Skip to main content
MathOverflow

Return to Answer

replaced the dead link
Source Link
Martin Sleziak
Martin Sleziak
  • 4.8k
  • 4
  • 39
  • 42

EDIT: Well, this was impressively wrong: not only did I somehow spend most of the answer thinking "ordered space" meant "linearly ordered space," I also missed the significance of the OP's restriction to universal Horn structures. Below is a heavily edited version of my original post. As it stands, this is not an answer, just a too-long comment. (My apologies to the OP.)

First, Wikipedia has the definition of ordered space: http://en.wikipedia.org/wiki/Partially_ordered_spacehttps://en.wikipedia.org/wiki/Partially_ordered_space.

As to first-order structures which are "compatible" with a given topology, let me say three things:

  • When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1https://projecteuclid.org/euclid.pl/1235417281) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

  • If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

  • I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

EDIT: Well, this was impressively wrong: not only did I somehow spend most of the answer thinking "ordered space" meant "linearly ordered space," I also missed the significance of the OP's restriction to universal Horn structures. Below is a heavily edited version of my original post. As it stands, this is not an answer, just a too-long comment. (My apologies to the OP.)

First, Wikipedia has the definition of ordered space: http://en.wikipedia.org/wiki/Partially_ordered_space.

As to first-order structures which are "compatible" with a given topology, let me say three things:

  • When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

  • If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

  • I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

EDIT: Well, this was impressively wrong: not only did I somehow spend most of the answer thinking "ordered space" meant "linearly ordered space," I also missed the significance of the OP's restriction to universal Horn structures. Below is a heavily edited version of my original post. As it stands, this is not an answer, just a too-long comment. (My apologies to the OP.)

First, Wikipedia has the definition of ordered space: https://en.wikipedia.org/wiki/Partially_ordered_space.

As to first-order structures which are "compatible" with a given topology, let me say three things:

  • When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (https://projecteuclid.org/euclid.pl/1235417281) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

  • If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

  • I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

Completely overhauled answer
Source Link
Noah Schweber
Noah Schweber
  • 19.6k
  • 10
  • 121
  • 370

(Here I assume that all spaces are Hausdorff EDIT: Well, since Priestley spaces are compact and at least sometimes that includes being Hausdorff.)

I'mthis was impressively wrong: not sure exactly what you're asking, but let me try to address partonly did I somehow spend most of it:

Firstthe answer thinking "ordered space" meant "linearly ordered space, note that" I also missed the notionsignificance of compatible you define makes sense for arbitrary first-order structures. I don't see that anything is gained by specializingthe OP's restriction to universal Horn structures, though I may be wrong.

However, note that many intuitively compatible structures are ruled out by your approach: for example, the usual topology on $\mathbb{R}$ is not compatible with the first-order structure $(\mathbb{R}; <)$, since $<$ Below is open (not closed) as a subsetheavily edited version of $\mathbb{R}$my original post. Conversely, if we demand that relations be openAs it stands, then the usual topology on $\mathbb{R}$ is not compatible with $(\mathbb{R}; \le)$.

The right response to this, I think, is to allow relations to be either open or closed. This plays nicely with negationnot an answer, even though it does seemjust a bit ad hoc. You can also go "higher," and ask that the relations merely be Borel, projective, etctoo-long comment. The point is, we probably want the "legal" topological complexities of relations(My apologies to be closed under complementthe OP.)

Then there's the specific question of what "ordered space" means. I think that the definition of "ordered space" being used inFirst, Wikipedia has the definition of Priestleyordered space demands that the ordering $<$ be open, or equivalently that the ordering $\le$ be closed; this is backed up by: http://en.wikipedia.org/wiki/Partially_ordered_space, as well as the classical definition of order topology (in the linear case). So, I think the definition of Priestley space is just:

A compact space $(X, <, \tau)$ is \it Priestley \rm if $<$ is an open subset of $X^2$ in the product topology from $\tau$, and the Priestley separation axiom is satisfied.

Also, Priestley's paper is available: http://plms.oxfordjournals.org/content/s3-24/3/507.full.pdf. From what I read, it seems that he beginsAs to first-order structures which are "compatible" with the ordering and then considers two topologies generated by it, the "upper" and "lower" topologies; these would both be coarser than the topologya given by requiring that $<$ be open.

Let me add that sometimes, no compatibility requirement at all is introduced. See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics; alsotopology, I believe this is the approach taken by Chang/Keisler in "Continuous model theory."let me say three things:

  • When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

  • If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

  • I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

(Here I assume that all spaces are Hausdorff, since Priestley spaces are compact and at least sometimes that includes being Hausdorff.)

I'm not sure exactly what you're asking, but let me try to address part of it:

First, note that the notion of compatible you define makes sense for arbitrary first-order structures. I don't see that anything is gained by specializing to universal Horn structures, though I may be wrong.

However, note that many intuitively compatible structures are ruled out by your approach: for example, the usual topology on $\mathbb{R}$ is not compatible with the first-order structure $(\mathbb{R}; <)$, since $<$ is open (not closed) as a subset of $\mathbb{R}$. Conversely, if we demand that relations be open, then the usual topology on $\mathbb{R}$ is not compatible with $(\mathbb{R}; \le)$.

The right response to this, I think, is to allow relations to be either open or closed. This plays nicely with negation, even though it does seem a bit ad hoc. You can also go "higher," and ask that the relations merely be Borel, projective, etc. The point is, we probably want the "legal" topological complexities of relations to be closed under complement.

Then there's the specific question of what "ordered space" means. I think that the definition of "ordered space" being used in the definition of Priestley space demands that the ordering $<$ be open, or equivalently that the ordering $\le$ be closed; this is backed up by http://en.wikipedia.org/wiki/Partially_ordered_space, as well as the classical definition of order topology (in the linear case). So, I think the definition of Priestley space is just:

A compact space $(X, <, \tau)$ is \it Priestley \rm if $<$ is an open subset of $X^2$ in the product topology from $\tau$, and the Priestley separation axiom is satisfied.

Also, Priestley's paper is available: http://plms.oxfordjournals.org/content/s3-24/3/507.full.pdf. From what I read, it seems that he begins with the ordering and then considers two topologies generated by it, the "upper" and "lower" topologies; these would both be coarser than the topology given by requiring that $<$ be open.

Let me add that sometimes, no compatibility requirement at all is introduced. See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics; also, I believe this is the approach taken by Chang/Keisler in "Continuous model theory."

EDIT: Well, this was impressively wrong: not only did I somehow spend most of the answer thinking "ordered space" meant "linearly ordered space," I also missed the significance of the OP's restriction to universal Horn structures. Below is a heavily edited version of my original post. As it stands, this is not an answer, just a too-long comment. (My apologies to the OP.)

First, Wikipedia has the definition of ordered space: http://en.wikipedia.org/wiki/Partially_ordered_space.

As to first-order structures which are "compatible" with a given topology, let me say three things:

  • When studying topological structures, we don't always care about any sort of compatibility. See See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics. Obviously, this isn't what you're talking about, but I think it's worth mentioning - especially if you start looking around for topological model theory, just so one is aware that frequently no such compatibility is required.

  • If the complexity of the structure is not restricted, then we don't always want relations to be closed; sometimes we want relations to be open. This is the approach taken by Robinson (http://matwbn.icm.edu.pl/ksiazki/fm/fm81/fm81115.pdf). On the other hand, we could simply demand that every predicate be closed; the price of that would be that some reasonable topologies on some natural first-order structures would be ommitted.

  • I don't know of a structure $M$ with a natural topology $\tau$ where $M$ is universal Horn and $\tau$ does not make all predicates closed; however, this doesn't mean that such structures aren't interesting. As an example, given a universal Horn structure $\mathcal{M}$ with a topology $\tau$, it is reasonable to consider adjoining a predicate $U$ to $\mathcal{M}$ naming a specific open subset of $\mathcal{M}$. The resulting expansion $\mathcal{M}^+$ might still be universal Horn, but now it's not clear to me that it's natural to want $U^{\mathcal{M}^+}$ to now be clopen. I think even in the universal Horn cases, demanding all predicates be closed might kill off some neat objects.

Source Link
Noah Schweber
Noah Schweber
  • 19.6k
  • 10
  • 121
  • 370

(Here I assume that all spaces are Hausdorff, since Priestley spaces are compact and at least sometimes that includes being Hausdorff.)

I'm not sure exactly what you're asking, but let me try to address part of it:

First, note that the notion of compatible you define makes sense for arbitrary first-order structures. I don't see that anything is gained by specializing to universal Horn structures, though I may be wrong.

However, note that many intuitively compatible structures are ruled out by your approach: for example, the usual topology on $\mathbb{R}$ is not compatible with the first-order structure $(\mathbb{R}; <)$, since $<$ is open (not closed) as a subset of $\mathbb{R}$. Conversely, if we demand that relations be open, then the usual topology on $\mathbb{R}$ is not compatible with $(\mathbb{R}; \le)$.

The right response to this, I think, is to allow relations to be either open or closed. This plays nicely with negation, even though it does seem a bit ad hoc. You can also go "higher," and ask that the relations merely be Borel, projective, etc. The point is, we probably want the "legal" topological complexities of relations to be closed under complement.

Then there's the specific question of what "ordered space" means. I think that the definition of "ordered space" being used in the definition of Priestley space demands that the ordering $<$ be open, or equivalently that the ordering $\le$ be closed; this is backed up by http://en.wikipedia.org/wiki/Partially_ordered_space, as well as the classical definition of order topology (in the linear case). So, I think the definition of Priestley space is just:

A compact space $(X, <, \tau)$ is \it Priestley \rm if $<$ is an open subset of $X^2$ in the product topology from $\tau$, and the Priestley separation axiom is satisfied.

Also, Priestley's paper is available: http://plms.oxfordjournals.org/content/s3-24/3/507.full.pdf. From what I read, it seems that he begins with the ordering and then considers two topologies generated by it, the "upper" and "lower" topologies; these would both be coarser than the topology given by requiring that $<$ be open.

Let me add that sometimes, no compatibility requirement at all is introduced. See Ziegler's article "Topological model theory" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?handle=euclid.pl/1235417281&view=body&content-type=pdf_1) in Model-Theoretic Logics; also, I believe this is the approach taken by Chang/Keisler in "Continuous model theory."