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Does descendant or ascendant Löwenheim-Skolem fail for $\mathcal{L}_{\omega_1\omega}$ -logic?

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    $\begingroup$ The question could be answered by looking at any standard reference, and as it is currently worded it shows little sign of effort being placed into writing it. I downvoted it for this reason. $\endgroup$
    – Carl Mummert
    Sep 28, 2012 at 20:30

2 Answers 2

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The downward Löwenheim-Skolem theorem continues to hold. The upward one does not; for example the infinitary formula $\bigvee_{n \in \omega} (x = n)$ has no uncountable models. This implies that the compactness theorem also fails, because it implies the upward Löwenheim-Skolem theorem.

This is all described in the usual references. You might start with the article on Infinitary logic at the Stanford Encyclopedia.

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    $\begingroup$ Compactness fails, but completeness still holds. Right? $\endgroup$
    – Asaf Karagila ♦
    Sep 28, 2012 at 20:43
  • $\begingroup$ @Asaf: the completeness theorem holds in a certain sense. Compared to first order logic, the deductive system is expanded and allows deductions of countable length rather than finite length. Second, the set of non-logical premises being assumed has to be countable. With these modifications, $L(\omega_1, \omega)$ has a completeness theorem. The SEP article talks about this. $\endgroup$
    – Carl Mummert
    Sep 29, 2012 at 0:59
  • $\begingroup$ Doesn't downward L-S fail also since the ordered field of reals can be characterized up to isomorphism by an uncountable set of $\mathcal{L}_{\omega_{1}, \omega}$-sentences? $\endgroup$
    – Quinn Culver
    Sep 30, 2012 at 4:07
  • $\begingroup$ @Quinn Culver: see the article I linked. Even for first-order logic the D-L theorem only says that for a theory $T$ you can get a model of size $|T| + \aleph_0$. $\endgroup$
    – Carl Mummert
    Sep 30, 2012 at 11:51
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I think, if ultrafilters everywhere in the proof are replaced by $\omega_1$-ultrafilters (those which are closed under intersection of $\omega$ many sets), then it is still valid in that form.

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  • $\begingroup$ You do know that the existence of a $\sigma$-complete ultrafilter requires a measurable cardinal, right? $\endgroup$
    – Asaf Karagila ♦
    Sep 28, 2012 at 17:12

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