Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $V_1$ be a generic extension of $V\models GCH$ obtained by adding $\aleph_{\omega}-$many Cohen reals. Then we have the following:

1- In $V_1$ there are $\aleph_{\omega+1}-$many reals,

2- In $V_1$ there are only $\aleph_{\omega}-$many Cohen reals.

What can we say about the other reals in $V_1$? Are they generic for some forcing notion over $V$?

share|improve this question
1  
I'm not sure that your logic holds about how many Cohen reals are in the extension. If you add only one Cohen real, the extension has $\aleph_0$ many Cohen reals and not just one. –  Asaf Karagila Jun 7 '12 at 7:15
1  
Asaf, there are actually continuum many Cohen reals in the extension, not merely $\aleph_0$ many, for the reason I explain in my answer. –  Joel David Hamkins Jun 7 '12 at 10:18

1 Answer 1

up vote 13 down vote accepted

In your extension $V_1$, there are actually continuum many, that is $\aleph_{\omega+1}$ many $V$-generic Cohen reals. In general, whenever you add even a single Cohen real, then the extension wills have continuum many $V$-generic Cohen reals, because if $c$ is a $V$-generic Cohen real and $x$ is any real in the ground model, then the bit-wise sum $c\oplus x$ will be a $V$-generic Cohen real, since this induces an automorphism of the forcing in the ground model, and these are all different. (In your case, as Goldstern mentions in the comments, we may view the forcing as adding all but one of the Cohen reals, and then a final Cohen real, to achieve $\aleph_{\omega+1}$ many $V$-generic Cohen reals by this reasoning.)

But it is true that not every new real of $V[G]$ is a $V$-generic Cohen real. For example, one may easily construct reals that obey some regular pattern, repeating their digits in pairs, for example, which prevent them from being Cohen reals, even if they are not in the ground model.

Nevertheless, if $V[G]$ is a forcing extension obtained by adding any number of Cohen reals and $z$ is any real in $V[G]$, then I claim that $z\in V[c]$ for some $V$-generic Cohen real in $V[G]$. This is because the countable chain condition of the forcing means that one needs only countably much information from $G$ in order to construct $z$, and restricting the generic sequence of Cohen reals to any countable domain is isomorphic again to adding a single Cohen real.

In general, if $V\subset V[G]$ is any forcing extension and $W$ is a model in between $V\subset W\subset V[G]$, such as $W=V[z]$ for some real $z$, then $W$ is also a forcing extension of $V$ by a complete subalgebra of the Boolean algebra giving rise to $G$.

In the case of adding a Cohen real, forcing which has a countable dense set, every subalgebra of its Boolean algebra also has a countable dense set, and all nontrivial forcing notions with a countable dense set are isomorphic to adding a Cohen real. In this sense, every real added in a Cohen real forcing extension (and this includes every real in your model by my observations above) is generic over $V$ for forcing that is isomorphic to the forcing to add a Cohen real.

So the answer to your final question is that yes, these extra reals are $V$-generic for some forcing notion, and that forcing notion is isomorphic to the forcing to add a single Cohen real!

Let me conclude with an interesting tidbit:

Theorem. In the forcing extension $V[c]$ obtained by adding a single $V$-generic Cohen real, there is a family of continuum many pairwise mutually generic Cohen reals. Indeed, there is a perfect set $P$ in $V[c]$, all of whose finite subsets are mutually $V$-generic Cohen reals.

Proof. Consider the forcing to add such a perfect set $P$. We want to force to create a tree, all of whose branches are $V$-generic Cohen reals, and such that any finitely many branches are mutually generic Cohen reals. Let conditions be finite binary trees, ordered by end-extension. It is dense for the leaves to be extended into any given dense subset of Cohen forcing. And for finite products of Cohen forcing with itself, it is dense to extend the tree so that all pairs (or triples etc.) of branches are inside any given dense set in the product forcing. Thus, this forcing will create such a tree and hence such a perfect set.

Finally, observe that our tree forcing has only countably many conditions, and thus it is isomorphic to adding a single Cohen real. So the forcing extension $V[c]$ already has such a pefect set. QED

Of course, the branches through the perfect set will not be $V$-generic for the forcing to add continuum many Cohen reals, since that forcing is not a subalgebra of the forcing to add only one. The reals in the perfect set are only mutually generic when taken finitely many at a time, but not fully mutually generic for infinite collections. For example, the perfect set contains reals that are the limits of other of its elements, and this violates mutual genericity for those infinite families.

share|improve this answer
    
Implicit in your first paragraph is the statement that adding $\aleph_\omega$ many Cohen reals is the same as first adding $\aleph_\omega$ many (blowing up the continuum to $\aleph_{\omega+1}$), then adding one more; this last one adds $\aleph_{\omega+1}$ many Cohens. –  Goldstern Jun 7 '12 at 10:25
    
Yes, thanks for this; I have edited to explain it. I also added the tidbit at the end, which always fascinated me. –  Joel David Hamkins Jun 7 '12 at 10:34

Your Answer

 
discard

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.