Collapsing cardinals before the first inaccessible 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?
 A: There are several small mistakes in your question, and one big mistake, leading to your confusion.
First, the small mistakes:


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*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.
A: 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 $\lambda = \aleph_1 = 2^{\aleph_0}$. 
