For a logic $\mathcal{L}$, say that a cardinal $\kappa$ is $\mathcal{L}$-correct iff every satisfiable $\mathcal{L}$-theory of size $<\kappa$ has a model of size $<\kappa$. First-order correctness is of course boring, but quite quickly we enter the realm of strong large cardinal properties (see e.g. here).

I'm interested in a kind of "iterated correctness:" repeatedly add to a given logic the ability to quantify over cardinals which are correct for it (or rather, for the previous iteration of this process). This doesn't make obvious sense in general, but for logics with reasonably nice syntax things are better. In particular, I'm interested in what happens when we iteratively "add correctness quantifiers" to second-order logic, as follows:

Let $\mathcal{L}^2_0$ be usual second-order logic, and let $\mathcal{L}^2_{n+1}$ be $\mathcal{L}_n^2$ augmented with a quantifier $$\mathsf{C}_nx\varphi(x)\equiv\mbox{$\vert\{x: \varphi(x)\}\vert$ is $\mathcal{L}^2_n$-correct.}$$ So $\mathcal{L}^2_k$, in addition to the usual logical symbols, has $k+2$ different types of quantifier: the first-order quantifiers, the second-order quantifiers, and $k$-many correctness quantifiers. These quantifiers can alternate however is desired.

My question is:

Do any of the "usual" large cardinal properties imply $\mathcal{L}^2_n$-correctness for every $n$?

This is not at all obvious to me. On the other hand, I can't pin down a concrete obstacle to e.g. supercompactness having the above property.

  • $\begingroup$ Whatever it is, it looks like it should be at most an extendible. $\endgroup$
    – Asaf Karagila
    Sep 29 '20 at 0:48
  • $\begingroup$ @AsafKaragila I'd love it if that were the case, but I don't see how to get that bound. (Maybe it's obvious and beer isn't helping my set theory.) $\endgroup$ Sep 29 '20 at 2:13
  • $\begingroup$ If beer isn't helping, try another one. In any case, argue by induction that if $\delta$ is extendible, then the $n$th logic still takes place below it. If it wasn't, for every theory, pick an elementary embedding that moves $\delta$ high enough, and then by elementarity we can pull back to below $\delta$. But that was just off the cuff thought before falling asleep. I must be missing something. $\endgroup$
    – Asaf Karagila
    Sep 29 '20 at 3:48
  • $\begingroup$ Why can't we consider $C_n$ to be just a unary predicate which is evaluated as the set of all $\mathcal{L}^2_n$-correct cardinals? I don't understand why $C_n$ is a quantifier and not simply an additional predicate. $\endgroup$
    – Yair Hayut
    Sep 29 '20 at 12:41
  • 1
    $\begingroup$ @YairHayut I see the issue, I was unclear in my definition (and comment): I mean that the cardinality of the set in question is appropriately correct. So it's not coding-dependent and is genuinely a quantifier. $\endgroup$ Sep 29 '20 at 20:58

Assume ZFC. Let $n$ be a meta-integer. Then $\kappa$ is $\mathcal{L}^2_n$-correct iff $\kappa$ is $\Sigma_{n+2}$-reflecting, i.e. $V_\kappa\preccurlyeq_{n+2}V$.

Proof: For $n=0$, i.e. 2nd order logic, we have: If $V_\kappa\preccurlyeq_2 V$ then easily $\kappa$ is $\mathcal{L}^2_0$-correct. Suppose now that $\kappa$ is $\mathcal{L}^2_0$-correct. Now for each $\alpha<\kappa$, if $|V_\alpha|<\kappa$ then $|V_{\alpha+1}|<\kappa$, as is easily seen using $\mathcal{L}^2_0$-correctness. So there is a limit ordinal $\eta\leq\kappa$ such that $|V_\eta|=\kappa$. Now observe that $V_\eta\preccurlyeq_2 V$ by using $\mathcal{L}^2_0$-correctness. It follows that $\eta=\kappa$ (if $\eta<\kappa$, take $\alpha<\eta$ such that $\eta\leq\beta=|V_\alpha|<\kappa$, and using $V_\eta\preccurlyeq_2 V$, show $\beta<\eta$ for a contradiction). So $V_\kappa\preccurlyeq_2 V$ as desired.

For $n=1$: Suppose $V_\kappa\preccurlyeq_3V$. Let $T\in V_\kappa$ be a satisfiable $\mathcal{L}^2_1$-theory. Then there is an ordinal $\eta$ such that $V_\eta\preccurlyeq_2 V$ and $M\in V_\eta$ such that $V_\eta\models$"$M\models T$", where the latter statement uses $V_\eta$'s own notions to compute $\Sigma_2$-elementarity, and hence the $C_0$ quantifier. But this is a $\Sigma_3$ statement, so $V_\kappa$ models it, and note that this really gives a true model of $T\in V_\kappa$.

Conversely, suppose $\kappa$ is $\mathcal{L}^2_1$-correct. So in particular it is $\mathcal{L}^2_0$-correct, and hence $V_\kappa\preccurlyeq_2V$. We want to see $V_\kappa\preccurlyeq_3V$. Let $x\in V_\kappa$ and suppose $V\models\exists w\varphi(x,w)$ where $\varphi$ is $\Pi_2$. Let $\alpha<\kappa$ with $x\in V_\alpha$. Consider the $\mathscr{L}^2_0$-theory theory $T$ in parameters in $V_\alpha$, which describes a rank segment $V_\eta$ of $V$ such that $\eta$ is a cardinal and $V_\eta\preccurlyeq_2V$ (by using the $C_0$-quantifier to require $\eta$ to be $\mathcal{L}^2_0$-correct), and $V_\eta\models\exists w\varphi(w,x)$. Note this is satisfiable, and hence satisfiable in cardinality ${<\kappa}$, and hence $V_\kappa\models\exists w\varphi(w,x)$.

Now proceed in this manner.

So in particular, least $\mathcal{L}^2_0$-correct cardinal is $<$ least strong and least supercompact, these are $<$ least $\mathcal{L}^2_1$-correct $<$ least extendible < least $\mathcal{L}^2_2$-correct.


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