At least on a continuum, the binary operations of intersection and union are Vietoris-continuous. But the Vietoris topology only applies the the collection of NONEMPTY closed subsets, and this means our lattice would lack a 0 element.
Is there any meaningful topology for the collection of ALL closed subsets of a space, such that if forms a topological lattice under the Boolean set operations? What restrictions must be made on the space under consideration, in order to allow this?
Let $(X,\tau)$ be a topological space. Then you can topologize the powerset $\mathcal{P}(X)$ with the topology $\tau_{\mathcal{P}}$ generated by the subbasis
$\{\mathcal{S} \subseteq \mathcal{P}(X): (\bigcup \mathcal{S}) \in \tau\}$.
Then the operations $\cap, \cup$ are continous operations on $(\mathcal{P}(X), \tau_{\mathcal{P}})$, and of course these operations stay continuous if you restrict yourself to the lattice of all closed subsets of $X$ (including $\emptyset$).
One does see (in some of the literature) the empty set as an element of the hyperspace $H(X)$, and in that case the $\emptyset$ is an isolated point when we use the Vietoris topology. This is in many cases unwanted: for non-empty closed sets we have (assuming $X$ is $T_1$) that $X$ connected iff $H(X)$ connected, but if $H(X)$ always has the isolated point, we'd have to modify that a bit to $H(X) \setminus \{\emptyset\}$ is connected, etc.
But $\emptyset$ being isolated might spoil continuity of $\cap$, but I don't think it does (for normal spaces at least) : if $(A,B)$ is a pair of closed sets with $A \cap B = \emptyset$, then we might have that $A = \emptyset$ or $B = \emptyset$, but then $\{\emptyset\} \times H(X)$, or $H(X) \times \{\emptyset\}$ resp. is then a neighbourhood of $(A,B)$ in $H(X) \times H(X)$ that maps into $\{\emptyset\}$, and otherwise $A,B$ are non-empty closed and disjoint sets, so in a normal space $X$ we find disjoint open sets $U \supseteq A$, $V \supseteq B$. And then $\langle U \rangle \times \langle V \rangle$ is a basic product neighbourhood (where $\langle O \rangle = \{F \in H(X): F \subseteq O, F \cap O \neq \emptyset \}$ is the usual Vietoris open set) in $H(X) \times H(X)$ that also maps into $\{\emptyset\}$. So at all new pairs (compared to the non-empty situation) the $\cap$ operation is also continuous.
