MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some help with this theorem: if $P_\beta=\langle P_\alpha,\dot{Q}_\alpha:\alpha\leq\beta\rangle$, $\beta<\omega_2$, is a CSI of proper forcings, $P_\alpha\Vdash \lvert \dot{Q}_\alpha\rvert\leq\aleph_1$, and CH holds in the ground model, then $P_\beta$ forces the CH in the generic extension.

A proof of this fact appears in the Handbook, but I need a diferent one. Just for simplicity, suppose $\beta=\omega$, then the problem is how to prove that $P_\omega$ forces the CH. Can someone give a clue or an idea of how to proceed?


share|cite|improve this question
I wouldn't mind seeing a more direct proof myself if there is one, but why do you need a different proof? – Amit Kumar Gupta Jan 27 '11 at 8:22
I agree with Gupta, seeing another would be nice. – Michael Blackmon Jan 27 '11 at 8:37

Here is a sketch. We may assume that each $\dot Q_\alpha$ has $\omega_1$ as its universe; in which case the underlying set of $P_{\alpha+1}$ can be taken to be $P_\alpha\times\omega_1$ and the ordering defined as usual. This shows that, by induction each $P_n$ has cardinality $\aleph_1$ (at most) and, by CH, so does $P_\omega$. As $P_\omega$ is proper for every name $\dot x$ of a real and every $p\in P_\omega$ there are $q\le p$ and a countable subset $\dot y$ of $P_\omega\times\omega$ (which then acts as a name of a subset of $\omega$) such that $q$ forces $\dot x=\dot y$. This means that those countable names produces the power set of $\omega$, so that, by CH again, in $V^{P_\omega}$ there are $\aleph_1$ many subsets of $\omega$.

share|cite|improve this answer
I'd believe it if you said, "As $P_{\omega}$ is ccc" instead of "proper," how does properness tell us there is some $q \leq p$ and a countable name $\dot{y}$ such that $q \Vdash \dot{x} = \dot{y}$? – Amit Kumar Gupta Jan 27 '11 at 10:28
Given $p$ and $\dot x$ (a subset of $P_\omega\times\omega$), take a countable elementary substructure $M$ of $H(\kappa)$ ($\kappa$ large enough) that contains $P_\omega$, $p$ and $\dot x$; take $q\le p$ that is $(M,P_\omega)$-generic. Then $q$ forces that $\dot x$ and $M\cap \dot x$ are the same real. – KP Hart Jan 27 '11 at 11:18
I am concerned about the cardinality; I think even $P*\dot{Q}$ might have cardinality $2^{\aleph_1}$. It's true we can take the underlying set of $\dot{Q}_\alpha$ to be forced to be $\omega_1$. But $P*Q$ consists of pairs $p,q$ where $q$ is a name for an element of $Q$ (ie of $\omega_1$). If $P$ has cardinality $\aleph_1$ there might be $2^{\aleph_1}$ such names. – Justin Palumbo Jan 27 '11 at 19:21
@KP, Yes I see, thanks. @Justin, $p \Vdash \dot{q} \in \omega_1$ then there is $p' \leq p$ and an ordinal $\alpha \in \omega_1$ such that $p' \Vdash \dot{q} = \alpha$. So $P \times \omega_1$ is dense in $P \ast \dot{Q}$, so we can just assume we're adding copies of $\omega_1$ at each stage of the iteration. – Amit Kumar Gupta Jan 27 '11 at 20:02
@Justin: For such an assumption, it is of course essential that $\aleph_1$ is not collapsed throughout the iteration (assuming $\dot{\aleph}_1$ is used rather than $\check{\aleph}_1$). This is of course guaranteed by the properness of the forcing so maybe $\aleph_1$-preserving is sufficient for your conclusion? I'd have to think more about this. @Amit: Yes, your argument does not require properness; so long as $\kappa$ is inaccessible and the forcing at each stage has hereditary size less than $\kappa$, the bounded support iteration of length $\kappa$ should always be $\kappa$-c.c. – Jason Jan 28 '11 at 10:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.