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I'm currently brushing up my topology using Willard's General Topology. Currently I'm working through the chapters 11 and 12 on nets and filters. Chapter 12 deals extensively with ultrafilters and proves (Theorem 12.12) that every filter is contained in an ultrafilter using Zorn's lemma.

Theorem 12.17 and the exercises connect nets and filters. In this way, I see a proof that:

Every net has a subnet which is an ultranet.

For reference:

A subnet $x_\mu$ of $x_\lambda$ is an increasing cofinal mapping $\phi: M\to\Lambda$ composed with $x: \Lambda\to X$.

An ultranet is a net $x: \Lambda \to X$ such that: $$\forall E \subseteq X: \exists \lambda_0: \forall \lambda\ge \lambda_0: x_\lambda \in E \lor \forall \lambda\ge \lambda_0: x_\lambda\notin E$$

However, this very statement also occurs as Exercise 11B.2. This suggests an easier proof. After one and a half week of intermittent attempts, I concede and humbly ask your help.

I would love to see "natural proofs", as opposed to deus ex machina constructions. Thanks in advance. (It should be noted that some choice is necessary, but even in choice proofs, some are more natural than others.)

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Lord_Farin
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  • Is there any indication that this is not somehow equivalent to some weak choice principle, thus eliminating "natural proofs"? Or is Zorn's llama a natural proof? – Asaf Karagila May 01 '15 at 19:40
  • Even though Zorn is choice, the order on which to apply Zorn is natural for the filter proof. But the point is that this exercise is here, before filters are even mentioned. – Lord_Farin May 01 '15 at 19:48
  • Well, was Zorn's lemon introduced just before talking about filters? – Asaf Karagila May 01 '15 at 19:50
  • It wasn't introduced to begin with. There was only a remark after the fact (but a remark stating that choice is necessary is in Chapter 11). – Lord_Farin May 01 '15 at 19:53
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    The easiest argument is by way of filters and ultrafilters; even Kelley essentially does it that way, though he does it before defining filters. Willard’s 11B.2 is a hard exercise at that point. (By the way, Willard uses the most restrictive of the three standard notions of subnet, and in many ways the hardest to work with. See my answer to this question for a discussion and references.) – Brian M. Scott May 01 '15 at 21:18
  • Not seeing a nice direct proof, I decided to look how Pedersen (Analysis Now) proved the existence of universal subnets [Pedersen calls universal nets what Willard dubs ultranets]. And whaddayaknow, he does it via filters and ultrafilters (not quite using that terminology, that is introduced in the exercises after). So if there is a nice direct way, it's at least not well known. – Daniel Fischer May 01 '15 at 21:19

1 Answers1

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Let $f: I \to X$ be net in $X$, where $(I, \le)$ is a directed set. To build a universal subnet: let $\mathcal{F}$ be the filter of tails on $X$, so the filter generated by all $T(i) = \{f(j): j \ge i\}, i \in I$, cf. definition 12.15 in Willard, and (the non-constructive part) let $\mathcal{U}$ be an ultrafilter on $X$ extending $\mathcal{F}$. Then let $J = \{(i, U) \in I \times \mathcal{U}: f(i) \in U \}$ be a set with order $(i, U_1) \le (i' ,U_2)$ iff $i \le i'$ and $U_2 \subseteq U_1$, which is clearly a directed set. Define a net $g: J \to X$ by $g(j) = g((i,U)) = x_i$. This is a subnet of $f$ (use the projection as the connecting map). One only has to show that $g$ is indeed a universal net.

If you get stuck on that part my note here might help...

All published proofs that I know of the existence of ultranets use ultrafilters or ultrafilters in disguise (as a maximal of family of subsets of certain type). I think this has to do with the fact that there is no set of subnets of a given net; as the index set of a subnet can be any directed set, the collection of subnets form a proper class in general, and so we cannot easily apply maximum principles like Zorn’s lemma.

Henno Brandsma
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    Thanks for the effort, but I don't really understand your answer. As I mentioned, I know proofs using filters and I completely follow chapter 12. However this is an exercise in chapter 11 and I'd like a direct proof, without filters... However I appreciate the clear writing! – Lord_Farin Jan 31 '18 at 07:33
  • @Lord_Farin what is it that you don’t understand? – Henno Brandsma Jan 31 '18 at 16:50
  • I wrote "Theorem 12.17 and the exercises connect nets and filters. In this way, I see a proof [of the statement]". So I want to know proofs without using the connection of nets and filters. But now it seems to me that your proof also uses this connection. So I don't really understand how you are answering the question (regardless of the clear and straightforward proof). – Lord_Farin Jan 31 '18 at 17:03
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    @Lord_Farin all proofs I have seen use filters. There is no set of subnets of a given net (they’re a class) so we cannot easily apply a maximum principle there. – Henno Brandsma Jan 31 '18 at 17:06
  • If you would be so kind to elaborate on these points in your answer I would be happy to cast a well-deserved upvote and accept. – Lord_Farin Jan 31 '18 at 20:07
  • @Lord_Farin I now do mention it.. – Henno Brandsma Feb 01 '18 at 22:47