Let $P$ be a polyhedron in $\mathbb{R}^n, a\in\mathbb{R}^n$ a vector, and $b\in\mathbb{R}$ a scalar. Consider the set $Q=\{x\in P:a^tx\le b\}$
$a)$ Prove that $Q$ is a polyhedron
$b)$ Is it always possible to choose $b$ in such way so that $Q=P$? Explain your answer
Definition of polyhedron
A polyhedron is the intersection of finitely many halfspaces: $$P=\{x∈\mathbb{R}^n,a^tx≤b\}$$
$a)$
Proof
Let $Q:=\{x\in P:a_0^tx\le b\}$
Show that $Q$ is the intersection of finitely many half-spaces
Since $P$ is a polyhedron we have $$\exists a_1^t\dots a_n^t\in\mathbb{R}^n,c_1\dots c_n\in\mathbb{R},s.t.$$
$$P=\{x\in\mathbb{R}^n:a_1^tx\le c_1\}\cap\dots\cap\{x\in\mathbb{R}^n:a_n^tx\le c_n\}$$
$$=\{x∈\mathbb{R}^n:a_1^tx≤c_1\wedge\dots\wedge a_n^tx\le c_n\}$$
Then
$$Q=\{x\in P:a_0^tx\le b\}=\{x\in \mathbb{R}^n:x\in P\wedge a_0^tx\le b\}$$
$$=\{x\in \mathbb{R}^n:a_0^tx\le b\wedge a_1^tx≤c_1\wedge\dots\wedge a_n^tx\le c_n\}$$
$$=\{x\in \mathbb{R}^n:a_0^tx\le b\}\cap\{x\in\mathbb{R}^n:a_1^tx\le c_1\}\cap\dots\cap\{x\in\mathbb{R}^n:a_n^tx\le c_n\}$$
Which is the intersection of finite many half-spaces. $\tag*{$\square$}$
$b)$
No, consider case $n=1$ in $\mathbb{R}^2$, if $P=\{x∈\mathbb{R}^2:x_1+x_2≤0\},Q=\{x\in P:x_1-x_2\le b\}$
Where $b\in\mathbb{R}$
Then
$$Q=\{x∈\mathbb{R}^2:x_1-x_2\le b\wedge x_1+x_2≤0\}$$
For example $\left(\frac{b}{2},-\frac{b}{2}-1\right)$ is in $P$ but not in $Q$.
Since $\frac{b}{2}-\frac{b}{2}-1≤0$ that $\left(\frac{b}{2},-\frac{b}{2}-1\right)\in P$
But $\frac{b}{2}-\frac{b}{2}-1≤0\wedge\frac{b}{2}-(-\frac{b}{2}-1)\le b$ is false that $\left(\frac{b}{2},-\frac{b}{2}-1\right)\not\in Q$ for any real $b$.
That proves sometimes$$\forall b\in\mathbb{R},Q\neq P$$$\tag*{$\square$}$
$\dots$ Is my proof correct ? Any suggestions would be appreciated.
Also please tell me if there is a better method to prove it.
Thanks for your help.