3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I think the point of the question is that a countable-support proper iteration of length $>\omega_2$ can collapse cardinals (even if the length is only $\omega_2+1$), so the $\aleph_2$ of the extension could be larger than that of the ground model. The OP wants hypotheses that prevent this, or at least prevent $\aleph_2$ from jumping up beyond $|\alpha|$. Unfortunately, even Sacks forcing (which morally "adds a real while doing as little other damage as possible") can collapse cardinals at stage $\omega_2$, so I'm not optimistic about getting a good hypothesis to prevent that. $\endgroup$ – Andreas Blass Aug 22 '13 at 15:16
  • 3
    $\begingroup$ You probably want that the $Q_\beta$ (at least most of them) do not add reals. (Necessary, not sufficient.) If all $Q_\beta$ add reals, then in stage $\kappa+\omega_1$ you will have that $|\kappa|\le\aleph_1$ for all cardinals $\kappa$. In particular, all cardinals $<\alpha$ will have size $\aleph_1$ at most, by the time you reach stage $\alpha$. $\endgroup$ – Goldstern Aug 22 '13 at 15:55
  • 3
    $\begingroup$ Similarly, you also need that most $Q_\beta$ are $\aleph_2$-c.c., since otherwise you will be generically collapsing the $\aleph_2$ of $V[G_\kappa]$ by stage $\kappa+\omega_1$. $\endgroup$ – Joel David Hamkins Aug 22 '13 at 17:06
  • 1
    $\begingroup$ @Theo, I think the notation used in the question is completely standard, and will be readily understandable to anyone familiar with iterated forcing. Let me add that I find the votes to close this question to be inappropriate. $\endgroup$ – Joel David Hamkins Aug 23 '13 at 18:43
  • 1
    $\begingroup$ @Theo, in this post, $\alpha$ refers to an ordinal (rather than a cardinal), which will be the length of the forcing iteration, and $\text{cf}(\alpha)$ means the cofinality of $\alpha$, which is the cardinality of the smallest unbounded subset of $\alpha$. Iterated forcing constructions are usually covered in an advanced graduate set theory class, such as Math 235 at Berkeley (or at least that is what it was called 20 years ago). This particular post uses the notation of Kunen's book Set Theory (amazon.com/…). $\endgroup$ – Joel David Hamkins Sep 18 '13 at 13:49
2
$\begingroup$

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 $|\alpha| > \aleph_2$.

$\endgroup$
  • 1
    $\begingroup$ Since your forcing is countably closed, it adds no reals, and so you are forcing with $\text{Add}(\omega_1,1)^V$ at each stage $\beta$. Is your overall iteration the same as $\text{Add}(\omega_1,\alpha)$? $\endgroup$ – Joel David Hamkins Sep 18 '13 at 13:42
  • $\begingroup$ Does $Add(\omega_1, \alpha) = \{f : f$ is a countable partial function from $\omega_1 \times \alpha$ into $\{0, 1\}\}$?. $\endgroup$ – Zoorado Sep 18 '13 at 14:48
  • $\begingroup$ Yes, the forcing to add $\alpha$ many Cohen subsets to $\omega_1$. $\endgroup$ – Joel David Hamkins Sep 18 '13 at 14:55
  • $\begingroup$ I think it is, but I don't have a proof yet. I shall give it some thought later. $\endgroup$ – Zoorado Sep 18 '13 at 18:25
  • 1
    $\begingroup$ I think it is the same. Your iteration doesn't add reals, and so the iterated forcing construction is really just the same as the product forcing to add $\alpha$ many Cohen subsets to $\omega_1$, which under CH will preserve all cardinals. $\endgroup$ – Joel David Hamkins Sep 18 '13 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.