# The Unapologetic Mathematician

## 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