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Is $ \text{Con}(\mathsf{ZFC}) $ absolute for transitive models of $ \mathsf{ZFC} $? It appears that $ \text{Con}(\mathsf{ZFC}) $ is a statement only about logical syntax. Taking any $ \in $-sentence $ \varphi $, we can write $ \text{Con}(\mathsf{ZFC}) $ as $ \mathsf{ZFC} \nvdash (\varphi \land \neg \varphi) $, which appears to be an arithmetical $ \in $-sentence.

If this is true, then I think one can get a quick proof of $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \nvdash \langle \text{There exists a transitive model of $ \mathsf{ZFC} $} \rangle, $$ assuming that $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is consistent.

Proof If $$ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) \vdash \langle \text{There exists a transitive model of $ \mathsf{ZFC} $} \rangle, $$ then let $ M $ be such a transitive model. By the absoluteness of $ \text{Con}(\mathsf{ZFC}) $, we see that $ M \models \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $. Hence, $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ proves the consistency of $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $. By Gödel’s Second Incompleteness Theorem, $ \mathsf{ZFC} + \text{Con}(\mathsf{ZFC}) $ is therefore inconsistent. Contradiction. $ \blacksquare $

Question: Is $ \text{Con}(\mathsf{ZFC}) $ absolute for transitive models, and is the above proof correct?

Thanks for any clarification.

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That's distinctly clearer, post edit -- so I've deleted my answer. Cheers, and sorry if I was being dense! (Too much Christmas imbibing dissolving the grey cells?) –  Peter Smith Dec 26 '12 at 16:20
    
But NB what you say about Con(ZFC) in the second sentence isn't right. $Con(ZFC)$ is the arithmetization of a claim about what can't be proved in $ZFC$. –  Peter Smith Dec 26 '12 at 16:27
    
By Arithmetical relations, I mean a definable subset of $HF = H(\aleph_0)$ with parameters from $HF$. –  William Dec 26 '12 at 16:33
    
If $ZFC$ is inconsistent, then for any $\varphi$, $ZFC \vdash \neg(\varphi \land \neg\varphi)$, and your $Con(ZFC)$ would still come out true. Not what you mean, surely. You want something along the lines of $ZFC \nvdash (\varphi \land \neg\varphi)$ or the usual $ZFC \nvdash \bot$ no? –  Peter Smith Dec 26 '12 at 16:38
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This proof is correct, but there is another way of proving it. If there is any transitive set model $M$ of ZFC, by taking the constructible universe $L^M$ inside $M$ we find that $L_\alpha$ is a transitive model of ZFC for some ordinal $\alpha = \text{Ord}^M$. Then, if $\beta$ is minimal such that $L_\beta$ satisfies ZFC, we have that $L_\beta$ satisfies "there is no transitive model of ZFC". –  Carl Mummert Jul 21 '13 at 19:46
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up vote 3 down vote accepted

Yes, $\text{Con}(\mathsf{ZFC})$ is an arithmetic statement ($\Pi^0_1$ in particular, because it says a computer program that looks for an inconsistency will never halt) so it is absolute to transitive models, and your proof is correct.

By the way, there are a couple of ways you can strengthen it. First, arithmetic statements are absolute to $\omega$-models (models with the standard integers, which may nevertheless have nonstandard ordinals) so $\text{Con}(\mathsf{ZFC})$ does not prove the existence of an $\omega$-model of $\mathsf{ZFC}$. Second, the existence of an $\omega$-model of $\mathsf{ZFC}$ does not prove the existence of a transitive model of $\mathsf{ZFC}$, because the existence of an $\omega$-model of $\mathsf{ZFC}$ is a $\Sigma^1_1$ statement, and $\Sigma^1_1$ statements are absolte to transitive models.

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