# Cohen forcing and bounds on GCH violation

I recall (though it might be my faulty memory) an exercise in some book (perhaps Jech?) that was along the lines

"Suppose $\mathrm{GCH}$ holds in $V$ and $P$ is a Cohen forcing that adds $\kappa$ subsets to $\lambda$, then in $V[G]$ there exists some $\beta$ such that $\mathrm{GCH}$ holds above $\aleph_\beta$".

Of course there is an assumption that $P$ is a set, otherwise Easton's theorem tells us otherwise.

If this is indeed from Jech's Set Theory then it is most likely correct, and I cannot see the reason why; if it is a result of some bad sectors in my brain, I still cannot come up with a counterexample.

(Also, if it is from some book I would be glad to have the reference!)

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I'm 30+ years rusty, but doesn't $P$ have the $(2^{<\lambda})^+$ chain condition and therefore preserve cofinalities and cardinals $\ge (2^{<\lambda})^+$? –  Brian M. Scott Jul 2 '11 at 20:56
@Brian: That seems like a very good argument, what about singular cardinals of cofinalities smaller than the chain condition's size? –  Asaf Karagila Jul 2 '11 at 21:04

A counting argument works in this case. Without much loss of generality $P$ is actually a complete Boolean algebra. In this case, names for subsets of a cardinal $\mu$ in $V[G]$ can be equated with ground model functions $f:\mu \to P$, where $f(\xi)$ is the Boolean value that $\xi$ is in the set in question. There are $(|P|^{\mu})^V$ such functions, so when $(|P|)^V \leq (2^{\mu})^V = (\mu^+)^V$ there are only $(\mu^+)^V$ such functions in the ground model. Since every subset of $\mu$ in the generic extension $V[G]$ has at least one name, we conclude that $(2^\mu)^{V[G]} \leq (\mu^+)^V$. Since we cannot have fewer than $(\mu^+)^{V}$ subsets of $\mu$, we conclude that $(\mu^+)^V = (\mu^+)^{V[G]} = (2^\mu)^{V[G]}$. Therefore, the GCH holds for all cardinals $\mu$ in $V[G]$ such that $(|P|)^V \leq (2^\mu)^V$.
Many thanks!${}$ –  Asaf Karagila Jul 3 '11 at 5:54