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

This is related to my previous question "Reals added after Cohen forcing" Reals added after Cohen forcing

The fact is that what I had in mind by $\lambda-$many Cohen reals was different from what was considered in answers. To be more precise by $\lambda-$many Cohen reals over $V$ I mean a sequence $(r_{\alpha}: \alpha < \lambda)$ which is $V-$generic for the Cohen forcing $Add(\omega, \lambda)$ for adding $\lambda-$many new Cohen reals (thus it is different from saying that we have $\lambda-$many reals where each of them is $V-$generic for the Cohen forcing $Add(\omega, 1)$ for adding a new Cohen real). Similarly for other forcing notions for adding reals.

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.

Question. Can we find $\aleph_{\omega+1}-$many new reals which are generic for some forcing notion over $V$?

share|improve this question
3  
A variant of this question: Does the generic extension $V[G]$ (for a generic $G \subseteq Add(\omega, \aleph_{\omega})$ contain a generic $H \subseteq Add(\omega, \aleph_{\omega+1}$? And if yes, can $H$ be found such that $V[G]=V[H]$? (This looks like it should be known, but I do not see it.) –  Goldstern Sep 13 '12 at 22:19
    
That is a very nice question. At first I posted a partial answer to it, but now I think I've got the whole thing, and have edited my answer accordingly. –  Joel David Hamkins Sep 14 '12 at 1:47
add comment

1 Answer

It is a general fact that every intermediate model $M\models\text{ZFC}$ sitting between a model and a forcing extension $V\subset M\subset V[G]$ is itself a forcing extension. Indeed, if $G\subset\mathbb{B}$ is $V$-generic for the complete Boolean algebra $\mathbb{B}$, then there is a complete subalgebra $\mathbb{B}_0\subset\mathbb{B}$ such that $M=V[G\cap \mathbb{B}_0]$. (This is proved in Jech's book.)

Thus, if $z$ is any real or indeed any set of ordinals in a forcing extension $V[G]$, then we may form the model $V[z]$, which is a model of ZFC, and by the general fact above it is obtained by forcing with a complete subalgebra of the forcing used with $G$.

In your case, every real $z\in V_1$ has a hereditarily countable name, and thus $z\in V[g_0]$ for some $V$-generic Cohen real in $V_1$. Since every nontrivial subalgebra of $\text{Add}(\omega,1)$ has a countable dense set, it follows that $V[z]=V[g_0]$ for some Cohen real $g_0\in V_1$. In short, every real in $V_1$ that is not in $V$, and this means all $\aleph_{\omega+1}$ of them, is added by the forcing to add a single Cohen real. (These reals are not themselves Cohen reals, but they have names for the forcing to add a Cohen real which interpret to them by some actual Cohen real in $V_1$.)

But I'm not sure if I have understood your question in the sense that you may have intended it. In particular, if you meant that you wanted a single forcing notion to add an entire $\aleph_{\omega+1}$ sequence of reals, then I would say that you already have it, namely, the forcing $\text{Add}(\omega,\aleph_\omega)$ itself was already adding all those extra reals.

Edit. Let me answer Goldstern's follow-up question in the comments, a question I find very interesting.

First, I claim that the $V[G]=V[H]$ situation he mentions is impossible, where $G$ is $V$-generic for adding $\aleph_\omega$ many Cohen reals and $H$ is $V$-generic for adding $\aleph_{\omega+1}$ many Cohen reals. The reason is that if these two extensions were equal, then it would follow that the two forcing notions $\text{Add}(\omega,\aleph_\omega)$ and $\text{Add}(\omega,\aleph_{\omega+1})$ are isomorphic below respective conditions and hence simply isomorphic. But the former has a dense set of size $\aleph_\omega$ and the latter has dense sets only of size at least $\aleph_{\omega+1}$, since any smaller set than this could not mention enough points in the domains of the conditions to be dense.

Second, a similar argument shows now that we cannot even have that $V[G]$ contains a $V$-generic filter $H$ for $\text{Add}(\omega,\aleph_{\omega+1})$, for in this case we would have that $\text{Add}(\omega,\aleph_{\omega+1})$ is isomorphic to a complete subalgebra of $\text{Add}(\omega,\aleph_\omega)$. But since this latter Boolean algebra is $\aleph_\omega$-dense, it follows that every complete subalgebra of it is also (at most) $\aleph_\omega$-dense. But $\text{Add}(\omega,\aleph_{\omega+1})$ has no dense set of size less than $\aleph_{\omega+1}$, for the reason described in the previous paragraph. This is a contradiction, and so $V[G]$ contains no $V$-generic filter for $\text{Add}(\omega,\aleph_{\omega+1})$.

More generally, essentially the same argument shows that adding $\theta$ many Cohen reals can never add a generic filter for adding $\lambda$ many Cohen reals, if $\theta\lt\lambda$.

But meanwhile, as my answer at the other question shows, $V[G]$ has a family of $\aleph_{\omega+1}$ many pairwise (and finitely-wise) mutually generic Cohen reals. But these do not rise to the level of mutual genericity necessary to form a generic for $\text{Add}(\omega,\aleph_{\omega+1})$.

share|improve this answer
    
But it seems that I said most of this over at the other question already. Could you clarify the question if this doesn't answer it? –  Joel David Hamkins Sep 13 '12 at 14:15
    
To see that a complete subalgebra $\mathbb{C}$ of a $\kappa$-dense complete Boolean algebra $\mathbb{B}$ is also at most $\kappa$-dense, fix a dense $D\subset\mathbb{B}$ of size $\kappa$, and for each $d\in D$ let $c_d$ be the infimum of the elements of $\mathbb{C}$ above $d$. This has size at most $\kappa$ and is dense in $\mathbb{C}$, because any $c\in\mathbb{C}$ has some $d\in D$ below it, and hence $c_d\leq c$. –  Joel David Hamkins Sep 14 '12 at 2:35
add comment

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.