Preservation of closed sets under linear transformation. Let $f:\mathbb{R^n} \rightarrow \mathbb{R^m}$ and let $A$ be a closed set in $\mathbb{R^n}$. I would like to know if $f(A)$ is a closed set.
I know this question is pretty much the same as this one: A linear transform of a closed set is closed. But the answer given in that question is wrong as the image of $E$ given is said to not be closed when I would claim that it actually is closed. Just take the compliment of $f(E)$ and find that it is open.
As for the correct answer, my intuition would tell me that linear transformations do in fact preserve closedness. To be clear: I am looking for a better counterexample to disprove this fact or indeed a proof for it.

For posterity, a necessary and sufficient when $A$ is an arbitrary subset of $\mathbb{R}^n$ and $f: \mathbb{R}^n \to \mathbb{R}^m$ is linear, is:
$$
f(A) \textrm{ is closed} \iff A + \textrm{ker}(f) \textrm{ is closed.}
$$
This is a corollary of the infinite dimensional version which can be found, for example, in Holmes' Geometric Functional Analysis and Its Applications (Lemma 17.H).
