Work in $V$. Let $P = \text{Col}(\omega, \omega_1)$ and suppose that $G$ is generic for $P$ over $V$. Then $V[G]\models |\omega_1^V|=\aleph_0$ and $\omega_2^V=\aleph_1$. In particular, $V[G]\models\omega_1^V$ is an ordinal between $\omega$ and $\omega_1^{V[G]}$. What ordinal is it? I would guess that the answer depends on $G$, and the best that we can say is that it is a limit ordinal.

  • $\begingroup$ It is the same ordinal it always was. No ordinals are created or change their order by collapsing. Can you make more precise what you mean? $\endgroup$ Aug 13, 2012 at 5:09
  • $\begingroup$ Yes, I was afraid that my question was malformed; I'll try to be clearer. In $V[G]$ I'd like to write $\omega_1^V$ in Cantor normal form. Another way of asking this: if we momentarily forget that we are in a forcing extension, and view $V[G]$ as our universe, what ordinal between $\omega$ and $\omega_1$ was our old $\omega_1$ collapsed to? You wouldn't call it $\omega_1^V$. $\endgroup$
    – Jambalaya
    Aug 13, 2012 at 5:15

1 Answer 1


I think that you're missing the point of the Levy collapse.

Forcing [over transitive models] does not add ordinals. It does not remove ordinals either. What the collapse does is to add a bijection between $\omega$ and $\omega_1$.

Note that all the "definable" ordinals ($\varepsilon_0$, etc.) and so are very small, and $\omega_1^V$ is far far beyond them.

And to your comment, yes. In fact this is what we would call it: $\omega_1^V$. If $\alpha$ is a countable ordinal and we know that there is an inner model $M$ in which $\alpha=\omega_1^M$ then we immediately know two things:

  1. $\alpha$ is quite a large countable ordinal; and
  2. $\omega_1^M$ is an excellent way to name it.

In the case of forcing we actually start with the inner model.

For the comment:

  1. Definability is a strong word. If the ground model was $L$, certainly $\omega_1^L$ is a definable ordinal. It is the least ordinal that there is no bijection between him and $\omega$ which satisfies the constructibility axiom. Furthermore, we now know that the ground model is definable with parameters. This means that $\omega_1^V$ is definable from parameters in $V[G]$ by the same trick.

  2. Note that ordinal arithmetics are not changed by forcing. This means that $\omega_1^V$ is an $\varepsilon$ number in $V[G]$ since $\omega^{\omega_1}=\omega_1$ in $V$; furthermore it is a fixed point of $\varepsilon$ numbers, for the same reasons. Namely if $\alpha=\omega_1^V$ then $\alpha=\varepsilon_\alpha$, which in $V$ is the $\omega_1$-th and in $V[G]$ is not.

  3. Since $\omega_1^V$ is an $\varepsilon$ number its Cantor normal form is in fact $\omega_1^V$, so there is no simpler way of writing it.

  • $\begingroup$ Thank you for your response. I apologize the question wasn't clearly stated, I am aware that we don't add or destroy ordinals. However, I think your reply narrows down what the issue is for me. Are you saying that this ordinal is not definable in $V[G]$? Could you say a bit about what you mean by that? Thank you. $\endgroup$
    – Jambalaya
    Aug 13, 2012 at 6:52
  • $\begingroup$ @Jambalaya: I edited to answer your questions. I hope it helps. $\endgroup$
    – Asaf Karagila
    Aug 13, 2012 at 6:59
  • $\begingroup$ Thank you Asaf, it helps a lot! A followup to help drive it home: Suppose that we start with the universe V, and then we are told that actually V was obtained as a forcing extension from some inner model, by collapsing $\omega_1$. However, we are not told what the inner model was. So we know there is some countable ordinal $\alpha$ that was $\omega_1$ of an inner model. But at best we could say it must be a countable fixed point of the $\epsilon$ numbers, correct? Thank you! $\endgroup$
    – Jambalaya
    Aug 13, 2012 at 7:29
  • $\begingroup$ @Jambalaya: We can do a bit more. Any "nice" ordinal function similar to $\varepsilon$ will behave in a similar fashion. We can say a bit more, too. It will never be below $\omega_1^L$. $\endgroup$
    – Asaf Karagila
    Aug 13, 2012 at 7:34
  • $\begingroup$ Interesting! I want to know more but don't want to take up any more of your time. Could you suggest any papers or keywords I might look up to learn more? Thanks again for all your help. $\endgroup$
    – Jambalaya
    Aug 13, 2012 at 7:54

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.