In their paper "On Löwenheim-Skolem-Tarski numbers for extensions of first order logic", Magidor and Väänänen make the following statement:

"For second order logic, $LS(L^{2})$ [the Löwenheim-Skolem number for second order logic--my comment] is the supremum of $\Pi_{2}$-definable ordinals..., which means that it exceeds the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinal if they exist ["the Löwenheim-Skolem number $LS(L^{2})$ of second order logic $L^{2}$ is the smallest cardinal $\kappa$ such that if a theory $T$$\subset$$L^{2}$ has a model, it has a model of cardinality$\le$max($\kappa$,$|T|$)", and "$L^{2}$ extends first order logic with quantifiers of the form $\exists$$R$$\phi$($R$,$x_0$,...,$x_{{n}-1}$), where the second order variable $R$ ranges over n-ary relations on the universe for some fixed n"--my comment also but substantially quoting the authors]".

Assume that the first measurable, the first $\kappa^{+}$-supercompact $\kappa$, and the first huge cardinals exist. What type of large cardinal, then, is $LS(L^{2})$? If the answer is known, please provide the reference.

  • $\begingroup$ Magidor has results on this. $\endgroup$ – Monroe Eskew Jul 19 '15 at 3:49
  • 2
    $\begingroup$ Why did this get a -1? Seems a fine question to me. $\endgroup$ – Noah Schweber Jul 19 '15 at 4:23
  • 5
    $\begingroup$ Thomas, in your definition of the Löwenheim-Skolem number, shouldn't it be $\kappa\leq\text{max}(\kappa,|T|)$, rather than $<$? $\endgroup$ – Joel David Hamkins Jul 19 '15 at 12:20
  • 1
    $\begingroup$ Strange. Wikipedia defines it as I suggest, and that definition makes sense to me: en.wikipedia.org/wiki/L%C3%B6wenheim_number#Extensions. $\endgroup$ – Joel David Hamkins Jul 20 '15 at 1:10
  • 1
    $\begingroup$ @JoelDavidHamkins is right. You can't reasonably expect to get models of $T$ that are smaller than $|T|$. $\endgroup$ – Andreas Blass Jul 20 '15 at 15:10

The following is due to Magidor:

Theorem 1. Is $\kappa$ is a strong cardinal, then $LS(L^2) < \kappa.$

The proof if easy. Let $T \subseteq L^2$ be a theory and let $A$ be a model for $T$. e may assume the universe is some cardinal $\delta.$ Take some cardinal $\beta > \beth_{\omega}(\delta)$, and let $j: V \to M$ witness $\kappa$ is $\beta$-strong. It is easily seen $M\models$``$A \models T$'', so $M \models \exists B( B \models T, |B| < j(\kappa))$. By elementarity in $V$, $T$ has a model of size $< \kappa.$

Also note that for any theory $T \subseteq L^2,$ there is a least $\delta_T$ such that if $T$ has a model, then it has a model of size $< \delta_T.$ Then $LS(L^2)=\sup \{\delta_T: T$ as above $\}$, so $LS(T^2)$ can be singular, even though it can be above some very large cardinals.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.