Countable chain condition and uncountable sets

Question

In the proof of Theorem 1.13 (a) here: http://www.math.psu.edu/jech/preprints/stat.pdf

Jech talks about a basic forcing fact which I can not prove:

If P is ccc and A is an uncountable subset of $\aleph_1$ in $V[G]$, then there is B in V which is an uncountable subset of A.

Could some body show me how to prove this? Thank you.

Answer 1

Unless I'm missing something, this statement is just false.

Consider the forcing $\mathbb{P}$ consisting of finite partial functions from $\omega_1$ to $2$ - basically, we're building a generic subset of $\omega_1$ similarly to how Cohen forcing builds a generic subset of $\omega$. By the $\Delta$-system lemma, $\mathbb{P}$ has c.c.c. Now let $g$ be $\mathbb{P}$-generic and $A=g^{-1}(1)$; then by genericity it's not hard to show that $A\not\supseteq X$ for any uncountable (even, any infinite!) $X\subseteq\omega_1$, but also that $A$ is unbounded.

What is true - and this is sufficient for Jech's purposes, and I think this is what he meant to write - is that any club $A\subseteq\omega_1$ in $V[G]$ has a club subset in $V$. (Note, incidentally, that all clubs in $V$ are still clubs in $V[G]$, so we're saved some possible confusion.)

This is a nice argument. Suppose $\mathbb{P}$ is a c.c.c. forcing, and $\nu$ is a name for a club subset of $\omega_1$. Let $\mu$ be the name for the principal function of $\nu$ - that is, $\mu[G](\eta)$ is the $\eta$th element of $\nu[G]$ in increasing order. We now argue as follows.

By c.c.c., for each $\eta$ the set $$Poss(\eta)=\{\alpha<\omega_1: \exists p\in\mathbb{P}(p\Vdash \mu(\eta)=\alpha)\}$$ is countable; we have $\sup(Poss(\eta))<\omega_1$ (since $Poss(\eta)$ is countable), and we also have that $\min(Poss(\eta))\ge\eta$ (since $\mu$ enumerates the elements of $\nu$ in increasing order.

We've "trapped" the possible elements of $\nu$ into relatively small blocks; we now use the fact that $\nu$ is a name for a club to show that this lets us know lots of the elements of $\nu$. Namely, observe that:

If $\beta_0<\beta_1<...$ with $\sup(Poss(\beta_i))<\beta_{i+1}$, then $\Vdash\sup\{\beta_i: i\in\omega\}\in\nu$.

This suggests a club subset of $A=\nu[G]$ in the ground model:

Let $B=\{\beta: \forall\gamma<\beta(\beta>\sup(Poss(\gamma)))\}$.

Clearly $B\in V$, $B$ is closed, and $B\subseteq A$ by the consideration above, so it's enough to show that $B$ is unbounded. But this is easy to do: for each $\beta<\omega_1$, define a sequence of ordinals as:

• $\beta_0=\beta$

• $\beta_{i+1}=\sup(Poss(\beta_i))+1$

and take $\alpha=\sup\{\beta_i: i\in\omega\}$; then $\alpha\in B$.