# Iterated forcing and CH

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?

Thanks.

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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 $\omega$.
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