TITLE: Convergence of Sets? (Topology on a Powerset of a Set?)
QUESTION [8 upvotes]: Given a sequence of sets, is there some well-defined notion of a limit of a set? 
In other words, given some universe set $U$, I am wondering if there is a topology on $2^U$ (the powerset of $U$) such that the usual intersection and the union limits converge in that topology.
As an explicit example, let $U=\mathbb{N}$, 
$S_n = \{x\in \mathbb{N} | n< x \le 2n \}$, 
 $T_n = \{n\}$.
The limit of both sequences above should be the empty set by the following argument:
\begin{align}
S_n &\subset (n,\infty) \\\\
\lim_{n\to\infty} S_n &\subset \lim_{n\to\infty} (n,\infty) = \cap_{n\in\mathbb{N}} (n,\infty) = \emptyset 
\end{align}
(I'm not sure how to justify passing a set inclusion to the limit.)

REPLY [12 votes]: The natural topology on $2^U$ is the compact-open topology, which here is the product topology.  This is precisely the topology of pointwise convergence of indicator functions $U \to 2$.  Thus a sequence $S_1, S_2, ...$ of sets converges in this topology if and only if, for every $u \in U$, either all but finitely many $S_i$ contain $u$ (so that $u$ is in the limit set) or all but finitely many $S_i$ do not contain $u$ (so that $u$ is not in the limit set).  So both of the sequences you describe have limit the empty set as desired.
Equivalently (I think), one can define a sequence of sets to converge if its liminf and limsup (defined in the usual way) converge to the same set.