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.)


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.

  • 1
    $\begingroup$ 1) Yes; 2) You can always take the discrete topology on $U$. $\endgroup$ – Asaf Karagila Jan 24 '11 at 19:17
  • 2
    $\begingroup$ @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). $\endgroup$ – Arturo Magidin Jan 24 '11 at 19:25
  • 2
    $\begingroup$ @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 $\mathcal{P}(U)$. $\endgroup$ – Arturo Magidin Jan 24 '11 at 19:28
  • 2
    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Jan 24 '11 at 19:44
  • 2
    $\begingroup$ @Zhen: in a category with finite coproducts and a terminal object, 2 is sometimes used to denote the coproduct of the terminal object with itself. In Top, this is the 2-point space with the discrete topology. (Alternately, the forgetful functor from Top to Set has two adjoints: its left adjoint gives a set the discrete topology and its right adjoint gives a set the indiscrete topology. Which adjoint is appropriate for your purposes depends, of course, on the purposes.) $\endgroup$ – Qiaochu Yuan Jan 24 '11 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.