# Prove the (path-) connectedness of the graph of a compact- and convex-valued upper hemi-continuous correspondence.

Let $$\Gamma: [0, 1] \to \mathbb{R}$$ be a compact- and convex-valued, upper hemi-continuous correspondence. Prove that the graph of $$\Gamma$$ is a connected set. Is it path-connected?

This is what I have so far:

Proof $$\space$$ We first show that $$Gr(\Gamma)$$ is connected. Assume to the contrary that $$Gr(\Gamma)$$ is not connected. Then $$Gr(\Gamma) = A \cup B$$ where $$\overline{A} \cap B = \phi$$ and $$A \cap \overline{B} = \phi$$, $$A \neq \phi$$, and $$B \neq \phi$$. Since $$\Gamma$$ is compact- and convex-valued, $$\Gamma(x)$$ us a closed interval or a singleton for all $$x \in [0, 1]$$. Since $$\Gamma$$ is upper hemi-continuous, $$Gr(\Gamma)$$ is closed. Let $$(x, y)$$ be a limit point of $$Gr(\Gamma)$$. Without loss of generality, assume $$(x, y) \in A$$, where $$y \in \Gamma(x)$$. Then $$(x, y) \notin \overline{B}$$. Take a sequence $$\{x_n\}$$ such that $$x_n \to x$$. Then there exists a subsequence $${x_{n_k}}$$ such that $$(x_{n_k}, y_{n_k}) \in \overline{B}$$, $$y_{n_k} \in \Gamma(x_{n_k})$$, and $$x_{n_k} \to x$$. However, no subsequence of $$\{y_{n_k}\}$$ can converge to $$y$$, for otherwise $$(x, y)$$ would be in $$\overline{B}$$. This contradicts the sequential characterization of upper hemi-continuous correspondence. Therefore, $$Gr(\Gamma)$$ is connected.

I would really appreciate it if someone could check if my proof of the connectedness is correct.

In addition, I have difficulties to determine whether $$Gr(\Gamma)$$ is path-connected or not. I tried to use the $$Gr(\Gamma)$$ = Deleted Comb Space as a counterexample, but it turns out that $$Gr(\Gamma)$$ is not closed, because its closure is the Comb Space. By the closed-graph theorem, if $$\Gamma$$ is upper hemi-continuous, $$Gr(\Gamma)$$ would be closed. Hence $$\Gamma$$ is not upper hemi-continuous, a contradiction.

This is a question about set-valued analysis and topological spaces. I would like to provide some related definitions and results first, because they are usually not covered in a standard math course in analysis or topology:

Definition $$\space$$ Let $$X$$ and $$Y$$ be two sets. If with each element $$x$$ of $$X$$ we associate a subset $$\Gamma(x)$$ of $$Y$$, we say that the correspondence $$x \to \Gamma(x)$$ is a mapping of $$X$$ into $$Y$$; the set $$\Gamma(x)$$ is called the image of $$x$$ under the mapping $$\Gamma$$.

Definition $$\space$$ Let $$\Gamma$$ be a mapping of a topological space $$X$$ into a topological space $$Y$$, and let $$x_0$$ be any point of $$X$$. We say that $$\Gamma$$ is lower hemi-continuous at $$x_0$$ if for each open set $$G$$ meeting $$\Gamma(x_0)$$, there is a neighborhood $$U(x_0)$$ such that $$$$x \in U(x_0) \Longrightarrow \Gamma(x) \cap G \neq \phi.$$$$ We say that $$\Gamma$$ is upper hemi-continuous at $$x_0$$ if for each open set $$G$$ containing $$\Gamma(x_0)$$ there exists a neighborhood $$U(x_0)$$ such that $$$$x \in U(x_0) \Longrightarrow \Gamma(x) \subset G.$$$$ We say that the mapping $$\Gamma$$ is continuous at $$x_0$$ if it is both lower and upper hemi-continuous at $$x_0$$.

Definition $$\space$$ We say that $$\Gamma$$ is lower hemi-continuous in $$X$$ (l.h.c. in $$X$$) if it is lower hemi-continuous at each point of $$X$$. We say that $$\Gamma$$ is upper hemi-continuous in $$X$$ (u.h.c. in $$X$$) if it is upper hemi-continuous at each point of $$X$$ and if, also, $$\Gamma(x)$$ is a compact set for each $$x$$. If $$\Gamma$$ is both lower hemi-continuous in $$X$$ and upper hemi-continuous in $$X$$, then it will be called continuous in $$X$$.

Definition $$\space$$ The correspondence $$\Gamma$$ is closed-valued if for each $$x \in X$$, $$\Gamma(x)$$ is closed in $$Y$$.

Definition $$\space$$ The correspondence $$\Gamma$$ is compact-valued if for each $$x \in X$$, $$\Gamma(x)$$ is compact in $$Y$$.

Definition $$\space$$ The correspondence $$\Gamma$$ is convex-valued if for each $$x \in X$$, $$\Gamma(x)$$ is a convex set in $$Y$$.

Definition $$\space$$ The graph of the correspondence $$\Gamma$$ is the set $$Gr(\Gamma) = \{(x, y) \in X \times Y | y \in \Gamma(x)\}$$.

Lemma $$\space$$ [Sequential characterization of lower hemi-continuous] A correspondence $$\Gamma: X \to Y$$ is lower hemi-continuous at $$x_0 \in X$$ if and only if, for any sequence $$\{x_m\}$$ converges to $$x_0$$ and any $$y \in \Gamma(x_0)$$, there exists a sequence $$\{y_m\}$$ converges to $$y$$ such that $$y_m \in \Gamma(x_m)$$ for all $$x_m$$.

Lemma $$\space$$ [Sequential characterization of upper hemi-continuous] Let $$\Gamma: X \to Y$$ be a correspondence. If, for every sequence $$\{x_m\}$$ in $$X$$ that converges to $$x_0 \in X$$ and for every sequence $$\{y_m\}$$ such that $$y_m \in \Gamma(x_m)$$, there exists a subsequence of $$\{y_m\}$$ that converges to a point in $$\Gamma(x_0)$$, then $$\Gamma$$ is upper hemi-continuous at $$x_0$$. If $$\Gamma$$ is compact-valued, then the converse is true.

Theorem $$\space$$ [Closed graph theorem] Let $$\Gamma: X \to Y$$ be a correspondence. If $$\Gamma$$ is closed-valued and upper hemi-continuous, then $$Gr(\Gamma)$$ is closed.

• Btw it might be nice to put \newcommand{\gr}{\operatorname{Gr}} at the start and then whenever you write \gr later in the post you get a nicer render (of $\operatorname{Gr}$) Jul 4, 2023 at 10:12
• And \emptyset works for $\emptyset$, rather than $\phi$ Jul 4, 2023 at 10:43
• Your proof the graph is connected doesn't make sense to me. It seems unjustified that there are $(x_{n_k})_k\in\overline{B}$ convergent to $x$, e.g. this is false if $(x,y)$ is an interior point of $A$ Jul 4, 2023 at 11:42
• @FShrike That makes sense. Thank you! But could you please help me fix it, coz I do have some difficulties figuring out how to construct a contradiction in that case. I really appreciate it! Jul 4, 2023 at 14:07

$$\newcommand{\gr}{\operatorname{Gr}}$$ It is not true in general that $$\gr(\Gamma)$$ is path-connected. Consider: $$\Gamma:[0,1]\ni x\mapsto\begin{cases}[-1,1]&x=0\\\{\sin x^{-1}\}&0Then $$\gr(\Gamma)$$ is the closed topologist's sine curve, which is well known to be connected but not path connected (nor locally connected nor locally path connected). We should also check $$\Gamma$$ is complex and convex valued as well as upper hemicontinuous, but this is nearly obvious.

So, why should $$\gr(\Gamma)$$ be connected? As commented, I don't like your proof since the convergent sequence $$(x_{n_k})_{k\in\Bbb N}$$ comes from nowhere. Firstly let's note that:

$$\Gamma:[0,1]\to\Bbb R$$ is compact and convex valued

Is a roundabout way of saying: $$\Gamma(x)$$ is a closed and bounded interval for every $$x$$. By a simple compactness argument (utilising upper hemicontinuity) we can conclude $$\gr(\Gamma)$$ is bounded. Using this closed graph theorem, $$\gr(\Gamma)$$ is then compact.

Suppose there exists a disconnection of $$\gr(\Gamma)$$ into $$A$$ and $$B$$. It follows that $$A,B$$ are both compact. Let $$\pi:[0,1]\times\Bbb R\to[0,1]$$ be the projection: then, $$\pi A$$ and $$\pi B$$ are both compact hence closed. Evidently $$\pi A\cup\pi B=[0,1]$$ with neither set being empty. Note that $$\Gamma(x)$$ is connected for all $$x$$, so $$\{x\}\times\Gamma(x)$$ must be fully contained in either $$A$$ or $$B$$ for any $$x$$. It follows that $$\pi A\cap\pi B=\emptyset$$. We conclude that $$\pi A$$ and $$\pi B$$ form a disconnection of $$[0,1]$$, contradicting the fact that $$[0,1]$$ is connected.

Therefore no $$A,B$$ can exist; therefore $$\gr(\Gamma)$$ is connected.

Generalising, if $$\Gamma:X\to Y$$ is connected-valued and $$X$$ is a connected Hausdorff space, then if $$\gr(\Gamma)$$ is compact we get to conclude $$\gr(\Gamma)$$ is connected.