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 $V$?

  • 3
    $\begingroup$ 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.) $\endgroup$
    – Goldstern
    Sep 13, 2012 at 22:19
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2012 at 1:47

1 Answer 1


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 $\text{Add}(\omega,\aleph_{\omega+1})$.

  • $\begingroup$ 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? $\endgroup$ Sep 13, 2012 at 14:15
  • $\begingroup$ 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 $c_d\leq c$. $\endgroup$ Sep 14, 2012 at 2:35

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