I have a question to solve that says.
Let $F \colon X \rightarrow Y$ be a correspondence and $a \in X$. Determine whether $F$ is lower semi continuous at $a$ if $\bar{F}$ (closure of $F$) is lower semi-continuous at $a$.
Now I know for sure $F$ being l.s.c. at $a$ implies that $\bar{F}$ is l.s.c. at $a$ too. However I believe that the converse is not true since the intersection with the open set $G$ in $Y$ $x \in \bar{F}(a) \cap G$ could be at the $\delta F(a)$ and $F(a)$ can be an open set. To find a counter-example I tried this.
Let $X = Y = \mathbb{R}$, and define $F\colon \mathbb{R} \rightarrow \mathbb{R}$ by $F(x) = [0,1]$ for $x \in \mathbb{Q}$ and $F(x) = [2,3]$ for $x \in \mathbb{R} \setminus \mathbb{Q}$.
Then $\bar{F}(x) = [0,1] \cup [2,3]$ for all $x \in \mathbb{R}$, which is a closed and hence lower semi-continuous correspondence.
Now my question is that. Would just saying that $F(x)$ is not closed suffice or do I need any further steps in here. I need a hint to show me whether I am on the right path. Thanks in advance.