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Specifically, the closed graph theorem I am referring to is:

Let $f : X \rightarrow Y$ exist and $Y$ be compact and Hausdorff. Then $f$ is continuous if and only if the graph of $f$ denoted by $G_f = \{(x,y) | f(x)=y\}$ is closed in $X \times Y$

More specifically, I am attempting to prove the "right-to-left" part of the implication which only needs compactness:

If $G_f$ is closed in $X \times Y$ and $Y$ is compact, then $f$ is continuous.

I appear to have a "proof" here that doesn't invoke compactness on Y, so what about this is invalid?

Let $x$ be arbitrary and let $V$ be an open set in $Y$ containing $f(x)$. let $U = f^{-1}(V)$.

Pick a point $(x, y)$ in $X \times Y$ such that $x \in U$ and $y \in V$ BUT $f(x) \neq y$. Note that $(x, y) \notin G_f$. Since $G_f$ is closed, it's complement is open and we can produce an open neighborhood around $(x, y)$ denoted $N=U'\times V'$ which also does not intersect $G_f$.

$U'$ is an open subset of $U$ and furthermore an open neighborhood of $x$ such that $f(U)\subset V$. Since this construction holds for any arbitrary subset of $Y$, $f$ must be continuous.

I don't appear to invoke compactness, but I'm missing what's wrong here.

cemulate
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    Your set $N$ is open in $X \times Y$, but that doesn't necessarily mean that $N = U \times V$. For example, in $\mathbb{R} \times \mathbb{R}$, the complement of the unit disk is open, but cannot be written in this form. – Thomas Credeur Apr 10 '14 at 23:03
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    Why is $U^\prime$ a subset of $U$? (Thomas: I don't think that's a big problem. $N$ need not be a product of open sets, but certainly contains one). – Ben Millwood Apr 10 '14 at 23:04
  • I was just making a comment. My topology professor made sure to drill this into our heads so much that I felt like sharing. – Thomas Credeur Apr 10 '14 at 23:05
  • Sure, it's worth remembering :) – Ben Millwood Apr 10 '14 at 23:07
  • Thanks; I see now there's nothing guaranteeing that $N$ can be be written as $U' \times V'$ with $U'$ definitely a subset of $U$. Can I fix this approach or am I off altogether? I could use some hints going forward because I'm pretty stumped. – cemulate Apr 11 '14 at 00:41
  • You want to show $U$ is open for continuity, but you do no such thing. – Henno Brandsma Jun 03 '18 at 08:57

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Let $p: X \times Y \to X$ be the continuous projection. It is well-known that $p$ is a closed map iff $Y$ is compact.

Then if $G_f$ is closed and $C \subseteq Y$ is closed, by the definitions we can check that

$$f^{-1}[C] = p[G_f \cap (X \times C)]$$

and note that $X \times C$ is closed in $X \times Y$ and as $G_f$ is closed too, we see that $f^{-1}[C]$ is the image of a closed set under $p$, and so $f^{-1}[C]$ is closed. This implies that $f$ is continuous.

Henno Brandsma
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