4
$\begingroup$

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 \begin{equation} x \in U(x_0) \Longrightarrow \Gamma(x) \cap G \neq \phi. \end{equation} 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 \begin{equation} x \in U(x_0) \Longrightarrow \Gamma(x) \subset G. \end{equation} 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.

$\endgroup$
4
  • $\begingroup$ 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 $\operatorname{Gr}$) $\endgroup$
    – FShrike
    Jul 4, 2023 at 10:12
  • $\begingroup$ And \emptyset works for $\emptyset$, rather than $\phi$ $\endgroup$
    – FShrike
    Jul 4, 2023 at 10:43
  • 1
    $\begingroup$ 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 $A$ $\endgroup$
    – FShrike
    Jul 4, 2023 at 11:42
  • $\begingroup$ @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! $\endgroup$
    – Beerus
    Jul 4, 2023 at 14:07

1 Answer 1

2
$\begingroup$

$\newcommand{\gr}{\operatorname{Gr}}$ It is not true in general that $\gr(\Gamma)$ is path-connected. Consider: $$\Gamma:[0,1]\ni x\mapsto\begin{cases}[-1,1]&x=0\\\{\sin x^{-1}\}&0<x<1\end{cases}\subseteq\Bbb R$$Then $\gr(\Gamma)$ 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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .