$\{K∈K(X):K⊆U\}$ for $U$ open in $X$ generates $\textbf{B}(K(X))$

Question

Let $X$ be a Polish space.

The family of set

$(i)$ $\{K∈K(X):K⊆U\}$

$(ii)$ $\{K∈K(X):K∩U≠∅\}$

for $U$ open in $X$ generates $\textbf{B}(K(X))$ where $K(X)$ is the space of all compact subsets of $X$.?

Any contribution'll be very grateful.

Answer 1

By the definition of the Vietoris topology the sets from (i) and (ii) form a subbase for the topology of the hyperspace $H(X)$ of $X$ (all non-empty closed subsets of $X$), and so also for its subspace $K(X)$. IIRC the Hausdorff metric $d_H$ on $K(X)$ (not generally defined on $H(X)$) generates this topology as well. I suppose you already know this; $(K(X), d_H)$ is a Polish space whenever $X$ is. E.g. the Kechris textbook covers this.

As $X$ is second countable (being Polish), the base from this subbase generates the topology as a $\sigma$-algebra as well, so this also holds for the subbase. So we know that the $\sigma$-algebra generated by the subbase is indeed the same as the $\sigma$-algebra generated by the topology, i.e. the Borel sets.