Is the preimage of a closed set closed?

Question

Let $A\in\mathbb{R}^{m\times n}$. Can I prove if $S\subseteq \mathbb{R}^m$ is closed then so is $A^{-1}(S) =\{ x \in \mathbb{R}^n : Ax \in S\}$

What I am thinking is that I can't, based on my knowledge that if $x \subseteq \mathbb{R}^m$ is closed, its linear transformation $ Ax \in \mathbb{R}^n$ is not always closed, because what I assume is when dealing with the first case, $x \in A^+S$, where $A^+$ is the pseudo-iverse of matrix $A$, which is another linear transformation.

Answer 1

Yes, it is closed since $A$ is continue and the inverse image of a closed subset by a continuous map is closed.

Answer 2

For a function $A\to B$, the preimage of a closed set is closed provided that function is continuous. Every linear transformation $\Bbb R^n\to\Bbb R^m$ is continuous, and has a matrix representation. You can conclude that the preimage of any closed set under a linear transformation is closed.