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I am studying set-valued analysis, and I saw three version of this Closed Graph Theorem.

Version 1: (what I was taught in class)

Let $\Gamma: X \to Y$ be a correspondence. If $\Gamma$ is closed-valued and upper hemi-continuous, then $Gr(\Gamma)$ is closed.

Version 2: (from Set-Valued Analysis by Jean-Pierre Aubin, Hélène Frankowska)

Let $\Gamma: X \to Y$ be an upper hemi-continuous correspondence with closed domain and is closed-valued, then $Gr(\Gamma)$ is closed.

Version 3: (from Infinite Dimensional Analysis: A Hitchhiker's Guide by Charlambos Aliprantis and Kim C. Border)

A correspondence with compact Hausdorff range space is closed if and only if it is upper hemi-continuous and closed-valued.

I want to know if they are equivalent to each other. I'm not sure how to prove it or give a counterexample. Could someone please help me with it? Thanks so much in advance!

Here is a reference about related definitions and results that might be helpful: Prove the (path-) connectedness of the graph of a compact- and convex-valued upper hemi-continuous correspondence.

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    $\begingroup$ Looking at this post from MathOverflow, it seems like compactness of $Y$ is only required for the "only if" direction, thus without compactness the theorems are almost identical if you only consider the "if" direction for "Version $3$", apart from the closed domain requirement for "Version $2$" (where by domain they mean the set $\{x \mid \Gamma(x) \neq \emptyset\}$). $\endgroup$
    – Bruno B
    Jul 3, 2023 at 23:19
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    $\begingroup$ Let $X=Y=\mathbb R$ and $gr(\Gamma):=\{ (x,y): xy=1\}$. Then $\Gamma$ has closed values and is upper hemicontinuous, but only version 1 allows to conclude that $gr(\Gamma)$ is closed. $\endgroup$
    – daw
    Jul 4, 2023 at 7:29
  • $\begingroup$ @BrunoB That makes sense, thank you so much! I would add an answer-proof. But I am still a bit confused about the Hausdorff condition. It seems that the other two versions implicitly assumes the Hausdorff condition for any topological spaces. Is my understanding correct? $\endgroup$
    – Beerus
    Jul 4, 2023 at 14:17
  • $\begingroup$ @daw I appreciate it! So you mean the domain in your example is $\mathbb{R}$ \ $\{0\}$, and it is not closed because its complement $\{0\}$ is not open. But we still have that $Gr(\Gamma)$ is closed, for $(𝑥,𝑦) \mapsto 𝑥𝑦$ is continuous and $Gr(\Gamma)$ is the preimage of the closed set {1}, so is closed by continuity. So should we conclude that the requirement for "closed domain" is redundant or false? $\endgroup$
    – Beerus
    Jul 4, 2023 at 14:30
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    $\begingroup$ It seems that version 1 is stronger than version 2. This might depend on the assumptions on $X,Y$. No idea, why they wrote version 2 like this. $\endgroup$
    – daw
    Jul 4, 2023 at 18:22

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