I made a proposition about models of ZFC, which says that every countable model of ZFC is really countable in larger countable model:

For every countable transitive model $M$ of ZFC there is a countable transitive model $N\supset M$ of ZFC such that $M\in N$ and $(|M|=\aleph_0)^N$.

Of course, ZFC + "there is no transitive model of ZFC" implies above statement trivially. However I wonder about non-trivial case. I prove that if there is a proper class of worldly cardinal, then the proposition holds. Here is my proof: For a model $M$ and a bijection $f:M\to\omega$, there is a worldly cardinal $\alpha$ such that $M, f\in V_\alpha$. Taking a Skolem closure to $\{M,f\}\cup M$ over $(V_\alpha,\in)$ and collapsing it gives a countable transitive model $N$ of ZFC. (We can find the transitive collapse because $\in$ is well-founded.) $M$ and $f$ are in the transitive collapse $N$ because the model before collapsed contains every elements in $M$ and $M$ is transitive.

Is my proof correct? I also wonder the consistency strength of the proposition. Thanks for any help.

  • 3
    $\begingroup$ The condition that $(|M|=\aleph_0)^N$ can actually be ignored, since as soon as you have an $N$ such that $M\in N$ you can make $M$ countable by forcing. $\endgroup$ – Eric Wofsey Jul 22 '16 at 4:06
  • 3
    $\begingroup$ Note that since $M$ is countable and transitive, $M\in V_{\omega_1}$. So we only need one worldly cardinal for your argument to work. Also, I think the right formulation for your proposition (to avoid trivialities) is "Any countable transitive set $A$ is contained in a countable transitive $M$ such that $M\models ZFC+\vert A\vert=\aleph_0$." $\endgroup$ – Noah Schweber Jul 22 '16 at 4:29
  • 3
    $\begingroup$ Since we're piling things you can say, this is the same as just saying that every real is an element of a transitive model. $\endgroup$ – Asaf Karagila Jul 22 '16 at 6:27

Your Answer

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

Browse other questions tagged or ask your own question.