This is again a question about forcing. Start in $L$, the constructible universe. CH holds. Let $\lambda$ be an inaccessible cardinal, also let $\lambda$ > $\aleph_0$. For each $\alpha < \lambda$, let $P_\alpha$ be the set of all functions such that $dom(p_\alpha) \subset \aleph_0$, $|dom(p_\alpha)|<\aleph_0$ and $ran(p_\alpha) \subset \alpha$. $p_\alpha$ is stronger than $q_\alpha$ iff $p_\alpha$ extends $q_\alpha$.

Now consider $(P,<)$ be the $\kappa$-product of the $P_\alpha$, $\alpha<\lambda$. Now the conditions of $P$ are functions taking their argument on the set of all subsets of $\lambda \cdot \aleph_0$ such that the cardinality of the domain of $p$ is strictly smaller than $\aleph_0$ and such that $p(\alpha,\xi)<\alpha$ for each pair $(\alpha,\xi) \in dom(p)$.

If $G$ is a generic set of conditions then let for each $\alpha$, $G_\alpha$ be the projection of $G$ on each $P_\alpha$, each $G_\alpha$ is a generic filter, let $f_\alpha= \bigcup G_\alpha$ is a function fro, $\aleph_0$ onto $\alpha$ for every $\alpha < \lambda$ we have $|\alpha| \leq \aleph_0$.

Since the forcing is $<\aleph_0$-closed so cardinals and cofinalities are preserved and it satisfies the $\lambda$-chain condition so $\lambda$ is a cardinal in the generic extension $L[G]$.

Now we have $\lambda=\aleph_0^+$. But in light of the basic fact (that I had overlooked in my previous post), the continuum can't be strong limit. So the above is clearly false. Can you help me point my mistakes?

  • $\begingroup$ Why do you think $\lambda$ will still be strong limit? $\endgroup$
    – Kiochi
    Jul 17, 2010 at 22:29
  • $\begingroup$ $\lambda$ is an inaccessible cardinal but since the forcing preserves cardinals and cofinalities (if I'm not mistaken, in light of François's comment) it is still inaccessible in the generic extension. $\endgroup$ Jul 17, 2010 at 22:47
  • $\begingroup$ All cardinals and cofinalities below $\lambda$ are collapsed to $\omega$, so they are not preserved. $\endgroup$ Jul 17, 2010 at 22:52
  • $\begingroup$ This is true François but $\lambda$ will still be inaccessible, right? $\endgroup$ Jul 17, 2010 at 23:15

2 Answers 2


There are several small mistakes in your question, and one big mistake, leading to your confusion.

First, the small mistakes:

  • You say that $\lambda$ is inaccessible and also $\lambda>\aleph_0$. This is redundant, since the usual definition has that $\lambda$ is an inaccessible cardinal if and only if it is an uncountable regular strong limit cardinal.

  • You say that $P$ is the $\kappa$-product of the $P_\alpha$, but what you really mean is that $P$ is the finite support $\lambda$-product of the $P_alpha$ (and this is indeed how you specify it just afterwards, by insisting that conditions have finite support). If you used full support, then you won't get the $\lambda$-chain condition, and $\lambda$ itself would be collapsed.

Second, the big mistake:

  • You say "Since the forcing is $\lt\aleph_0$-closed so cardinals and cofinalities are preserved...," but this is completely wrong. As François mentions, every partial order is $\lt\omega$ closed, since this is just saying that finite descending sequences have lower bounds, which is trivial. And in fact, the forcing $P_\alpha$ clearly collapses $\alpha$ to $\omega$, and so $P$ collapses all $\alpha\lt\lambda$ to $\omega$. The cardinal $\lambda$ itself is not collapsed, because the forcing $P$ is $\lambda$-c.c., a fact which can be proved by a combinatorial $\Delta$-system argument. This fact is surely the key to understanding the Levy collapse, which is how your forcing is known.

The Levy collapse collapses all cardinals below $\lambda$ to $\omega$ and preserves $\lambda$ itself, thereby making $\lambda$ into the $\omega_1$ of the forcing extension. There is no need to start in $L$ with this forcing, or with CH. It works quite generally. Furthermore, one needn't collapse to $\omega$; by collapsing all ordinals $\alpha\lt\lambda$ to some other fixed cardinal $\delta$, one forces $\lambda$ to be $\delta^+$ in the extension. For example, one can make $\lambda$ equal to $\aleph_2$ or $\aleph_3$ in the extension.

Finally, let me point out that $\lambda$ is indeed the continuum in the extension. A chain condition argument shows that every real of $V[G]$ is added by some stage $V[G_\alpha]$, where we cut off the forcing at $\alpha\lt\lambda$, and so the reals of $V[G]$ are the union of the reals of $V[G_\alpha]$, each of which are countable in $V[G]$. So the continuum of $V[G]$ has size $\lambda$, which is now $\omega_1$. In particular, $V[G]$ satisfies CH, even if the ground model $V$ does not. But of course, $\lambda$ is no longer inaccessible in $V[G]$, since it is a successor cardinal there, and so the objection that the continuum cannot be $\lambda$ in $V[G]$ evaporates.

  • $\begingroup$ Thank you for this enlightening answer. Also, I was thinking that $\lambda$ would stay the same in the new model as it was in the old model which, I realize actually, is not the same as saying that it is just not collapsed to $\omega$ (it can be $\aleph_1$ or $\aleph_2$ as you point out). $\endgroup$ Jul 17, 2010 at 23:50

This product forcing (known as the Lévy collapse) is not ${<\kappa}$-closed for any $\kappa > \omega$ since the individual factors $P_\alpha$ are not ${<\kappa}$-closed. The forcing is $\lambda$-c.c., so it preserves all cardinals and cofinalities from $\lambda$ and above, but not those below $\lambda$. In the extension, we have $\lambda = \aleph_1 = 2^{\aleph_0}$.

  • $\begingroup$ Sorry, there was a typo, I meant $<\aleph_0$-closed not $<\kappa$-closed $\endgroup$ Jul 17, 2010 at 22:24
  • $\begingroup$ But still, every finite descending chain of functions will have a lower bound and every countable descending chain of functions will have a lower bound since $B(P)=Col(\aleph_0,<\lambda)$ is a complete Boolean algebra. Maybe I am missing something...? $\endgroup$ Jul 17, 2010 at 22:44
  • $\begingroup$ Every forcing is $<\omega$-closed. Not every countable sequence in the Boolean completion of $P$ will have a nonzero lower bound. $\endgroup$ Jul 17, 2010 at 22:50
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    $\begingroup$ Since you mention the Boolean algebra, it is interesting to note (a common confusion for forcing beginners) that infinite complete Boolean algebras are never countably closed (in the nonzero lower bound sense), because if $A$ is an infinite antichain, one can take the supremum and then strip off successive elements one at a time. Thus, even if $P$ is a highly closed forcing partial order, its Boolean algebra is not. $\endgroup$ Jul 17, 2010 at 23:55

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