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Let $f:X\rightarrow Y$ be any map. The graph of f is the set $\Gamma_f=\{(x,f(x))| x\in X\}\subset X \times Y$.

Assume that Y is compact. Prove that if $\Gamma_f \subset X \times Y$ is closed then $f$ is continous.

I was able to prove the converse (where it was also given that Y is Hausdorff), but I'm having trouble proving this direction.

Drew
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1 Answers1

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Assume $F\subseteq Y$ is a closed set. Then $(X\times F)\cap\Gamma_f$ is closed in $X\times Y$ as a finite intersection of closed sets. Now define the projection $\pi:X\times Y\to X$ by $(x,y)\to x$. Since $Y$ is compact this is a closed map. Hence $f^{-1}(F)=\pi((X\times F)\cap\Gamma_f)\subseteq X$ is closed in $X$. So we proved that the inverse image of any closed set is closed and this is equivalent to continuity.

Mark
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  • Can you explain why the projection is a closed map due to Y being compact? – Drew May 14 '19 at 15:52
  • I was stuck just there, too. – ajotatxe May 14 '19 at 15:52
  • @ajotatxe See https://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map – YuiTo Cheng May 14 '19 at 15:59
  • @Drew This is a known theorem which was proved on this forum before. You can see a proof here: (not the accepted answer, but the answer written by hengxin for example): https://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map/673505 – Mark May 14 '19 at 16:00