# $< \aleph_1-$support Product of Cohen forcings

Suppose $\kappa$ is an inaccessible cardinal, and let $P$ be the $< \aleph_1-$support product of $Add(\alpha^{++}, 1)$ for singular cardinals $\alpha < \kappa.$

1- Does this forcing preserve cardinals?

2-(A weaker question) Does $\kappa$ remain inaccessible in the generic extension?

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Your forcing notion will collapse all uncountable cardinals below $\kappa$ to $\aleph_1$. To see this, fix any uncountable $\gamma\lt\kappa$. Consider the first $\aleph_1$ many cardinals after $\gamma$ at which forcing occurs. For the $\xi^{th}$ such cardinal $\alpha$, we are adding a subset to $\alpha^{++}$. Consider the first ordinal in the set added at this stage. This can be any ordinal up to $\alpha^{++}$, including any ordinal below $\gamma$. Because you are using countable support, any condition specifies nontrivial sets only on a countable number of these cardinals, and so it is dense that any particular $\beta\lt\gamma$ appears at least once in such a way. Thus, in the generic extension, there will be a surjective map from $\aleph_1$ onto $\gamma$, and so $\gamma$ is collapsed.
The cardinal $\kappa$ itself is not collapsed, by a $\Delta$-system argument (and neither is any cardinal above $\kappa$ collapsed), and so this forcing makes $\kappa$ the $\aleph_2$ of the extension. In particular, it does not preserve the inaccessibility of $\kappa$.
Another issue is that you will need a GCH-type assumption in order even to know that individual factors $\text{Add}(\alpha^{++},1)$ do not collapse cardinals. After all, this partial order will force $2^{\alpha^+}=\alpha^{++}$, which will collapse cardinals unless this was true originally. If you start with GCH, then perhaps you may want to use Easton support, rather than countable support, and then things will be much better for you. – Joel David Hamkins May 18 '11 at 12:20