Set Valued maps

Question

Let $X$ and $Y$ be two topological spaces $\psi: X \rightarrow Y $ be a correspondence (or) set valued map. Assume that the map $\psi$ is lower hemicontinuous at $x_{0}$. Define a correspondence $\psi: X \rightarrow Y $ by $\phi(x)=\left[\psi(x)\right]^{c}$ (i.e) complement of the correspondence $\psi$. My question: Is it true that $\psi$ is lower hemicontinuous at $x_{0}$? Is there any result (or) theorem talks about the lower hemi continuity of the complement of the correspondence map? Please help in this regard.

Answer 1

The answer is "no".

A correspondence $\psi$ is lower hemicontinuous at $x_0 \in X$ if for any open $V \subset Y$ such that $V \cap \psi(x_0) \ne \emptyset$ there exists a neighbourhood $U$ of $x_0$ such that $V \cap \psi(x) \ne \emptyset$ for all $x \in U$.

Define $\psi : \mathbb{R} \to \mathbb{R}$ by $\psi(0) = \emptyset$, $\psi(x) = \mathbb{R}$ for $t \ne 0$. Then $\psi$ is lower hemicontinuous at $0$ (simply because there does not exist any open $V \subset \mathbb{R}$ such that $V \cap \psi(0) \ne \emptyset$). On the other hand $\psi^c(0) = \mathbb{R}$, $\psi^c(t) = \emptyset$ for $t \ne 0$. $\psi^c(0)$ has nonempty intersection with any nonempty open $V \subset \mathbb{R}$, but no other $\psi^c(t)$ has a nonempty intersection with $V$. This shows that $\psi^c$ is not lower hemicontinuous at $0$.

Answer 2

Here is another example.

Let $X$ be any space having a closed non-open single-point subspace $A = \lbrace x_0 \rbrace$. Note that this implies $A \subsetneqq X$.

Define $\psi : X \to X$ by $\psi(x_0) = A$, $\psi(x) = X-A$ for $x \ne x_0$. Then $\psi$ is lower hemicontinuous at $x_0$ because each open $V \subset X$ which intersects $\psi(x_0)$ (i.e. contains $x_0$) also intersects $X-A$.

On the other hand $\psi^c(x_0) = X-A$, $\psi^c(x) = A$ for $x \ne x_0$. $\psi^c(x_0)$ has nonempty intersection with $X-A$, but no other $\psi^c(x)$ has a nonempty intersection with $X-A$. This shows that $\psi^c$ is not lower hemicontinuous at $x_0$.

Let us moreover observe

(1) All $\psi(x)$ are nomempty.

(2) All $\psi^c(x)$ are nomempty.

(3) $\psi$ is lower hemicontinuous in all $x \in X$ (in $x \ne x_0$ this follows from the fact that $X-A$ is open and $\psi$ is constant on $X-A$).