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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.

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

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  • $\begingroup$ I know that $\Bbb P_\omega$ adds many $\omega$-sequences of ordinals, but what is it doing to the PCF of the measurable cardinals? $\endgroup$ – Asaf Karagila Nov 12 '18 at 12:09
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    $\begingroup$ It collapses it, probably to be of size less or equal to the continuum. $\endgroup$ – Yair Hayut Nov 12 '18 at 13:27
  • $\begingroup$ @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. $\endgroup$ – Mohammad Golshani Nov 13 '18 at 4:34

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