leq 0,\,\, x_2 - x_3 \leq 0,\,\, x_3 - x_4 \leq 0,\,\, \ldots, \,\, x_{n-1} - x_n \leq 0, \\ x_2 - 2x_1 \leq 0, \,\, x_3 - 2x_2 \leq 0, \,\, \ldots, \,\, x_n - 2 x_{n-1} \leq 0. $$
The first line corresponds to the matrix $$ A_1 = \begin{pmatrix} -1 & 0 & \cdots & 0 \\ 1 & 0 & \cdots & 0\end{pmatrix} \in \mathbb R^{2 \times n} $$ the second one to $$ A_2 = \begin{pmatrix} 1 & -1 \\ & 1 & -1 \\ & & 1 & -1 \\ & & & \ddots & \ddots \\ & & & & 1 & -1 \end{pmatrix} \in \mathbb R^{(n-1)\times n} $$ and the third one to $$ A_3 = \begin{pmatrix} -2 & 1 \\ & -2 & 1 \\ & & -2 & 1 \\ & & & \ddots & \ddots \\ & & & & -2 & 1 \end{pmatrix} \in \mathbb R^{(n-1) \times n} $$ such that $X = P(A, b)$ with $$ A = \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix} \in \mathbb R^{2n \times n}, \quad b = \begin{pmatrix} -1/2 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \in \mathbb R^{2n}. $$ Now for an extreme point, I need $n$ linear independent rows of $A$... and that's where I'm stuck. There are so many possibilities to choose $n$ linear independent row that there has to be a "smart" solution or some "reduction" before... does anyone have an idea?
Your $A$ and $b$ are correct.
You have $n$ pairs of inequalities, and at most one inequality in each pair can be active, so there are $2^n$ extreme points, which you can obtain by independently selecting $$x_1\in\{1/2,1\}, x_2\in\{x_1,2x_1\}, x_3\in\{x_2,2x_2\},\dots,x_n\in\{x_{n-1},2x_{n-1}\}.$$