# 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.

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Posted by John Armstrong | Point-Set Topology, Topology

## 6 Comments	»

1. hi, i was reading about nets recently and became quite excited about how so many thing in topology become so much more natural when thinking in terms of sets(like definitons of continuity, compactness, product topology, tychonoff theorem). it would be nice to have a definition of a topological space just in terms of nets. but all the characterizations of topological spaces in terms of nets prove that the map from topological spaces to
the class {sets with information about which nets converge to which points} is an injective map. But it would a nice to know what what subset of the above class come from topological spaces,
for instance, one obvious requirement is that the set of accumulation points of a subnet is a subset of the accumulation points of the net and that isomorphic nets have
the same accumulation points…
but these conditions aren’t enough, there could be no topological structure on the set which has the same info about accumulation points of nets.
If there were some sufficient conditions though, then one get rid entirely of the open sets approach and start with nets from the begining and that would be a very nice thing.

Comment by harsha | November 20, 2007 | Reply

2. harsha, I don’t know the answer exactly. Something similar goes on here to what happened when I talked about neighborhoods. That is, any collection of filters gives a topology, but many different collections will give rise to the same topology. In effect, there’s a sort of “closure” going on.

Here instead we have the fact that no matter what accumulation points we assign to each net, we get some closure operator, but not every such collection of data arises from a closure operator. So the process of going from data to operator to data is a sort of “closure” on the class of such data.

So here’s how I’d suggest going about it: try to figure out how this process works. That is, if you start with an arbitrary collection of accumulation points for each net, use that data to define a topology, and see what accumulation points each net now has. Then determine what the fixed points of this process are. Those will be your answers.

First, though, I’d ask around your department. I know at least one venerable-mathematician-who-knows-everything sort of topologist who hangs around there you might try. If nothing else you might be able to verify that this actually hasn’t been done, and then it would make an excellent thesis problem

Comment by | November 20, 2007 | Reply

3. [...] and Subbases We’ve defined topologies by convergence of nets, by neighborhood systems, and by closure operators. In each case, we saw some additional hypothesis [...]

Pingback by | November 22, 2007 | Reply

4. I just aksed myself the same question harsha brought up here, and hence found this webpage. Is there the notion of a basis for a space’s nets? I only have a vague notion of what it should be: A basis in terms of nets is a full subcategory of the category of nets in some set X, together with their accumulation points such that the topology that arises from this subcategory agrees with the original one.
For instance, if a space is 1st countable, then instead of considering all nets and their accumulation points it suffices to take all sequences into account. There are various other examples.
Still, the difficulty is what properties needs the accumulation point map to have to give rise to a topology?

Comment by olf | March 11, 2008 | Reply

5. Olf, I really wish I knew. But when it comes down to it I’m not a point-set topologist and I just haven’t spent a lot of time on this problem. I’d love to hear if you turn up anything, though.

Comment by | March 11, 2008 | Reply

6. [...] The first thing to note about is that it’s a closed set. That is, it should contain all its accumulation points. So let be such an accumulation point and assume that is isn’t in , so . So there must exist [...]

Pingback by | December 9, 2009 | Reply

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