# When collapsing a cardinal, what ordinal does it become?

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.

• 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? Aug 13, 2012 at 5:09
• 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 $\omega_1^V$. Aug 13, 2012 at 5:15

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.

• 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. Aug 13, 2012 at 6:52
• @Jambalaya: I edited to answer your questions. I hope it helps. Aug 13, 2012 at 6:59
• 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! Aug 13, 2012 at 7:29
• @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 $\omega_1^L$. Aug 13, 2012 at 7:34
• 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. Aug 13, 2012 at 7:54