Continuity and Inverse Preservation of Boundedness Implies Preservation of Closed Sets Sorry about the rather long title - wasn't sure what else to call it!
Here is my question:

Let $f : \Bbb R^n \rightarrow \Bbb R^m$ be continuous and such that $f^{-1}(F)$ is bounded whenever $F$ is bounded. Is it then true that $f(E)$ is closed whenever $E$ is closed? [We are not assuming uniform continuity.]

(This is a kind of reverse direction of the standard "continuous if and only if inverse image preserves closedness/openness.)
If it is not true, then please don't just give an example with no motivation! :)
Thank you in advance! =D
 A: Yes, this is true. If $y_n$ is a sequence in $f(E)$ for some closed set $E\subset \mathbb{R}^n$ which converges to $y$, then the set $M:=\{y_n\}$ is bounded. Hence $f^{-1}(M)$ is bounded and therefore there exists a bounded(!) sequence $x_n$ in this set with $f(x_n)=y_n$. Since $\{x_n\}$ is bounded, we find a converging subsequence $\{x_{n_k}\}$ converging to some $x$ (because we are in finite dimension). Because $E$ is close we have $x\in E$. Now, $f$ is continuous, thus $f(x_{n_k})$ converges to $f(x)$, but at the same time it converges to $y$. Consequently, $y=f(x)\in f(E)$.
A: Let $\{x_n\}_{n\in \mathbb{N}}$ be a sequence of elements of $f(E)$ converging to $x\in \overline{f(E)}$ - we wish to show that $x\in f(E)$. 
Choose a sequence $\{y_n\}_{n\in\mathbb{N}}$ of elements of $E$ such that $f(y_n)=x_n$. We know that $\{y_n\}_{n\in\mathbb{N}}$ is a bounded sequence in $\mathbb{R}^n$ and therefore has a convergent subsequence $\{y_{n_k}\}_{k\in\mathbb{N}}$. Since $E$ is closed, the limit $y$ of this convergent sequence is an element of $E$.
Since $f$ is continuous,
$f(y)$
$=f(\lim_{k\to\infty} y_{n_k})$
$=\lim_{k\to\infty} f(y_{n_k})$
$=\lim_{k\to\infty} x_{n_k}$
$=x$.
Therefore, $x\in f(E)$, and the proof is complete.
Hope this helps!
