Is the Vietoris topology a semigroup topology with the union action?

Question

I read a question from users about the Vietoris topology, but now I was faced with the problem.

Let $X$ be a topological space and $\operatorname{H}(X)$ be the set of all closed non-empty subsets of $X$.

If $U, V_{1}, V_{2}\ldots V_{n}$ are non-empty open subsets in $X$, define:

$$ \langle U , V_{1}, V_{2}, \ldots V_{n} \rangle = \{ F \in \operatorname{Exp}(X)\mid F\subseteq U, (\forall 1 \leq i \leq n)( F\cap V_{i} \neq \emptyset )\}.$$

The family $B$ consisting of all sets of the form $\langle U, V_{1}, V_{2}, ......V_{n} \rangle$ is a basis for a topology on $H(X)$, and this topology is called the Vietoris topology.

Question: Is this space a semigroup topology with the union action?

Answer 1

It seems like yes to me. Suppose we have two points $F_1,F_2 \in H(X)$ and we want to check continuity at $(F_1, F_2)$ of $f: H(X) \times H(X) \rightarrow H(X): f(A,B) = A \cup B$.

So let $O = \langle U; V_1, \dots V_n \rangle$ be a basic neighbourhood of $F_1 \cup F_2$, in particular this means that $(F_1 \cup F_2) \cap V_i \neq \emptyset$ for all $i$. This intersection must lie in $F_1$ or $F_2$, so if we define $I_1 = \{i \in \{1,\ldots,n\}: F_1 \cap V_i \neq \emptyset\}$ and $I_2 = \{i \in \{1,\ldots,n\}: F_2 \cap V_i \neq \emptyset\}$ we know that $I_1 \cup I_2 = \{1, \ldots n\}$. Also $F_i \subset F_1 \cup F_2 \subset U$ for $i=1,2$.

Now define $O_1 = \langle

So we have found $O_1, O_2$ such that $(F_1,F_2) \in O_1 \times O_2$ and $f[O_1 \times O_2] \subset O$, where $O$ was any basic open neighbourhood of $f(F_1,F_2)$. So $f$ (the union) is continuous at every point of $H(X) \times H(X)$.