## Nets, Part II

Okay, let’s pick up our characterization of topologies with nets by, well, actually using them to characterize a topology. First we’re going to need yet another way of looking at the points in the closure of a set.

Here goes: a point is in if and only if every open neighborhood of has a nonempty intersection with . To see this, remember that the closure of is the complement of the interior of the complement of . And we defined the interior of the complement of as the set of points that have at least one open neighborhood completely contained in the complement of . And so the closure of is the set of points that have no open neighborhoods completely contained in the complement of . So they all touch somewhere. Cool?

Okay, so let’s kick it up to nets. The closure consists of exactly the accumulation points of nets in . Well, since we know that every accumulation point of a net is the limit of some subnet, we can equivalently say that consists of the limit points of nets in . So for every point in we need a net converging to it, and conversely we need to show that the limit of any convergent net in lands in .

First, let’s take an . Then every open neighborhood of meets in a nonempty intersection, and so we can pick an . The collection of all open neighborhoods is partially ordered by inclusion, and we’ll write if . Also, for any and we have and . Thus the open neighborhoods themselves form a directed set. The function is then a net in . And given any neighborhood of there is a neighborhood contained in . And then for any we have , so our net is eventually in . Thus we have a net which converges to .

Now let’s say is a net converging to . Then is eventually in any open neighborhood of . That is, every open neighborhood of meets in at least one point, and thus .

So for any , the closure is the collection of accumulation points of all nets in . And now we can turn this around and define a closure operator by this condition. That is, we specify for each net the collection of its accumulation points, and from these we derive a topology with this closure operator.

Let’s see that we really have a closure operator. First of all, clearly for all because we can just take constant nets. Even easier is to see that there are no nets into , and so its closure is still empty. Any accumulation point of a net in is the limit of a subnet, which we can pick to lie completely within either or , and so .

To finish, we must show that this purported closure operator is idempotent. For this, I’ll use a really nifty trick. A point in is the limit of some net , and each of the points is the limit of a net . So let’s build a new directed set by taking the disjoint union and defining the order as follows: if and for and in , then if , or if and in . Then combining the nets we get a net from this new directed set, which clearly has an accumulation point at the limit point of , and which is completely contained within . This completes the verification that is indeed a closure operator, and thus defines a topology.