It's the first time I'm posting here so I don't know if I really should put this question here... I tried to post it on math.stackexchange, but a friend told me I would get better results by posting it here. So I decided to give it a try. The original posting is here. https://math.stackexchange.com/questions/668929/consistency-of-p-1-on-kunen [I have deleted it, it had no answers]

I'm finally finished studying Kunen's "Introduction to Independence proofs" but I'm having trouble on understanding something in the last theorem of the book (!!!).

I think that in order to show that $2^{\omega_1}\leq \kappa$ in $M[G]$ I must show that $|\mathbb P_\xi| < \kappa$ if $\xi<\kappa$ by induction (as in the proof of MA), but I can't. Can someone give me some advice?

In $M$, assume CH, and assume that $\kappa>\omega_1$, $\kappa$ is regular and $2^{<\kappa}=\kappa$; then there is a $\mathbb P \in M$ such that $\mathbb P$ is $\omega_1$-closed and $\omega_2$-c.c. in $M$, and whenever $G$ is $\mathbb P$-generic over $M$, $2^{\omega_1}=\kappa$ and $P_1$ hold in $M[G]$.

$P_1$ is the statement that whenever $\mathscr A \subset \mathscr P(\omega_1)$, $|\mathscr A|<2^{\omega_1}$ and $\forall F \subset \mathscr A(|F|<\omega_1 \rightarrow |\omega_1 \setminus \bigcup F|=\omega_1$ there exists $d \subset \omega_1$ such that $|d|=\omega_1$ and $\forall x \in \mathscr A(|x \cap d|<\omega_1)$.

What I have done so far:

Let $f \in M$ such that $f: \kappa \rightarrow \kappa \times \kappa$ is onto and $\forall \xi, \eta, \gamma<\kappa(f(\xi)=\langle \eta, \gamma\rangle \rightarrow \eta \leq \xi)$.

We shall build within $M$ an iterated forcing construction of the form $\langle \langle \mathbb P_\xi: \xi \leq \kappa\rangle, \langle \pi_\xi: \xi < \kappa\rangle\rangle$ such that for all $\xi$, $\pi_\xi$ is the standard $\mathbb P_\xi$-name for $\mathbb Q_\tau$ for a $\tau$ such that $|\tau|\leq \kappa$ and $1\Vdash(\tau \subset \mathscr P(\check {\omega_1}))$.

This will imply that for each $\xi\leq\kappa$, $|\mathbb P_\xi|\leq \kappa$. By induction: Step 0 is trivial. For the induction step, $\pi_\xi=\{\langle op(\check s, \sigma), 1\rangle: s \subset \omega_1 \wedge |s|\leq \omega_1 \wedge 1 \Vdash(\sigma \subset \tau \wedge |\sigma|<\omega_1) \wedge \sigma$ is a $\mathbb P_\xi$ nice name for a subset of $\tau\}$. By CH within $M$, we have $\omega_1^\omega=\omega_1$ possibilities for $s$. Within M, $\kappa^{\omega_2}=\kappa$. By the previous theorems of the book, all the $\mathbb P_\xi$ will have $\omega_2$-c.c., therefore we will have $\kappa^{\omega_2}=\kappa$ antichains and since $|\tau|\leq \kappa$ we will have we will have at most $\kappa^\kappa$ nice names for subsets of $\tau$. Well, here is where I am stuck. I'm not able to conclude that $|\pi_\xi| \leq \kappa$. IF I had $|\tau|<\kappa$ I would be able to conclude that, since $\kappa^\lambda=\kappa$ if $\lambda<\kappa$. Supposing $|\mathbb P_\xi|<\kappa$ whenever $\xi<\kappa$ makes no difference at this point.

So I tried to workaround supposing that $|\tau|<\kappa$. Then the argument works and the limit step follows easy. So let's try to continue the construction.

Suppose we have already constructed $\mathbb P_\xi$ as above. Let $\langle \sigma_\gamma^\xi: \gamma < \kappa\rangle$ enumerate all $\mathbb P_\xi$-names $\sigma$ such that for some $\lambda<\kappa$, $\sigma$ is a nice name for a subset of $(\lambda \times \omega_1\check)$ and let $\tau^\xi_\gamma$ be such that $$1 \Vdash \tau^\xi_\gamma=\{x \subset \omega_1: \exists \mu < \check \kappa(x=\{v:\langle \mu, \nu\rangle \in \sigma^\xi_\gamma)\}$$

Now I need to find a $\tau^\xi_\gamma$ such that $|\tau^\xi_\gamma|<\kappa$. Therefore I can't use the maximal principle. I tried something like $$\tau^\xi_\gamma=\{\langle \sigma, p\rangle: \sigma \text{ is a nice name for a subset of }\check \omega_1 \wedge p \Vdash \exists \mu < \check \kappa(\sigma=\{v: \langle\mu, \nu\rangle \in \sigma^\xi_\gamma)\}$$

So, for this having chance to work, I have to assume in my previous induction that $|\mathbb P_\xi|<\kappa$. But I can't guarantee that there are less than $\kappa$ such nice names. So I'm really stuck again.

  • 2
    $\begingroup$ Don't worry -- you came to the right place. $\endgroup$
    – Todd Trimble
    Feb 9, 2014 at 23:01

1 Answer 1


Note that in fact it suffices to consider those $\tau$'s which have size $<\kappa.$ The reason is simply as follows:

Suppose $M[G]$ is the final extension and we are going to show that $P_1$ holds in it. So we consider some $A\subset P(\omega_1)$ of size $<\kappa.$ As the forcing has $\omega_2-$c.c., $A$ has a name of size $<\kappa,$ and in fact by our book-keeping function it is some $\tau$ which is appeared during the iteration.

So from the beginning we can just consider those $\tau$'s which have size $<\kappa,$ and if some $\tau$ has size $\kappa,$ we can do the trivial iteration in that step.


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