$f:D\to\Bbb R^m$ is proper iff it pushes the boundary to infinity Let $D\subset\subset\Bbb R^n$, i.e. $D$ is relatively compact  and open in $\Bbb R^n$, and consider $f:D\to\Bbb R^m$ continuous.
I conjectured that

$f$ is proper iff $\exists p\in\partial D$ s.t. $f(x)\stackrel{x\to p}{\longrightarrow}\infty_{\Bbb R^m}$

$\Rightarrow$ if BC $f(x)\stackrel{x\to p}{\longrightarrow}\infty_{\Bbb R^m}$ is false $\forall p\in\partial D$, we get that $f$ is bounded on $D$.
So we can write $M:=\|f\|_D<+\infty$ and consider $K:=B(0,M]\Subset\Bbb R^m$. Now $f^{-1}(K)=D$ which is NOT compact in $R^n$, thus $f$ cannot be proper.
$\Leftarrow$ Take $K\Subset\Bbb R^m$; since $f$ is continuous, $f^{-1}(K)$ is clearly closed. Moreover, $f^{-1}(K)\subseteq D$, thus it is bounded too, so $f^{-1}(K)$ is compact, thus $f$ is proper.
Is this correct? I am suspicious because in the latter part, the hypotesis is not taken into account.
Is the proof correct? How do I have to modify the conjecture to make it true?Thanks
 A: No, it is not correct, because it is not true that $f^{-1}(K)$ is closed in $\overline{D}$. Take for example, $D=(0,1)\subset\mathbb{R}$ and consider the function $f(x)=\sin\frac{1}{x}$. $D$ is relatively compact, the function $f$ is continuous and $f(0,1)=[-1,1]$, so with $K=[-1,1]$ you have $f^{-1}(K)=(0,1)$ which is not compact in $\mathbb{R}$. It is true that this $f$ does not satisfy also your conjectured condition (the limit at the boundary point zero does not exist, and the limit at the other boundary point is some finite number), but still, it does show that your second argument is incorrect.
A: Let us first make precise that  your characterizing condition means the following:
There exists $p \in \partial D$ such that $\lim_{x \in D, x \to p} \lVert f(x) \rVert = \infty$.
$\Leftarrow$ is not true.
Let $D = (0,1) \times (0,1) \subset \mathbb R^2$ and $f : D \to \mathbb R^2, f(x,y) = (e^{1/x},y)$. Then $\lim_{(x,y) \to (0,0)} \lVert f(x,y)\rVert = \infty$, but $f^{-1}(\{e^{-2}\} \times [0,1]) = \{1/2\} \times (0,1)$ is not compact.
However, $\Rightarrow$ is true since $\partial D \ne \emptyset$. In fact, we have the following result:
$f$ is proper $\Leftrightarrow$ For all $p \in \partial D$ one has  $\lim_{x \in D, x \to p} \lVert f(x) \rVert = \infty$.


*

*Let $f$ be proper and $A_r \subset \mathbb R^n$ be the closed ball with radius $r$ centered at $0$. Since  $A_r$ is compact, $B_r = f^{-1}(A_r) \subset D$ is compact. Now let $p \in \partial D$ and $r > 0$. Since $B_r$ is compact, $\overline{D} \setminus B_r$ is an open neighborhood of $p$ in $\overline{D}$. For $x  \in D \setminus B_r$ we have $\lVert f(x) \rVert > r$ which shows that for each $r > 0$ there exists $\delta > 0$ such that if $x \in D$ with $\lVert x - p \rVert < \delta$, then $\lVert f(x) \rVert > r$. Thus $\lim_{x \in D, x \to p} \lVert f(x) \rVert = \infty$.

*Let $f$ have the property that for all $p \in \partial D$ one has  $\lim_{x \in D, x \to p} \lVert f(x) \rVert = \infty$. Assume that $f$ is not proper. Then there exists a compact $K \subset \mathbb R^n$ such that $K' = f^{-1}(K)$ is not compact. Since $K'$ is closed in $D$, it must have a limit point $p \in \overline{D} \setminus D = \partial D$. Let $(x_k)$ be a sequence in $K'$ converging tp $p$. Then $f(x_k) \in K$, thus $(f(x_k))$ is bounded. This contradicts the above condition on $f$. Thus $f$ must be proper.
