# Convergence of Sets? (Topology on a Powerset of a Set?)

Given a sequence of sets, is there some well-defined notion of a limit of a set?

That is, given some universe set $U$, is there topology on $2^U$ (powersets of $U$) such that the usual intersection and the union limits converge to what they normally converge to in that topology?

For instance, if I were to take the following sequence of sets for $U=\mathbb{N}$,

$S_n = \{x\in \mathbb{N} | n< x \le 2n \}$,

or even simpler, something like:

$T_n = \{x\in \mathbb{N} | n\le x < n+1 \} =\{n\}$.

I'm wondering whether there is a notion of convergence that can say whether the limit of such a sequence makes sense or not.

On one hand, it feels like the limit of both sequences above should be the empty set, by the following type of 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}

And of course, the same for $T_n$.

On the other hand, I don't see why I should be able to pass a set inclusion to the limit. (I feel like I'm just declaring that this should be a property of limits of sets...)

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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.
1) Yes; 2) You can always take the discrete topology on $U$. – Asaf Karagila Jan 24 '11 at 19:17
@Braindead: The set $2=\{0,1\}$ has the discrete topology; the set $2^U$ is naturally bijectable with a direct product of copies of $2$ indexed by $U$, and the product has a natural topology (the product topology) since it is a product of topological spaces. As Qiaochu is noting, if you do this, then it turns out to coincide with the compact-open topology if you give $U$ the discrete topology (which makes all maps $U\to 2$ continuous). – Arturo Magidin Jan 24 '11 at 19:25
@Braindead: If you choose the indiscrete topology on $U$ instead, then the only continuous maps are the constant maps; but to identify $2^U$ with $\mathcal{P}(U)$ you need all maps, not just the constant ones. So if you place that topology on $U$, you are essentially just looking at $\emptyset$ and $U$ instead of all of $\mathcal{P}(U)$; similarly if you place other topologies: you are restricting yourself to a subset of $\mathcal{P}(U)$. The discrete topology is the only one that gives you all of $\mathcal{P}(U)$. – Arturo Magidin Jan 24 '11 at 19:28
@Braindead: Not at all: for the compact-open topology to be defined, you need a topological structure on both sets. But if you have "all functions from $X$ to $Y$", and $Y$ is a topological space, then you can take the product topology on $Y^X$, which happens to coincide with the compact-open topology if you give $X$ the discrete topology. It's not a silly question, and the coincidence of the two topologies is something you may want to prove for yourself, just to make sure and get more comfortable with both. – Arturo Magidin Jan 24 '11 at 19:44