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I had to examine the closed graph theorem under the following circumstances:

$X, Y$ metric spaces with $Y$ compact. Does the theorem also hold if Y is not compact?
(Assuming compactness in the context of metric spaces means complete and bounded. → link)

I have an idea of how to proceed (draft below), but I cannot visualize—and hence incorporate into the proof—why $Y$ would need to be compact, but not $X$.

For example, $f:\mathbb R \to \mathbb R, x \mapsto x$, is continuous, while $\mathbb R$ is not compact. What am I missing?

$(\Rightarrow)$ by contradiction:

Assume $f$ is continuous, but $G(f)$ is not closed. Then $$\exists (x_n,y_n)_n\in G(f): \lim_{n \to \infty} (x_n, y_n)\not\in G(f).$$ But since the graph only consists of tuples where $y_n = f(x_n)$ and $f$ is continuous that leads to a contradiction.

$(\Leftarrow)$ worked out in a similar fashion: If there is a sequence $x_n \in X$ with limit $c \in X$ such that the limit of $f(x_n)\not = f(c)$, then the graph would have at least one point—$(c,f(c))$—afflickting the graph's closedness.

355durch113
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  • In the context of metric spaces, being complete and bounded is not enough for $X$ to be compact. Take $B[0,1]\subset \mathcal{l}^2$ The closed ball of center $0$ and radius $1$ in the hilbert space $\mathcal{l}^2$ is complete and bounded but it is not compact. However the result holds if $X$ is complete and totally bounded – Joaquin Liniado Jun 04 '15 at 20:40
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    Your proof for $\impliedby$ does not actually show $G(f)$ is not closed. $G(f)$ may still be closed with this point not being continuous. For example, $f(x)=1/x$ is a function not continuous at $x=0$ but, $G(f)$ is closed. The example given was not linear but you did not use that assumption in your proof. Hope this helps, goodluck! – Eoin Jun 04 '15 at 21:30
  • @Eoin Wouldn't 1/x be continuous since it's not defined at x=0? But one could still let f equal some arbitrary c for x=0. And thank you for your input! – 355durch113 Jun 05 '15 at 15:01
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    I’m voting to close this question because there are multipile closed graph theorems with different hypothesis. Since OP did not specify what are the assumptions on codomain/domain, the question does not make any sense. – Clemens Bartholdy Sep 02 '24 at 08:01

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In fact, this theorem holds if both $X,Y$ are F-spaces, or metric spaces with a complete invariant metric.

The reason why $Y$ is assumed compact is because this is a necessary condition for every projection $\pi_1:X\times Y\rightarrow X$ to be a closed map. This is discussed in the post here. However, this claim can be omitted using the fact that $f$ is a linear function.

To prove this using only the fact $X,Y$ are F-spaces and $f:X\rightarrow Y$ linear, and $G(f)$ closed:

Since $f$ is linear, $G(f)$ is a subspace of $X\times Y$ under the metric $d((x_1,y_1),(x_2,y_2))=d_X(x_1,x_2) + d_Y(y_1,y_2)$. Since $G(f)$ is assumed closed, it is also an F-space. Take the mappings $\pi_1:G\rightarrow X$ and $\pi_2:X\times Y\rightarrow Y$ by $\pi_1(x,f(x))=x$ and $\pi_2(x,y)=y$.

The map $\pi_1$ is an injective continuous linear mapping of $G$ onto $X$. So it has an inverse $\pi_1^{-1}$ which is continuous by the open mapping theorem.

Also, $f=\pi_2\circ \pi_1^{-1}$ and $\pi_2$ is continuous. So $f$ is continuous.

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