Prove that if $A\subseteq Y$ is compact, and $f:X\to Y$ is continuous, where $X$ is compact, then $f^{-1}(A)$ is compact.

I am asked to show that the preimage of a compact set is compact.

I am given $(X,d)$ compact metric space, arbitrary metric space $Y$, compact set $A$ $\subset$ $Y$, and continuous function $f$: X$\rightarrow$Y. I need to show $f^{-1}$(A) compact. My textbook gives a solution by showing using the closed set of compact set is compact theorem. I would like to test an alternative approach. Can you please check it?

Attempt:

Since A is compact, there exists a finite subcover $U$ such that $A$ $\subset$ $U$. We know that $U$ is an open set since it is the union of finite open sets. Since $F$ continuous and $A$ open, we know that $f^{-1}$($U$) open. Recall $A$ $\subset$ $U$, then $f^{-1}$($A$) $\subset$ $f^{-1}$($U$). Thus, $f^{-1}$($A$) has a finite open cover of $f^{-1}$($U$) and $f^{-1}$($A$) compact.

• Stop and think about the first sentence of your argument: a finite subcover of what? In any case, you’re starting in the wrong place. To show that $f^{-1}[A]$ is compact, you must start with an arbitrary open cover of $f^{-1}[A]$, not of $A$, and show that it has a finite subcover. And as @LASV’s answer shows, it need not. – Brian M. Scott Dec 8 '13 at 22:40
• I strongly encourage changing the title of the question. – LASV Dec 8 '13 at 22:43
• I will do it now – kpz Dec 8 '13 at 22:44

There is no need of considering open covers. Just use these three simple facts (the second one is actually proved with open covers, of course).

1. A compact subset of a metric space is closed

2. A closed subset of a compact metric space is compact

3. The inverse image of a closed subset via a continuous map is closed

Your alternative approach is flawed. You want to show that if $A\subset Y$ is compact, then $B=f^{-1}(A)$ is compact. To show this, you have to take an arbitrary open cover of $B$ and try extracting from it a finite subcover.

Say $\mathcal{U}$ is the open cover. Then, since $B$ is closed, $\mathcal{U}\cup\{X\setminus B\}$ is an open cover of $X$. Therefore, there exists a finite subcover: $X=U_1\cup\dots\cup U_n\cup (X\setminus B)$. Obviously

$$B\subset U_1\cup\dots\cup U_n$$

so we have found a finite subcover of $B$ from $\mathcal{U}$. But this is exactly the proof of statement 2 above.

• I'm guessing this is the proof the OP's book gives. He's asking about an alternate approach. – Santiago Canez Dec 8 '13 at 22:43
• I see - this should be the general strategy then? To take an open cover of the entire metric space and narrow it? – kpz Dec 8 '13 at 22:54
• @user2853043 No; one has to take an open cover of the set $B$, which can be turned into an open cover of the space by adding $X\setminus B$ to it. But an arbitrary open cover, not a finite one, to begin with. – egreg Dec 8 '13 at 23:00

Please change the title of your question, it is misleading. For example $f:\mathbb{R}\to\{0\}$ yields a counterexample to the statement.

Use the fact that preimages of closed sets are closed, the fact that compact sets of metric spaces are closed, and the fact that a closed subset of a compact set is compact.

• $X$ is compact, in the question. I don't think $\mathbb{R}$ is, at least with the usual topology. – egreg Dec 8 '13 at 22:39
• The domain is supposed to be compact. – Santiago Canez Dec 8 '13 at 22:40
• Ah, then the question should be changed. Then use the fact that preimages of closed sets are closed, the fact that compact sets of metric spaces are closed, and the fact that a closed subset of a compact set is compact. – LASV Dec 8 '13 at 22:41