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I'm studying a forcing argument which produces a generic extension in which GCH holds, but I am, somewhat embarrassingly, stuck on a minor detail. I hope someone can point out the thing I'm missing.

Let $P_\alpha=\mathrm{Fn}(\beth_\alpha^+,\beth_{\alpha+1},\beth_\alpha^+)$ be the Lévy collapsing notions. Denote by $P$ their Easton product over all ordinals $\alpha$; each element of $P$ is a function $p$, defined on some subset of the ordinals, so that $p(\alpha)\in P_\alpha$ for all $\alpha$ and $$|\{\alpha<\gamma;p(\alpha)\neq\emptyset\}|<\gamma$$ holds for all regular cardinals $\gamma$. Order $P$ coordinate-wise. Also, for any $\alpha$ define $$P^{>\alpha}=\{p|_{\{\beta;\beta>\alpha\}};p\in P\}$$ as the class of restrictions of elements of $P$ to ordinals, greater than $\alpha$ (don't be alarmed that these things are proper classes; it doesn't really matter at this point).

I am told that $P^{>\alpha}$ is $\beth_{\alpha+1}$-closed (as Jech would say, or $\leq\beth_{\alpha+1}$-closed, as Kunen would say). So let's take $\mu\leq\beth_{\alpha+1}$ and a descending $\mu$-sequence $(q_\xi)_{\xi<\mu}$ and define $q$ on $\bigcup_{\xi<\mu}\mathrm{dom}(q_\xi)$ by $q(\beta)=\inf_{\xi<\mu} q_\xi(\beta)$, where the existence of the infimum in $P_\beta$ is guaranteed by the fact that $P_\beta$ is $\beth_\beta$-closed. For $q$ to be a lower bound for our sequence, we have to check the support condition mentioned above. Taking a regular $\gamma$, we have $$|\{\beta<\gamma;q(\beta)\neq\emptyset\}|= \left|\bigcup_{\xi<\mu}\{\beta<\gamma;q_\xi(\beta)\neq\emptyset\}\right|$$

This is where I get confused. Certainly, $\gamma$ can be taken greater than $\alpha$. Also, since $\gamma$ is regular, if we also have $\mu<\gamma$, the above union has cardinality strictly less than $\gamma$ and we get what we want. What worries me is what happens when $\gamma\leq\mu$. This doesn't seem at all right, so I suspect I messed up something in setting up the proof.

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Since $q_\xi$ is a descending sequence, $\bigcup_{\xi<\mu}\{\beta < \gamma\ |\ q_\xi(\beta) \neq \emptyset\}$ is an increasing union. If $\gamma < \mu$ then this union must clearly stabilize, and thus it has size less than $\gamma$. We need to see that if $\mu = \gamma$ the increasing union still stabilizes. If $\mu$ is weakly Mahlo then this is easy to see by a pressing down argument. But what about the general case for $\gamma = \mu$? – Amit Kumar Gupta Jun 14 '12 at 8:23
Wait, is this even true? Assume GCH, and let $\alpha = 0$, $\mu = \gamma = \beth_{\alpha+1} = \beth_1 = \aleph_1$. For $\xi < \omega_1$, let $q_\xi$ be empty on every coordinate except $q_\xi(\beta) = \{(0,0)\}$ for $\omega < \beta < \omega+\xi$. Doesn't this give a counterexample? – Amit Kumar Gupta Jun 14 '12 at 8:59
@AmitKumarGupta I considered basically the same example you brought up and I couldn't figure out why it shouldn't disprove the claim. On the other hand, I found out that more or less the same approach is given as an exercise in Jech, where he also claims that $P^{>\alpha}$ is closed. I'm confused. – Miha Habič Jun 14 '12 at 12:36
where in Jech?${}$ – Amit Kumar Gupta Jun 15 '12 at 3:13
@AmitKumarGupta It's exercise 15.15 of the Third Millenium edition. – Miha Habič Jun 15 '12 at 5:52
up vote 3 down vote accepted

(Compiled from comment thread above)

If you look at Jech's hint, you'll see the Easton-like requirement on the product is only for inaccessible $\gamma$. Also, to verify $\beth_{\alpha+1}$-closure one need only consider $\mu = \beth_{\alpha+1}$. So now, if $\gamma < \beth_{\alpha+1}$, the increasing union $\bigcup_{\xi < \beth_{\alpha+1}}\{\beta < \gamma | q_\xi(\beta)\neq \emptyset\}$ must stabilize, thus it has size less than $\gamma$. $\gamma = \beth_{\alpha+1}$ never happens because $\gamma$ is inaccessible. So that'll do it.

The Easton requirement here is different from the usual one. The best way to see the analogy between this Easton requirement and the usual one is to re-index the forcing notions here: Define $P_{\beth_\alpha}$ to be the Levy collapse of $\beth_{\alpha+1}$ and $P_\beta$ to be trivial if $\beta$ is not a $\beth$-number. Then the appropriate Easton requirement here is the usual one, but the only $\gamma$ for which this requirement does anything is when $\gamma$ is inaccessible.

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