Let $cf(\alpha) > \omega$, and $P\_\alpha := \langle P\_{\beta}, \dot{Q}\_{\beta} : \beta < \alpha \rangle$ be a countable support iterated forcing construction (so for each $\beta$, $P\_{\beta} = P\_{\beta - 1} \ast \dot{Q}\_{\beta}$ if $\beta$ is a successor ordinal, $P\_{\beta} =$ inverse limit of $\langle P\_{\gamma} : \gamma < \beta \rangle$ if $cf(\beta) = \omega$, and $P\_{\beta} =$ direct limit of $\langle P\_{\gamma} : \gamma < \beta \rangle$ otherwise). Let $\Vdash\_{P\_{\beta}}$ "$\dot{Q}\_{\beta}$ is non-trivial", for all $\beta$. What can we specify for the $\dot{Q}\_{\beta}$ (and $\alpha$) so that $\Vdash\_{P\_\alpha} |\alpha| > \aleph\_2$?


Thanks in advance.



Countable support iterated forcing of length $\alpha$ which forces $|\alpha| > \aleph_2$
Let each $\dot{Q}\_{\beta}$ be $\{f | f : \gamma \rightarrow \{0, 1\}$ and $\gamma < \omega\_1\}$, ordered by inclusion, as defined in $V[G\_{\beta}]$. Let the length of the iteration, $\alpha$, be $> (2^{\omega})^{++}$, with $cf(\alpha) > \omega$. Then $P\_{\beta}$ forces $|\dot{Q}\_{\beta}| = \aleph\_1$ for all $\beta < \alpha$, if CH holds. Now each $\dot{Q}\_{\beta}$ is forced to be $\aleph\_1$-complete and thus proper. By a theorem of Shelah, $\aleph\_1$-completeness of the $\dot{Q}\_{\beta}$ means $P\_{\alpha}$ has the $\aleph\_2$-c.c., if CH holds. But $P\_1$ forces CH and forcing with $P\_{\alpha}$ is the same as forcing with $P\_1$ then with $P\_{\alpha}$ defined in $V[G\_1]$, since $P\_1$ is proper. Thus $P\_{\alpha}$ forces $|\alpha| > \aleph\_2$.


