# The GCH in a reverse Easton support iteration

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 $V[G_\kappa]$?

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.