# Question about the equivalence of three versions of Closed Graph Theorem

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.

• 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 $\{x \mid \Gamma(x) \neq \emptyset\}$). Jul 3, 2023 at 23:19
• 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.
– daw
Jul 4, 2023 at 7:29
• @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? Jul 4, 2023 at 14:17
• @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? Jul 4, 2023 at 14:30
• 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.
– daw
Jul 4, 2023 at 18:22