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On page 232 of Jech's Set Theory (3rd Edition, 2003), we have the following statement of Easton's theorem.

Theorem 15.18 (Easton). Let $M$ be a transitive model of ZFC and assume that the Generalized Continuum Hypothesis holds in $M$. Let $F$ be a function (in $M$) whose arguments are regular cardinals and whose values are cardinals, such that for all regular $\kappa$ and $\lambda$:

  1. $F(\kappa) > \kappa$

  2. $F(\kappa) \leq F(\lambda)$ whenever $\kappa \leq \lambda$

  3. $\mathrm{cf} F(\kappa) > \kappa$

Then there is a generic extension $M[G]$ of $M$ such that $M$ and $M[G]$ have the same cardinals and cofinalities, and for every regular $\kappa$,

$$M[G] \models 2^κ = F(κ).$$

This doesn't make sense to me. In particular, assume (in the ambient set theory) both GCH and the existence of an inaccessible $\iota$, we can take $M = V_\iota.$ But doesn't this imply that $M$ has no (non-trivial) forcing extensions?

What am I missing here?

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  • $\begingroup$ Why wouldn't $V_\kappa$ have any forcing extensions? What is your preferred formalization of forcing? $\endgroup$ – Miha Habič Aug 13 '14 at 13:40
  • $\begingroup$ @MihaHabič, Boolean valued models, I guess, not that I really understand it. But anyway, I thought $V_\kappa$ wouldn't have any forcing extensions because we're not allowed to ordinals (this is part of the definition, is it not?) and because, well, how are we going to add more elements to $V_\kappa$ without adding ordinals? $\endgroup$ – goblin Aug 13 '14 at 13:45
  • $\begingroup$ The fact that forcing extensions don't add ordinals isn't a part of the definition of the extension but rather a consequence of the construction itself. As for your point about adding elements, how do you then force to add an element to any model at all? Wouldn't your argument just say that there are no forcing extensions of anything at all (which is absurd)? $\endgroup$ – Miha Habič Aug 13 '14 at 14:05
  • $\begingroup$ Ha. Exactly three hours ago one of the guys here started giving a lecture based exactly on that. $\endgroup$ – Asaf Karagila Aug 13 '14 at 14:22
  • $\begingroup$ @AsafKaragila, coincidence, or something more sinister? $\endgroup$ – goblin Aug 13 '14 at 14:28
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First of all, you can always assume that $M$ is countable, by taking an elementary submodel. And in particular $F$ "reflects down" to the submodel (since it is definable by a formula in $M$, and we can assume that the parameters are in the submodel).

So if $V_\kappa$ is a model of set theory, then by taking a countable elementary submodel, and collapsing the submodel, we get a countable transitive model with the exact same properties, as far as first-order logic is concerned.

Secondly, it's not true that if $V_\kappa$ is a model of $\sf ZFC$ then it has no nontrivial generic extensions. It is true that you can't add sets which have rank ${<}\kappa$ (since you are taking sets from the universe, in which case you can't find any sets of rank ${<}\kappa$ which are not in $V_\kappa$ already), but you might add classes to $V_\kappa$ by forcing, meaning you'd make subsets of $\kappa$ which weren't definable in $(V_\kappa,\in)$ definable in $(V_\kappa[G],\in)$.

In either case, if you force "over the universe" you can always talk about Boolean-valued models, and so on, since you are really looking to prove a consistency result here.

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  • $\begingroup$ I'm having a really hard time trying to understand what this question is asking and I think your third paragraph confuses the issue a bit. The way I read it it seems to say that I can't add a Cohen real to $V_\kappa$. I would suggest an edit, but I don't understand the question well enough to know what the edit should be. $\endgroup$ – Miha Habič Aug 13 '14 at 22:09
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    $\begingroup$ @Miha: If you force in $V$, you can't add a Cohen real to $V_\kappa$. Recall that the von Neumann rank of all the reals in $V$ is at most $\omega+1$, therefore all the reals are already in $V_\kappa$. Forcing a Cohen real over $V_\kappa$ requires you either to consider a countable submodel, or force "over the universe". $\endgroup$ – Asaf Karagila Aug 13 '14 at 22:24
  • $\begingroup$ @Miha: The point is that even if you force using Boolean-valued models, you don't realized them to actual sets in the universe. You can't. Even if you do this over $V_\kappa$ as a set. All this shows you is that the truth value of some statement (which you wanna prove) is not $1_\Bbb B$, and therefore it is consistently true/false. But again, you don't add sets, you don't extend $V_\kappa$ in a meaningful way in $V$. $\endgroup$ – Asaf Karagila Aug 13 '14 at 23:01
  • $\begingroup$ "You can always assume that $M$ is countable, by taking an elementary submodel." Sure: if all we want is to prove independence results, then we might as well assume that $M$ is countable. But suppose we're interested in models of ZFC themselves, independently of what sentences of first-order logic they satisfy. Then we definitely do not want to assume that $M$ is countable. So I am focusing on your third paragraph. I don't get it! Let $\iota$ be an inaccessible to avoid confusion. Then how can we make new subsets of $V_\iota$ definable by forcing without adding elements of rank $<\iota$? $\endgroup$ – goblin Aug 14 '14 at 3:02
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    $\begingroup$ @goblin: And I want a fanless PSU and a passive cooling system without paying a fortune. If you want to work with models of ZFC of arbitrary size, there are some limitations you cannot avoid. To the actual question, this would be a class forcing over the model. $\endgroup$ – Asaf Karagila Aug 14 '14 at 8:28
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If $\mathbb{Q}$ is any atomless and separative poset in $M = V_\iota$ (where $\iota$ is an inaccessible cardinal), then adding a generic subset of it (over $M$) enlarges the whole universe. This follows from these two facts:

  1. If $\mathbb{P} \in M$, then $\mathcal{P}(\mathbb{Q})$, the powerset of $\mathbb{Q}$ (which is the same for $M$ and $V$ since $\iota$ is an inaccessible) is in $M$, so all of the dense open sets of $\mathbb{Q}$ are in $M$;
  2. For such a $\mathbb{Q}$, there is no fully generic set in $V$. This is because for such a $\mathbb{Q}$, if we had a fully generic set in $V$, then we can define a new dense open set (in $V$) which is completely disjoint from the generic set (here we would use that the poset is atomless and separative).

So, adding a new generic set, even if over $M$, adds a new set to $V$ (since, by the first point, a set is generic over $M$ if and only if it is generic over $V$), and notions such as 'rank' need to be recalculated. Put another way, there are no generic sets for partial orders in $M$ in $V$.

In particular, after forcing over $M$ with a partial order in $M$, you do ''add elements of rank $<\iota$'' to $V$(think of the poset to add a new real for example), but 'rank' in the new model and in $V$ might mean different things. For example, consider the partial order to make an ordinal $\alpha$ which is the $\omega_1$ of $V$ countable. Then, the rank of $\alpha$ changes from uncountable in $V$ to countable in the new model.

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