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.
 A: Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be the projection $(x,y) \mapsto x$.  Next, let $\{q_n\}_{n=1}^\infty$ be a list of all the rational numbers indexed by $\mathbb{N}$.  Note that the set $A = \Big\{(q_k, k) \ | \ k \in \mathbb{N} \Big\}$ is closed in $\mathbb{R}^2$ because all of its points are isolated (any two distinct points have at least $1$ unit of vertical separation).  Now consider $f(A)$.  

As an aside, any function between two topological spaces that takes closed sets to closed sets is called a closed map.
A: This is not true in general.  For a counterexample, consider $T : \mathbb{R}^2 \to \mathbb{R}$ given by $T(x, y) = x$, and the closed subset $A := \{ (x, y) \in \mathbb{R}^2 \mid xy = 1 \}$ of $\mathbb{R}^2$.  Then $T(A) = \mathbb{R} \setminus \{ 0 \}$ is not closed.
A: 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).
