Let $X$ be a topological space and $\mathcal{B}$ be a basis of the space. The Chabauty-Fell topology basis of the hypertopology on $\mathcal{C}(X)$, or whatever Miss-Hit topology this is, is comprised of collections of the form
$$ \mathcal{U}(F,\mathcal{O}):=\{ Y\in\mathcal{C}(X): Y\cap F=\emptyset, Y\cap O\neq \emptyset \; \text{for all} \; O\in \mathcal{O} \} $$
where $F\subseteq X$ is closed and $\mathcal{O}$ is a finite collection of open subsets in $X$. My question is whether it is enough to consider only collections of the form
$$ \mathcal{U}(X\setminus B,\mathcal{E}):=\{ Y\in\mathcal{C}(X): Y\cap (X\setminus B)=\emptyset, Y\cap B_i\neq \emptyset \; \text{for all} \; 1\leq i\leq k \} \tag{*} $$
where $B\in \mathcal{B}$ and $\mathcal{E}=\{ B_i \}_{i=1}^k \subseteq \mathcal{B}$. I suspect that this is indeed the case, but I was wondering whether sets of the form (*) also constitute a basis for the Chabauty-Fell topology. I know that substituting $\mathcal{O}$ with $\mathcal{E}$ works, but I\m not sure about discussing complements of base elements instead of general closed sets.
I have a feeling this should work, but my attempts thus far have not really yielded anything.
Best Answer
See the definition of the Vietoris topology from my comment above:
Your sets $\mathcal{U}(F,\mathcal{O})$ are just special cases of Vietoris basic sets and conversely all Vietoris basic sets can be written as $\mathcal{U}(X\setminus \bigcup_{i=1}^n O_i, \{O_1, \ldots, O_n\})$.
Your idea about creating a smaller base by using just finite subfamilies from the base and their complements (as first component) does not quite work:
Example: $\mathcal{C}(\Bbb N)$, where $\Bbb N$ has its standard discrete topology, has base $\mathcal{B}$ consisting of all singletons $\{n\}, n \in \Bbb N$. We do not get the Vietoris topology when we restrict your $\mathcal{U}$-base sets as you suggest. This cannot be as $\mathcal{C}(\Bbb N)$ is well-known to be non-$T_4$ but $T_3$ (see the exercises on hyperspaces in Engelking) while a second countable $T_3$ space that is second countable would be normal, even metrisable. So the restricted basic sets (which form a countable base for some topology on $\mathcal{C}(\Bbb N)$) generate a different topology altogether.