Countable transitive model of ZFC?

Question

If there is countable transitive model of ZFC, this model cannot capture all ordinals of ZFC. But we use it for stuffs like forcing.

But this seems to violate ZFC's axioms - for example, power set axiom (take, $\omega = \aleph_0$ and apply power set operator.).

So how can we use countable transitive model of ZFC, then?

score 13 · Accepted Answer · edited Apr 13 '17 at 12:21

13

You are mixing internal and external point of views.

If $M$ is a countable transitive model of ZFC then it doesn't know that it is countable, and it certainly don't know that all its members are countable. That is to say, there are many sets $x\in M$ such that $M\models\aleph_0<|x|$. So while that in the full universe there is a bijection between $x$ and $\omega$ it is not an element of $M$.

When we use c.t.m for forcing we work internally when we define the forcing poset, and we argue externally when we prove that there is a generic set, and that the construction results in another countable transitive model of ZFC.

Somewhat related:

edited Apr 13 '17 at 12:21

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answered Feb 23 '13 at 14:47

Asaf Karagila

393,674

A countable transitive model of ZFC means that it obeys axiom of regularity, which says that all sets are well-founded. The question is, by Lowenheim-Skolem, we can guarantee that a countable model exists. But can we use Mostowski collapse lemma to guarantee that countable transitive model exists? – Mark Zwazingker Feb 23 '13 at 15:11
Mark, the existence of transitive countable models is stronger than the existence of countable models. Internally all models of ZFC are well-founded, but we can only collapse those which are really well-founded; those which are not well-founded of course are not aware of this fact. – Asaf Karagila Feb 23 '13 at 15:12
Just to make sure, I understand that internal and external views can differ. I was just asking an additional question. Thanks for all your help :) – Mark Zwazingker Feb 23 '13 at 15:15
So do we just have to blindly assume that such countable transitive model, which it internally obeys ZFC axioms, exists? Is there any way to show this type of model can exist? – Mark Zwazingker Feb 23 '13 at 15:17
Mark, ZFC itself cannot even prove it has any model at all. If we want to work with models then we need to assume they exist. For forcing we prefer [countable] transitive models. Of course, if we assume large cardinals then this becomes trivial because large cardinals prove the existence of transitive models, then we can use Skolem-Lowenheim and the Mostowski collapse and prove there are countable transitive models. – Asaf Karagila Feb 23 '13 at 15:21
But Mostowski collapse lemma can only be used if model is really well-founded externally - if so, how can we use Mostowski collapse lemma even if we assume large cardinals? – Mark Zwazingker Feb 23 '13 at 15:26
Mark, if you have an inaccessible cardinal $\kappa$ then $V_\kappa$ is a transitive model of ZFC, take a countable elementary submodel, since its $\in$ relation is the true $\in$ this model must be well-founded, so we can collapse it. Now we have a countable transitive model. But do note that large cardinals are much much stronger in consistency strength than the existence of transitive models. – Asaf Karagila Feb 23 '13 at 15:28
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Perhaps it is convenient to point out that the talk of countable transitive models is unnecessary, but facilitates exposition. We can use the language of Boolean valued models to do forcing arguments without the need for any assumptions beyond ZFC. – Andrés E. Caicedo Feb 23 '13 at 17:23
@Andres: Yes, that thought has crossed my mind in the writing of the one of the comments. Thank you. – Asaf Karagila Feb 23 '13 at 17:25
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The other comment to add is that, by the reflection theorem, without additional assumptions, we do have countable transitive models of any finite fragment of ZFC, which in practice suffices for forcing arguments. – Andrés E. Caicedo Feb 23 '13 at 17:28
@Mark, you may want to read the comments of Andres; Andres: on the second remark I did not think... :-) – Asaf Karagila Feb 23 '13 at 17:30

score 5 · Answer 2 · answered Feb 23 '13 at 14:48

The set that serves within the model as the power set of a set would not contain all subsets of the set, but only all subsets that belong to the model.

The theorem that the power set of an infinite set is uncountable would be true within the model simply because such a set would not be "internally countable", i.e. no enumeration of the set would be a member of the model.

score 5 · Answer 3 · answered Feb 23 '13 at 14:48

Let $M$ be a countable transitive model of $\mathsf{ZFC}$. Obviously $\wp(\omega)^M$, the set that $M$ ‘thinks’ is the power set of $\omega$, is countable in $V$, since it’s a subset of $M$, but there is no $f\in M$ such that $M\models\text{'}f\text{ is a bijection between }\omega\text{ and }\wp(\omega)^M\text{'}$. Thus, from $M$’s point of view $\wp(\omega)^M$ is uncountable.

It’s hard at first, but you have to keep straight what $V$ ‘thinks’ and what $M$ ‘thinks’.

Countable transitive model of ZFC?

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