Let $A$ be a $m\times n$ matrix and $b\in\mathbb{R}^m$ a vector, which define the set $$ S(A,b):=\{x\in\mathbb{R}^n\::\:Ax\leq b\}. $$
- Let $A_k\to A$ and $b_k\to b$, w.r.t. some norm, such that all the sets $S(A_k,b_k)$ are non-empty. Can the non-emptyness of S(A,b) be concluded?
- If $S(A,b)$ is non-empty, and $A'$ and $b'$ are "near enough" $A$ and $b$, respectively, can we say something about the non-emptyness of $S(A',b')$?
Edit: Item 1 is very simple when considering "$=$" instead of "$\leq$". The answer for item 1 is "no" in the equality case: is enough to consider $A_k=1/k$, $b_k=1$, so each solution set $\{x\::\:A_kx=b_k\}=\{k\}$ is non-empty but the "limit" solution set is $\{x\::\:0\cdot x=1\}=\emptyset$. But for the inequality case I couldn't find a counter example. I believe item 1 is true, here is my attempt to prove it: first observe that $S(A,b)\neq\emptyset$ if and only if $$ \forall\,y\geq 0,\,y^{\top}b\geq 0\text{ or }A^{\top}y\neq 0. $$ This is the "Feasibility Theorem" found here (I actually found this reference by having a look at this question).
Second, for $y\geq 0$, denote $$ I(y)=\{k\in\mathbb{N}\::\:y^{\top}b_k\geq 0\}\text{ and }J(y)=\{k\in\mathbb{N}\::\:A_k^{\top}y\neq 0\}. $$ Since $S(A_k,b_k)\neq \emptyset$ for all $k$, we have that $\mathbb{N}=I(y)\cup J(y)$, for all $y\geq 0$.
Now, given $y\geq 0$, then either $I(y)$ or $J(y)$ is infinite. If $I(y)$ is infinite, I can take the limit to the inequality $y^{\top}b_k\geq 0$ and obtain $y^{\top}b\geq 0$. However I cannot do the same when $J(y)$ is infinite, as I cannot conclude that $A^{\top}y\neq 0$ when $A_k^{\top}y\neq 0$ for $k\in J(y)$.