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Question:

There is a continuous bijection $f: C \rightarrow D \subset \Bbb R^m$,$\,C\subset \Bbb R^n$ is a compact set. Then prove$\,f^{-1}$ is continuous.

The question is that I know the proof using topological tools by the definition of continuous in topological version. However, I do not know how to only use the metric to prove it (although one can use the usual metric on $\Bbb R^m$ to induce a topology and prove it in a topological way).

Could someone tell me a hint to prove by the continuity at every point using $\epsilon-\delta$ language?

1 Answers 1

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Assume for a contradiction that $f^{-1}$ is not continuous.

This means that there is an $y_0\in D$, $\varepsilon>0$ such that there are $y$s arbitrarily close to $y_0$ such that $f^{-1}(y)$ differ from $f^{-1}(y_0)$ by at least $\varepsilon$.

Choose a sequence $(y_1,y_2,\ldots)$ of such $y$s that converge towards $y_0$.

The corresponding sequence of $x_n=f^{-1}(y_n)$ lives in $C$, which is assumed compact. Therefore it has a convergent subsequence. Suppose it converges towards $x'_0$. Then $x'_0\ne f^{-1}(y_0)$ because all of the $x_n$s are at least $\varepsilon$ away from $f^{-1}(y_0)$.

But because $f$ is continuous, we must have $f(x'_0)=y_0$, but we also have $f(f^{-1}(y_0))=y_0$, contradicting the assumption that $f$ was bijective.