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.

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 '17 at 19:33infinitarylogic,geometriclogic, andsecond orderlogic 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 '17 at 19:36Lindstrom's characterization theoremfor first-order logic. Briefly, you have to take one more semantical property, in addition to completeness, in order tocharacterizefirst-order logic. With only one property, thereareproper 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 '17 at 19:47sentencesof $L_{\omega_1,\omega}$ 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. $\lnot\sigma$ isnotprovable) if and only if $\sigma$ has a model. $\endgroup$ – Alex Kruckman May 31 '17 at 23:11