# Are proper maps compact?

Recently I learned about the notion of a proper map in metric spaces. Namely, if $X$, $Y$ are metric spaces, then a map $f:X\rightarrow Y$ is called proper iff for every compact set $K\subseteq Y$ the set $f^{-1}(K) \subseteq X$ is compact. I wondered whether all proper maps between metric spaces are closed as well. All the proofs I found on the net (e.g. http://vmm.math.uci.edu/PalaisPapers/WhenProperMapsAreClosed.pdf or When is the image of a proper map closed?) use at some point that the image of a compact set is compact. I thought that the following is a counterexample: Endow $\mathbb{R}$ with the Euclidian metric then

$f: \mathbb{R} \rightarrow \mathbb{R}, \ f(x)= \begin{cases} \vert x \vert, &\vert x \vert\geq 1, \\ \frac{1}{\vert x \vert}, &0<\vert x \vert<1, \\ -1, &x=0. \end{cases}$

is proper, $[-1,1]$ is compact, but $f([-1,1])=[1,\infty)$ is not compact. What am I doing wrong?

• Proper continuous maps between metric spaces are closed. – Jack Lee Apr 15 '16 at 20:26
• Thanks for your response. I was aware of the proof in the case when the map is continous. But Ambrosetti, Prodi claim in "A Primer of Nonlinear Analysis" on p. 46 that proper maps are closed and they do not assume continuity. Those it hold in such generality? – Severin Schraven Apr 15 '16 at 20:35
• Well, as your counterexample shows, it's just not true that every proper map is closed without assuming continuity. Are you sure Ambrosetti & Prodi aren't operating under some implicit assumption of continuity? – Jack Lee Apr 15 '16 at 20:44
• Unfortunately, my example just shows that proper maps are not compact, it is in fact closed. Ambrosetti & Prodi make the remark immediately after the definition and state in every theorem explicitely when the map is assumed to be continuous. – Severin Schraven Apr 15 '16 at 20:56
• Oh, sorry -- I didn't look carefully enough at your example. I'll see if I can cook up one of my own. – Jack Lee Apr 15 '16 at 21:04

Let $X,Y$ be metric spaces, and $f\colon X\to Y$ be a proper map. Suppose $A\subset X$ is any closed set. We need to show that $f(A)$ is closed in $Y$.
Let $y$ be any limit point of $f(A)$. Then there is a sequence $(y_i)$ in $f(A)$ such that $y_i\to y$, and we may assume that all of the $y_i$'s are distinct from each other and from $y$. This implies that for each $i$, there is a point $x_i\in A$ such that $f(x_i)=y_i$. The fact that the $y_i$'s are distinct implies that the $x_i$'s are distinct as well.
Now the set $K = \{y_i: i\in \mathbb Z^+\} \cup \{y\}$ is compact, so $f^{-1}(K)$ is compact. The sequence $(x_i)$ is contained in $f^{-1}(K)$, so it has a subsequence $x_{i_j}$ that converges to a point $x\in f^{-1}(K)$. Because $A$ is closed, we also have $x\in A$. Because $f(x)\in K$, we must either have $f(x)=y$ or $f(x) = y_{i_0} = f(x_{i_0})$ for some ${i_0}$. If $f(x)=y$, we are done, because this means $y = f(x)\in f(A)$; so we just have to rule out the possibility that $f(x)=y_{i_0}$. If this is the case, let $\widetilde K= K \smallsetminus \{y_{i_0}\}$, which is still compact. The set consisting of all $x_i$'s except $x_{i_0}$ lies in the compact set $f^{-1}(\widetilde K)$, so its limit point $x$ also lies in $f^{-1}(\widetilde K)$. But this is a contradiction, because $f(x) = y_{i_0}\notin \widetilde K$.