Let $X,Y$ be metric spaces, $x^*\in X$
We define two multifunctions $F_1:X\rightrightarrows Y$,$F_2:X\rightrightarrows Y$.
We recall the upper-semi-continuity in Berge's sense :
A multifunction $F:X\rightrightarrows Y$ is said to be u.s.c-B in $x^*$ if $$\forall \mathcal O\subset Y: F(x^*)\subset \mathcal O, \exists \delta>0, \forall x\in X : x\in B(x^*,\delta) \implies F(x^*)\subset \mathcal O$$
Now we observe the following theorem:
Theorem :
If $F_1$ is closed in $x^*$, and $F_2$ is u.s.c-B and the set $F_2(x^*)$ is compact, then the application $F_1\cap F_2$ is u.s.c-B.
The demonstration of this theorem is clear and it relies on the finite coverage of open sets of the compact set $F_2(x^*)$.
My question is the following :
Under the weakened hypothesis :
$F_1$ is closed in $x^*$
$F_2(x^*)$ is bounded, not closed, and is u.s.c-B in $x^*$.
How can we show that $F_1\cap F_2$ is not u.s.c-B in $x^*$ ? I tried the direct approach but couldn't show the desired result.
I am looking for a clue on how initiate my proof OR an example which illustrates such case.
