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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeJun 5th 2011
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI've started [[relational beta-module]]. It would be lovely if somebody who really grasps it could fill in the abstract definition and maybe check (or even show how to derive) the concrete one, which I extracted from [this blog post by Todd Trimble](http://topologicalmusings.wordpress.com/2008/07/18/ultrafilter-convergence-compactness-and-the-spectrum-of-a-boolean-algebra/). (Hey, maybe Todd could check it!) This started when I realised that being infinitely close is a uniform (not topological) property in nonstandard analysis, which is hinted at by the very bottom of the page (as it is now).

   I’ve started relational beta-module. It would be lovely if somebody who really grasps it could fill in the abstract definition and maybe check (or even show how to derive) the concrete one, which I extracted from this blog post by Todd Trimble. (Hey, maybe Todd could check it!)

   This started when I realised that being infinitely close is a uniform (not topological) property in nonstandard analysis, which is hinted at by the very bottom of the page (as it is now).

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJun 6th 2011
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   Author: Mike Shulman
   Format: MarkdownItexThanks! I added links back and forth to [[generalized multicategory]] and to [[pseudotopological space]]. I also added a remark about a different generalization of this that is known to handle uniform spaces.

   Thanks! I added links back and forth to generalized multicategory and to pseudotopological space. I also added a remark about a different generalization of this that is known to handle uniform spaces.

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeJun 7th 2011
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   Author: TobyBartels
   Format: MarkdownItexThanks. This helped me see that much of the abstract business has already been given by Todd at [[ultrafilter monad]], so I'm going to look at that (and maybe make it into its own page too).

   Thanks. This helped me see that much of the abstract business has already been given by Todd at ultrafilter monad, so I’m going to look at that (and maybe make it into its own page too).

4. • CommentRowNumber4.
   • CommentAuthorTim Campion
   • CommentTimeNov 23rd 2015
   • PermaLink
   Author: Tim Campion
   Format: MarkdownItexIn the current version, I think there's a problem with Lemma 1 (which comes a little way down from [here](http://ncatlab.org/nlab/show/relational+beta-module#proof_of_main_theorem), right before Theorem 3). In order for the conclusion to be true, we should need to assume that ξ is a relational β-module (which suffices anyway for our purposes). The error comes in the second sentence, which claims that "¬A is a neighborhood of x if and only if ∀F:βS F⇝x⇒¬A∈F, but this is incorrect, because in order for ¬A to be open, this criterion needs to apply not only at x, but at _all_ y∈¬A. Philosophically, it seems to me that the lemma as stated shouldn't be correct because in order to get the left-to-right implication, we need some way of building filters F such that F⇝x, but if ξ is an arbitrary relation, there simply is no "introduction rule" for such F's that we can apply. What this lemma really is trying to accomplish is to provide an alternate description of the topology τ(ξ) which we can apply when ξ is a relational β-module, and thus it seems to be more or less equivalent to Barr's Proposition 3.6. I haven't carefully read his proof, but in reconstructing it I found that I did indeed need to use both relational β-module inequalities (unit and associativity), constructing an ultrafilter on βR in the process in order to apply associativity. This ultrafilter is kind of interesting: whereas later in the proof of Theorem 3 we extend a subfilter on R with the "S"-coordinate fixed, in this case we need to extend a subfilter on R with the "βS"-coordinate fixed. This duality is kind of comforting, actually -- it indicates that we're really using the full power of the associativity axiom for relational β-modules, as we must have to. The reason I'm interested is that I think this glossed-over point is exactly where the story will break down if we try to replace β by β<sub>ω</sub>, the monad of ultrafilters which contain a countable set. The hope would be that relational β<sub>ω</sub> - modules might be the same as [countably generated spaces](https://en.wikipedia.org/wiki/Countably_generated_space), but I think what happens is that although countably generated spaces are indeed β<sub>ω</sub> - modules, the converse does not hold. I think this is more-or-less stated in a relatively recent paper of [Manes](http://www.sciencedirect.com/science/article/pii/S0166864109005331). I'll try to write up this part of the proof in the article when I get the chance. I've been attempting to get my head around this whole business on and off for awhile now, and this article has been a big help -- thanks everyone who's worked on it so far!

   In the current version, I think there’s a problem with Lemma 1 (which comes a little way down from here, right before Theorem 3). In order for the conclusion to be true, we should need to assume that ξ is a relational β-module (which suffices anyway for our purposes). The error comes in the second sentence, which claims that “¬A is a neighborhood of x if and only if ∀F:βS F⇝x⇒¬A∈F, but this is incorrect, because in order for ¬A to be open, this criterion needs to apply not only at x, but at all y∈¬A.

   Philosophically, it seems to me that the lemma as stated shouldn’t be correct because in order to get the left-to-right implication, we need some way of building filters F such that F⇝x, but if ξ is an arbitrary relation, there simply is no “introduction rule” for such F’s that we can apply.

   What this lemma really is trying to accomplish is to provide an alternate description of the topology τ(ξ) which we can apply when ξ is a relational β-module, and thus it seems to be more or less equivalent to Barr’s Proposition 3.6. I haven’t carefully read his proof, but in reconstructing it I found that I did indeed need to use both relational β-module inequalities (unit and associativity), constructing an ultrafilter on βR in the process in order to apply associativity. This ultrafilter is kind of interesting: whereas later in the proof of Theorem 3 we extend a subfilter on R with the “S”-coordinate fixed, in this case we need to extend a subfilter on R with the “βS”-coordinate fixed. This duality is kind of comforting, actually – it indicates that we’re really using the full power of the associativity axiom for relational β-modules, as we must have to.

   The reason I’m interested is that I think this glossed-over point is exactly where the story will break down if we try to replace β by β_ω, the monad of ultrafilters which contain a countable set. The hope would be that relational β_ω - modules might be the same as countably generated spaces, but I think what happens is that although countably generated spaces are indeed β_ω - modules, the converse does not hold. I think this is more-or-less stated in a relatively recent paper of Manes.

   I’ll try to write up this part of the proof in the article when I get the chance. I’ve been attempting to get my head around this whole business on and off for awhile now, and this article has been a big help – thanks everyone who’s worked on it so far!

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 23rd 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexMight you be misreading things? I didn't say $\neg A$ was open. I said $\neg A$ was a neighborhood of $x$, meaning that $x$ belongs to the interior of $\neg A$. A set is open if it is a neighborhood of every point it contains.

   Might you be misreading things?

   I didn’t say $\neg A$ was open. I said $\neg A$ was a neighborhood of $x$, meaning that $x$ belongs to the interior of $\neg A$.

   A set is open if it is a neighborhood of every point it contains.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 23rd 2015
   • (edited Nov 23rd 2015)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexMaybe it will help if I say the following things: for a topological space, we say that an ultrafilter $F$ *converges to* $x$ if the neighborhood filter is contained in $F$. Now any filter is equal to the intersection of all ultrafilters containing it; this is a consequence of the ultrafilter theorem. So we can retrieve the neighborhood filter from the convergence relation (attached to a topological space) by considering the intersection of all ultrafilters that converge to $x$. Thus, a set $B$ is a neighborhood of $x$ iff $$\forall_{F: \beta S} F \rightsquigarrow x \Rightarrow B \in F.$$ To define a topology from a relation $R$ from $\beta S$ to $S$, we turn it around and use the preceding observations as our springboard. First we say that $B$ is a *neighborhood* of $x$ if $$\forall_{F: \beta S} R(F, x) \Rightarrow B \in F.$$ Then we say that $B$ is *open* if it is a neighborhood of all the points it contains. In logical symbolism, this becomes $$\forall_{F: \beta S} \forall_{x \in S} x \in B \wedge R(F, x) \Rightarrow B \in F$$ which is the definition of the topology $\tau(R)$ stated somewhere near the beginning of the article.

   Maybe it will help if I say the following things: for a topological space, we say that an ultrafilter $F$ converges to $x$ if the neighborhood filter is contained in $F$.

   Now any filter is equal to the intersection of all ultrafilters containing it; this is a consequence of the ultrafilter theorem. So we can retrieve the neighborhood filter from the convergence relation (attached to a topological space) by considering the intersection of all ultrafilters that converge to $x$. Thus, a set $B$ is a neighborhood of $x$ iff

   $$\forall_{F: \beta S} F \rightsquigarrow x \Rightarrow B \in F.$$

   To define a topology from a relation $R$ from $\beta S$ to $S$, we turn it around and use the preceding observations as our springboard. First we say that $B$ is a neighborhood of $x$ if

   $$\forall_{F: \beta S} R(F, x) \Rightarrow B \in F.$$

   Then we say that $B$ is open if it is a neighborhood of all the points it contains. In logical symbolism, this becomes

   $$\forall_{F: \beta S} \forall_{x \in S} x \in B \wedge R(F, x) \Rightarrow B \in F$$

   which is the definition of the topology $\tau(R)$ stated somewhere near the beginning of the article.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 23rd 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexIt's still possible however that I did overlook a subtlety. Given a relation $\xi: Rel(\beta S, S)$, my previous comment gives a definition of neighborhood of a point $x$ relative to $\xi$, and then goes on to define open sets in terms of this neighborhood relation. But I'm not seeing immediately how it follows that a neighborhood of $x$ is the same thing as a set containing an open about $x$, on the basis of those definitions (relative to $\xi$). It's a little late here and maybe I'm too tired to think about it clearly.

   It’s still possible however that I did overlook a subtlety. Given a relation $\xi: Rel(\beta S, S)$, my previous comment gives a definition of neighborhood of a point $x$ relative to $\xi$, and then goes on to define open sets in terms of this neighborhood relation. But I’m not seeing immediately how it follows that a neighborhood of $x$ is the same thing as a set containing an open about $x$, on the basis of those definitions (relative to $\xi$). It’s a little late here and maybe I’m too tired to think about it clearly.

8. • CommentRowNumber8.
   • CommentAuthorTim Campion
   • CommentTimeNov 23rd 2015
   • (edited Nov 23rd 2015)
   • PermaLink
   Author: Tim Campion
   Format: MarkdownItexRight, I think this distinction is exactly the subtlety it turns on. It started bothering me at some point that I couldn't see this equivalence and I think it's because you need $\xi$ to be a relational $\beta$-module for this to hold. I think I have a counterexample. Consider a three-element set $S = \{x,y,z\}$, with $\mathrm{prin}(x) \rightsquigarrow_\xi y$, $\mathrm{prin}(y) \rightsquigarrow_\xi z$, and no other convergence relations. A set $B$ is open iff the implications $z \in B \implies y \in B$ and $y \in B \implies x \in B$ hold, so the open sets are $\emptyset, \{x\}, \{x,y\}, \{x,y,z\} = S$. So the neighborhood filter for $z$ is just $\{S\}$. But $\{y,z\}$ _does_ have the property that $F \rightsquigarrow_\xi z$ implies $\{y,z\} \in F$. For a simpler counterexample to the statement of the lemma, take $\xi$ to be empty (with $S$ nonempty). Then every set is open, and hence every set is closed. But it's certainly not the case that every element of every set has an ultrafilter $\xi$-converging to it!

   Right, I think this distinction is exactly the subtlety it turns on. It started bothering me at some point that I couldn’t see this equivalence and I think it’s because you need $\xi$ to be a relational $\beta$-module for this to hold.

   I think I have a counterexample. Consider a three-element set $S = \{x,y,z\}$, with $\mathrm{prin}(x) \rightsquigarrow_\xi y$, $\mathrm{prin}(y) \rightsquigarrow_\xi z$, and no other convergence relations. A set $B$ is open iff the implications $z \in B \implies y \in B$ and $y \in B \implies x \in B$ hold, so the open sets are $\emptyset, \{x\}, \{x,y\}, \{x,y,z\} = S$. So the neighborhood filter for $z$ is just $\{S\}$. But $\{y,z\}$ does have the property that $F \rightsquigarrow_\xi z$ implies $\{y,z\} \in F$.

   For a simpler counterexample to the statement of the lemma, take $\xi$ to be empty (with $S$ nonempty). Then every set is open, and hence every set is closed. But it’s certainly not the case that every element of every set has an ultrafilter $\xi$-converging to it!

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 23rd 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexThanks for the counterexamples! So you're right that Lemma 1 is not correct as stated. I expect that it wouldn't be too hard to patch things up, but how much time I have available to work it out myself in the near future is somewhat in question. So go ahead and revise if you are clear on how to go about this.

   Thanks for the counterexamples! So you’re right that Lemma 1 is not correct as stated.

   I expect that it wouldn’t be too hard to patch things up, but how much time I have available to work it out myself in the near future is somewhat in question. So go ahead and revise if you are clear on how to go about this.

10. • CommentRowNumber10.
    • CommentAuthorTim Campion
    • CommentTimeNov 23rd 2015
    • (edited Nov 24th 2015)
    • PermaLink
    Author: Tim Campion
    Format: MarkdownItexI've just entered revisions. I split the statement into an easy part which I labeled a remark, and a hard part which I labeled a lemma. Of course, feel free to make any further changes you like.

    I’ve just entered revisions. I split the statement into an easy part which I labeled a remark, and a hard part which I labeled a lemma. Of course, feel free to make any further changes you like.

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 24th 2015
    • (edited Nov 24th 2015)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexThanks for this, Tim! Your arguments seem to check out; nice work. I expect that overall the arguments could be condensed further; right now they feel slightly repetitive to me (I mean the proofs of lemma 1 and the wrap-up of the main theorem). But again I'm not sure what time I have available in the near future to look into this. I never really bothered looking at Barr's paper; I had worked out most of the proofs by myself, which shows in the mistake you caught. It's possible that his treatment is better in places than what we now have at the nLab; someone should check at some point. Edit: Oh, actually Tim's comment #4 apparently explains why the proofs looked "repetitive". Clearly Tim you've put some thought into this!

    Thanks for this, Tim! Your arguments seem to check out; nice work.

    I expect that overall the arguments could be condensed further; right now they feel slightly repetitive to me (I mean the proofs of lemma 1 and the wrap-up of the main theorem). But again I’m not sure what time I have available in the near future to look into this.

    I never really bothered looking at Barr’s paper; I had worked out most of the proofs by myself, which shows in the mistake you caught. It’s possible that his treatment is better in places than what we now have at the nLab; someone should check at some point.

    Edit: Oh, actually Tim’s comment #4 apparently explains why the proofs looked “repetitive”. Clearly Tim you’ve put some thought into this!

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 26th 2015
    • (edited Nov 26th 2015)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexTim Campion has just answered at [MO](http://mathoverflow.net/a/224536/2926) with an analysis alluded to in comment #4.

    Tim Campion has just answered at MO with an analysis alluded to in comment #4.

13. • CommentRowNumber13.
    • CommentAuthorTim Campion
    • CommentTimeDec 1st 2015
    • PermaLink
    Author: Tim Campion
    Format: MarkdownItexTodd, you caught my secret motivation for learning about this stuff! This lemma saying $\overline{A} = \{x \in X \mid \exists F \in \beta X, A \in F, F \rightsquigarrow x\}$ is a sort of "idempotency" property. It turns out that this idempotency _does_ generalize to the $\beta_\omega$ case, contrary to what I surmised in #4 (which is a relief!). But there's another case where it doesn't work -- in a sequential space, the operation $A \mapsto \{ x \in X \mid \exists (x_n)_n \subseteq A, (x_n)_n \rightsquigarrow x\}$ may need to be iterated transfinitely many times before it stabilizes and gives you the closure of $A$. A space where this operation is actually idempotent is sometimes called a [Fréchet-Urysohn space](https://en.wikipedia.org/wiki/Sequential_space#Fr.C3.A9chet.E2.80.93Urysohn_space). I currently have a write-up about countably-generated spaces and $\beta_\omega$, but talk about repetitive -- it basically involves going through all the proofs in the relational $\beta$-modules article and checking some extra conditions at each stage.

    Todd, you caught my secret motivation for learning about this stuff!

    This lemma saying $\overline{A} = \{x \in X \mid \exists F \in \beta X, A \in F, F \rightsquigarrow x\}$ is a sort of “idempotency” property. It turns out that this idempotency does generalize to the $\beta_\omega$ case, contrary to what I surmised in #4 (which is a relief!). But there’s another case where it doesn’t work – in a sequential space, the operation $A \mapsto \{ x \in X \mid \exists (x_n)_n \subseteq A, (x_n)_n \rightsquigarrow x\}$ may need to be iterated transfinitely many times before it stabilizes and gives you the closure of $A$. A space where this operation is actually idempotent is sometimes called a Fréchet-Urysohn space.

    I currently have a write-up about countably-generated spaces and $\beta_\omega$, but talk about repetitive – it basically involves going through all the proofs in the relational $\beta$-modules article and checking some extra conditions at each stage.

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 1st 2015
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexTim, I'm very glad to have gotten all your input on all this, and it gives me pleasure that the technical details of the relational beta-module article were of use to you after all. Your observations at MO look publishable, and I hope you're considering doing that.

    Tim, I’m very glad to have gotten all your input on all this, and it gives me pleasure that the technical details of the relational beta-module article were of use to you after all.

    Your observations at MO look publishable, and I hope you’re considering doing that.