I'm attempting the following proof while working through the math appendix of my economics textbook.

Prove that $f:D \rightarrow \mathbb{R}^n$ is continuous if and only if $f^{-1}(T)$ is compact in the domain $D \subset \mathbb{R}^m$ for every compact set $T$ in the range $\mathbb{R}^n$.

I've found numerous examples on here of showing that if $S$ in the domain is compact then $f(S)$ in the range is also compact, but nothing that goes the other way around that I could understand.

One of the problems I have with this proof is that it seems like there are counterexamples. Take $f(x) = sin(x)$ where $D$ is the non-negative reals. Let $T=[0,1]$, then $f^{-1}(T)=[0,\pi]\cup[2\pi,3\pi]\cup[4\pi,5\pi]....$. This seems to be an unbounded set in $D$.

So does the wording of this proof imply that $D$ itself must be compact in $\mathbb{R}^m$?

Secondly, it seems fairly obvious that if $f$ is continuous and $T$ is closed, then $f^{-1}(T)$ is a closed set. But I'm not quite sure how to show that the same applies for boundedness. The topology section of my textbook does not mention the concept of the cover of set, so the only definition of boundedness we have is that a set $S$ is bounded if and only if there exists an $\epsilon > 0$ such that $S \subset B_\epsilon (x)$ for some $x \in D$.

I've also tried manipulating the definition of Cauchy continuity, ie that $\forall \epsilon>0, \exists \delta>0$ such that $f(B_{\delta}(x) \cap D) \subset B_{\epsilon}(f(x))$, but I get stuck.

I also note that if $T$ is a closed ball in $\mathbb{R}^n$, then $f^{-1}(T)$ will be closed in $D$. Likewise if $T$ is an open ball in $\mathbb{R}^n$, then $f^{-1}(T)$ will be open in $D$. I wondered whether there was a way to combine these two facts, ie if $T$ is bounded, then the complement of $T$ in $\mathbb{R}^n$ will be both open and closed, hence I can show that the compliment of $f^{-1}(T)$ in $D$ is both open and closed, and hence $f^{-1}(T)$ must be bounded? Is this avenue of reasoning likely to be fruitful?

I'd be most grateful for any hints. This is for my own off-season preparation, and not part of any assignment.


  • 2
    $\begingroup$ They simply forgot to assume compactness of the domain $D$ of $f$ and of the codomain of $f$. Otherwise, there are counter examples in both directions. $\endgroup$ Feb 1 at 23:59

1 Answer 1


As I said in the comment, they should have assumed that $f: D\to K$ is a map of compact subsets. Then the claim holds: it amounts to saying that the preimage of each closed subset is closed, which is a standard characterization of continuity. Without the assumption of compactness of $D$ the simplest example to use is the zero map of $R^n$: the preimage of $\{0\}$ is noncompact. To get a counter- example in the opposite direction take $D=\{0, 1/n: n\in \mathbb N\}$ and $f(1/n)=n$, $f(0)=0$. Preimages of compacts are compact but the map is discontinuous.


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