I consider $X$ a metric space and $F_1,F_2$ two disjoint subsets of $X$. Let $T : X\rightrightarrows\mathbb{R}$ be a multi valued function defined by :
$T(x) =\{0\}$ on $F_1$
$T(x)=\{1\}$ on $F_2$
$T(x) = (0,1)$ on $(F_{1}\cup F_2)^c$
I would like to prove or disprove that under the assumption that $F_1$ and $F_2$ are closed $T$ is lower semi continuity.
Since $T$ is not compact valued we will use the definition of lower semi continuity in terms of neighborhoods.
Let $U$ be an open set of $\mathbb{R}$ and $x$ such that $T(x)\cap U\neq\emptyset$.
First case : $x\in F_1$ so $T(x)=\{0\}$ which implies that $0\subset U$, $U$ being open it means there exists $\delta>0$ such that $B(0,\delta)\subset U$.
From this, we observe that if $(0,1)\subset U$ we have no problem to conclude that there exists a neighborhood $V_x$ of $x$ such that $T(x’)\cap U\neq\emptyset$ for all $x’\in V_x$. However, for an open set as $B= (-\frac{1}{4}, \frac{1}{4})$ we will not be able to find such a neighborhood even if $0\in B$ (for example the case of $x\in\partial F_1$).
This disproves that $T$ is lower semi continuous.
I would like to know if it is correct please and if it is the case, under what supplementary assumptions could we get the lower semi continuous please.
Thank you