Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a set and let ${\mathcal N}$ be a collection of nets on $X.$

I've been told by several different people that ${\mathcal N}$ is the collection of convergent nets on $X$ with respect to some topology if and only if it satisfies some axioms. I've also been told these axioms are not very pretty.

Once or twice I've tried to figure out what these axioms might be but never came up with anything very satisfying. Of course one could just recode the usual axioms regarding open sets as statements about nets and then claim to have done the job. But, come on, that's nothing to be proud of.

Has anyone seen topology axiomatized this way? Does anyone remember the rules?

share|improve this question
2  
Axiomatizing via filters is a much more fruitful endeavor. For that I suggest Bourbaki's "General Topology". Filters have a number of really nice applications to logic, set theory, and even category theory (filtered limits and colimits!). –  Harry Gindi Mar 25 '10 at 13:40
    
@fpqc: Filters are in some exact sense "dual" to nets, right? –  Theo Johnson-Freyd Mar 25 '10 at 19:28
    
I'm having a hard time giving you an answer that is completely elementary, so don't take this as me using overly complicated examples just to be a jerk: I would say that a net is to a filter what a pseudofunctor is to a fibered category. The filter encodes all equivalent nets, and getting a net from a filter just requires you to make choices (similar to choosing a cleavage, for example). So they're not really dual, but rather, related by something similar to the grothendieck construction. –  Harry Gindi Mar 26 '10 at 4:57
2  
@TJF: Section 6 of the notes referred to in my answer below makes explicit the "correspondence" between nets and filters. It is not a perfect correspondence, and fpqc's description [I have ignored the categorical jargon] seems correct to me: a net is fundamentally an extrinsic object to the set in question, whereas a filter is intrinsic to the set. The assignment net |-> filter is canonical; the other one requires choices. Note that extrinsic constructions have their merits: I discuss this in Section 7. In particular, I think Bourbaki uses "too many filters". –  Pete L. Clark Mar 26 '10 at 5:06
1  
Theo, maybe you're thinking of: "filters are dual to ideals". Which is exactly true. –  Fabrizio Polo Mar 26 '10 at 12:14
show 3 more comments

2 Answers 2

up vote 21 down vote accepted

Yes. This is given in Kelley's General Topology. (Kelley was one of the main mathematicians who developed the theory of nets so that it would be useful in topology generally rather than just certain applications in analysis.)

In the section "Convergence Classes" at the end of Chapter 2 of his book, Kelley lists the following axioms for convergent nets in a topological space $X$

a) If $S$ is a net such that $S_n = s$ for each $n$ [i.e., a constant net], then $S$ converges to $s$.
b) If $S$ converges to $s$, so does each subnet.
c) If $S$ does not converge to $s$, then there is a subnet of $S$, no subnet of which converges to $s$.
d) (Theorem on iterated limits): Let $D$ be a directed set. For each $m \in D$, let $E_m$ be a directed set, let $F$ be the product $D \times \prod_{m \in D} E_m$ and for $(m,f)$ in $F$ let $R(m,f) = (m,f(m))$. If $S(m,n)$ is an element of $X$ for each $m \in D$ and $n \in E_m$ and $\lim_m \lim_n S(m,n) = s$, then $S \circ R$ converges to $s$.

He has previously shown that in any topological space, convergence of nets satisfies a) through d). (The first three are easy; part d) is, I believe, an original result of his.) In this section he proves the converse: given a set $S$ and a set $\mathcal{C}$ of pairs (net,point) satisfying the four axioms above, there exists a unique topology on $S$ such that a net $N$ converges to $s \in X$ iff $(N,s) \in \mathcal{C}$.

I have always found property d) to be unappealing bordering on completely opaque, but that's a purely personal statement.

Addendum: I would be very interested to know if anyone has ever put this characterization to any useful purpose. A couple of years ago I decided to relearn general topology and write notes this time. The flower of my efforts was an essay on convergence in topological spaces that seems to cover all the bases (especially, comparing nets and filters) more solidly than in any text I have seen.

http://math.uga.edu/~pete/convergence.pdf

But "even" in these notes I didn't talk about either the theorem on iterated limits or (consequently) Kelley's theorem above: I honestly just couldn't internalize it without putting a lot more thought into it. But I've always felt/worried that there must be some insight and content there...

share|improve this answer
4  
+1 for mentioning the oft-overlooked "General Topology". –  Steve D Mar 25 '10 at 8:51
1  
An interesting addendum: Suppose we make a tiny change (So tiny that some have done it inadvertantly by mistake.) We allow a directed set to be empty. Keep these axioms. Define open set, closed set, etc., from convergence as usual. Almost everything is the same as usual. But the whole space need not be open. So we end up in another recent question at MO. –  Gerald Edgar Mar 25 '10 at 12:22
1  
@Andrew L: could you be a little more specific in your reference? –  Pete L. Clark Mar 26 '10 at 4:31
1  
The "Handbook of Analysis and its Foundations" by Eric Schechter contains an extremely extensive and thorough discussion of all notions of convergence, including the relationship between nets and filters, a classification of various notions of subnets and similar material. I would highly recommend the book to anyone interested in the connection between filters and nets. –  Michael Greinecker Jul 11 '10 at 18:46
2  
Interesting -- property (d) is exactly the sort of thing I would expect, since the characterization of topologies in terms of (ultra)filter convergence has a similar axiom on ultrafilters of ultrafilters, as does the characterization of sequential topologies in terms of convergence of sequences. It's especially nice in the ultrafilter case, because that definition is equivalent to saying that a topological space is a lax algebra for the ultrafilter monad on Rel, and then property (d) is just associativity for the lax algebra structure. –  Mike Shulman Aug 23 '10 at 0:21
show 5 more comments

(too long for a comment to Pete's answer)

Garrett Birkhoff was my Ph.D. advisor. Let me provide a few remarks of a historical nature.

From a 25-year-old Garrett Birkhoff we have: Abstract 355, "A new definition of limit" Bull. Amer. Math. Soc. 41 (1935) 636. (Received September 5, 1935)

According to the report of the meeting (Bull. Amer. Math. Soc. 42 (1936) 3) the paper was delivered at the AMS meeting in New York on October 26, 1935.

In the abstract we find what would nowadays be called convergence of a filter base. (See also Definition 4 in Birkhoff's 1937 paper.) Birkhoff remarked to me once that Bourbaki never acknowledged his (Birkhoff's) priority.

It seems that some time after Birkhoff's talk, his father (G. D. Birkhoff) remarked that it reminded him of a paper of Moore and Smith. So young Garrett read Moore and Smith, and in the end adopted their system for the subsequent paper, calling it "Moore-Smith convergence in general topology". Since that Annals of Mathematics paper was received April 27, 1936, one can only imagine young Garrett working furiously for 6 months converting his previous filter-base material into the Moore-Smith setting!

share|improve this answer
    
Did Bourbaki acknowledge anyone 's priority? I remember his thanking tits in a footnote for the exercises on Coxeter groups &c. –  Mariano Suárez-Alvarez Jul 11 '10 at 23:18
    
In Garrets Paper I think he uses a definition for convergence of nets using neighborhoods. So, this doesn't give nets as sth fundamental rather it just imposes indirectly axioms on what is known as a neighborhood ...plz correct me if I'm wrong –  Freeze_S Jan 24 at 3:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.