

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

If we start with a model of $\sf ZFC$, $M$ and $(P,\le)\in M$ is a notion of forcing, $G\subseteq P$ a generic filter, then in $M[G]$ we can define some generic object from $G$. For example, if $P$ is the Levy collapse of $\omega_1$ to $\omega$ then $G$ defines a new function $f\colon\omega\to\omega_1$ which is bijective.

Now suppose that we have a product forcing $P=\prod P_i$ in $M$, then the generic filter $G$ can be projected on every coordinate and $G_i$ (its projection) is a generic filter over $P_i$, which defines some generic object. Then a priori we can think that $G$ defines some generic collection $\{g_i\}$ such that $g_i$ is the generic object defined by $G_i$.

So for example, if we take the product of two Cohen-like forcings, one adding a subset of $\omega$ and the other adding a subset of $\omega_1$ - we can think of the collection as the pair of the new subsets.

In Jech Set Theory, 3rd Millennium edition, in the relevant chapter (Ch. 15) Jech discusses this very shortly, proving some basic theorems about this. However in the exercises there is only one problem related to this issue:

Let $P$ be the notion of forcing (15.1) that adjoins $\kappa$ Cohen reals. Then $P$ is (isomorphic to) the product of $\kappa$ copies of the forcing for adding a single Cohen real (Example 14.2).

This means, that we can think of the product of $\kappa$ Cohen forcings as adding $\{g_i\mid i<\kappa\}$ as a set of $\kappa$ new Cohen reals, just like we would think at first.

However, there is no mention of this being true or false in a general framework. So to my question:

Suppose $P=\prod P_i$ is the product of $\kappa$ copies of some $P'$ a fixed notion of forcing, can we automatically assume that $G\subseteq P$, a generic filter, adds a set of $\kappa$ new generic elements, each defined by a generic filter, $G_i$ over $P'$?

If this is true, then we can ask even further:

Suppose $P=\prod P_i$ is a product of $\kappa$ notions of forcings, can we say that $G\subseteq P$, a generic filter, adds a set of generic objects each defined solely by $G_i$?

share|cite|improve this question
Dear anonymous downvoter: If you have something to say to me, please say it in words so I can understand you. – Asaf Karagila Nov 9 '13 at 22:46
Since this question seems to be a target for some many downvotes, I can only assume that those who read it has fierce objections to its content. Since the mathematical content seems to me as fine (and I have received no remarks on that), I can only conclude that the objections were to minor grammatical slips. I have tried my best to correct them, but I'd be glad to hear further points for improvements! Thank you very much downvoters for bringing these issues to my attention! (Because otherwise, I'd just assume these downvotes represent some disagreement over meta issues, which would be silly.) – Asaf Karagila Jul 15 '14 at 12:30
up vote 6 down vote accepted

For your first question, yes, it is true in complete generality. If $G$ is a $V$-generic filter on the product $\Pi_i P_i$, then the projection of $G$ onto each factor, that is, the set $G_j$ consisting of the $j^{\rm th}$ coordinates of the conditions in $G$, is a $V$-generic filter for $P_j$. This is because if $D\subset P_j$ is any dense subset of $P_j$ in $V$, then the set of conditions $p\in \Pi_i P_i$ that have their $j^{\rm th}$ coordinate in $D$ is dense in the product forcing, and thus it is met by $G$, and so $G_j$ meets $D$.

In particular, if the product consists of $\kappa$ many copies of a single nontrivial forcing notion $P'$, then the product forcing will add $\kappa$ many $V$-generic filters for $P'$. If $P'$ is nontrivial in the sense that there are incompatible conditions below any given condition (that it, it is splitting), then it is dense in the product that the generic filters $G_j$ added on each factor are distinct, since for any pair $i,j$ the set of conditions in the product for which the $i^{\rm th}$ coordinate is incompatible with the $j^{\rm th}$ coordinate is dense in the product.

Conversely, we can reconstruct the full product generic $G$ from the projections $G_j$, since a condition is in $G$ if and only if its projection on coordinate $j$ is in $G_j$ for every $j$. Perhaps this is the what you ask in your second question?

But not every collection of $V$-generic filters $G_j$ for $P_j$ will give rise to a generic for the product forcing. For example, we can't have them be all the same on every coordinate for the reasons mentioned above. The additional property that the factor filters form a generic filter for the product is called mutual genericity.

share|cite|improve this answer
Thanks for the detailed answer, Joel, I do have a question on that last part. Starting with a model of GCH, there are only $\aleph_1$ many possible generic filters on a Cohen forcing; however if we add $\omega_2$ many reals we have that many of them must be defined by the same filter. No? – Asaf Karagila Dec 11 '11 at 16:55
No, the Cohen reals determine and are determined by their filters. You seem to be counting the filters in the ground model, but they should be counted in the forcing extension, where they live. In the extension after adding $\omega_2$ many Cohen reals, you have at least $\omega_2$ many different Cohen reals. (In fact, after adding just one $V$-generic Cohen real, then regardless of GCH there are continuum many $V$-generic Cohen reals in $V[c]$.) – JDH Dec 11 '11 at 17:03
No, in $V[c]$ there is a continuum size set of pairwise mutually generic Cohen reals. (But just pairwise, or finitely-wise, is all I am claiming here.) To see this, consider a Cohen real as laid out on the nodes of the tree $2^{\lt\omega}$, and consider the family of reals obtained by the paths through this tree. – JDH Dec 11 '11 at 17:13
Asaf, I'm not saying that adding one Cohen real is forcing equivalent to adding continuum many; that would not be true. I'm just saying that if you add one or more Cohen reals, then in the extension you've got continuum many such Cohen reals. But you may not have a generic filter for the forcing to add $\kappa$ many for any uncountable $\kappa$. This just shows that, ``the number of $V$-generic Cohen reals,'' is not the same as the number of Cohen reals that you forced to add, since the former ignores the issues of (infinitary) mutual genericity. – JDH Dec 11 '11 at 18:44
I think I've got what you mean. Regardless to how many new reals I've added - there will be continuum many new generic reals (which makes sense regardless to a forcing argument), however the amount of mutually generic reals depends on how we forced them (that is how many were forced upon the model). – Asaf Karagila Dec 11 '11 at 19:03

Your Answer


