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I have a function $f \colon X \rightarrow Y$, where $X,Y$ are both compact, Hausdorff spaces, and I need to prove that if $\mathcal{G}(f)$ (the graph of $f$) is closed, then $f$ is continuous.

I am aware about the Closed Graph Theorem, but I need to prove this result without using the Open Image Theorem (I think this is possible because of the extra hypotheses about the space $X$), and all the proofs I found about the CGT uses the OIT.

What I have from now on is that if $(x_{\lambda})_{\lambda \in \Lambda}$ is a net converging to $x \in X$, then the net $(f(x_{\lambda})_{\lambda \in \Lambda}$ has $f(x)$ as one of its accumulation points. How can I proceed with that?

Martin Sleziak
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Klaramun
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  • You can use the fact that for all accumulation points $y$ of $(f(x_{\lambda}))$ there is a subnet such that $f(x_{\mu}) \to y$. But you can also use that the projection to the first coordinate is a continuous bijection $\mathcal{G}(f) \to X$. – Daniel Fischer Oct 24 '15 at 11:53
  • Yeah I thought about the subnet converging there, but I need to prove that the whole net converges to $y$, and that is where I am stuck. Nevertheless, I will try thinking about the projection map, thank you. – Klaramun Oct 24 '15 at 11:56
  • A net $(p_{\lambda})$ converges to $q$ if and only if every subnet $(p_{\mu})$ has a further subnet $(p_{\nu})$ such that $p_{\nu} \to q$. – Daniel Fischer Oct 24 '15 at 11:58
  • So tell me if I am correct about that: if in a Hausdorff space we have a net which has $x$ as an accumulation point, then the whole net converges to $x$? If that is not true, I cannot see the relation about what you are telling and finishing the proof. – Klaramun Oct 24 '15 at 12:02
  • No, that's not correct. There are nets in Hausdorff spaces with more than one accumulation point. The relation between what I'm telling you and the proof is in the penultimate sentence of your question. Try to make a connection between that and my comments. – Daniel Fischer Oct 24 '15 at 12:05

1 Answers1

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Let $\pi_X: X \times Y \to X$ be the continuous projection.

Then let $C$ be closed in $Y$. Then $$f^{-1}[C] = \pi_X[\mathcal{G}(f) \cap (X \times C)]$$

which is closed in $X$ as the continuous image of the compact set $\mathcal{G}(f) \cap (X \times C)$. So $f$ is continuous (inverse image of a closed set is closed).

Henno Brandsma
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