9
$\begingroup$

$ZFC + \exists V_\alpha$ model of $ZFC \vdash Con(ZFC + \exists$ transitive standard model of $ZFC)$

and then

$ZFC + \exists$ transitive standard model of $ZFC \vdash Con(ZFC + \exists \omega-model$ of $ZFC)$

For the first one :

We can always find a countable extentional $M \subset V_\alpha$ elementary equivalent to $V_\alpha$. Let $M'$ be the mostowski collapse of $M$. $M' \approx M$ so $M'$ is model of ZFC. And because $M'$ is countable and transitive then $M' \in V_\alpha$ (since $H_{\omega_1} \subset V_{\omega_1}$ and $\alpha$ is surely far larger than $\omega_1$).

So $V_\alpha$ is the model of '$\exists$ a standard transitive model of ZFC'.

For the second one :

I don't really know how to do it... Does anyone have an idea ?

$\endgroup$
2
  • $\begingroup$ I'm not quite sure about the question. E.g. for (1), are you trying to prove that $(ZFC + (\exists a)(V_\alpha \vDash ZFC)) \vdash \mathrm{Con}(ZFC + (\exists x)(\text{x is a countable transitive model of ZFC}))$, or are you trying to prove in the metatheory that any model of $ZFC + (\exists a)(V_\alpha \vDash ZFC)$ can be used to create a model of $ZFC + (\exists x)(\text{x is a countable transitive model of ZFC})$? The former is done, essentially, by formalizing the proof of the latter in ZFC. $\endgroup$
    – Carl Mummert
    Apr 18, 2011 at 21:49
  • $\begingroup$ sorry, It was confusing, I fixed it. I want to do what you mention first. $\endgroup$
    – Archimondain
    Apr 18, 2011 at 22:12

1 Answer 1

Reset to default
10
$\begingroup$

Part 1 is correct. [And note that we get more: For example, your $M'$ in fact is a model of Con(ZFC+there is a transitive model of ZFC).]

For part 2: The statement "There is an $\omega$-model of ZFC" is $\Sigma^1_1$: Note that if there is an $\omega$-model, there is a countable one (take a countable elementary substructure), and now we can express this by saying that "there is a real $x$ coding a model of ZFC, and there is a real $y$ coding an order isomorphism of $\omega$ onto the natural numbers of the model coded by $x$".

Mostowski's absoluteness theorem gives us that any transitive model of ZFC is correct about $\Sigma^1_1$ statements (see Section 13 of Kanamori's book, for example). In particular, your transitive model is a model of the statement that there is an $\omega$-model.

$\endgroup$
13
  • $\begingroup$ I don't understand what you mean, but I ll have a look at what $\Sigma_1^1$ means. Somehow I managed something else (probably equivalent) : Any standard model is also an $\omega$-model, so assuming $Con(ZF)$, any $\omega$-model is also a model of $Con(ZF)$ (otherwise we would have a standard proof of 0=1, which is impossible assuming Con(ZF)). So by completness theorem, our standard model $M$ is model of $\exists (N,E) \vDash ZFC$. I just don't see why $(N,E)$ would be an $\omega$-model, but this is a good start I think ;) $\endgroup$
    – Archimondain
    Apr 18, 2011 at 22:03
  • $\begingroup$ Archimondian: the problem with your approach is the following: Assume only that there is an $\omega$-model $M$. Exactly the argument you suggest gives you that $M$ is a model of Con(ZFC), so it contains models of ZFC. But certainly we cannot expect those models to be $\omega$-models, or we contradict second incompleteness. So this approach cannot succeed. We need in the case you are interested in to use some some additional absoluteness fact that allows us to conclude that we have not just a model but an $\omega$-model. $\Sigma^1_1$-absoluteness does the trick in this case. $\endgroup$
    – Andrés E. Caicedo
    Apr 18, 2011 at 22:38
  • $\begingroup$ Kanamori's book is a good place to look this up; I think Jech also explains this material carefully. The argument is relatively simple: $\Sigma^1_1$ statements are statements of the form "There is a real $x$ such that $\phi(x)$", where $\phi$ is recursive. The key insight is that this is equivalent to saying that certain trees have branches, but ill-foundedness of trees is absolute. $\endgroup$
    – Andrés E. Caicedo
    Apr 18, 2011 at 22:40
  • 1
    $\begingroup$ @Andrés It's sometime later, maybe I get the argument now: by downward-LS if there is an $\omega$-model, then there is a (hereditarily) countable one, and then you can use reals to code things as you indicate in your answer (using en.wikipedia.org/wiki/Code_(set_theory)). Then by Mostowski absoluteness, since standard models are $\omega$-models (and we have a standard model), in every transitive model of ZFC the sentence "there exists an $\omega$-model" is true. In particular so in the standard transitive model we assumed to exist (...) $\endgroup$
    – Jori
    Apr 26, 2020 at 16:19
  • 1
    $\begingroup$ So that model is a model of ZFC + "there exists an $\omega$-model". And so by soundness: ZFC + "there exists an $\omega$-model" is consistent. $\endgroup$
    – Jori
    Apr 26, 2020 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.