I am trying to understand the proof that the GCH can first fail at a weakly compact cardinal. We assume the GCH and that there exists a weakly compact cardinal $\kappa$, and we construct a reverse Easton support iteration $\{\mathbb P\_\alpha\}\_{\alpha\leq \kappa}$ adding a Cohen subset to each inaccesible cardinal, and then we add $\kappa^{++}$ Cohen subsets to $\kappa$. We get a generic extension $V[G\_\kappa][H]$ where $\kappa$ remains weakly compact.


My problem is why the GCH holds below $\kappa$. More specifically, let $\{\mu\_n\}\_{n\in\omega}$ be the sequence of the first inaccesible cardinals, and let $\mu$ be its supremum. The extensions $V=V[G\_{\mu\_0}]\subset V[G\_{\mu\_1}]\subset \cdots$ have the same cardinals and satisfy the GCH. Moreover, each $\mu\_n$ is a cardinal in $V[G\_\kappa]$ because the iteration can be factored as $\mathbb P\_\kappa\cong \mathbb P\_{\mu\_n}\*\pi^{\mu\_n}$ and $\Vdash \pi^{\mu\_n}$ is $\mu\_n$-directed closed. Hence $\mu$ is also a cardinal in $V[G\_\kappa]$.


Now, the poset $\mathbb P\_\mu$ is the inverse limit of the previous ones. We have $|\mathbb P\_\mu|=\mu^+$, but it may fail satisfying the $\mu^+$-c.c., so the usual bounds on the number of nice names for subsets of $\mu$ provide the bound $2^\mu\leq \mu^{++}$ in $V[G\_\mu]$, and this is not enough. What am I missing? Does not this construction guarantees the GCH in $V[G\_\kappa]$?



The GCH in a reverse Easton support iteration
Every subset of $\mu$ in the extension is determined by the
$\omega$-sequence of its initial segments, that is, an
$\omega$-sequence of subsets of $\mu\_n$ as $n$ increases. Each
such subset of $\mu\_n$ has a nice $\mathbb{P}\_{\mu\_n+1}$-name. The
forcing altogether is $\leq\omega$-closed, and so we may find a
sequence in the ground model of such nice $\mathbb{P}\_{\mu\_n+1}$-names, that
collectively name the initial segments of our given subset of
$\mu$. But for subsets of a given $\mu\_n$, there are strictly
fewer than $\mu$ many such nice names, and so there are at most
$\mu^\omega=\mu^+$ many such $\omega$-sequences of such names in $V$.
So we still have at most $\mu^+$ many subsets of $\mu$ in the
extension.


A similar argument works at other singular limits of inaccessible
cardinals, simply by working above the cofinality of that
cardinal.


