Assume that $A\subseteq\mathbb{R}^n$ is compact and that $\mathbf{f}:A\to\mathbb{R}^m$ is continuous and injective. Let $B=\mathbf{f}(A)\subseteq\mathbb{R}^m$ and show that $\mathbf{f}^{-1}(B)\rightarrow A$ is continuous.

I want to use the sequential definition of continuity.

Let $\mathbf{y} \in B$ and $(\mathbf{y}_n) \rightarrow \mathbf{y}$ with $\mathbf{y}_n \in B$ for each $n\in \mathbb{N}$.

Then since $B=\mathbf{f}(A)$, there are $\mathbf{x} , \mathbf{x}_n \in A$ such that $\mathbf{f}(\mathbf{x}) =\mathbf{y}$ and $\mathbf{f}(\mathbf{x}_n) = \mathbf{y}_n$ for each $n$.

Since $A$ is compact, it is closed and bounded. So $(\mathbf{x}_n)$ is bounded. We show that $(\mathbf{x}_n)$ converges.

It's enough to show that any convergent subsequence of $(\mathbf{x}_n)$ converges to the same point in $\mathbb{R}^n$. So let $(\mathbf{x}_{n_k})$ be a convergent subsequence that converges to $\mathbf{a}$. Since $A$ is closed, $\mathbf{a} \in A$. By continuity, $\mathbf{f}(\mathbf{x}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. Therefore, $(\mathbf{y}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. But since $(\mathbf{y}_{n_k})$ is a subsequence of $(\mathbf{y}_n)$, we get that $\mathbf{y} =\mathbf{f}(\mathbf{a})$. Thus, $\mathbf{f}^{-1}(\mathbf{y})=\mathbf{a} =\lim_{k\to \infty} \mathbf{x}_{n_k}$. And every convergent subsequence of $ (\mathbf{x}_{n})$ converges to $\mathbf{f}^{-1}(\mathbf{y}):=\mathbf{a}$, so that $(\mathbf{x}_n)\rightarrow \mathbf{a}$.

EDITED QUESTION: Does this proof work? I'm a bit concerned that I haven't used the fact that $\mathbf{f}({A})$ is a compact set. Overall, I'm not sure if I made a mistake somewhere.

  • $\begingroup$ Do you have to show it with sequences? It is much easier in a general topology setting. $\endgroup$ – Jakob Elias Apr 25 '17 at 22:37
  • $\begingroup$ @JakobElias Yes, with sequences. $\endgroup$ – CuriousKid7 Apr 25 '17 at 22:37
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    $\begingroup$ It is a long time I did any general topology, so correct me if I am wrong. A continuous function from $R^n$ to $R^m$ maps compact set to compact set (to see this, consider the sub sequence criterion, a set is compact iff. every sequence contains converging sub-sequence). Now a closed subset of compact is compact, and a compact subset in $R^m$ is closed. In a nutshell, $f$ maps closed sets to closed sets. $\endgroup$ – Ran Wang Apr 25 '17 at 22:50

Start from your sequence $(\mathbf{y_n})\to \mathbf{y}$ and consider an arbitrary subsequence $(\mathbf{y_{n_k}})$ of $(\mathbf{y_n})$. By what you proved, you can find a subsequence $(\mathbf{y_{n_{k_j}}})$ of $(\mathbf{y_{n_k}})$ such that $(f^{-1}(\mathbf{y_{n_{k_j}}}))\to f^{-1}(\mathbf{y})$. This implies that the entire sequence $(f^{-1}(\mathbf{y_{n}}))\to f^{-1}(\mathbf{y})$ since otherwise, you could find a subsequence $(\mathbf{y_{n_k}})$ such that $\vert f^{-1}(\mathbf{y_{n_{k}}})-f^{-1}(\mathbf{y})\vert\ge \varepsilon>0$ for all $k$, which would be a contradiction. I am using this fact theorem on subsequences Does this answer your question?

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  • $\begingroup$ I think this is correct~ $\endgroup$ – Ran Wang Apr 25 '17 at 22:51
  • $\begingroup$ I actually was really unclear. I edited my question. I actually am just looking for verification that my proof works as is. $\endgroup$ – CuriousKid7 Apr 25 '17 at 23:11
  • $\begingroup$ I was making a small correction to your proof. When you say "It's enough to show that any convergent subsequence of ", if you actually try to prove the "It's enough to show" you will see that you need to make the change I suggested. $\endgroup$ – Gio67 Apr 26 '17 at 1:17
  • $\begingroup$ Ok, to be clear: I have already proven a theorem beforehand that says that if every convergent subsequence of a bounded sequence $(x_n)$ has the same limit $l$, then $(x_n)$ converges to $l$ as well. So I have already justified the "it's enough..." part. Having said this, do you see any issue with my proof as it stands now? I never used the fact that $f(A)$ is compact and want to be sure that my proof is valid anyway. $\endgroup$ – CuriousKid7 Apr 26 '17 at 1:51
  • $\begingroup$ yes, you do not need to use the fact that $f(A)$ is compact since you start with a sequence which is already convergent in $f(A)$ $\endgroup$ – Gio67 Apr 26 '17 at 1:55
  1. Since $f$ is continuous, $f(A)$ is compact and therefore closed.

  2. Let $y_{n}\in f(A)$ a convergent sequence. By 1., we can supose $y_{n}=f(x_{n})\to f(x)=y$.

  3. We want to show that $f^{-1}(y_{n})\to f^{-1}(y)$, i.e. $x_{n}\to x$. If $x_{n}\not\to x$, then there exists a subsequence such that $|x_{n_{k}}-x|\geq \epsilon$. Then we can extract a convergent subsequence $x_{n_{k_{j}}}$ of $x_{n_{k}}$ to, say, $x'$. But then we would have $f(x')\leftarrow f(x_{n_{k_{j}}})=y_{n_{k_{j}}}\to y=f(x)$. So, $f(x)=f(x')$ and then $x=x'$, a contradiction.

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