$\sf LS(\aleph_0)$: Every model $M$ of a first order theory $\sf T$ with countable signature has an elementary submodel $N$ which is at most countable.

Now this theorem is equivalent over axioms of $\sf ZF$ to the axiom of dependet choice $``\sf DC"$.

Now does this mean that in $\sf ZF + \neg DC$, we can have a consistent theorey in a countable signature that doesn't have a countable model?


No. If a theory is well-orderable, then it has a well-orderable model satisfying all the usual properties in $\sf ZFC$.

To see why, note that you can code all the thing into a set of ordinals $A$, then in $L[A]$ your theory exists and it is a model of $\sf ZFC$, so the theory has a countable model. But this is upwards absolute to $V$.

  • 1
    $\begingroup$ That's a massively overkill argument - just observe that the usual model construction goes through without choice if the theory is well-orderable. (+1 of course, though.) $\endgroup$ Jul 23 at 17:59
  • $\begingroup$ Sure, but that's a neater argument. $\endgroup$
    – Asaf Karagila
    Jul 23 at 18:00
  • $\begingroup$ can we have first order theories in countable signature that are not well orderable? $\endgroup$
    – Zuhair
    Jul 23 at 20:05
  • 1
    $\begingroup$ @Zuhair: No. If a theory is countable, the signature is countable. If a signature is countable, there are countably many sentences from whence a theory is born. If a signature is well-orderable, then there are well-orderably many sentences. If you just add symbols, but never ever use them, what's the point of having them? $\endgroup$
    – Asaf Karagila
    Jul 23 at 20:06

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