6
$\begingroup$

Let $M$ be a model of ZFC and take the forcing notion $P(\omega,\omega_2)$ where: $P(\omega,\omega_2)=\{p|p \space is \space a \space function \space and \space \exists n \space s.t. (dom(p)=n) \space n \space and \space ran(p)\subset \omega_2 \}$. ($G$ is a $P$-generic filter)

I am trying to show that in $M[G]$ we can collapse $\omega_2$ to a smaller cardinal.

Any idea how this can be done?

Thank you

EDITenter image description here

$\endgroup$
4
  • 2
    $\begingroup$ What do you mean by "we can collapse"? Did you mean "we collapse" (as in: $\omega_2^{M[G]}<\omega_2$) or something else? The phrasing suggests that what you want is to show that in $M[G]$ there is a further forcing poset such that, forcing with it over $M[G]$, will collapse $\omega_2^{M[G]}$. But maybe you mean something different... $\endgroup$ Mar 1, 2015 at 22:58
  • 1
    $\begingroup$ To possibly add to Andres's question. Which $\omega_2$. When you talk about forcing, particularly multiple forcings, you really need to qualify where things live. If I'm reading your first forcing right and you mean to say $n\in\omega$ then you can't collapse the base model $\omega_2^{M}$ any further since the first forcing collapsed it to $\omega^{M}$. On the other hand if you mean the extensions $\omega_2^{M[G]}$ then it seems a strange question since you can just force again with $P(\omega^{M[G]},\omega_2^{M[G]})$ to collapse the $\omega_2$. $\endgroup$
    – DRF
    Mar 2, 2015 at 7:03
  • 1
    $\begingroup$ I have added the source for my question $\endgroup$
    – user135172
    Mar 2, 2015 at 8:39
  • $\begingroup$ You're using the term "collapse" wrong. You can never use the Mostowski collapse to collapse an ordinal, because the Mostowski collapse is the identity for transitive sets, in particular for ordinals. When we say that we collapse a cardinal in a generic extension, what we mean to say is that it is no longer a cardinal; or that we collapsed some interval below that cardinal (e.g. "The inaccessible is collapsed to be the new $\omega_1$" means that we collapsed the cardinals below the inaccessible, but not the inaccessible itself). $\endgroup$
    – Asaf Karagila
    Mar 2, 2015 at 10:29

1 Answer 1

3
$\begingroup$

For each $\alpha<\omega^M_2$, it is easy to show that $\{p\in P: \alpha\in rng(p)\}\in M$ is dense in $P$. So $rng(\cup G) = \omega^M_2$ and $M[G]\vDash \omega = |\omega^M_2|$.

$\endgroup$
4
  • $\begingroup$ Sorry for the "stupid question", but, why is it important that this set will be dense? density will give me that every generic filter intersects it. So, it means I get a function which is onto? Why can't we do the same thing with $P(\omega_2 \times \omega,2)$? $\endgroup$
    – user135172
    Mar 2, 2015 at 15:15
  • 1
    $\begingroup$ That's right; you get a function from $\omega$ which is onto $\omega^M_2$. Why would you think that the same thing could be done with $P(\omega_2\times\omega, 2)$? $\endgroup$
    – user104955
    Mar 2, 2015 at 15:40
  • 2
    $\begingroup$ I see. Thank you. The thing is that I am not sure why something like this can not be done. I am not sure why it is true in example 20 above that we get not only ${2^{\omega}}^{M}={\omega_2}^{M}$ but also, ${2^{\omega}}^{M[G]}={\omega_2}^{M[G]}$. I can't put my finger on the reason that allows me to move from what is true in $M$ to what is true in $M[G]$. $\endgroup$
    – user135172
    Mar 2, 2015 at 15:54
  • 1
    $\begingroup$ Well, the question why $M[G]\vDash 2^\omega = \omega_2$ is different, and I take it what example 20 is intending to show. The bit underlined in red is just pointing out that it won't in general be true that $\omega_2^M = \omega^{M[G]}_2$. In particular, it won't be if we force over $P(\omega, \omega_2)$ for the reason I outlined. $\endgroup$
    – user104955
    Mar 2, 2015 at 22:07

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.