I understand (at least I think I understand) the notion behind these topics.
In my class we use the following criterion:
A correspondence $\Gamma: X \rightrightarrows Y$ is LHC at $x$ if whenever $x$ is in the lower inverse of an open set $E \subset Y$, there is an open set $F$ containing $x$ that satisfies $F \subset \Gamma^l[E]$
and
A correspondence $\Gamma: X \rightrightarrows Y$ is UHC at $x$ if whenever $x$ is in the upper-inverse of an open set $E \subset Y$, then there is an open set $F$ containing $X$ such that $F \subset \Gamma^U[E]$
Where the upper inverse and lower inverse are defined as:
$\Gamma^U[E]$ is the $\textbf{Upper inverse}$ of $E$, defined as:
$$\Gamma^U[E] = \{ x \in X | \Gamma(x) \subset E\}$$
and
$\textbf{The lower inverse}$ of $E \subset Y$, denoted by $\Gamma^l[E]$ is the set:
$$\Gamma^l[E] = \{ x \in X | \Gamma(x) \cap E \neq 0\} $$
But I can never score full points on a question, and can't seem to find any rigorous proofs, only pictures and one sentence arguments.
For instance, consider the two correspondences:
\begin{equation*}
\Gamma(x) = \begin{cases}
[0, \frac{1}{x}] & \text{if } x > 0 \\
\{0\} & \text{if } x = 0
\end{cases}
\end{equation*}
Or: $\phi: \mathbb{R}_+ \rightrightarrows \mathbb{R}$ defined as:
$$\phi(x) = \{x^{0.5}, -x^{0.5}\} $$
Yes, looking at the graph I can immediately tell that $\Gamma(x)$ is not UHC at $x=0$, and is LHC at $x=0$, and $\phi(x)$ is UHC LHC everywhere, but I just don't know how to rigorously prove it. How would you go about the proof, using the above definitions, to a blind person who had a text-to-voice translator (but obviously couldn't see the graph)?
Best Answer
Well, you just go through the definitions step by step. Let's prove $\Gamma$ is not UHC at $x=0$. Negating the definition of UHC, this means we must find an open set $E$ such that $0\in \Gamma^U(E)$, but there is no open set $F$ such that $0\in F$ and $F\subseteq\Gamma^U(E)$. In order to get $0\in\Gamma^U(E)$, we need $\Gamma(0)=\{0\}$ to be a subset of $E$, so let's just pick $E=(-1,1)$, say.
Now we need to prove this choice of $E$ works, so let $F\subseteq\mathbb{R}_+$ be any open set containing $0$. We wish to show $F\not\subseteq\Gamma^U(E)$, so we want to find an element $x\in F$ that is not in $\Gamma^U(E)$, in other words such that $\Gamma(x)\not\subseteq (-1,1)$. Looking back at our definition of $\Gamma$, this just means we need $x\neq 0$ and $1/x\geq 1$, or $x\leq 1$. But since $F$ is open and contains $0$, it contains an interval $[0,\epsilon]$ for some $\epsilon>0$, and then $x=\min(1,\epsilon)$ is an element of $F$ that works.
Let's now show $\Gamma$ is LHC at $0$. By the definition, this means we start with an arbitrary open set $E$ such that $0\in \Gamma^l(E)$, and wish to find an open set $F$ containing $0$ such that $F\subseteq\Gamma^l(E)$. What does $0\in\Gamma^l(E)$ tell us? It means that $\Gamma(0)\cap E\neq \emptyset$, or just that $0\in E$ since $\Gamma(0)=\{0\}$.
What does the desired condition $F\subseteq\Gamma^l(E)$ mean? It means for each $x\in F$, $\Gamma(x)\cap E\neq\emptyset$. Aha, but we know that $0\in E$, and looking back at our definition of $\Gamma$, we see that actually $0\in \Gamma(x)$ for all $x$! So $\Gamma(x)\cap E\neq \emptyset$ is true for all $x$, and so we can take $F$ to be any open set containing $0$ at all.