Does descendant or ascendant Löwenheim-Skolem fail for $\mathcal{L}_{\omega_1\omega}$ -logic?
Does Löwenheim-Skolem fail for $\mathcal{L}_{\omega_1\omega}$?
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logic
model-theory
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1The 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. – 2012-09-28
2 Answers
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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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0@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 $|T| + \aleph_0$. – 2012-09-30
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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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0You do know that the existence of a $\sigma$-complete ultrafilter requires a measurable cardinal, right? – 2012-09-28