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The closed map lemma says that if $f : X \to Y$ is a continuous function, $X$ is compact and $Y$ is Hausdorff, then $f$ is a closed map.

How can I prove this ?

Here is my attempt so far:

Suppose for contradiction that $f$ is not a closed map. Then there exists a closed subset $V$ of $X$ whose image $f(X)$ is not closed in $Y$. This means that there exists a convergent sequence $(w_n)_{n \in \mathbb{N}}$ in $f(V)$ whose limit is not contained in $f(V)$.

Now by compactness of $X$, we know the inverse image $f^{-1}((w_n)_{n \in \mathbb{N}})$ of the sequence contains a limit point, denote it by $z$, which lies in $V$ (as $V$ is closed). But then, by continuity of $f$, we also know that $f(z)$ must be a limit point of the sequence $(w_n)_{n \in \mathbb{N}}$. By the fact that the latter converges, we deduce that this limit point must be the unique limit. Thus we have

\begin{equation} V \ni z \mapsto f(z) \notin f(V) \end{equation}

which is a contradiction.

Here is my question: Where does the Hausdorff condition for $Y$ enter the proof ? Since it is not mentioned in my above attempt I must have a mistake somehwhere .. what did I miss ?

Many thanks for your help !!

harlekin
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    I think when you said "unique limit point" – David Mitra May 11 '12 at 19:09
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    In a non-Hausdorff space a sequence can converge to more than one point. – Brian M. Scott May 11 '12 at 19:09
  • AHH - sure - that's what I missed .. very embarassing since it is the very essence of the property of being Hausdorff .. THANKS ! – harlekin May 11 '12 at 19:11
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    Also, you should replace "sequence" with "net" since you don't assume metrizability. The argument works all the same. – Xabier Domínguez May 11 '12 at 19:11
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    To extend Xabier's, compactness is generally not equivalent to sequentially compactness. However it is true that in a compact space every infinite set has a limit point (so there is a net converging to it). – Asaf Karagila May 11 '12 at 19:13
  • hm .. ok, I have to look into the concept of a net. So far I have touched upon topology only as a sidestep from my real analysis courses - it seems I am working way too much under the assumption that my spaces are metrizable. Thanks very much for your comments! – harlekin May 11 '12 at 19:15
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    A shorter argument would use: 1) Closed subsets of compact spaces are compact. 2) Continuous images of compact sets are compact. 3) Compact sets in Hausdorff spaces are closed. – David Mitra May 11 '12 at 19:20

2 Answers2

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The theorem is true in topological spaces in general, not just spaces in which sequences determine the topology (like metric spaces, for instance), so it can be proved without any use of sequences. In the general setting it's nets and filters that generalize sequences, but you don't need them here. If you want to learn more about them, a very good starting point is these notes by Pete L. Clark, who also posts here.

Suppose that $H\subseteq X$ is closed, and let $K=f[H]\subseteq Y$. Let $\mathscr{U}$ be a family of open subsets of $Y$ such that $K\subseteq\bigcup\mathscr{U}$; that is, $\mathscr{U}$ is an open cover of $K$ in $Y$. Let $\mathscr{V}=\{f^{-1}[U]:U\in\mathscr{U}\}$; since $f$ is continuous, $\mathscr{V}$ is an open cover of $H$ in $X$. But $H$, being a closed subset of the compact space $X$, is compact, so some finite subset $\{V_1,\dots,V_n\}$ of $\mathscr{V}$ covers $H$. For $k=1,\dots,n$ there is $U_k\in\mathscr{U}$ such that $V_k=f^{-1}[U_k]$, and it's clear that $\{U_1,\dots,U_n\}$ covers $K$. Thus, every open cover of $K$ has a finite subcover, and $K$ is compact. Now just use (or prove) the result that every compact subset of a Hausdorff space is closed.

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Brian M. Scott
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  • In short: take $E$ closed, it is compact as closed subset of a compact set, thus $f(E)$ is compact by continuity of $f$ (fact proven by Brian here), which in turn is closed as compact subset of a Hausdorff space. – Cryme Nov 06 '23 at 17:18
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I think that in Brian's proof he proved in the process that the continuous image of a compact set is compact.

Now in your proof above, you used sequences but you can prove it without using any of these The advantage of this is that your proof will almost generalise immediately to a general topological space. Here's how you use up all your hypotheses:

Let $B$ be a closed subset of our compact space $X$. Since a closed subset of a compact space is compact, we have that $B$ is a compact subset of $X$. Now by assumption $f$ is a continuous function so that $f(B)$ is compact. But then $Y$ is a Hausdorff space so that compact subsets of $Y$ are closed. It follows that $f(B)$ is closed and hence that $f$ is a closed map.