For a forcing notion $Q$, let $\dot{S^Q}$ be the $Q$-name for the class of ordinals $\{\kappa : \kappa = \omega_1^{V}$ $or$ $\kappa$ $is$ $a$ $regular$ $uncountable$ $cardinal \}$ in $V^Q$.

We say that $Q$ is $\dot{S^Q}$-semiproper iff for all $\lambda$ large enough, for all $N$ countable elementary submodel of $(H(\lambda), \in)$ such that $\{Q, \dot{S^Q}\} \in N$, for all $q \in Q \cap N$, there exists $p \in Q$, $p$ extends $q$ and $p$ satisfies the following:

for all regular uncountable cardinal $\kappa$ in $N$, for all $\dot{\beta}$ a $Q$-name for an ordinal in $\kappa$, $p \Vdash_Q $ $``if$ $\kappa \in \dot{S^Q},$ $then$ $\exists A \in N$ $(|A| < \kappa$ $and$ $\dot{\beta} \in A"$.

Now let $\alpha$ be a limit ordinal, $\bar{Q} = \langle P_i, \dot{Q_i} : i < \alpha \rangle$ be a revised countable support (RCS) iteration, and $P_{\alpha} = Rlim(\bar{Q}) =$ RCS limit of $\bar{Q}$.

Assume for all $i < j < \alpha$, $P_i \Vdash$ $``P_j/\dot{G_i}$ $is$ $\dot{S^{P_i/\dot{G_i}}}-semiproper"$. Is it true in general that there exists a $\dot{\xi}$, a $\bar{Q}$-name for an ordinal $< \alpha$ such that:

If $p \in P_{\gamma + 1}$, $p \Vdash_{\bar{Q}} \dot{\xi} = \gamma$, then $p \upharpoonright \gamma \Vdash_{P_{\gamma}} ``cf(\alpha) = \aleph_0$ $or$ $\forall k$ $(\gamma < k < \alpha \rightarrow$ $\Vdash_{P_k/\dot{G_{\gamma}}}$ $``cf(\alpha)^{V^{P_{\gamma}}}$ $is$ $regular$ $uncountable")"$?

  • $\begingroup$ When you say RCS, do you mean that as Shelah defined, or in the work of Viale and his students recently posted on arXiv (recently being the last year or so) which is based on the Dunder-Fuchs paper? $\endgroup$ – Asaf Karagila Sep 11 '14 at 5:56
  • $\begingroup$ A note: $\dot{S}^Q$ is clearly a proper class, so what does it mean to say $\dot{S}^Q\in N$? $\endgroup$ – Mohammad Golshani Sep 11 '14 at 6:11
  • $\begingroup$ Perhaps I am misunderstanding the question. Can't you just let $Q_0$ be an antichain of size $\alpha:=\aleph_2$, and then continue with an iteration which at the $Q_0$-generic coordinate makes $cf(\alpha)=\omega$ (and is otherwise trivial)? $\endgroup$ – Goldstern Sep 11 '14 at 6:12
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    $\begingroup$ Namba forcing may be semiproper in the traditional sense (i.e., $\{\aleph_1\}$-semiproper). If you let $\kappa=\aleph_3$, $\lambda=\aleph_4$, and you compose Namba with a forcing that collapses $\kappa$ to $\aleph_1$ with countable functions, then the whole forcing is $\lambda$-cc and hence satisfies your version of semiproperness, if I am not mistaken. Alternatively, let $\alpha$ be a measurable cardinal and use Prikry forcing. $\endgroup$ – Goldstern Sep 13 '14 at 17:40
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    $\begingroup$ Your revised version does not make sense. There is no demand on $\dot \xi$, other than being a name decided by $p$. $\endgroup$ – Goldstern Sep 13 '14 at 17:49

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