Closed graph implies $f$ continuous
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?
$\endgroup$ 51 Answer
$\begingroup$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).
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