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?



Collapsing cardinals before the first inaccessible
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.


