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1. • CommentRowNumber1.
   • CommentAuthorMike Shulman
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexStarted work on [[syntopogenous space]].

   Started work on syntopogenous space.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexWorking on... what? :-)

   Working on… what? :-)

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIs that a question addressed to me?

   Is that a question addressed to me?

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI was just joking... the term "syntopogenous" is one I'd never heard of, and it just *sounds* so esoteric. Have you known about them a long time?

   I was just joking… the term “syntopogenous” is one I’d never heard of, and it just sounds so esoteric. Have you known about them a long time?

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexNo, I just read about them recently. As far as I know, they aren't good for anything; I'm just amused that they exist.

   No, I just read about them recently. As far as I know, they aren’t good for anything; I’m just amused that they exist.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexAnd that the "underlying topology" functor $Unif \to Top$ (and others) can be described as a coreflection.

   And that the “underlying topology” functor $Unif \to Top$ (and others) can be described as a coreflection.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeAug 28th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI think I'm done with [[syntopogenous space]] now, by the way.

   I think I’m done with syntopogenous space now, by the way.

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 27th 2014
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexShould there be a clause in condition 2 of topogenous relation that $A \cup B \delta C$ iff $A \delta C$ or $B \delta C$? (What's there involves unions only in the second variable.) It seems you'd need this to justify the claim that the opposite of a topogenous relation is also topogenous. Is it correct that inserting this clause doesn't affect any of the results on this page?

   Should there be a clause in condition 2 of topogenous relation that $A \cup B \delta C$ iff $A \delta C$ or $B \delta C$? (What’s there involves unions only in the second variable.) It seems you’d need this to justify the claim that the opposite of a topogenous relation is also topogenous.

   Is it correct that inserting this clause doesn’t affect any of the results on this page?

9. • CommentRowNumber9.
   • CommentAuthorTobyBartels
   • CommentTimeJun 27th 2014
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexYes, definitely. I\'ll fix it. >Is it correct that inserting this clause doesn't affect any of the results on this page? On the contrary, inserting this clause is necessary to make many of them true!

   Yes, definitely. I'll fix it.

   Is it correct that inserting this clause doesn’t affect any of the results on this page?

   On the contrary, inserting this clause is necessary to make many of them true!

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeJul 18th 2014
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexSome changes made, mainly to mention $\ll$ (and $\bowtie$) earlier and to clarify the role of nontriviality/reflexivity.

    Some changes made, mainly to mention $\ll$ (and $\bowtie$) earlier and to clarify the role of nontriviality/reflexivity.

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 28th 2014
    • (edited Dec 28th 2014)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexBehind the scenes there has been some further discussion with Victor Porton, including this [discussion at MO](http://mathoverflow.net/questions/191372/is-certain-topology-related-set-a-distributive-lattice/191444) with an illuminating response from Mamuka Jibladze. This has produced what I regard as (potentially) the most useful way of regarding [topogenous relations](http://ncatlab.org/nlab/show/syntopogenous+space#topogenous_relations) on a set $X$: they can be identified with closed subsets $C \hookrightarrow \beta X \times \beta X$ containing the diagonal, that is to say with reflexive relations from $\beta X$ to itself in the pretopos of compact Hausdorff spaces. I have added this remark to [[syntopogenous space]]. (Briefly: a map $P X \times P X \to \mathbf{2}$ which preserves finite joins in separate arguments amounts to a join-preserving map $P X \otimes P X \to \mathbf{2}$ where the tensor product is just that of join-semilattices viewed as (idempotent) commutative monoids. This tensor product gives the coproduct of commutative rigs, and this coproduct is one of Boolean algebras as well. By Stone duality, this corresponds to the product $\beta X \times \beta X$, where the poset of join-preserving maps as above is dual to the poset of ideals in $CompHaus(\beta X \times \beta X, \mathbf{2})$, i.e., is dual to the topology of the Zariski spectrum $\beta X \times \beta X$. This leads to the identification with closed subsets $C \subseteq \beta X \times \beta X$. Details can be found at my article [topogeny](http://ncatlab.org/toddtrimble/published/topogeny).) I have not tried to work out what a syntopogeny is from this point of view. But it seems like potentially useful point of view, since so much is known about $\beta$.

    Behind the scenes there has been some further discussion with Victor Porton, including this discussion at MO with an illuminating response from Mamuka Jibladze. This has produced what I regard as (potentially) the most useful way of regarding topogenous relations on a set $X$: they can be identified with closed subsets $C \hookrightarrow \beta X \times \beta X$ containing the diagonal, that is to say with reflexive relations from $\beta X$ to itself in the pretopos of compact Hausdorff spaces. I have added this remark to syntopogenous space.

    (Briefly: a map $P X \times P X \to \mathbf{2}$ which preserves finite joins in separate arguments amounts to a join-preserving map $P X \otimes P X \to \mathbf{2}$ where the tensor product is just that of join-semilattices viewed as (idempotent) commutative monoids. This tensor product gives the coproduct of commutative rigs, and this coproduct is one of Boolean algebras as well. By Stone duality, this corresponds to the product $\beta X \times \beta X$, where the poset of join-preserving maps as above is dual to the poset of ideals in $CompHaus(\beta X \times \beta X, \mathbf{2})$, i.e., is dual to the topology of the Zariski spectrum $\beta X \times \beta X$. This leads to the identification with closed subsets $C \subseteq \beta X \times \beta X$. Details can be found at my article topogeny.)

    I have not tried to work out what a syntopogeny is from this point of view. But it seems like potentially useful point of view, since so much is known about $\beta$.

12. • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeDec 29th 2014
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    Author: Mike Shulman
    Format: MarkdownItexInteresting! How does the correspondence act concretely? E.g. if $\delta$ is a topogenous relation, what does $A\delta B$ mean in terms of the corresponding closed subset of $\beta X \times \beta X$?

    Interesting! How does the correspondence act concretely? E.g. if $\delta$ is a topogenous relation, what does $A\delta B$ mean in terms of the corresponding closed subset of $\beta X \times \beta X$?

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 29th 2014
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    Author: Todd_Trimble
    Format: MarkdownItexI believe it means this: given a topogenous relation $\delta$, consider the set $$\{(\mathcal{U}, \mathcal{V}) \in \beta X \times \beta X: (\forall (A, B) \in P X \times P X)\; A \in \mathcal{U} \wedge B \in \mathcal{V} \Rightarrow (A, B) \in \delta\}.$$ This is a closed subset of $\beta X \times \beta X$. For example, take $\delta$ to be the minimal topogenous relation, consisting of $(A, B)$ such that $A \cap B \neq \emptyset$. I claim the corresponding closed subset is the diagonal of $\beta X \times \beta X$. Any point $(\mathcal{U}, \mathcal{U})$ certainly belongs to the subset, since $A \in \mathcal{U}$ and $B \in \mathcal{U}$ implies $A \cap B \in \mathcal{U}$, so necessarily $A \cap B \neq \emptyset$. On the other hand, if $\mathcal{U}$ and $\mathcal{V}$ are distinct ultrafilters, then we can find an $A \in \mathcal{U}$ for which $A \notin \mathcal{V}$, so that $A \in \mathcal{U}$ and $\neg A \in \mathcal{V}$. Then if $(\mathcal{U}, \mathcal{V})$ belongs to the closed set, we reach the conclusion $(A, \neg A) \in \delta$ or $A \cap \neg A \neq \emptyset$, a contradiction. (I'll probably add more to this response in a while; have to run now.)

    I believe it means this: given a topogenous relation $\delta$, consider the set

    $$\{(\mathcal{U}, \mathcal{V}) \in \beta X \times \beta X: (\forall (A, B) \in P X \times P X)\; A \in \mathcal{U} \wedge B \in \mathcal{V} \Rightarrow (A, B) \in \delta\}.$$

    This is a closed subset of $\beta X \times \beta X$. For example, take $\delta$ to be the minimal topogenous relation, consisting of $(A, B)$ such that $A \cap B \neq \emptyset$. I claim the corresponding closed subset is the diagonal of $\beta X \times \beta X$. Any point $(\mathcal{U}, \mathcal{U})$ certainly belongs to the subset, since $A \in \mathcal{U}$ and $B \in \mathcal{U}$ implies $A \cap B \in \mathcal{U}$, so necessarily $A \cap B \neq \emptyset$. On the other hand, if $\mathcal{U}$ and $\mathcal{V}$ are distinct ultrafilters, then we can find an $A \in \mathcal{U}$ for which $A \notin \mathcal{V}$, so that $A \in \mathcal{U}$ and $\neg A \in \mathcal{V}$. Then if $(\mathcal{U}, \mathcal{V})$ belongs to the closed set, we reach the conclusion $(A, \neg A) \in \delta$ or $A \cap \neg A \neq \emptyset$, a contradiction.

    (I’ll probably add more to this response in a while; have to run now.)

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 29th 2014
    • (edited Dec 29th 2014)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexContinuing the previous post: let's check that the subset $K_\delta$ of $\beta X \times \beta X$ corresponding to a topogenous relation $\delta$ is closed, as claimed. Suppose $(\mathcal{U}, \mathcal{V}) \notin K_\delta$, so that there exist $A, B \in P X$ such that $A \in \mathcal{U}$ and $B \in \mathcal{V}$ and $(A, B) \notin \delta$. Then, letting $[A]$ denote the basic clopen $\{\mathcal{U}: A \in \mathcal{U}\}$, it seems to be trivially true that the open set $[A] \times [B]$ contains $(\mathcal{U}, \mathcal{V})$ and does not intersect $K_\delta$, so the subset is closed. So far we have not said anything about the conditions for $\delta$ to be a topogenous relation; we do know that $K_\delta$ contains the diagonal if $\delta$ contains all pairs $(A, B)$ that intersect in a nonempty set. So let's go in the opposite direction: starting from a closed reflexive relation $K \subseteq \beta X \times \beta X$, produce a corresponding topogenous relation $\delta_K$. It seems clear what to do: define $$\delta_K \coloneqq \{(A, B) \in P X \times P X: ([A] \times [B]) \cap K \neq \emptyset\}.$$ Let's see: the map $P X \to P\beta X$ mapping $A$ to $[A]$ is a Boolean algebra map and therefore preserves joins, and the operations $- \times [B]$ and $- \cap K$ also preserve joins, and one pretty quickly concludes from there that the conditions for $\delta_K$ to be a topogenous relation are met. We should go on and check that $\delta \mapsto K_\delta$ is inverse to $K \mapsto \delta_K$, giving a covariant bijection between topogenous relations and reflexive relations on $\beta X$. I think I'll skip this for now (I'm reasonably confident).

    Continuing the previous post: let’s check that the subset $K_\delta$ of $\beta X \times \beta X$ corresponding to a topogenous relation $\delta$ is closed, as claimed. Suppose $(\mathcal{U}, \mathcal{V}) \notin K_\delta$, so that there exist $A, B \in P X$ such that $A \in \mathcal{U}$ and $B \in \mathcal{V}$ and $(A, B) \notin \delta$. Then, letting $[A]$ denote the basic clopen $\{\mathcal{U}: A \in \mathcal{U}\}$, it seems to be trivially true that the open set $[A] \times [B]$ contains $(\mathcal{U}, \mathcal{V})$ and does not intersect $K_\delta$, so the subset is closed.

    So far we have not said anything about the conditions for $\delta$ to be a topogenous relation; we do know that $K_\delta$ contains the diagonal if $\delta$ contains all pairs $(A, B)$ that intersect in a nonempty set. So let’s go in the opposite direction: starting from a closed reflexive relation $K \subseteq \beta X \times \beta X$, produce a corresponding topogenous relation $\delta_K$. It seems clear what to do: define

    $$\delta_K \coloneqq \{(A, B) \in P X \times P X: ([A] \times [B]) \cap K \neq \emptyset\}.$$

    Let’s see: the map $P X \to P\beta X$ mapping $A$ to $[A]$ is a Boolean algebra map and therefore preserves joins, and the operations $- \times [B]$ and $- \cap K$ also preserve joins, and one pretty quickly concludes from there that the conditions for $\delta_K$ to be a topogenous relation are met.

    We should go on and check that $\delta \mapsto K_\delta$ is inverse to $K \mapsto \delta_K$, giving a covariant bijection between topogenous relations and reflexive relations on $\beta X$. I think I’ll skip this for now (I’m reasonably confident).

15. • CommentRowNumber15.
    • CommentAuthorMike Shulman
    • CommentTimeDec 30th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexOk, this is starting to make sense. If we think of points of $\beta X$ as "ideal points" of $X$, then $K_\delta$ is the "ideal set" of all points $(x,y)$ such that every pair of sets $A\ni x$ and $B\ni y$ are $\delta$-related. Except that we have to allow $x,y$ to be in $\beta X$ rather than $X$ for this to retain enough information. That makes me wonder what topogenous relations look like in nonstandard analysis. The obvious thing would be to consider the analogous subset of ${^*X} \times {^*X}$, the pairs $(x,y)$ such that if $x\in {^*A}$ and $y\in {^*B}$ then $A\delta B$. This gives a reflexive (external) relation $\approx_\delta$ on nonstandard points of $X$. Does this relation retain all the information about $\delta$? Can one characterize the external relations which arise from topogenous relations?

    Ok, this is starting to make sense. If we think of points of $\beta X$ as “ideal points” of $X$, then $K_\delta$ is the “ideal set” of all points $(x,y)$ such that every pair of sets $A\ni x$ and $B\ni y$ are $\delta$-related. Except that we have to allow $x,y$ to be in $\beta X$ rather than $X$ for this to retain enough information.

    That makes me wonder what topogenous relations look like in nonstandard analysis. The obvious thing would be to consider the analogous subset of ${^*X} \times {^*X}$, the pairs $(x,y)$ such that if $x\in {^*A}$ and $y\in {^*B}$ then $A\delta B$. This gives a reflexive (external) relation $\approx_\delta$ on nonstandard points of $X$. Does this relation retain all the information about $\delta$? Can one characterize the external relations which arise from topogenous relations?

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 30th 2014
    • (edited Dec 30th 2014)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexSo okay, I might as well go ahead and establish the bijection mentioned in my last message. It will simplify things a bit if we drop the reflexivity from both sides, and just concentrate on conditions 2 and 3 for a topogenous relation (I call such a thing a *topogeny*), and similarly on the $\beta$ side drop the reflexivity. It will also work more smoothly if I set up a Galois connection between *complements* of topogenies (call them antitopogenies) and closed sets of $\beta X \times \beta X$. I'll remark that if $\delta$ is an antitopogeny, then $(A, B) \in \delta$ and $(A', B) \in \delta$ together implies $(A \cup A', B) \in \delta$, and similarly in the second argument; also sets $(\emptyset, B)$ and $(A, \emptyset)$ belong to $\delta$. So, we set up a relation $R \subseteq P X \times P X \times \beta X \times \beta X$ where $(A, B, \mathcal{U}, \mathcal{V}) \in R$ if $\neg A \in \mathcal{U}$ or $\neg B \in \mathcal{V}$. This induces a Galois connection in the usual way, where given a subset $\delta \subseteq P X \times P X$ we consider $$\delta \multimap R \coloneqq \{(\mathcal{U}, \mathcal{V}): (\forall (A, B) \in \delta)\; R(A, B, \mathcal{U}, \mathcal{V})\}$$ which is an intersection of basic closed sets $\neg ([A] \times [B])$ with $(A, B)$ ranging over $\delta$, hence closed. And given a subset $K \subseteq \beta X \times \beta X$ we consider $$K \multimap R \coloneqq \{(A, B): (\forall (\mathcal{U}, \mathcal{V}) \in K)\; R(A, B, \mathcal{U}, \mathcal{V})\}$$ which is an intersection of antitopogenous relations $\neg (\mathcal{U} \times \mathcal{V})$, hence antitopogenous. With the Galois connection infrastructure in place, it's enough to show that (1) every closed set of $\beta X \times \beta X$ is of the form $K_\delta$ for some $\delta \subseteq P X \times P X$, and (2) every antitopogeny is of the form $\delta_K$ for some $K \subseteq \beta X \times \beta X$. (1) is trivial since every closed $K$ is an intersection of basic closed sets of the form $\neg ([A] \times [B])$ (i.e., the basic open sets are $[A] \times [B]$), so just take those $(A, B)$ we need to form the set $\delta$. (2) is an application of the ultrafilter lemma. I won't go through all the details, but if $\delta \subseteq P X \times P X$ is an antitopogeny, then the collection of sets $(A, 0) \in \delta$ is an ideal and is contained in a maximal ideal, and furthermore one has sufficiently many such maximal ideals (i.e., every ideal is the intersection of the maximal ideals $\neg \mathcal{U}$ containing it). Similarly sets of the form $(0, B)$ form an ideal which is the intersection of sufficiently many maximal ideals, and one can work out that any antitopogeny is an intersection of sets of the form $\{(A, B): A \notin \mathcal{U} \;\text{or}\; B \notin \mathcal{V}\}$ where $\mathcal{U}, \mathcal{V}$ range over sufficiently many ultrafilters. In fact, I'm going to add the remark now to [[syntopogenous space]] that every topogeny is a union of "basic topogenies" which are those of the form $\mathcal{U} \times \mathcal{V} \subset P X \times P X$, a cartesian product of ultrafilters.

    So okay, I might as well go ahead and establish the bijection mentioned in my last message. It will simplify things a bit if we drop the reflexivity from both sides, and just concentrate on conditions 2 and 3 for a topogenous relation (I call such a thing a topogeny), and similarly on the $\beta$ side drop the reflexivity. It will also work more smoothly if I set up a Galois connection between complements of topogenies (call them antitopogenies) and closed sets of $\beta X \times \beta X$. I’ll remark that if $\delta$ is an antitopogeny, then $(A, B) \in \delta$ and $(A', B) \in \delta$ together implies $(A \cup A', B) \in \delta$, and similarly in the second argument; also sets $(\emptyset, B)$ and $(A, \emptyset)$ belong to $\delta$.

    So, we set up a relation $R \subseteq P X \times P X \times \beta X \times \beta X$ where $(A, B, \mathcal{U}, \mathcal{V}) \in R$ if $\neg A \in \mathcal{U}$ or $\neg B \in \mathcal{V}$. This induces a Galois connection in the usual way, where given a subset $\delta \subseteq P X \times P X$ we consider

    $$\delta \multimap R \coloneqq \{(\mathcal{U}, \mathcal{V}): (\forall (A, B) \in \delta)\; R(A, B, \mathcal{U}, \mathcal{V})\}$$

    which is an intersection of basic closed sets $\neg ([A] \times [B])$ with $(A, B)$ ranging over $\delta$, hence closed. And given a subset $K \subseteq \beta X \times \beta X$ we consider

    $$K \multimap R \coloneqq \{(A, B): (\forall (\mathcal{U}, \mathcal{V}) \in K)\; R(A, B, \mathcal{U}, \mathcal{V})\}$$

    which is an intersection of antitopogenous relations $\neg (\mathcal{U} \times \mathcal{V})$, hence antitopogenous. With the Galois connection infrastructure in place, it’s enough to show that (1) every closed set of $\beta X \times \beta X$ is of the form $K_\delta$ for some $\delta \subseteq P X \times P X$, and (2) every antitopogeny is of the form $\delta_K$ for some $K \subseteq \beta X \times \beta X$. (1) is trivial since every closed $K$ is an intersection of basic closed sets of the form $\neg ([A] \times [B])$ (i.e., the basic open sets are $[A] \times [B]$), so just take those $(A, B)$ we need to form the set $\delta$. (2) is an application of the ultrafilter lemma. I won’t go through all the details, but if $\delta \subseteq P X \times P X$ is an antitopogeny, then the collection of sets $(A, 0) \in \delta$ is an ideal and is contained in a maximal ideal, and furthermore one has sufficiently many such maximal ideals (i.e., every ideal is the intersection of the maximal ideals $\neg \mathcal{U}$ containing it). Similarly sets of the form $(0, B)$ form an ideal which is the intersection of sufficiently many maximal ideals, and one can work out that any antitopogeny is an intersection of sets of the form $\{(A, B): A \notin \mathcal{U} \;\text{or}\; B \notin \mathcal{V}\}$ where $\mathcal{U}, \mathcal{V}$ range over sufficiently many ultrafilters.

    In fact, I’m going to add the remark now to syntopogenous space that every topogeny is a union of “basic topogenies” which are those of the form $\mathcal{U} \times \mathcal{V} \subset P X \times P X$, a cartesian product of ultrafilters.

17. • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTimeDec 17th 2016
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexAt [[syntopogenous space]] I wrote as the definition: * For any $\delta\in \mathcal{O}$, there exists a $\delta'\in\mathcal{O}$ such that if $A,B\subseteq X$ have the property that whenever $C\cup D = X$, either $A\;\delta'\; C$ or $B\;\delta'\; D$, then $A\;\delta\; B$. But now that seems wrong to me: instead of $B\;\delta'\; D$ it should be $D\;\delta'\; B$. That's what we have in the corresponding condition at [[proximity space]], and that's the only way I can see to make (for instance) the identification of topological spaces with simple perfect syntopogenies true. So I've fixed it. I no longer have Császár's book in front of me, but I suspect I translated incorrectly from his use of $\ll$ instead of $\delta$. (Actually, it would be nice to rewrite the page treating $\delta,\bowtie,\ll$ on an equal footing the way we do at [[proximity space]].)

    At syntopogenous space I wrote as the definition:

    • For any $\delta\in \mathcal{O}$, there exists a $\delta'\in\mathcal{O}$ such that if $A,B\subseteq X$ have the property that whenever $C\cup D = X$, either $A\;\delta'\; C$ or $B\;\delta'\; D$, then $A\;\delta\; B$.

    But now that seems wrong to me: instead of $B\;\delta'\; D$ it should be $D\;\delta'\; B$. That’s what we have in the corresponding condition at proximity space, and that’s the only way I can see to make (for instance) the identification of topological spaces with simple perfect syntopogenies true. So I’ve fixed it. I no longer have Császár’s book in front of me, but I suspect I translated incorrectly from his use of $\ll$ instead of $\delta$. (Actually, it would be nice to rewrite the page treating $\delta,\bowtie,\ll$ on an equal footing the way we do at proximity space.)

18. • CommentRowNumber18.
    • CommentAuthorMike Shulman
    • CommentTimeDec 29th 2016
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexI added some more remarks to [[syntopogenous space]] about $\bowtie$ and $\ll$ and how things behave constructively.

    I added some more remarks to syntopogenous space about $\bowtie$ and $\ll$ and how things behave constructively.

19. • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 6th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexComing back and re-reading Todd's [page on topogenies](https://ncatlab.org/toddtrimble/published/topogeny), it's clear from the positioning of negations that constructively, the definition using $\ll$ is the most natural. This was also the way Császár originally phrased it; I now regret that I originally wrote [[syntopogenous space]] in terms of $\delta$ relations instead. Specifically, what Császár calls "topogenous orders" $\ll$ are what Todd calls "flat profunctors", whose composites are simply relational composites. Moreover, a simple syntopogeny (i.e. a quasi-proximity) is just a comonoid in the bicategory of such profunctors, and an arbitrary syntopogeny should be obtainable similarly from its local ind-completion. Why a *comonoid*? It seems related to the observation that [[apartness relations]] are the open complements, in a discrete locale, of closed equivalence relations, so that their "comparison" axiom is more like a comultiplication than a multiplication. So I am wondering whether the characterizations of topogenies and [[relational beta-modules]] in terms of the ultrafilter space $\beta$ could constructively be rephrased using the localic Stone-Cech compactification, whose frame of opens is directly related to ideals and filters without needing to pass through the ultrafilters (its points, of which constructively there may not be enough). Perhaps this would yield the corresponding "apartness" notions. However, I'm currently a bit stuck on the constructive counterpart of Todd's Proposition 4.10 with its slick use of Stone duality. I think if we remove the dualizations, then the result ought to be that opens in $\beta X_1\times\dots\times \beta X_n$ correspond to separately join-preserving maps $P X_1 \times \cdots \times P X_n \to \Omega^{op}$ (or perhaps some subclass of these satisfying a "decomposability" property dual to complete regularity). The latter in the case $n=2$ would be the $\bowtie$ version of a topogenous relation, so that its complement "$\delta$" would be a closed sublocale. Separately join-preserving maps $P X_1 \times \cdots \times P X_n \to \Omega^{op}$ should still correspond to ideals in the distributive lattice $P X_1 \otimes \cdots \otimes P X_n$, but where do we go from there?

    Coming back and re-reading Todd’s page on topogenies, it’s clear from the positioning of negations that constructively, the definition using $\ll$ is the most natural. This was also the way Császár originally phrased it; I now regret that I originally wrote syntopogenous space in terms of $\delta$ relations instead. Specifically, what Császár calls “topogenous orders” $\ll$ are what Todd calls “flat profunctors”, whose composites are simply relational composites. Moreover, a simple syntopogeny (i.e. a quasi-proximity) is just a comonoid in the bicategory of such profunctors, and an arbitrary syntopogeny should be obtainable similarly from its local ind-completion.

    Why a comonoid? It seems related to the observation that apartness relations are the open complements, in a discrete locale, of closed equivalence relations, so that their “comparison” axiom is more like a comultiplication than a multiplication. So I am wondering whether the characterizations of topogenies and relational beta-modules in terms of the ultrafilter space $\beta$ could constructively be rephrased using the localic Stone-Cech compactification, whose frame of opens is directly related to ideals and filters without needing to pass through the ultrafilters (its points, of which constructively there may not be enough). Perhaps this would yield the corresponding “apartness” notions.

    However, I’m currently a bit stuck on the constructive counterpart of Todd’s Proposition 4.10 with its slick use of Stone duality. I think if we remove the dualizations, then the result ought to be that opens in $\beta X_1\times\dots\times \beta X_n$ correspond to separately join-preserving maps $P X_1 \times \cdots \times P X_n \to \Omega^{op}$ (or perhaps some subclass of these satisfying a “decomposability” property dual to complete regularity). The latter in the case $n=2$ would be the $\bowtie$ version of a topogenous relation, so that its complement “$\delta$” would be a closed sublocale. Separately join-preserving maps $P X_1 \times \cdots \times P X_n \to \Omega^{op}$ should still correspond to ideals in the distributive lattice $P X_1 \otimes \cdots \otimes P X_n$, but where do we go from there?

20. • CommentRowNumber20.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 6th 2017
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    Author: Mike Shulman
    Format: MarkdownItexHmm, now I am getting a glimmer. There are actually two, or maybe three, different compactifications hiding under the name $\beta$. Namely, for any locale $X$, we can construct: * The locale $K_0(X)$ with $O(K_0(X)) = Idl(X)$, the frame of ideals in $X$ (which is the free frame on the underlying distributive lattice of $X$) * The locale $K_2(X)$ with $O(K_2(X))$ the frame of "completely regular" ideals (those for which any $U\in I$ is $\ll$ some $V\in I$, where $\ll$ is the "very inside" relation of complete regularity) * The locale $K_1(X)$, which is like $K_2(X)$ but using "strongly regular" ideals instead, for which $\ll$ is the finest interpolating relation contained in the "well-inside" relation of regularity Countable choice gives $K_1(X) = K_2(X)$. Moreover, if $O(X)$ is a Boolean algebra (such as if excluded middle holds and $X$ is discrete), then $K_0(X) = K_1(X) = K_2(X)$ too, since in a Boolean algebra every element is very-inside itself. In general, $K_1(X)$ or $K_2(X)$ is the "Stone-Cech compactification" $\beta$; but I suspect that when applying $\beta$ to *discrete* spaces/locales as we do when defining relational $\beta$-modules and so on, constructively we may be more interested in $K_0(X)$. In particular, while $K_1$ and $K_2$ are reflections into the full subcategories of compact regular and compact completely-regular locales, $K_0$ is a non-idempotent monad on $Loc$; it is dual to the comonad on $Frm$ induced by the free-forgetful adjunction to $DLat$. So it seems more likely to give an interesting notion of generalized multicategory.

    Hmm, now I am getting a glimmer. There are actually two, or maybe three, different compactifications hiding under the name $\beta$. Namely, for any locale $X$, we can construct:

    • The locale $K_0(X)$ with $O(K_0(X)) = Idl(X)$, the frame of ideals in $X$ (which is the free frame on the underlying distributive lattice of $X$)
    • The locale $K_2(X)$ with $O(K_2(X))$ the frame of “completely regular” ideals (those for which any $U\in I$ is $\ll$ some $V\in I$, where $\ll$ is the “very inside” relation of complete regularity)
    • The locale $K_1(X)$, which is like $K_2(X)$ but using “strongly regular” ideals instead, for which $\ll$ is the finest interpolating relation contained in the “well-inside” relation of regularity

    Countable choice gives $K_1(X) = K_2(X)$. Moreover, if $O(X)$ is a Boolean algebra (such as if excluded middle holds and $X$ is discrete), then $K_0(X) = K_1(X) = K_2(X)$ too, since in a Boolean algebra every element is very-inside itself. In general, $K_1(X)$ or $K_2(X)$ is the “Stone-Cech compactification” $\beta$; but I suspect that when applying $\beta$ to discrete spaces/locales as we do when defining relational $\beta$-modules and so on, constructively we may be more interested in $K_0(X)$.

    In particular, while $K_1$ and $K_2$ are reflections into the full subcategories of compact regular and compact completely-regular locales, $K_0$ is a non-idempotent monad on $Loc$; it is dual to the comonad on $Frm$ induced by the free-forgetful adjunction to $DLat$. So it seems more likely to give an interesting notion of generalized multicategory.

21. • CommentRowNumber21.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 9th 2017
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    Author: Mike Shulman
    Format: MarkdownItexTo start with, here is a constructive proof of Todd's theorem that topogenies from $X$ to $Y$ correspond to closed subspaces of $\beta X \times \beta Y$. Constructively, I will dualize both the topogenies and the subspaces, and prove that topogenous *apartnesses* from $X$ to $Y$ (relations $\bowtie$ between subsets of $X$ and $Y$ such that $A\cup B \bowtie C \iff (A\bowtie C \wedge B\bowtie C)$ and $\emptyset \bowtie C$ and dually) correspond to *open* sublocales of $K_0 X \times K_0 Y$, where $O(K_0(X)) = Idl(P(X))$ as above. (Note that the strongly closed sublocales are exactly the duals of the open ones, though this is not true for spaces constructively.) By definition of locale products as suplattice tensor products, the opens in $K_0 X \times K_0 Y$ are formal joins of tensor products of an ideal in $P(X)$ with an ideal in $P(Y)$, modulo an equivalence relation making the tensor product distribute over joins in both variables. In one direction, given a topogenous apartness $\bowtie$, we send it to the join of tensor products of principal ideals $\bigvee_{A\bowtie B} \downarrow(A) \otimes \downarrow(B)$. In the other direction, given an open $\bigvee_i (J_i\otimes K_i)$, we define $A\bowtie B$ to mean $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ where for each $j,k$ there is an $i$ such that $A_j \in J_i$ and $B_k \in K_i$. To show that these are inverses, starting from $\bowtie$, if $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ and for each $j,k$ there exist principal ideals $\downarrow(C)$ and $\downarrow(D)$ with $A_j \subseteq C$ and $B_k \subseteq D$ and $C\bowtie D$, then also $A_j \bowtie B_k$ by isotony, hence $A\bowtie B$ by finite additivity of $\bowtie$. (And obviously, if $A\bowtie B$ then $A\in \downarrow(A)$ and $B\in \downarrow (B)$.) On the other side, since any ideal is the directed join of the principal ideals it contains (directed joins of ideals are just unions) and $\otimes$ distributes over joins, any open is equivalent to a join of tensor products $\bigvee_i \downarrow(C_i) \otimes \downarrow (D_i)$ of tensor products of principal ideals. Then each $C_i \bowtie D_i$ in the resulting apartness $\bowtie$, hence $\downarrow(C_i) \otimes \downarrow (D_i)$ appears in the open reconstructed from it. Conversely, if $A\bowtie B$, then $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ and for each $j,k$ there exists an $i_{j,k}$ with $A_j \subseteq C_{i_{j,k}}$ and $B_k \subseteq D_{i_{j,k}}$. Then by definition of the finite join of ideals, the ideal $J = \bigvee_{j,k} \downarrow(C_{i_{j,k}})$ contains $A$ and the ideal $K = \bigvee_{j,k} \downarrow(D_{i_{j,k}})$ contains $B$, and the tensor product $J\otimes K$ might as well appear in the given open since $\otimes$ distributes over joins, hence so might as well also $\downarrow(A)\otimes \downarrow(B)$. A similar, but simpler, argument shows that opens in $X\times K_0(Y)$ correspond to point-set apartness relations $x\bowtie A$ for $x\in X$ and $A\subseteq Y$ satisfying isotony and finite additivity on the right.

    To start with, here is a constructive proof of Todd’s theorem that topogenies from $X$ to $Y$ correspond to closed subspaces of $\beta X \times \beta Y$. Constructively, I will dualize both the topogenies and the subspaces, and prove that topogenous apartnesses from $X$ to $Y$ (relations $\bowtie$ between subsets of $X$ and $Y$ such that $A\cup B \bowtie C \iff (A\bowtie C \wedge B\bowtie C)$ and $\emptyset \bowtie C$ and dually) correspond to open sublocales of $K_0 X \times K_0 Y$, where $O(K_0(X)) = Idl(P(X))$ as above. (Note that the strongly closed sublocales are exactly the duals of the open ones, though this is not true for spaces constructively.)

    By definition of locale products as suplattice tensor products, the opens in $K_0 X \times K_0 Y$ are formal joins of tensor products of an ideal in $P(X)$ with an ideal in $P(Y)$, modulo an equivalence relation making the tensor product distribute over joins in both variables. In one direction, given a topogenous apartness $\bowtie$, we send it to the join of tensor products of principal ideals $\bigvee_{A\bowtie B} \downarrow(A) \otimes \downarrow(B)$. In the other direction, given an open $\bigvee_i (J_i\otimes K_i)$, we define $A\bowtie B$ to mean $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ where for each $j,k$ there is an $i$ such that $A_j \in J_i$ and $B_k \in K_i$.

    To show that these are inverses, starting from $\bowtie$, if $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ and for each $j,k$ there exist principal ideals $\downarrow(C)$ and $\downarrow(D)$ with $A_j \subseteq C$ and $B_k \subseteq D$ and $C\bowtie D$, then also $A_j \bowtie B_k$ by isotony, hence $A\bowtie B$ by finite additivity of $\bowtie$. (And obviously, if $A\bowtie B$ then $A\in \downarrow(A)$ and $B\in \downarrow (B)$.)

    On the other side, since any ideal is the directed join of the principal ideals it contains (directed joins of ideals are just unions) and $\otimes$ distributes over joins, any open is equivalent to a join of tensor products $\bigvee_i \downarrow(C_i) \otimes \downarrow (D_i)$ of tensor products of principal ideals. Then each $C_i \bowtie D_i$ in the resulting apartness $\bowtie$, hence $\downarrow(C_i) \otimes \downarrow (D_i)$ appears in the open reconstructed from it. Conversely, if $A\bowtie B$, then $A = A_1 \cup \cdots \cup A_n$ and $B = B_1 \cup \cdots \cup B_m$ and for each $j,k$ there exists an $i_{j,k}$ with $A_j \subseteq C_{i_{j,k}}$ and $B_k \subseteq D_{i_{j,k}}$. Then by definition of the finite join of ideals, the ideal $J = \bigvee_{j,k} \downarrow(C_{i_{j,k}})$ contains $A$ and the ideal $K = \bigvee_{j,k} \downarrow(D_{i_{j,k}})$ contains $B$, and the tensor product $J\otimes K$ might as well appear in the given open since $\otimes$ distributes over joins, hence so might as well also $\downarrow(A)\otimes \downarrow(B)$.

    A similar, but simpler, argument shows that opens in $X\times K_0(Y)$ correspond to point-set apartness relations $x\bowtie A$ for $x\in X$ and $A\subseteq Y$ satisfying isotony and finite additivity on the right.

22. • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeFeb 9th 2017
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    Author: Todd_Trimble
    Format: MarkdownItexI haven't read through this properly yet, but this sounds awesome! I never tried to consider the constructive aspects while writing the topogeny stuff up, because the stuff I was trying to reinterpret seemed so inundated with excluded middle in the first place (and I felt sure the guy I was communicating with at the time would not be the least interested). So I'm glad you're taking this up.

    I haven’t read through this properly yet, but this sounds awesome!

    I never tried to consider the constructive aspects while writing the topogeny stuff up, because the stuff I was trying to reinterpret seemed so inundated with excluded middle in the first place (and I felt sure the guy I was communicating with at the time would not be the least interested). So I’m glad you’re taking this up.

23. • CommentRowNumber23.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 10th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexPerhaps it's better to consider the opens in $K_0 X \times K_0 Y$ as (-1)-sheaves for a coverage, the way Johnstone does it. Then they are sets $\mathcal{J}$ of tensor products $J\otimes K$, where $J\in Idl(P(X))$ and $K\in Idl(P(Y))$, such that 1. if $J_2\otimes K_2\in \mathcal{J}$ and $J_1 \subseteq J_2$ and $K_1\subseteq K_2$, then $J_1\otimes K_1 \in \mathcal{J}$ (presheaf) 2. if $J_i \otimes K \in \mathcal{J}$ for all $i$ then $(\bigvee_i J_i)\otimes K \in \mathcal{J}$ and dually, where $\bigvee_i J_i = \{ A_1 \vee \cdots \vee A_n \mid $ each $A_i$ belongs to some $J_i \}$ (sheaf) With this representation, the apartness relation corresponding to an open $\mathcal{J}$ says that $A\bowtie B$ if $\downarrow(A) \otimes \downarrow(B) \in \mathcal{J}$, and the open corresponding to an apartness relation is $\{ J\otimes K \mid \forall A\in J, \forall B\in K, A\bowtie B\}$. The bijection between the two is then even easier.

    Perhaps it’s better to consider the opens in $K_0 X \times K_0 Y$ as (-1)-sheaves for a coverage, the way Johnstone does it. Then they are sets $\mathcal{J}$ of tensor products $J\otimes K$, where $J\in Idl(P(X))$ and $K\in Idl(P(Y))$, such that

    1. if $J_2\otimes K_2\in \mathcal{J}$ and $J_1 \subseteq J_2$ and $K_1\subseteq K_2$, then $J_1\otimes K_1 \in \mathcal{J}$ (presheaf)
    2. if $J_i \otimes K \in \mathcal{J}$ for all $i$ then $(\bigvee_i J_i)\otimes K \in \mathcal{J}$ and dually, where $\bigvee_i J_i = \{ A_1 \vee \cdots \vee A_n \mid$ each $A_i$ belongs to some $J_i \}$ (sheaf)

    With this representation, the apartness relation corresponding to an open $\mathcal{J}$ says that $A\bowtie B$ if $\downarrow(A) \otimes \downarrow(B) \in \mathcal{J}$, and the open corresponding to an apartness relation is $\{ J\otimes K \mid \forall A\in J, \forall B\in K, A\bowtie B\}$. The bijection between the two is then even easier.

24. • CommentRowNumber24.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 10th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexNow I claim that just like apartness relations are exactly closed equivalence relations on a discrete locale, *decomposable [[proximity space|proximal]] (set-set) apartnesses* are exactly closed equivalence relations on $K_0 X$ for a discrete locale $X$ (and similarly decomposable quasi-proximal apartnesses correspond to closed preorders). The above argument shows that closed sublocales of $K_0 X \times K_0 X$ correspond to set-set relations $\bowtie$ satisfying the appropriate isotony and additivity properties, and the correspondence of symmetry is clear; so it suffices to consider reflexivity and decomposable transitivity. For reflexivity, the diagonal is contained in a closed sublocale $\mathsf{C}U$ of $K_0 X \times K_0 X$ iff $U$ is contained in the *exterior* of the diagonal, which is the largest open whose restriction along the diagonal is $0$. As a (-1)-sheaf, this exterior is the set of all $J\otimes K$ such that $J\cap K = 0$. Thus, $\mathsf{C}U$ contains the diagonal iff $A\bowtie B$ implies $A\cap B = 0$, which is the proximal reflexivity axiom. Transitivity of $\mathsf{C}U$ means that the pullback $\pi_2^* U$ is contained in $\pi_1^* U \vee \pi_3^* U$ as open subsets of $K_0 X \times K_0 X \times K_0 X$. If $U$ corresponds to $\bowtie$, then we have $$ \pi_1^* U = \{ J\otimes K\otimes L \mid \forall B\in K, \forall C\in L, B\bowtie C \} $$ $$ \pi_2^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall C\in L, A\bowtie C \} $$ $$ \pi_3^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall B\in K, A\bowtie B \} $$ The tricky thing is $\pi_1^* U \vee \pi_3^* U$, since it is not just the union of these sets but the (-1)-sheafification of the union. However, I claim that $$\pi_1^* U \vee \pi_3^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall B\in K, \forall C\in L, \exists D,E, (A\bowtie D \wedge E\bowtie C \wedge B = D\cup E \} $$ It's easy to see that this is a (-1)-sheaf, and that it contains $\pi_1^*U$ and $\pi_3^*U$. Moreover, every $J\otimes K\otimes L$ in this set can be obtained from those in $\pi_1^*U$ and $\pi_3^*U$ by three steps of gluing: $$\begin{aligned} J\otimes K \otimes L &= \bigvee_{A\in J} (\downarrow(A) \otimes K\otimes L)\\ &= \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K \otimes \downarrow(C))\\ &= \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K_A \otimes \downarrow(C)) \vee \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K_C \otimes \downarrow(C)) \end{aligned} $$ where $K_A = \{ D\in K \mid A\bowtie D\}$ and $K_C = \{ E \in K \mid E\bowtie C \}$, so that $\downarrow(A) \otimes K_A \otimes \downarrow(C) \in \pi_3^*U$ and $\downarrow(A) \otimes K_C \otimes \downarrow(C)\in \pi_1^*U$. The final decomposition as a binary join uses the assumed property of $J\otimes K\otimes L$ and the fact that joins of ideals are computed by taking finite joins of their elements. Thus, the above set of triples of ideals is a (-1)-sheaf containing $\pi_1^*U$ and $\pi_3^*U$ and contained in the sheafification of their union, so it must be their join as (-1)-sheaves. Now transitivity $\pi_2^* U\subseteq \pi_1^* U \vee \pi_3^* U$ is equivalent to the special case that each triple of the form $\downarrow(A) \otimes \downarrow(X) \otimes \downarrow(C)$, with $A\bowtie C$, is an element of the above set. This means exactly that whenever $A\bowtie C$ there are sets $D,E$ such that $A\bowtie D$ and $E\bowtie C$ and $X = D\cup E$, the decomposability axiom for a proximal apartness.

    Now I claim that just like apartness relations are exactly closed equivalence relations on a discrete locale, decomposable proximal (set-set) apartnesses are exactly closed equivalence relations on $K_0 X$ for a discrete locale $X$ (and similarly decomposable quasi-proximal apartnesses correspond to closed preorders). The above argument shows that closed sublocales of $K_0 X \times K_0 X$ correspond to set-set relations $\bowtie$ satisfying the appropriate isotony and additivity properties, and the correspondence of symmetry is clear; so it suffices to consider reflexivity and decomposable transitivity.

    For reflexivity, the diagonal is contained in a closed sublocale $\mathsf{C}U$ of $K_0 X \times K_0 X$ iff $U$ is contained in the exterior of the diagonal, which is the largest open whose restriction along the diagonal is $0$. As a (-1)-sheaf, this exterior is the set of all $J\otimes K$ such that $J\cap K = 0$. Thus, $\mathsf{C}U$ contains the diagonal iff $A\bowtie B$ implies $A\cap B = 0$, which is the proximal reflexivity axiom.

    Transitivity of $\mathsf{C}U$ means that the pullback $\pi_2^* U$ is contained in $\pi_1^* U \vee \pi_3^* U$ as open subsets of $K_0 X \times K_0 X \times K_0 X$. If $U$ corresponds to $\bowtie$, then we have

    $$\pi_1^* U = \{ J\otimes K\otimes L \mid \forall B\in K, \forall C\in L, B\bowtie C \}$$ $$\pi_2^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall C\in L, A\bowtie C \}$$ $$\pi_3^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall B\in K, A\bowtie B \}$$

    The tricky thing is $\pi_1^* U \vee \pi_3^* U$, since it is not just the union of these sets but the (-1)-sheafification of the union. However, I claim that

    $$\pi_1^* U \vee \pi_3^* U = \{ J\otimes K\otimes L \mid \forall A\in J, \forall B\in K, \forall C\in L, \exists D,E, (A\bowtie D \wedge E\bowtie C \wedge B = D\cup E \}$$

    It’s easy to see that this is a (-1)-sheaf, and that it contains $\pi_1^*U$ and $\pi_3^*U$. Moreover, every $J\otimes K\otimes L$ in this set can be obtained from those in $\pi_1^*U$ and $\pi_3^*U$ by three steps of gluing:

    $$\begin{aligned} J\otimes K \otimes L &= \bigvee_{A\in J} (\downarrow(A) \otimes K\otimes L)\\ &= \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K \otimes \downarrow(C))\\ &= \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K_A \otimes \downarrow(C)) \vee \bigvee_{A\in J} \bigvee_{C\in L} (\downarrow(A) \otimes K_C \otimes \downarrow(C)) \end{aligned}$$

    where $K_A = \{ D\in K \mid A\bowtie D\}$ and $K_C = \{ E \in K \mid E\bowtie C \}$, so that $\downarrow(A) \otimes K_A \otimes \downarrow(C) \in \pi_3^*U$ and $\downarrow(A) \otimes K_C \otimes \downarrow(C)\in \pi_1^*U$. The final decomposition as a binary join uses the assumed property of $J\otimes K\otimes L$ and the fact that joins of ideals are computed by taking finite joins of their elements. Thus, the above set of triples of ideals is a (-1)-sheaf containing $\pi_1^*U$ and $\pi_3^*U$ and contained in the sheafification of their union, so it must be their join as (-1)-sheaves.

    Now transitivity $\pi_2^* U\subseteq \pi_1^* U \vee \pi_3^* U$ is equivalent to the special case that each triple of the form $\downarrow(A) \otimes \downarrow(X) \otimes \downarrow(C)$, with $A\bowtie C$, is an element of the above set. This means exactly that whenever $A\bowtie C$ there are sets $D,E$ such that $A\bowtie D$ and $E\bowtie C$ and $X = D\cup E$, the decomposability axiom for a proximal apartness.

25. • CommentRowNumber25.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 10th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexNone of this actually requires $X$ to be discrete. However, if we allow $X$ to be non-discrete, then there should be some compatibility condition on a proximity, such as that every $A\in O(X)$ satisfies $A = \bigvee \{ B\in O(X) \mid \exists C, A\cup C = X \wedge B\bowtie C \}$. I'm not sure yet how to interpret that in terms of a closed equivalence relation. However, $K_0X$ has the interesting property that any map $f:X\to Y$ with $Y$ compact Hausdorff (i.e. compact regular) factors through a map $g:K_0 X \to Y$ defined by $g^*(W) = \{ A\in O(X) \mid A \triangleleft f^*W \}$, where $\triangleleft$ is the well-inside relation of regularity. I don't think $g$ is unique, and while $K_0 X$ itself is compact and contains $X$ densely, it is not Hausdorff --- but quotienting it by a closed equivalence relation seems like a natural way to make it Hausdorff! Certainly the kernel pair of a Hausdorff quotient is a closed equivalence relation, so that if an *effective* equivalence relation has a Hausdorff quotient then it is closed. Dually, if a closed equivalence relation is effective, then the quotient map is proper, and since closed sublocales descend along proper maps it follows that the quotient is Hausdorff. Thus, the effective closed equivalence relations on $K_0 X$ correspond bijectively to quotients of $K_0 X$ that are Hausdorff, giving a locale-theoretic approach to the standard bijective correspondence between proximities and compactifications. This suggests strongly that effectiveness should correspond to the above compatibility condition on proximities. However, I don't know how to show this because I don't know a characterization of which equivalence relations in $Loc$ are effective.

    None of this actually requires $X$ to be discrete. However, if we allow $X$ to be non-discrete, then there should be some compatibility condition on a proximity, such as that every $A\in O(X)$ satisfies $A = \bigvee \{ B\in O(X) \mid \exists C, A\cup C = X \wedge B\bowtie C \}$. I’m not sure yet how to interpret that in terms of a closed equivalence relation.

    However, $K_0X$ has the interesting property that any map $f:X\to Y$ with $Y$ compact Hausdorff (i.e. compact regular) factors through a map $g:K_0 X \to Y$ defined by $g^*(W) = \{ A\in O(X) \mid A \triangleleft f^*W \}$, where $\triangleleft$ is the well-inside relation of regularity. I don’t think $g$ is unique, and while $K_0 X$ itself is compact and contains $X$ densely, it is not Hausdorff — but quotienting it by a closed equivalence relation seems like a natural way to make it Hausdorff! Certainly the kernel pair of a Hausdorff quotient is a closed equivalence relation, so that if an effective equivalence relation has a Hausdorff quotient then it is closed. Dually, if a closed equivalence relation is effective, then the quotient map is proper, and since closed sublocales descend along proper maps it follows that the quotient is Hausdorff. Thus, the effective closed equivalence relations on $K_0 X$ correspond bijectively to quotients of $K_0 X$ that are Hausdorff, giving a locale-theoretic approach to the standard bijective correspondence between proximities and compactifications.

    This suggests strongly that effectiveness should correspond to the above compatibility condition on proximities. However, I don’t know how to show this because I don’t know a characterization of which equivalence relations in $Loc$ are effective.

26. • CommentRowNumber26.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 10th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexOh, that's silly -- the compatibility condition doesn't mean Hausdorffness of the quotient, it means that the composite map from $X$ to the quotient is a sublocale embedding. Maybe the quotient is always Hausdorff. I think I can show that it does look like other definitions of the compactification of a proximity locale.

    Oh, that’s silly – the compatibility condition doesn’t mean Hausdorffness of the quotient, it means that the composite map from $X$ to the quotient is a sublocale embedding. Maybe the quotient is always Hausdorff. I think I can show that it does look like other definitions of the compactification of a proximity locale.