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This question already has an answer here:

This is probably a pretty simple question, but I'm tying myself in knots over it.

We're all familiar with the Reflection Theorem, Lowenheim-Skolem Theorem, and Mostowski Collapse Lemma for getting countable transitive models of finite fragments of ZFC. You take a finite fragment $\Gamma$, use the reflection theorem to get that $\Gamma$ holds in some $V_\kappa$, Skolemise over $V_\kappa$, and then Collapse to a ctm.

My question. Is there (or could there be) a countable transitive model satisfying the same first-order truths as $V$? Obviously, by Tarski, such a countable transitive model is not first-order definable.

A possible route:

Suppose full second-order reflection is true of $V$. Let $A$ be a second-order parameter that contains a witness for every parameter-free sentence of first-order $ZFC$ true in $V$. Then, just reflect the sentence $x \in A$, Skolemise, and Collapse as normal (I feel like I might be pulling a fast-one here).

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marked as duplicate by Mario Carneiro, Community Oct 22 '15 at 17:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Can also do this with a truth predicate. Add it, add the Tarski axioms, expand Replacement for the new language, and then reflection will be provable for the expanded language. So there'll be a $V_\alpha$ which is an elementary substructure of $V$. Then proceed as above. $\endgroup$ – GME Oct 22 '15 at 17:05
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Well. There could be, or there might not be.

If there are no transitive models, then of course the answer is negative. But it could be positive, and in fact without evening increasing the consistency strength of $\sf ZFC$.

We augment the language of $\sf ZFC$ by adding a constant symbol $M$, and we add the following axioms:

  1. $M$ is countable and transitive.
  2. For every $\varphi$ in the language of set theory (without $M$, that is), $\varphi\leftrightarrow\varphi^M$.

If $\sf ZFC$ is consistent, then any finite fragment of this theory is consistent due to the reflection theorem. Simply find a large enough $\alpha$ such that $V_\alpha$ reflects whatever you wanted, and find an elementary equivalent countable transitive model.

If $V$ satisfies this theory, then $M$ is a countable transitive model and $\varphi\leftrightarrow\varphi^M$ holds there. So $M$ is in fact a countable model which is elementary equivalent to $V$.

(The entire thing is due to Feferman.)

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  • $\begingroup$ Boom. Ace as always Asaf. What's the source for that? $\endgroup$ – Neil Barton Oct 22 '15 at 17:02
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    $\begingroup$ I actually found that this is a duplicate, and the answer I wrote on the other answer has a reference. math.stackexchange.com/questions/811042/… $\endgroup$ – Asaf Karagila Oct 22 '15 at 17:03
  • $\begingroup$ Just to be 100% clear: The existence of a set of Skolem functions for V is inconsistent by Tarski? This looks obvious, but I've got this sort of thing mixed up before. $\endgroup$ – Neil Barton Oct 22 '15 at 18:29
  • $\begingroup$ Well, depends on what you mean by a Skolem function for V, because that would seem to me as each one of those is a proper class. And there is no set whose elements are proper classes. $\endgroup$ – Asaf Karagila Oct 22 '15 at 19:12
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    $\begingroup$ Well, since the statements you Skolemize don't live in V, but in the meta-theory, Tarski's theorem really ensures you that even if there is such set, V doesn't know that it is a set of Skolem functions, or what statement each function is Skolemizing. For example, if $\kappa$ is inaccessible, $V_\kappa$ knows the set of its Skolem functions and its truth and so on. It just doesn't know that those are in fact the Skolem functions and truth definition etc. $\endgroup$ – Asaf Karagila Oct 22 '15 at 19:16

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