*Previously asked and bountied without response at [MSE](https://math.stackexchange.com/q/4056043/28111).*




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This question is a companion to [this one](https://mathoverflow.net/questions/386478/how-many-steps-does-it-take-to-tarski-vaughtify-second-order-logic), about a tame(?) fragment of second-order logic with the standard semantics, $\mathsf{SOL}$, motivated by the Tarski-Vaught test. The question is:



> 
> Working in $\mathsf{ZFC}$ + "There is an extendible cardinal," can the strong compactness number$^1$ of $\mathsf{SOL}\_{TV}$ be strictly less than the smallest extendible cardinal?
> 
> 
> 


To keep things self-contained, here is the relevant definition. First, given structures $\mathfrak{A},\mathfrak{B}$ in the same signature $\Sigma$, say $\mathfrak{A}\trianglelefteq\mathfrak{B}$ iff $\mathfrak{A}\subseteq\mathfrak{B}$ and $\varphi^\mathfrak{B}\not=\emptyset \implies\varphi^\mathfrak{B}\cap \mathfrak{A}^{arity(\varphi)}\not=\emptyset$ for every second-order $\Sigma$-formula $\varphi$ with parameters from $\mathfrak{A}$. We then let $\mathsf{SOL}\_{TV}$ be the set of $\mathsf{SOL}$-formulas which are absolute with respect to $\trianglelefteq$, that is the set of $\varphi\in\mathsf{SOL}$ such that whenever $\mathfrak{A}\trianglelefteq\mathfrak{B}$ are structures in the signature of $\varphi$ we have $\varphi^\mathfrak{A}=\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}.$


Note that despite having the downward Lowenheim-Skolem property, $\mathsf{SOL}\_{TV}$ is much stronger than first-order logic. For example, well-foundedness is $\mathsf{SOL}\_{TV}$-characterizable$^2$; this means that there is an $\mathsf{SOL}\_{TV}$-sentence pinning down $\mathbb{N}$ up to isomorphism, which in turn implies that the $\mathsf{SOL}\_{TV}$-definable relations on $\mathbb{N}$ are exactly the $\mathsf{SOL}$-definable ones. Beyond this, not much is clear to me; the question above seems like it might be approachable.


At present I don't see anything preventing the strong compactness number of $\mathsf{SOL}\_{TV}$ from being *extremely* small - although of course since $\mathsf{SOL}\_{TV}$ has the downward Lowenheim-Skolem property it must be uncountable by Lindstrom's Theorem.




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$^1$The *strong compactness number* of a logic $\mathcal{L}$, if it exists at all, is the least cardinal $\kappa$ such that every unsatisfiable $\mathcal{L}$-theory has an unsatisfiable subtheory of cardinality $<\kappa$. Large cardinals imply the existence of strong compactness numbers; in particular, second-order logic has a strong compactness number iff there is an extendible cardinal, in which case its strong compactness number is the least extendible. As an amusing aside, [Vopenka's Principle](https://mathoverflow.net/a/29310/8133) turns out to be equivalent to the existence of strong compactness numbers for all logics [in a precise sense](https://www.jstor.org/stable/2273786).


$^2$Since well-foundedness is preserved downwards, it's enough to show (working in a language with a distinguished binary relation symbol $<$) that if $\mathfrak{A}\trianglelefteq\_{\mathsf{SOL}}\mathfrak{B}$ and $<^\mathfrak{B}$ is illfounded then $<^\mathfrak{A}=<^\mathfrak{B}\cap\mathfrak{A}$ is illfounded. Consider first the formula $\psi(x)\equiv$ "The initial segment of $<$ below $x$ is ill-founded." We have $\psi^\mathfrak{B}\not=\emptyset$, so there must be some $a\in\mathfrak{A}$ which is in the ill-founded part of $<^\mathfrak{B}$. But now for each $u\in\mathfrak{A}$ in the ill-founded part of $<^\mathfrak{B}$, consider the formula $\theta\_u(y)\equiv$ "$y<u$ and $y$ is in the ill-founded part of $<$." Applying $\triangleleft\_{\mathsf{SOL}}$-ness, we get that the set of elements of $\mathfrak{A}$ which are in the ill-founded part of $\mathfrak{B}$ has no $<^\mathfrak{B}$-minimal element, which (together with its above-established nonemptiness) implies that $<^\mathfrak{A}$ is ill-founded as desired.



Compactness number for a fragment of second-order logic
I think that if you have a logic $\mathcal{L}$ which has downward Lowenheim-Skolem (for theories of arbitrary cardinality, i.e. if $T$ has cardinality $\lambda$ and $N$ is a model $A$ of size $\theta\geq\lambda$, and $\lambda\leq\gamma\leq\theta$, then we can find a sub-model of size $\gamma$ which is elementary in $A$ (w.r.t. the relevant formulas of the logic)), then its strong compactness number is $\leq$ the least supercompact $\kappa$.


For suppose $T$ is a theory in $\mathcal{L}$ such that all subsets of $T$ of size ${<\kappa}$ have a model. Let $\lambda$ be the cardinality of $T$; we consider $T\subseteq\lambda$. Let $j:V\to M$ with $\mathrm{crit}(j)=\kappa$ and $\mathcal{P}(\lambda)\subseteq M$ and $j\upharpoonright\lambda\in M$. Then $M$ thinks that every sub-theory of $j(T)$ of size $<j(\kappa)$ has a model. But we have $T\in M$ and $T\subseteq\lambda$, and note that $T$ is equivalent to $j``T\in M$,
and this has size $\lambda<j(\kappa)$ in $M$. So $T$ has a model $B$ in $M$. But by Lowenheim-Skolem, then it has a model of size $\lambda$ in $M$. Since $\mathcal{P}(\lambda)\subseteq M$, this is truly a model in $V$.


Edit 4 (previous edits done over): Now regarding lower bounds (and cf. Noah's comments below)...


Say that a (possibly beyond 1st order) logic (I'm not sure of what the formal definition of "logic" is here, but it doesn't really matter; I only use it in a limited sense below which is clear enough) is $\kappa$-compact iff whenever $T$ is a set-sized theory in that logic and all subtheories of size ${<\kappa}$ have a model, then $T$ also has a model.


Now consider the logic $\mathcal{L}$ in the (first order) language of set theory, where we have available a proper class of constant symbols, and an extra predicate symbol (which can be taken to code multiple predicates/functions below),
together with the second-order statement ``I am wellfounded''. (Remark: I mistakenly said "one" constant symbol in an earlier version; but I used set-many. Actually for the purposes below we can restrict the number of constants available to something like the cardinality of $V\_{\kappa+2}$, where $\kappa$ is the cardinal in question.)


Claim: Let $\kappa$ be a cardinal and suppose that $\mathcal{L}$ as above is $\kappa$-compact. Then there is a measurable cardinal $\leq\kappa$.


Proof: First suppose that $\kappa$ is inaccessible.
Consider the $\mathcal{L}$-theory $T$ which includes the full first order theory of $V\_{\kappa+1}$ in parameters, the formulas “$\alpha < \dot{\mu} < \kappa$”, for each $\alpha < \kappa$, where $\dot{\mu}$ is a new constant, and the (2nd order) formula “I am wellfounded”. If $M$ is a (wellfounded) model of $T$, note that there is an elementary $j:V\_{\kappa+1}\to M$, the critical point $\mathrm{crit}(j)$ of $j$ exists, and is a measurable cardinal $\leq\kappa$. So we just need to see the small subtheories have models, but this follows easily from the inaccessibility of $\kappa$ (consider elementary substructures of $V\_{\kappa+1}$).


Now suppose $\kappa$ is singular.
Let $f:\eta\to\kappa$ be cofinal, where $\eta<\kappa$. Consider the theory $T$ consisting of the first order theory in parameters of $V\_\kappa$, and using function symbols $\dot{f},\dot{g}$, the statement "$\dot{f}:\eta\to\mathrm{Ord}$ is cofinal", the equations "$\dot{f}(\alpha)=\beta$" for each $\alpha,\beta$ such that $f(\alpha)=\beta$, and the statement "$\dot{g}$ is a surjection from some ordinal onto $\mathrm{Ord}$", and (2nd order) "I am wellfounded". Then for each $\theta<\kappa$, each fragment $\bar{T}$ of size $\theta$ is satisfiable, since we may assume $\eta<\theta$, and we can find an elementary substructure of $(X,f\cap X)\preceq(V\_\kappa,f)$ of cardinality $\theta$, with $\theta+1\subseteq X$, and then the transitive collapse $(M,\bar{f},g)$ is a model, where we take $g:\theta\to\mathrm{Ord}^M$ a surjection. So we get a wellfounded model $(M,f',g')$ of the full theory. Let $j:V\_\kappa\to M$ be the resulting elementary embedding. Note that $\mathrm{crit}(j)<\kappa$, because otherwise $f'=f$, so by cofinality, $\kappa=\mathrm{Ord}^M$, so $M=V\_\kappa$, so $g':\gamma\to\kappa$ is a surjection for some $\gamma<\kappa$, contradiction. But then $\mathrm{crit}(j)<\kappa$ is measurable.


Finally suppose that $\gamma<\kappa\leq 2^\gamma$. Then consider the theory consisting of the first order theory in parameters of $\mathcal{H}\_{(2^\gamma)^+}$, "I am wellfounded", and the statement "$\dot{f}:\mathcal{P}(\gamma)\to\mathrm{Ord}$ is surjective". Any sub-theory of size $<\kappa$ is satisfiable (in fact, any sub-theory of size $\leq 2^\gamma$ is satisfiable), so we have a model, and note it gives a measurable $\leq\gamma$.


(This leaves $\omega$, but it's easy to see this is impossible; also cf. Noah's comments in the question and below.)


