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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.

some_math_guy
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  • Does "compact" include Hausdorff? – Paul Frost Oct 22 '23 at 10:18
  • @PaulFrost uhmm Judging by the quote from my lecturer, it doesn't. I didn't know that some people may consider that, as it seems your question implies that. If it helps this problem is in the context of Lie groups. We mostly follow these lecture notes https://webspace.science.uu.nl/~ban00101/lecnotes/lie2010.pdf – some_math_guy Oct 22 '23 at 10:23
  • @PaulFrost To make it more confusing in the notes, section 13 a proper map is defined for locally compact Hausdorff (topological) spaces X and Y., but it seems like the lecturer changed this definition – some_math_guy Oct 22 '23 at 10:25
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    Search for "French school" in https://en.m.wikipedia.org/wiki/Compact_space. Many authors use this interpretation, but it is not a commonly accepted standard. – Paul Frost Oct 22 '23 at 10:32
  • Note that converging sequence in metric space with its limit is compact set. So $K\cap f^{-1}((s_n)\cup s)$ is closed quasi-compact set and its image is closed (by Hausdorffness of $Y$) compact set. From here, I think one can derive, $s\in f(K).$ – dsh Oct 22 '23 at 11:01
  • @PaulFrost I have confirmation that "compact" does not include Hausdorff for us – some_math_guy Oct 22 '23 at 11:24
  • @dsh I hope you are non using the definition that "compact" includes Hausdorff in this argument , as stated above we don't use that. (...Since you're are talking about quasi-compactness, which to to be frank I haven't heard of until now) – some_math_guy Oct 22 '23 at 11:25
  • @some_math_guy Yes, quasi-compact does not imply Hausdorfness (the case of $X$). I use compact in the case of $Y.$ Quasi-compact subset in Hausdorff space is compact and necessarily closed. – dsh Oct 22 '23 at 11:49
  • @some_math_guy Quasi-compact is from French school mentioned in the wiki, does not imply Hausdorfness to avoid confusion in cases like this (only existence of finite subcover of any open cover). – dsh Oct 22 '23 at 11:52
  • @some_math_guy We do not need Hausdorff ;-) – Paul Frost Oct 22 '23 at 12:11
  • @PaulFrost I have impression that compact includes Hausdorff in the algebro-geometric setting. – Mihail Oct 22 '23 at 16:01

2 Answers2

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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 $s.$
dsh
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  • Are you using the French school definitions of compactness and quasi-compactness? And why are you saying "Y is metrizable". Y is already metric by assumption. Are you changing the hypothesis of the problem? If so, please state clearly what is the statement of the problem you are solving – some_math_guy Oct 22 '23 at 13:41
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    @some_math_guy Yes, I used Bourbaki's definitions. No, I did not change hypothesis. Yes, every metric space is metrizable, but metrizable space may have different distance functions. And one may replace metric/metrizable space with Hausdorff and https://en.wikipedia.org/wiki/First-countable_space. – dsh Oct 22 '23 at 15:32
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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 $s \in f(K') \subset f(K)$.

Paul Frost
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  • Awesome! Thanks a lot. Why does "each open neighborhood of s contains all but finitely many $s_n$" means S is compact? I can't find that as one of the equivalent definitions of compactness in the wikipedia entry https://en.m.wikipedia.org/wiki/Compact_space and why is it useful to include $s \in S$? – some_math_guy Oct 22 '23 at 12:46
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    @some_math_guy In each open cover $\mathfrak U$ of $S$ one of the open sets, call it $U$, must contain $s$. This set contains all but finitely many $s_n$. The fnitely many $s_n$ not in $U$ are covered by finitely many members of $\mathfrak U$. Thus $\mathfrak U$ has a finite subcover. We have to include $s$ because the set ${s_n \mid n \in \mathbb N}$ is not compact in general. Take $s_n = 1/n$ as an example. – Paul Frost Oct 22 '23 at 13:03
  • @PaulFrost Why do we know that there exist ''suitable" $x_n$ s.t $x_n \in K$? – darkside Oct 24 '23 at 09:10
  • @darkside Because $s_n \in f(K)$. This means $s_n = f(x_n)$ for some $x_n \in K$. – Paul Frost Oct 24 '23 at 09:37
  • and why is $ f(K')$ closed in $X$? $ f(K')$ lives in $Y$ – darkside Oct 24 '23 at 09:45
  • I guess it was a typo, it should be closed in Y and then we used that a compact set in a Hausdorff space is closed, correct? – darkside Oct 24 '23 at 09:53
  • @darkside Yes, it was typo. Thank you for pointing this out. I made a correction. – Paul Frost Oct 24 '23 at 10:55
  • @PaulFrost If you have a minute , could you answer my comment in the answer of this question https://math.stackexchange.com/questions/4791946/if-a-g-times-m-rightarrow-m-is-a-proper-action-and-that-m-is-a-metric-sp ? – darkside Oct 24 '23 at 11:03