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.

  • $\begingroup$ I should mention that I'm willing to subject myself to any and all collapsing and cofinality adjustments that occur in the process. $\endgroup$ Dec 21, 2012 at 6:42
  • $\begingroup$ Relevant material by Friedman is here logic.univie.ac.at/~sdf/papers/class-forcing.pdf in section 3. $\endgroup$ Dec 21, 2012 at 7:13
  • $\begingroup$ 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 $\lambda$. $\endgroup$ Dec 21, 2012 at 13:50
  • $\begingroup$ Also, $\lambda^{< \lambda}=\lambda$ if and only if $\lambda$ is regular and $2^{< \lambda}=\lambda$. $\endgroup$ Dec 21, 2012 at 15:50
  • $\begingroup$ Yes, I did specify that $\lambda$ was regular, e.g. a successor. Thanks for clarifying. $\endgroup$ Dec 21, 2012 at 21:44

1 Answer 1


It's impossible to accurately answer this since you don't say what forcing you have in mind. However, in the "typical" case, the answer is yes.

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.

  • $\begingroup$ Note that I'm using "typical" in the sociological sense, not in the sense of "general". Every occurrence of "typical" means there is something to check there that probably fails in the "general" case. Still, this is the "typical" pattern for class forcing in the sense that it's the one you see most often in the literature. $\endgroup$ Dec 21, 2012 at 14:03
  • $\begingroup$ I did say Long Easton forcing. :-) $\endgroup$ Dec 21, 2012 at 21:49
  • $\begingroup$ OK. Is that a reverse type? If so, you'll need to use the extended form of the theorem since it won't split as a product. $\endgroup$ Dec 21, 2012 at 22:54
  • $\begingroup$ Hmm, not sure. What's the extended version of the theorem in any case? $\endgroup$ Dec 21, 2012 at 23:27
  • $\begingroup$ The extension of the theorem you stated for when the forcing splits as an iteration instead of as a product. $\endgroup$ Dec 21, 2012 at 23:30

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.