The motivation for this question is that I am working through an exercise to force the GCH (generalized continuum hypothesis) over a model of ZFC and obtain a model of ZFC where GCH holds.

The forcing is an ORD length iteration of $Add(\gamma^+, 1)$ at every cardinal $\gamma$ ($\gamma$ is a cardinal in $V^{P_{\gamma}})$, with Easton support. I'll just say here what Easton support is:

1) at regular $\gamma$, $P_{\gamma}$ is the direct limit of $P_{\alpha}$, $\alpha < \gamma$, meaning that for all $p \in P_{\gamma}$ there is a $\beta < \gamma$ such that $p\restriction \beta \in P_{\beta}$ and for all $\xi \ge \beta$, $p(\xi) = 1$

2) otherwise, $P_{\gamma}$ is an inverse limit, meaning for all $p \in P_{\gamma}$, the restriction of $p$ to any smaller cardinal is in that stage of the forcing, and there is no imposition that the coordinates of $p$ be trivial past some smaller stage.

Note that at successor cardinals $\gamma$, $P_{\gamma}$ is automatically a direct limit if we force at cardinal stages.

My question is about forcing at singular cardinal stages. These mathoverflow questions, answers, and the surrounding comments (what goes wrong, definition of easton product), were very helpful to me, but I still have a confusion.

Let me first explain what I do understand regarding forcing the GCH at singular cardinals and then my question.

Since $\gamma^+$ is regular for all $\gamma$, the important arguments to show that when we factor $P \cong P^{\le \gamma}*P^{> \gamma}$, the first factor satisfies the $\gamma^+$-chain condition and the second is $<\gamma^+$-closed, all go through, whether $\gamma$ is regular or singlular.

However, part of the above claim rests on a $\Delta$-system argument to show that $P^{\le \gamma}$ satisfies the $\gamma^+$-chain condition. And it is here where we use that any condition $p \in P^{\le \gamma}$ has a small domain compared to $\gamma^+$, and so a $\gamma^+$ sized set of domains of conditions in $P^{\le \gamma}$ forms a $\Delta$-system.

What can I say in the case of singular cardinals? How can the $\gamma^+$ sized set of domains of conditions in $P^{\le \gamma}$ form a $\Delta$-system if we used an inverse limit to support the iteration at that stage? Why not impose a direct limit at every stage of forcing?


Let me take the case of $\aleph_\omega$ as a specific example. Similar considerations apply at other singulars.

If you want $\sigma$-closure above $\omega_1$, you need the full support at $\aleph_\omega$; bounded support (i.e., direct limit) will not be closed enough.

Also: You need not worry about collapsing $\aleph_\omega$ (or changing its cofinality). If the elements of an unbounded set of cardinals below $\aleph_\omega$ stay cardinals, then $\aleph_\omega$ will stay a cardinal as well (and if a cobounded set of cardinals below $\aleph_\omega$ is collapsed to $\lambda$, then $\aleph_\omega$ will certainly not become $\lambda^+$ in the extension, and hence collapse as well. If $\kappa$ is a singular cardinal of cofinality $\lambda < \kappa$, then the cofinality of $\kappa$ will only change if the cofinality of $\lambda$ changes. This will be true no matter what support you take.

  • $\begingroup$ I'm starting to get the idea, but I don't understand fully. What you mean by the $\sigma$-closure above $\omega_1$ (that the second p.o. of the factoring at $\omega_1$ is $\omega_1$-closed?) and why we need full support at $\aleph_{\omega}$ to ensure this? $\endgroup$ – user10290 Jan 14 '13 at 20:57
  • $\begingroup$ Yes, I mean $<\omega_1$-closure in the second factor at $\omega_1$. We need full support because otherwise we could construct a decreasing $\omega$-sequence of conditions $(p_n)$, where $p_n$ has support bounded by $\aleph_n$, which will not have a limit. $\endgroup$ – Goldstern Jan 14 '13 at 21:36
  • $\begingroup$ Thanks. I see why we need full support at $\aleph_{\omega}$ in order to have a limit point for the sequence you described above, and see how to generalize to all singular cardinals. And, your last paragraph answers why we don't even need a $\Delta$-system argument on a singular cardinal for the proof since preserving regular cardinals (and applying a $\Delta$-system argument there) will ensure this. And, in the part of the proof where we show $Add(\lambda, 1)$ forces $2^{\lambda} = \lambda^+$ at stage $\lambda$, we only need closure on the later stages. $\endgroup$ – user10290 Jan 15 '13 at 18:13

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