# Prove that a continuous image of a closed subset of a compact space is a closed subset

Suppose $f$ is a continuous mapping from a compact metric space $X$ into a metric space $Y$. Prove that if $F$ is a closed subset of $X$, then $f[F]$ is a closed subset of $Y$.

Here is my idea for the proof: The continuous image of a connected space is connected. Use the intermediate value theorem to show that the image of every continuous real-valued function is an interval, and should return closed sets into closed sets.

Corrections are appreciated!

• I don't think this has something to do with intermediate value theorem, $\mathbb{R}$ is not mentionned in the problem. Here are some hints : in $X$ every closed set is compact (why), any continuous function from metric space to metric space sends compacts to compacts, in $Y$ every compact set is closed. – Clément Guérin Nov 5 '15 at 7:55
• Thank you. Would it be possible then to show there are sequences (and subsequences) in the compact space that sends compacts to compacts? Or am I still thinking too much in $\Bbb R$ ? – Iff Nov 5 '15 at 7:59
• I am not really sure about what you mean by "sequence that sends compacts to compacts". However if you suggest that to show compacity of some set $Z$, one only needs to show that sequences of $Z$ always admit a converging subsequence then you are right. Sequential compacity is equivalent to compacity in metric spaces (not only in $\mathbb{R}$). – Clément Guérin Nov 5 '15 at 8:05
• I apologize, I'm a bit sleepy. I was thinking of a theorem that states if a metric space is compact, then it has the Bolzano-Weierstrass property. So I thought if I could use sequences to show that it is closed - more properly I think I should use coverings and subcollection of coverings. I think my problem is I'm still confusing metric space properties and $\Bbb R$ properties. I appreciate your help! – Iff Nov 5 '15 at 8:12
• Remember that for a metric space it is the metric that outputs a real number to denote the distance between two objects in the space. The objects are not constrained to be of any particular kind. – user21820 Nov 5 '15 at 11:07

First let me tell you the flaws in your argument. 1) nobody said anything about $X$ being connected. 2) You never used compactness of $X$. 3) You are using a property of "real" valued functions, these are metric spaces, which are much more general than real spaces.

Here's how you do it. Take any open covering for $f(F)$ in $Y$, say $\{U_\alpha\}_{\alpha \in I}$ ($I$ possibly infinite index set). Since $f$ is continuous $f^{-1}(U_\alpha)$ is open in $X$ for every open set in the cover. Moreover $\{f^{-1}(U_\alpha)\}$ covers $F$. But $X$ is compact, so there is a finite subcover, say $\{f^{-1}(U_i)\}_{i=1, \cdots, n}$ covering the whole $F$, i.e. $F\subset \bigcup_{i=1}^n f^{-1}(U_i)$. Then $\{U_i\}_{i=1, \cdots, n}$ covers $f(F)$. So any open cover of $F$ has a finite subcover. Hence $f(F)$ is compact. Any compact subspace of a metric space is closed and bounded, so $f(F)$ is closed.

• I see, thank you very much. I am confusing my definitions. Is it correct to apply Bolzano-Weierstrass for the mention of subcovers or would you consider that superfluous? – Iff Nov 5 '15 at 8:15
• Bolzano–Weierstrass is just a fancy name for the fact that any subset of $\mathbb{R}^n$ is sequentially compact if and only if it is closed and bounded. So an easy generalization of it basically says that in metric spaces sequential compactness is equivalent to closed and bounded (which is another definition of compactness for metric spaces). So yes you can show the same thing using sequences (using Bolzano–Weierstrass). – Hamed Nov 5 '15 at 8:27
• Thank you for your insight! – Iff Nov 5 '15 at 15:59

You might want to proceed in the following manner:

1. Closed subsets of compact sets are compact, thus $F$ is compact;
2. Continuous image of a compact set is compact, hence $f(F)$ is compact in $Y$;
3. Compact sets in a metric space are closed, hence $f(F)$ is closed.
• Thank you. That's very helpful! – Iff Nov 5 '15 at 15:59

A slightly different answer using sequences. Take $F$ a closed set of $X$. Now take a sequence $(y_n)$ of elements in $f(F)$ converging to $y\in Y$. We want to show that $y\in f(F)$.

Because $y_n\in f(F)$ we can find $x_n\in F$ such that $f(x_n)=y_n$. Now $(x_n)$ is a sequence of $X$ (compact), so it admits a subsequence converging $(x_{\varphi(n)})$ to $x\in X$. Since $F$ is closed and $(x_{\varphi(n)})$ is a sequence of $F$ we know that actually $x\in F$.

Finally, because $f$ is continuous we have that $f(x_{\varphi(n)})$ converges to $f(x)$. In other words $(y_{\varphi(n)})$ converges to $f(x)$. Since $(y_{\varphi(n)})$ also converges to $y$, by unicity of the limit we get that $y=f(x)\in f(F)$.

We have shown that $f(F)$ is sequentially closed so it is closed (we are in a metric space).