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One of most classical and somehow striking result in classical model theory states:

A consistent first order theory $T$ has a model.

Few considerations are needed.

  • This result is not true for infinitary logics, such as geometric or second order. So it tells us something about the logic we are working with.

  • From the perspective of a topos theorist this result tells something about the elementary topos Set. In fact, if we change topos in which we take models, some first order theories have no models. An easy example is that there is no model of Peano arithmetic in FinSet.

    I want to focus on the topos theorist perspective. A natural question would be the following:

Characterize toposes $G$ such that any first order theory has a model in $G$

And strangely it turns out that this is linked to presentability of the topos.

A consistent coherent theory (using terminology of Sketches of an Elephant) has a model in any locally presentable elementary topos $G$.

Proof: There is a unique geometric morphism $\text{Set} \rightleftharpoons G $. Since a coherent theory is a first order theory, there is a model in Set, i.e. we have a geometric morphism $B(T) \rightleftharpoons \text{Set}$ so one can prolong this geometric morphism $$B(T) \rightleftharpoons \text{Set} \rightleftharpoons G $$ obtaining a model of $T$ in $G$.

Same proof shows the following refinement.

A first order theory which has a classifying topos has a model in any locally presentable elementary topos $G$.

It would be interesting to understand how far one can goes, in any of two direction:

  • Characterize toposes $G$ such that any first order theory $T$ which is classified by a topos has a model in $G$.
  • How big is the class of theories that have models in any Grothendieck topos?

And yet, former theorems are a partial answer:

  • Any Grothendick topos is ok.
  • This class contains at least first order theories which have a classifying topos.

Locally presentable elementary toposes (i.e. Grothendieck Toposes) have a beautiful notion of cardinality for a model, which is its presentability rank.

Find hypotesis on $B(T)$ such that the category $$\text{Mod}(T, G) = \text{Geom}(B(T),G) $$ is locally presentable or at least accessible.

Maybe one can hope in the following.

  • If $B(T)$ is locally presentable, than $\text{Geom}(B(T),G)$ is reflective in $G^{B(T)},$ so is locally presentable.

  • $B(T)$ is very often locally presentable.

    When $\text{Mod}(T, G)$ is locally presentable one can state freely Shealah conjecture for this categories and try to understand what happens. But first, what about Lowenheim-Skolem theorem?!

    • What about Lowenheim-Skolem theorem?! This result has a very partial answer in the book Topos Theory, by Johnstone. Also there are two papers from Zawadowski.
    • How does Shelah conjecture look like in these categories of models?

If these questions have not a trivial answer and someone finds them interesting, I would like to discuss them.

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    $\begingroup$ writing "not true for infinitary logics, such as geometric or second order" is misleading in several ways. First, using the plural "infinitary logics" misleadingly suggests that the completeness theorem you quote in your first box would fail for all (usual) infinitary logics---this is not quite so; a notable example is the very usual logic $L_{\omega_1,\omega}$: e.g. by a theorem of Carol Karp, for any $L_{\omega_1,\omega}$ sentence $\sigma$, $\sigma$ is provable if and only if $\sigma$ has a model. Second, writing "such as geometric or second order" conflates concepts. $\endgroup$
    – Peter Heinig
    May 31, 2017 at 19:33
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    $\begingroup$ Which concepts does it conflate? Well, infinitary logic, geometric logic, and second order logic simultaneously. This half-sentence conflates three concepts at once, each of which is a distinct concept. Would you please make the sentence less sweeping, e.g. by writing "not true for certain extensions of FOL. "? $\endgroup$
    – Peter Heinig
    May 31, 2017 at 19:36
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    $\begingroup$ It's quite clear that is not the point of the question, as you can read after. Anyway, you can answer to this question: math.stackexchange.com/questions/2276485/… I would appreciate. $\endgroup$
    – Ivan Di Liberti
    May 31, 2017 at 19:39
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    $\begingroup$ What you are getting at in "so it tells us something about the logic that [...]" is the circle of ideas around Lindstrom's characterization theorem for first-order logic. Briefly, you have to take one more semantical property, in addition to completeness, in order to characterize first-order logic. With only one property, there are proper extensions of FOL which still have this property. An example of a characterization, briefly, is: there does not exist an extension of FOL which still would have (0) a computable finitary syntax, (1) the Lowenheim--Skolem property and (2) be complete. $\endgroup$
    – Peter Heinig
    May 31, 2017 at 19:47
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    $\begingroup$ @PeterHeinig While the model existence theorem holds for sentences of $L_{\omega_1,\omega}$ and hence for theories in countable fragments of $L_{\omega_1,\omega}$, it does not hold for complete theories in the full logic $L_{\omega_1,\omega}$. Also, what you wrote isn't right: you meant that $\sigma$ is consistent (i.e. $\lnot\sigma$ is not provable) if and only if $\sigma$ has a model. $\endgroup$
    – Alex Kruckman
    May 31, 2017 at 23:11

1 Answer 1

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While I won't be able to give full answers to the questions, I would like to point to a few ideas which I believe are relevant.

I first would like to respond to your question: ``How big is the class of theories that have models in any Grothendieck topos?''. It was a little unclear whether you intended the question to only be about first order theories or not so I will consider the full case of (set sized) theories of $L_{\infty, \omega}$.

First note that if a theory It is worth pointing out that given any geometric theory $T$ which has a model in all Grothendieck toposes must also have a model in SET. Therefore any such $T$ must be classically consistent. However, if $T$ is any classically consistent theory (not just geometric) we can find an equivalent geometric theory by Morleyization. As such if there is some Grothendieck topos $G$ in which $T$ has no models then either:

  1. $T$ is classically inconsistent, or

  2. $T$ is classically consistent but has no models in SET.

It is worth noting that if $T$ is a first order theory or a sentence of $L_{\omega_1,\omega}$ (2) can't happen.

This is connected with Barr's Covering Theorem.

Next I want to talk about your question on the Lowenheim-Skolem theorem and your observation about in local presentability giving a notion of cardinality in a Grothendieck topos. While the notion of presentability does give a nice notion of size in a Grothendieck topos, once we are in this framework there are several other notions of size which are useful. For a discussion of several of them as well as several model theoretic results that lift to Grothendieck toposes (including the Lowenheim-Skolem theorem) you might look at:

On transferring model theoretic theorems of $L_{\infty, \omega}$ in the category of sets to a fixed Grothendieck topos.

This paper also discusses Morley's theorem on the number of models in a Grothendieck topos:

The Number Of Countable Models In Categories of Sheaves.

Also if you are interested in model theory in realizability toposes you can also check out:

The Number Of Countable Models In A Realizability Topos.

While the specific case of presentability rank isn't discussed in these papers it should be relatively straightforward to prove a Lowenheim-Skolem theorem with respect to presentability rank. Specifically, given a theory $T$ which has a model in a Grothendieck topos $G$ it should be relatively straight forward to find a model with presentability rank at most $|T| + |(C, J)|$ where $G \cong Sh(C, J).$ Roughly the argument should go something like the following:

  1. Let $X$ be a transitive set containing $T$ along with $(C, J)$ and with $|X| = |T| + |(C, J)|$.

  2. Let $V^*$ be a transitive elementary substructure of the universe $V$ containing $X$ with $|V^*| = |X|$.

  3. As $V^*$ is an elementary substructure of $V$ and $V$ thinks there is a model $M$ of $T$ in $Sh(C, J)$ we also have $V^*$ thinks there is a model of $T$ in $Sh(C, J)^{V^*}$.

  4. Because the satisfaction relation is absolute $M$ should also model $T$ in $Sh(C, J)^V \cong G$.

  5. But $M$ should have presentability rank no more than $|V^*| = |T| + |(C, J)|$.

You need to be a little careful here with what we mean by the relativization of a Grothendieck topos, but it is worked out in this paper:

Relativized Grothendieck topoi.

Finally by the Shelah conjecture, I assume you mean the "Shelah categoricity conjecture"? In general models of theories (even first order theories) inside a Grothendieck topos are like``models of a sentence of $L_{\infty, \omega}$ with a very little second order logic thrown in''. However to the best of my knowledge I don't think his eventual categoricity conjecture is even known for sentences of $L_{\omega_1, \omega}$ (although I could be wrong). If so this seems like a very hard problem in general.

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  • $\begingroup$ I don't understand "any geometric theory 𝑇 which has model in any Grothendieck topos must also have models in SET." Consider, for example, the propositional theory, using propositional variables $p_{n,\xi}$ for $n\in\omega$ and $\xi\in\omega_1$, with axioms $p_{n,\xi}\land p_{n,\eta}\to\bot$ for all $n$ and $\xi\neq\eta$, and $\top\to\bigvee_{n\in\omega}p_{n,\xi}$ for all $\xi$. This has no models in SET, because a model amounts to a partial function from $\omega$ onto $\omega_1$, but it has models in some Grothendieck topoi [continued in next comment] $\endgroup$
    – Andreas Blass
    Aug 21, 2019 at 17:53
  • $\begingroup$ namely some topoi of Boolean-valued sets, because one can force such a function (collapsing the cardinal $\omega_1$). More generally, any Grothendieck topos $E$ without points can be viewed as the classifying topos of a geometric theory that has no models in SET (because such a model would amount to a point of $E$) but has a model in the topos $E$. $\endgroup$
    – Andreas Blass
    Aug 21, 2019 at 17:55
  • $\begingroup$ If $T$ has models in any Grothendieck topos then it has models in SET as SET is a Grothendieck topos. $\endgroup$
    – Nate Ackerman
    Aug 22, 2019 at 19:01
  • $\begingroup$ The usual meaning of "any" is universal quantification with maximum dcope. In this case, that would be "for every Grothendieck topos $E$, if $T$ has a model in$E$ then it has one in SET." Similarly, "If anyone can solve this problem then Saharon can" is a nontrivial assertion about Saharon's problem-solving ability, not a trivial consequence of the fact that Saharon is someone. $\endgroup$
    – Andreas Blass
    Aug 22, 2019 at 19:29
  • $\begingroup$ I changed "any" to "all" so hopefully now it is clearer. Thanks for the suggestion. $\endgroup$
    – Nate Ackerman
    Aug 24, 2019 at 2:29

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