# Nets and Convergence: Why directed indices?

1.) Why do we require nets to be indexed by directed sets (apart from it simply works compared to filters and topology). Is there a reason w.r.t. to the notion of convergence?

So far there are those hints:

• Notion of a good Direction
• Uniqueness of Limit

2.) Moreover, why does it make sense to say that the following net converges even though it grows arbitrarily large on one side: $$\Lambda:=\{1,2\}\times\mathbb{N}:(i,n)\leq(i',n') :\Leftrightarrow i\leq i',n\leq n'\\ x_{(1,n)} :=n, x_{(2,n)} :=0$$ Thanks in advance. Cheers, Alex

• Note that the relation you describe isn't transitive. Probably you mean $(i,n)\le(i',n')\iff i\le i'\wedge n\le n'$, which makes this set the product of the directed sets $\{0<1\}$ and $\Bbb N$. – Stefan Hamcke Jan 8 '14 at 19:31
• Your example net is "eventually in" every neighborhood of $0$, which is what we define as convergence. It does not mean "all but finitely many in", however. Your example illustrates the difference. – GEdgar Jan 8 '14 at 20:44

Nets emerges from the work of Moore and Smith to understand integrals, i.e., integrals are the great example of nets. For localizing it, imagine you want to calculate

$$\int_a^bf(x)\,dx$$

for a sufficient good function $f$. Now, we consider all pairs of $(\{X_i\}_{i<n},\{\xi_i\}_{i<n})$, where $\{X_i\}_{i<n}$ is a partition in measurable set of $[a,b]$, i.e., a family of sets such that $$\bigcup_{i<n}X_i=[a,b]$$ and such that for each $i$, we can calc the length of $X_i$, $l(X_i)$ and where $\{\xi_i\}_{i<n}$ a family of points such that for all $i$, $\xi_i\in X_i$; and for each of these pairs $\mathcal{X}=(\{X_i\}_{i<n},\{\xi_i\}_{i<n})$ we consider the sum $$I(f,\mathcal{X})=\sum_{i<n} f(\xi_i)l(X_i)$$ which is an approximation of our desired integral. The problem is that we need now a way to understand how these sums converge to the desire integral and this can only be done via directed sets and nets. In this way directed sets arise in a natural manner as the only possible way to order these partitions.

In this particular case, the usual order is that $$\mathcal{X}=(\{X_i\}_{i<n},\{\xi_i\}_{i<n})\geq \mathcal{X'}=(\{X_i'\}_{i<n'},\{\xi_i'\}_{i<n'})$$ whenever for all $i<n$ there is some $j_i<n'$ such that $X_i\subseteq X_{j_i}'$, i.e., the first partition is a refinement of the second one. And here the usual axioms of directed sets are translated in the following way:

1. Every partition is a refinement of itself.
2. If $\mathcal{X}$ is a refinement of $\mathcal{X}'$ and $\mathcal{X}'$ a refinement of $\mathcal{X}''$, then $\mathcal{X}$ is a refinement of $\mathcal{X}''$.
3. Given any two partitions $\mathcal{X}$ and $\mathcal{Y}$, there is always a partition $\mathcal{Z}_{\mathcal{X},\mathcal{Y}}$ which refines both $\mathcal{X}$ and $\mathcal{Y}$.´

If you haven't, take the dfinition of net and check that $\{I(f,\mathcal{X})\}_\mathcal{X}$ converges effectively to the integral of $f$.

In this sense we can speak that the previous partial sums converge to the integral we want to calculate because there is a sense about what means all partitions that refines the one you take. Furthermore, if we take any partition we can refine it to be a refinement of the given one, which means that we know what means we don't miss any

So, why we need directed sets? Because we need to have a sense of direction to advance so as in the case of the partitions, we have to have a clear notions of what means advancing and this is not only possible with the third axioms. If you omit it, then it is not clear that convergence has the meaning of the net approaching to the point as it would happen if it were not possible for each pair of partitions to build a partition that refines both of them. This is the essence of the direction, giving the sense that the net advance towards to its limit points.

For your example, even though $$\lim_{n\to\infty} x_{(1,n)}=\infty\text{,}$$ it has sense to say that the net converges because when you are advancing whit one fixed, even if you "arrived" to $(1,\infty)$, you have not advance enough because there are points far beyond to advance. This means that by fixing one, you are not advancing all you can to the given direction because there are pints that you will never pass beyond in this way.

Note: If any inquiry, please tell me.

• The Notion of Advance sounds reasonable.<br> A plain relation gives the Notion of Direction: $n\geq m$ Now, what we want next is it to be transitive, namely, if sth. holds for all $n\geq M_0$ and there is a $n\geq M_0$ for some $m\geq n$ then it should hold for this $m$ as well. Reflexivity is not so important since one can switch back and forth between $\geq$ and $>$ without troubles. Antisymmetry resp. Asymmetry I'm not so sure yet. Last, will we always be able to advance all members so that $m\geq N_0$ and luckily, yes we can.<br> Thank you Tonelli-Cueto, thank you very much! =D – C-Star-W-Star Jan 8 '14 at 22:34
• @Alex Antysymmetry is not in my definition. For example, in Willard's General Topology you don't have that axiom for directed sets. – Josué Tonelli-Cueto Jan 8 '14 at 22:46

To define a notion of converge of a net $Φ$ you need some sort of order on the indexing set, and it should at least be reflexive and transitive. You can think of a net as many "lines" $α ≤ α_1 ≤α_2\le...$ and $β ≤ β_1 ≤ β_2 ≤ ...$. The third condition ensures that every two lines eventually meet. If a net were defined merely as a function from an arbitrary set $S$, then this $S$ would become irrelevant due to having no structure at all and we could just look at its image $Φ(S)$. A limit of $Φ$ could then at best be defined as a point $x$ where infinitely many points of $Φ(S)$ are gathering. This, however, would not be any different from $x$ being a limit point of the set $Φ(S)$ in the space $X$. In other words, there would be no difference between a net in $X$ and just an ordinary subset of $X$.

Another reason for directed sets is that we get a sort of equivalence between nets and filters. Suppose $\Phi$ is a net over the directed set $I$. You can get a filter "Filter$(Φ)$" by taking the tails $T_j=\{Φ_i\mid i\ge j\}$ as a filter base. The condition that the intersection of $T_j$ and $T_k$ contains another $T_l$ comes just from the property that for any $j,k$ there is another $l\ge k,j$.
On the other hand, if you have a filter $\cal F$, then define the directed set $$I_\cal F=(\{(x,F)\mid x\in F\in\mathcal F\},\ (x,F)\le (y,G)\iff G\subseteq F)$$ and let Net$(\cal F)_{(x,F)}=x$. Note that the tails of Net$(\cal F)$ are precisely the $F\in\cal F$. This implies that $\text{Filter(Net($\cal F$))}=\cal F$. With a reasonable notion of "subnet" we can show that $\text{Net(Filter($Φ$))}$ and $Φ$ are subnets of each other. This means that going from a net to a filter and back (or from a filter to net and back) yields the original object.

• Thanks for the hint to subsets ;-) The explanation of the interrelation between nets filters and topology was not asked. – C-Star-W-Star Jan 8 '14 at 20:21
• But I'm not talkin of a Subset without havin a partial ordering! – C-Star-W-Star Jan 8 '14 at 20:29
• @Alex: Sorry. I think I got distracted by the answer by Alex ... wait, isn't that the same user as you? Did not notice that until now. Well, in that answer you mentioned the possibility that a net is indexed by an arbitrary set. Maybe you should have made more clear that you mean directed sets in contrast to preordered sets. – Stefan Hamcke Jan 8 '14 at 20:36
• Oh yes, what I meant was preordered index sets. Thanks @Stefan Hamcke =) – C-Star-W-Star Jan 8 '14 at 20:40

1) For convergence you obviously need some sense of direction. Sequences give a simple of direction (towards $\infty$) but it turns out this is not good enough for dealing with arbitrary topological spaces. So people generalized sequences to nets by using arbitrary directed sets. There was no a priori reason this would work for arbitrary spaces either but it turned out it does work. Filters are equally good and are equivalent to nets.

2) The subnet $y$, given by restricting to the top row, is cofinal in $x$ and it obviously converges to $0$, so $x \to 0$ as well. Informally, the net eventually goes up. More formally, for every $(a,n) \in x$, there is $(b, m) \in y$ such that $(b, m) \geq (a,n)$. Just take $(b, m) := (2, n)$. If you reverted the vertical arrows, the resulting net would go bottom and wouldn't converge.

• 1.) Yes, I know that it simply works by generalizing sequences and yes, I know nets are sufficient and necessary to describe a topology. But that doesn't answer my question. Is there a reason w.r.t. convergence?<br> Some sort of direction seems to be a good starting point.<br> So one can talk about convergence in which one chooses either to say sth holds true eventually upwards or downwards. 2.) Yes, I know that it converges, as I wrote so. But that was not my question either. Why does it make sense not does it converge. – C-Star-W-Star Jan 8 '14 at 20:11
• @Alex: 1) I'm not sure what kind of answer you expect. Convergence is a phenomenon of getting near to something. So having sense of a direction is useful. I'm not sure if you can have more intuition than this since bad things happen in pathological spaces. 2) I've explained why it converges both formally and informally. What more do you need? For nets, the concept of direction is more complex than for sequences. Here, the net goes up and right and so for convergence issues only top matters and the net is constant there. So it's eventually close to $0$. I'm not sure what more do you want. – Marek Jan 8 '14 at 21:17

Collecting everything, finally we have:

• Notion of "Direction": Plain Order Relation
• Notion of "Eventually All": Transitive Order
• Notion of "Being Expandable": Directed Order

In detail, this means that:

• First of all, we must be able to talk of a direction which gives rise to the notion of convergence and thus distinguishes sequences from sober subsets. That is, for all $U_x$ there is some $\Lambda_0$ s.t. for all $\lambda\geq \Lambda_0$ it holds $x_\lambda \in U_x$.
• Next, we'd like to be sure that whenever sth. eventually holds it won't drop out anymore so we can say we're only getting closer. That is, if $\lambda_0\geq \Lambda_0$ then for every succesor $\lambda\geq \lambda_0$ it holds as well $\lambda\geq \Lambda_0$.
• Last, we want to be able to expand everything so that it holds as well. That is, for every $\lambda$ there is some $\lambda_0\geq\lambda$ s.t. as well $\lambda_0\geq\Lambda_0$.