MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a givien partial order, how many generic extensions might exist? In other words, for a boolean valued model class which dreams of a generic extension, how many unique generic objects exist for a given partial order (which gives rise to a boolean algebra naming such an extension)?

share|cite|improve this question
Enough for what? – Will Jagy Jul 1 '11 at 5:36
Enough to accept the complete reversal of ordering. As in adding a cohen real and considering "stronger" to be <= in the order, but stronger in the sense of a larger subset. of that. – Erin Carmody Jul 1 '11 at 5:42
I understand all the (3, so far) answers, but I still do not understand the question. What does "generous enough to reinterpret the order" mean? Perhaps you could give an example that is not "generous enough"? – Goldstern Jul 1 '11 at 21:52
I understand your comment, the wording is not technical. What I mean is a partial order that one would use to force. An example of a partial order which is not "generous enough" would be a partial order on finite binary sequences where p < q if p is a subset of q. A generic filter on this partial order does not seem to give rise to anything new. – Erin Carmody Jul 2 '11 at 18:27
up vote 5 down vote accepted

It is often the case that when we force with a partial order $\mathbb{P}$, then in the resulting forcing extension $V[G]$ there may be other $V$-generic filters and indeed many other $V$-generic filters $G'$, even giving rise to the same extension $V[G]=V[G']$. Thus, although from a the filter $G$ we may construct the forcing extension $V[G]$, we should not think of the of the forcing extension $V[G]$ as necessarily determining $G$; different filters can lead to the same forcing extension.

For example, if $\mathbb{P}$ has an automorphism $\pi:\mathbb{P}\to\mathbb{P}$ in $V$, then it is not difficult to see that $\pi"G$ is also $V$-generic, and furthermore these generic filters give rise to the same forcing extension $V[G]=V[\pi"G]$. Since many of the most-commonly-used forcing notions admit numerous automorphisms, it is quite common that we find ourselves in a forcing extension $V[G]$ for which there are numerous filters $G'\in V[G]$ for the same forcing notion $\mathbb{P}$ with $V[G]=V[G']$. And we can often count them in a precise way, as Stefan did in his answer. In particular, highly homogeneous forcing notions, having numerous automorphisms, will always lead to forcing extensions having many $V$-generic filters.

Meanwhile, there are also interesting cases where the forcing extension necessarily has fewer or sometimes only one generic filter for that order, which it seems is part of what you are asking about. For example, from $\Diamond$ one can construct Souslin trees with the unique branch property, meaning that the tree in the forcing extension has precisely one branch (and with a Souslin tree, the generic filters in any extension are precisely the branches), and one can view this property of a forcing notion as a strong kind of rigidity. This kind of situation with Souslin trees was the theme of my article with Gunter Fuchs on the Degrees of rigidity for Souslin trees, where we separate various strong notions of rigidity into a hierarchy.

Another example is self-encoding forcing, the forcing notion that adds a subset of some cardinal $\kappa$, and then codes this set into the GCH pattern (or whatever), and then codes these new sets into the GCH pattern a bit higher up, and so on. The process catches its tail in $\omega$ many steps, and the result is that in the corresponding forcing extension $V[G]$, there is only one $V$-generic filter for the poset, and it is therefore definable there from parameter $\kappa$.

When a nontrivial forcing notion $\mathbb{P}$ necessarily leads to a forcing extension in which $\mathbb{P}$ has exactly one generic filter, then one should think of $\mathbb{P}$ as being far from homogeneous.

share|cite|improve this answer
Thank you for each paragraph. Especially the Souslin tree example. – Erin Carmody Jul 1 '11 at 16:02

It is a bit difficult to understand the question. But are you asking for the number of different forcing extensions by a given partial order? From the perspective of the universe that we are living in? Or from the outside?

Let me be more specific. Let $M$ be a countable transitive model of set theory. In $M$, let $P$ be Cohen forcing, i.e., the partial order of functions from a natural number to $2$, ordered by reverse inclusion.
Now, $P$ is a countable set, both in the real world and from the perspective of $M$.

In the real world, there are $2^{\aleph_0}$ different $P$-generic filters over $M$.
($2^{\aleph_0}$ is clearly an upper bound, but it can also be shown that there are $2^{\aleph_0}$ different ones.) Now, some of these filters might actually give the same generic extension, for example if one filter is just a finite modification of the other. But for each generic filter $G$ the extension $M[G]$ is countable (in the real world). Hence, if we have $2^{\aleph_0}$ different generic filters that each yield only a countable extension of $M$, there must be $2^{\aleph_0}$ pairwise different generic extensions of $M$.

Now, if you are interested in the number of extensions from the perspective of $M$, that question does not really make sense, since $M$ doesn't have access to the generic filters over $M$.
For each cardinal $\kappa$ of $M$, there is a forcing extension of $M$ by iterated Cohen forcing in which $2^{\aleph_0}$ is at least $\kappa$. Note that $M$ and this large generic extension have the same cardinals. And that forcing extension contains, as substructures, $\kappa$ many different generic extensions of $M$ by single Cohen reals by the same argument as above.

Since $\kappa$ was an arbitrary cardinal of $M$, this shows that it is impossible to gauge the number of generic extensions from inside the ground model.

I hope this does at least tell you how the question should be phrased.

share|cite|improve this answer
Thank you for this answer, it was helpful indeed. – Erin Carmody Jul 1 '11 at 16:05

Let me add an easier observation about the number of $M$-generic filters, for a given poset $P\in M$, if we do the counting "from the outside". That is, suppose $M$ is a countable transitive model and we want the number of generic filters in the real world (which, in particular, sees a counting of $M$). As Stefan pointed out, there is an obvious upper bound of $2^{\aleph_0}$ because there are only that many subsets of $P$ altogether. On the other hand, this bound is attained by any non-trivial forcing $P$, where "non-trivial" means that there is some $q\in P$ such that every extension of $q$ has at least two incompatible extensions. (If this fails, then the generic extensions are just the ground model.) In this situation, the usual proof that generic filters exist (namely, list $M$'s dense subsets of $P$ in an $\omega$-sequence and build the generic filter in $\omega$ steps, hitting the $n$-th dense set at the $n$-th step) is easily modified to give $2^{\aleph_0}$ different generic filters, by inserting steps at which you choose arbitrarily one of two (previously selected) incompatible extensions of the condition chosen at the previous step. As Stefan pointed out, once you have $2^{\aleph_0}$ generic filters, you also have $2^{\aleph_0}$ generic extensions, because each of the extensions contains at most countably many of the generic filters.

share|cite|improve this answer
Thank you Andreas! Very detailed and interesting. – Erin Carmody Jul 1 '11 at 18:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.