I'm teaching myself bits and pieces of forcing at the moment, for the purposes of translating them into sheaf-theoretic versions. I'm trying to write down what I feel is a cleaner description of the Easton product of forcing posets, by which I mean a global description rather than one in terms of elements like $$ \left|{(\kappa,α,β) ∈ dom(p) : \kappa\leq \lambda}\right| \lt \lambda \qquad (1) $$ for $p$ in the Easton product $\prod^E P(\kappa)$, which is not very useful at the level of the category of sets if it is not the category of ZF(C)-sets.

I'm confident I understand what this condition does when you do the forcing, namely I think it puts a bound on how many sets are added to any given $\lambda$, else you might get silly things like a proper class of sets added to some set.

However, at the risk of embarrassing myself, and in the interest of educating others, here is my best guess for what this condition translates to. Consider, as Jech does (Set theory, 3rd edition), first the Easton product over some set $A$ of regular cardinals. For $\kappa\in A$, let $P(\kappa)$ be the set of functions $p:D(p)\to 2$ where $D(p)\subset \kappa\times B(\kappa)$ has cardinality less than $\kappa$. Here $B(\kappa)$ is some cardinal, not necessarily given by an Easton function on regular cardinals, and forcing using the 'usual order' on this poset will add $B(\kappa)$ subsets to each $\kappa\in A$. There is a distinguished element $\top$ of each $P(\kappa)$, namely the unique function $\emptyset \to 2$.

Let $P=\prod P(\kappa)$ be the product over $\kappa\in A$. The Easton product is the subset consisting of those $p\in P$ such that the support condition (1) holds. So what does this mean? An element $p\in P$ is a collection of functions $p_\kappa$, one for each $\kappa\in A$. Then let $supp(p) \subset A$ be the set of those $\kappa\in A$ such that $p_\kappa\not=\top$. Analogously, define $supp_\lambda(p) \subset A$ to be the set of those $\kappa\leq \lambda$ such that $p_\kappa\not=\top$. This last definition is where I am the most unsure, as the definition in Jech really involves $supp(p)\cap \lambda$, which doesn't make sense from a structural set theory point of view unless $\lambda$ is viewed as a subset of $A$ (even though it makes perfect sense in a material set theory such as ZFC).

Now assuming the definition of $supp_\lambda(p)$ is correct, the support condition on $p$ as first stated in Jech is that $$ \forall \text{ regular } \lambda, \left|supp_\lambda(p)\right| \lt \lambda.\qquad (2) $$ Apparently it's enough to enforce (2) whenever $\lambda$ is weakly inaccessible, (though Friedman just says 'inaccessible'), and that's fine by me. However, this doesn't look the same as the condition (1), which has me slightly worried. The condition (1) imposes a condition on the size of the domain of $p$ as well as (2), and in fact implies that the domain of $p_\kappa$ is smaller than $\kappa$ for all $\kappa$.

So where has my reasoning gone wrong? At this point I really just want to understand the set underlying the Easton product, rather than any intricacies of class forcing.

  • $\begingroup$ I don't get it. You seem to be worried that (2) does not imply that $dom(p_\kappa)$ is small. But if $p\in \prod_{k\in A}P(\kappa)$, then $p_\kappa\in P(\kappa)$, so by definition its domain is smaller than $\kappa$. $\endgroup$
    – Goldstern
    Dec 17, 2012 at 8:46
  • $\begingroup$ I realised something like this just before I went rock climbing, and couldn't note this here at the time. I knew I was missing something obvious. $\endgroup$ Dec 17, 2012 at 10:51

1 Answer 1


Here is a general way to think about product forcing, which may be helpful. One has forcing notions $\mathbb{Q}_\gamma$ for each $\gamma$ in a class $D$ of ordinals. (In your example, $D$ is the class of regular cardinals, and $\mathbb{Q}_\gamma=\text{Add}(\gamma,E(\gamma))$ for the Easton function $E$.) One wants to define the notion of product forcing $\Pi_\gamma\mathbb{Q}_\gamma$.

The idea is that fundamentally different product forcing notions are obtained by allowing different kinds of support in the product. This is exactly the same issue as the difference between direct sums and direct products in algebra. For each ideal $I$ on $D$, one may use that ideal as collection of allowed supports, so that the product $\Pi_\gamma\mathbb{Q}_\gamma$ with supports in $I$ consists of the functions $p:\gamma\mapsto p(\gamma)\in\mathbb{Q}_\gamma$, such that $\{\gamma\mid p(\gamma)\neq\top\}\in I$. That is, we insist that the support of a condition $p$ is in the ideal. It is very convenient for the collection of allowed supports to be an ideal, since we often want to manipulate conditions, either by weakening them on each coordinate or by combining two conditions that are compatible on each coordinate, and this leads us to want an ideal, which is closed under subsets and finite unions. Several natural cases arise:

  • If we use the ideal of all finite sets, then this product is called the finite-support product, and this corresponds to taking what in set theory is called the direct limit at each limit stage, and this can be characterized in terms of a nice universal property of how the product relates to the factors $\mathbb{Q}_\gamma$. This is like the direct sum.

  • If we use the ideal of all countable sets, the product is called a countable-support product.

  • If we use the (improper) ideal of all subsets of $D$, then the product is called the full-support product, and this corresponds to using inverse limits at every stage. This also has a nice characterization by a universal property, like the direct product.

  • The Easton support ideal consists of sets $A$, which are bounded below every inaccessible cardinal. This ideal corresponds to taking direct limits at every inaccessible cardinal stage, and inverse limits at all other limit stages.

This ideal perspective makes sense whether one speaks of product forcing or iterated forcing. Although Easton used his support with product forcing, there are now numerous applications of Easton forcing with iterated forcing.

Each of these ideals and others finds numerous applications in forcing. For example, the finite support iteration of ccc forcing is necessarily ccc. The countable support iteration of proper forcing remains proper.

Note that the Easton ideal is exactly the same as the improper ideal of all sets, if there should happen to be no inaccessible cardinals. Easton realized, however, that he could prove his theorem on the continuum function, without splitting into cases as to the existence of inaccessible cardinals, by using this hybrid support. Easton support is important because with it, one is often able to show that cardinals are preserved, when they would not be preserved with a full-support iteration. The insertion of direct limits at inaccessible stages enables one to prove better chain conditions, which are the means by which one shows that cardinals are not inadvertently collapsed by the forcing.

Lastly, let me say that the particular confusion about inaccessible versus regular cardinals in the Jech presentation is resolved when one realizes that he is forcing only at cardinal stages. And so if $\lambda$ is a successor cardinal $\lambda=\delta^+$, then there at most $\delta$ many cardinals below $\lambda$, and so the support will have size less than $\lambda$, for free. So the only operative restriction there occurs at inaccessible cardinals. (My belief is that this presentation in the book could be improved to explain this better, since numerous students have had exactly this confusion.)

  • $\begingroup$ So if there is just one inaccessible and we are using full support we will collapse it? (And so if there is a proper class of inaccessible the forcing would violate power set, or something similar?) $\endgroup$
    – Asaf Karagila
    Dec 17, 2012 at 9:53
  • $\begingroup$ If you use full support up to an inaccessible $\kappa$, then you can collapse $\kappa^+$. It is like the situation in mathoverflow.net/questions/114590/…. $\endgroup$ Dec 17, 2012 at 10:01
  • $\begingroup$ Thanks, Joel. The question was definitely sub-par, but this answer is such that I will gladly suffer the ignominy it causes to keep your answer around. $\endgroup$ Dec 18, 2012 at 23:31

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