# Iterated forcing with distributive forcing notions

Assume $$(\kappa_n| n<\omega)$$ is an increasing sequence of inaccessible cardinals with $$\kappa_\omega=\sup_{n<\omega}\kappa_n.$$. Let $$((\mathbb{P}_n| n \leq \omega), (\dot{\mathbb{Q}}_n | n<\omega))$$ be a full support iteration of forcing notions, where for each $$n< \omega$$, we have $$\Vdash_{\mathbb{P}_n}$$$$\dot{\mathbb{Q}}_n$$ has size < $$\kappa_{n+1}$$ and does not change $$\dot{V}_{\kappa_n}$$''.

Question (a) Does forcing with $$\mathbb{P}_\omega$$ add a new real$$?$$

(b) Does forcing with $$\mathbb{P}_\omega$$ collapse $$\kappa_\omega?$$

Remark. I assume each $$\kappa_n$$ is inaccessible but not Mahlo.

• This is why working with symmetric iterations is so much nicer. If the forcings are homogeneous, nothing bad happens. :-) – Asaf Karagila Nov 12 '18 at 12:10

One can get at least a counter example for (2) from countably many measurable cardinals.

Let $$\mu_n$$ be a measurable cardinal between $$\kappa_n$$ and $$\kappa_{n+1}$$. Let $$\kappa_{\omega} = \sup \kappa_n = \sup \mu_n$$. Let me assume that there is no inner model with a Woodin cardinal.

Let $$\mathbb{Q}_n$$ be the Prikry forcing for singularizing $$\mu_n$$ (as defined in the generic extension by $$\mathbb{P}_n$$). Let me claim that the full support iteration $$\mathbb{P}_\omega$$ collapses $$\kappa_{\omega}^+$$ and does not add reals.

Let $$\langle P_n \mid n < \omega\rangle$$ be the sequence of the generic Prikry sequences.

Claim: For $$r \in {}^\omega \omega$$ let $$g_r(n) = P_n(r(n))$$. Then $$\left(\prod \mu_n\right)^V$$ and $$\{g_r \mid r\in {}^\omega \omega\}$$ are interleaved.

Proof: First, let us note that $$\mathbb{P}_\omega$$ does not add reals. Indeed, let $$\dot{r}$$ be a name for a new real and let $$p \in \mathbb{P}_\omega$$. Let us define by induction a sequence of conditions $$p_n$$ such that for all $$n$$: $$p_n \restriction n \leq^* p_{n-1} \restriction n$$, $$p_n \leq p_{n-1}$$ and $$p_n$$ decides the value of $$n \in \dot{r}$$. This is done using the Prikry property of $$\mathbb{P}_n$$.

Next, let $$f\in \prod \mu_n \cap V$$. Using the Prikry Property of the finite iterations we can show that for every condition $$p$$ there is a direct extension $$p'$$ such that for all $$n$$, $$p'$$ decides the length of the stem of $$p'(n)$$. Let us extend $$p'$$ in all coordinates in order to obtain a function which dominates $$f$$ everywhere.

On the other hand, if $$r\in {}^\omega \omega$$, let $$p$$ be a sufficiently strong condition so that $$p$$ decides the length of its stems and $$\mathrm{len}\ \mathrm{stem}(p(n)) > r(n)$$. Now the the name for the ordinal $$g_r(n)$$ is a $$\mathbb{P}_n$$-name. Since this is a small forcing (of size $$<\mu_n$$), there is some $$\beta_n < \mu_n$$ such that $$\Vdash \dot{g}_r(n) < \beta_n$$. Let $$f(n) = \beta_n$$, then clearly $$p$$ forces that $$f$$ dominates $$g_r$$. QED

We conclude that the cofinality of $$\kappa_{\omega}^{+}$$ in the generic extension is at most the continuum. By the weak covering lemma, $$\kappa_\omega$$ cannot be a cardinal in the generic extension.

• I know that $\Bbb P_\omega$ adds many $\omega$-sequences of ordinals, but what is it doing to the PCF of the measurable cardinals? – Asaf Karagila Nov 12 '18 at 12:09
• It collapses it, probably to be of size less or equal to the continuum. – Yair Hayut Nov 12 '18 at 13:27
• @YairHayut Thanks a lot for the answer. I did not mention it, but I am mostly interested when there are no very large cardinals, say those incompatible with $V=L.$ In fact I am thinking on a question of Woodin, and the above question, was just a very start step that I asked myself. – Mohammad Golshani Nov 13 '18 at 4:34