# Can we weaken GCH in this class forcing?

I've just stumbled across the following theorem (here):

Theorem Let $P$ be a partial class order in $M$, a transitive model of ZF (resp. ZFC). Suppose for arbitrarily large cardinals $\kappa$ we have an isomorphism $P \simeq P^- \times P^+$ where $P^-$ is $\kappa^+$-cc and $P^+$ is $\leq\kappa$-closed. For $G$ an $M$-generic for $P$, $M[G]$ satisfies ZF (resp. ZFC).

Now Easton forcing constructs a particular $P_E$, and at least in Jech and S. Friedman's books GCH is assumed, I believe in order to show that for $\kappa$ any given regular cardinal we have $P_E = P^{\leq \kappa} \times P^{\gt\kappa}$ where $P^{\leq \kappa}$ is $\kappa^+$-cc. I think $\leq\kappa$-closedness of $P^{\gt\kappa}$ comes for free. My first question is this: shouldn't the existence of class-many regular cardinals $\lambda$ such that $\lambda^+ = 2^\lambda$ be enough to satisfy the hypotheses of the above theorem? Could we weaken this to $\lambda^{\lt\lambda} = \lambda$? (or is this equivalent to CH at $\lambda$?)

Secondly (for example in these notes) we have

Lemma The poset $\mathrm{Fn}(\eta,\lambda,\rho)$ of partial functions $\eta \rightharpoonup \lambda$ with domain smaller than $\rho$ is $(\lambda^{\lt\rho})^+$-cc.

My guess is that we need regularity of $\kappa$ so that $P^{\leq \kappa}$, which is a product of posets as in the lemma, is $\kappa^+$-cc. Is this correct?

I'm actually thinking of Long Easton forcing, so I don't want to assume Easton support. Does the above line of reasoning still work? Namely, assume the existence of class-many regular cardinals satisfying CH (or $\lambda^{\lt\lambda} = \lambda$, if that's weaker), and construct the partial class order so that we can fullfil the hypotheses of the theorem above.

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I should mention that I'm willing to subject myself to any and all collapsing and cofinality adjustments that occur in the process. –  David Roberts Dec 21 '12 at 6:42
Relevant material by Friedman is here logic.univie.ac.at/~sdf/papers/class-forcing.pdf in section 3. –  David Roberts Dec 21 '12 at 7:13
Regarding your question in the third paragraph, the assertion $2^\lambda=\lambda^+$ isn't equivalent to $\lambda^{\lt\lambda}=\lambda$, and neither generally implies the other, as can be seen at $\lambda=\omega_1$, since the situations of $2^\omega=2^{\omega_1}=\omega_2$ and also $2^\omega=\omega_1,2^{\omega_1}=\omega_3$ are both consistent. Meanwhile, $2^\lambda=\lambda^+$ is equivalent to $(\lambda^+)^{\lt\lambda^+}=\lambda^+$, that is, making the corresponding assertion at $\lambda^+$ rather than at $\lambda$. –  Joel David Hamkins Dec 21 '12 at 13:50
Also, $\lambda^{< \lambda}=\lambda$ if and only if $\lambda$ is regular and $2^{< \lambda}=\lambda$. –  Emil Jeřábek Dec 21 '12 at 15:50
Yes, I did specify that $\lambda$ was regular, e.g. a successor. Thanks for clarifying. –  David Roberts Dec 21 '12 at 21:44

The "typical" case is when the forcing poset $P$ comes from a "typical" iteration of set forcings. These are "typically" arranged so that for unboundedly many $\kappa$, the partial iteration $P_\kappa$ up to $\kappa$ consists of well-behaved $\kappa^+$-cc forcings (often enough these are small so that $|P_\kappa| \leq \kappa$), and the remaining posets in the iteration after $\kappa$ are $\kappa$-closed. The $\Delta$-System Lemma (or even $|P_\kappa| \leq \kappa$ in nice cases) can "typically" be used to show that $P_\kappa$ is $\kappa^+$-cc; the "typical" $\Delta$-System Lemma application (or counting argument) here requires $\kappa^{\lt\kappa} = \kappa$. Then a standard variation of the theorem you state shows that $P$ preserves ZFC.
This is the "typical" case but it is only "typical" because it is fairly straightforward to show that such $P$ are tame. Note that the theorem you state deals with a very special case where $P$ splits as a product at $\kappa$. This is not uncommon but in general $P$ will only split as an iteration, but the general idea of the argument "typically" still applies to the more general case.