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$ZFC + \exists V_\alpha$ 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 ?

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    sorry, It was confusing, I fixed it. I want to do what you mention first.2011-04-18

1 Answers 1

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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.

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    Well, I am a bit disappointed with the $\Sigma_1^1$ thing, it seems like a there is a big connection I did not suspected at all between 'set theory (independance proof)' and 'descriptive set theory'. (and so a lot of work to do in order to get all of it...). Thanks a lot for your answer. I'll have a look at kanamori's book.2011-04-18