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Lemma: Let $(X,d)$ be compact. Let $f: (X,d) \to (Y, \rho)$ be a continuous bijection. Then $f$ maps closed sets to closed sets.

Pf: Let $V$ be a closed set in $(X,d)$. We have to show that $f(V)$ is closed, i.e., $f(V) = \overline{f(V)}$. Let $y \in \overline{f(V)}$. Then $\{y_n\} \subset f(V)$ such that $y_n \to y$. So $\forall n \in \mathbb{N}, \exists x_n \in V$ such that $y_n = f(x_n)$. Then $\forall n \in \mathbb{N}, x_n \in f^{-1}(y_n) = x_n$. Now since $f^{-1}$ is continuous and $y_n \to y$, $f^{-1}(y_n) \to f^{-1}(y)$. $\Rightarrow f^{-1}(y) \in V$ since $V$ is closed. $\Rightarrow y \in f(V)$.

I have not used the fact that X is compact so I realised that my proof might be wrong. But I fail to see where the proof might break, logically speaking. Can someone help me out?

stange
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Meer Sisodia
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    In general, $f$ being continuous doesn't imply $f^{-1}$ is continuous. – Mark Jul 26 '24 at 16:23
  • So then would the proof be correct if I add the hypothesis of a continuous inverse? – Meer Sisodia Jul 26 '24 at 16:31
  • It would be correct, but with this assumption this becomes a pretty much trivial problem. No need to add this hypothesis, you just need to use compactness in order to prove that $f$ maps closed sets to closed sets. Which will by the way imply that the inverse is continuous. (try to understand why) – Mark Jul 26 '24 at 16:38
  • You have no reason to assume that $f$ has an inverse. – José Carlos Santos Jul 26 '24 at 16:58

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