If $Y$ is a metric space and $f: X \rightarrow Y$ is proper, $f$ is closed

Question

Def: Let $f : X \rightarrow Y$ be a continuous map. $X,Y$ Topological spaces. $f$ is called proper if $f^{-1}(K)$ is compact for every compact $K \subseteq Y$.

I want to prove that :

If $Y$ is a metric space and $f: X \rightarrow Y$ is proper, $f$ is closed

My try:

Let $K \subset X$ be closed. I want to prove that $f(K)\subseteq Y$. Since $Y$ is a metric space I can use that a closed set is one that contains all of its limit points:

Let $s$ be a limit point of $f(K) \subseteq X \iff \exists$ a sequence $ (s_n)\subseteq f(K) $ s.t. $s_n \rightarrow s$ I'd like to prove that $s \in f(K)$. Suppose it doesn't. Then $s \in f(K)^C$. Then there should exist a neighborhood $U $of $s$ that contains all but finitely many elements of the sequence. Now I am not sure if I can take this nbhd to be closed and or included in $f(K)^C$, If I could then $U$ would be compact because a compact set in a Hausdorff space ($Y $is metric, thus Hausdorff) is closed $f^{-1}(U)$ would be compact by properness of $f$ and closed by continuity of $f$. Still I am stuck here. Any help is welcome

Looking around it seems like people think we need more about X like local compactness , Hausdorfness.

Answer 1

Your approach (considering a sequence $(s_n)$ in $f(K)$ converging to $s \in X$) is fine, but we need additional arguments.

Let $S = \{s\} \cup \{s_n \mid n \in \mathbb N\}$. This is a compact subset of $X$ because each open neighborhood of $s$ contains all but finitely many $s_n$. Hence $f^{-1}(S)$ is compact. The set $K' = K \cap f^{-1}(S)$ is a closed subset of $f^{-1}(S)$, thus it is compact. Therefore $f(K')$ is a compact subset of $Y$. We have $s_n = f(x_n)$ for suitable $x_n \in K$. Since $x_n \in f^{-1}(s_n) \subset f^{-1}(S)$, we have $x_n \in K'$ and thus $s_n \in f(K')$. Since $f(K')$ is closed in $Y$ and $s_n \to s$, we see that $s \in f(K') \subset f(K)$.

Answer 2

Using quasi-compact notion for space which has finite subcover for any open cover. Compact means quasi-compact and Hausdorff. $f$ is has the property that for each $C\subseteq Y$ compact, $f^{-1}(C)$ is quasi-compact.

1. Closedness of $f(K)$ can be checked using sequences as $Y$ is metrizable. Consider $(s_n)\subseteq f(K)$ converging to $s\in Y.$
2. $S = (s_n)\cup \{s\}$ is compact in $Y$ by direct check.
3. $f^{-1}(S)$ is closed (by continuity) and quasi-compact by assumption on $f.$
4. $K_1 = K\cap f^{-1}(S)$ is closed subset of $f^{-1}(S)$ and thus quasi-compact.
5. $f(K_1)$ is also quasi-compact (check using covers). As $Y$ is Hausdorff, $K_1$ is closed.
6. $f(K_1)$ contains $(s_n)$ and by closedness, it contains $s.$