The Unapologetic Mathematician

Mathematics for the interested outsider

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 x is in \mathrm{Cl}(A) if and only if every open neighborhood of x has a nonempty intersection with A. To see this, remember that the closure of A is the complement of the interior of the complement of A. And we defined the interior of the complement of A as the set of points that have at least one open neighborhood completely contained in the complement of A. And so the closure of A is the set of points that have no open neighborhoods completely contained in the complement of A. So they all touch A somewhere. Cool?

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

First, let’s take an x\in\mathrm{Cl}(A). Then every open neighborhood U of x meets A in a nonempty intersection, and so we can pick an x_U\in U\cap A. The collection of all open neighborhoods is partially ordered by inclusion, and we’ll write U\geq V if U\subseteq V. Also, for any U_1 and U_2 we have U_1\cap U_2\geq U_1 and U_1\cap U_2\geq U_2. Thus the open neighborhoods themselves form a directed set. The function U\mapsto x_U is then a net in A. And given any neighborhood N of x there is a neighborhood U contained in N. And then for any V\geq U we have x_V\in V\subseteq U\subseteq N, so our net is eventually in N. Thus we have a net which converges to x.

Now let’s say \Phi:D\rightarrow A\subseteq X is a net converging to x\in X. Then \Phi is eventually in any open neighborhood U of x. That is, every open neighborhood of x meets A in at least one point, and thus x\in\mathrm{Cl}(A).

So for any A, the closure \mathrm{Cl}(A) is the collection of accumulation points of all nets in A. And now we can turn this around and define a closure operator by this condition. That is, we specify for each net \Phi 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 U\subseteq\mathrm{Cl}(U) for all U because we can just take constant nets. Even easier is to see that there are no nets into \varnothing, and so its closure is still empty. Any accumulation point of a net in U\cup V is the limit of a subnet, which we can pick to lie completely within either U or V, and so \mathrm{Cl}(U\cup V)=\mathrm{Cl}(U)\cup\mathrm{Cl}(V).

To finish, we must show that this purported closure operator is idempotent. For this, I’ll use a really nifty trick. A point in \mathrm{Cl}(\mathrm{Cl}(U)) is the limit of some net \Phi:D\rightarrow\mathrm{Cl}(U), and each of the points \Phi(d) is the limit of a net \Phi_d:D_d\rightarrow U. So let’s build a new directed set by taking the disjoint union \biguplus\limits_{d\in D}D_d and defining the order as follows: if a\in D_c and b\in D_d for c and d in D, then a\geq b if c\geq d, or if c=d and a\geq b in D_d. Then combining the nets \Phi_d we get a net from this new directed set, which clearly has an accumulation point at the limit point of \Phi, and which is completely contained within U. This completes the verification that \mathrm{Cl} is indeed a closure operator, and thus defines a topology.

November 20, 2007 Posted by | Point-Set Topology, Topology | 6 Comments