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  • $\begingroup$ I think you're right. For some reason I kept coming to the conclusion that GCH implies that for successor $\kappa$, we have that $|B(\kappa)| = \kappa^+.$ However, your argument looks impeccable. $\endgroup$
    – goblin GONE
    Commented May 13, 2014 at 7:39
  • $\begingroup$ I'd be interested in the proof that ZFC allows $dv(\kappa)>\kappa$, if you don't mind sharing it. $\endgroup$
    – Noah Schweber
    Commented May 13, 2014 at 7:40
  • 2
    $\begingroup$ Isn't it simpler to note that under GCH $2^{<\kappa} = \kappa$? $\endgroup$
    – Sam Roberts
    Commented May 13, 2014 at 8:15
  • 1
    $\begingroup$ Fair enough - it just looks to me a little over complicated to merely establish that $2^{<\kappa} = \kappa$. $\endgroup$
    – Sam Roberts
    Commented May 13, 2014 at 8:21
  • 1
    $\begingroup$ I don't know. I'd be inclined to give the following proof. Assume GCH below $\kappa$. Then $2^{<\kappa} = sup\{2^{\lambda}: \lambda<\kappa\} = sup\{\lambda^+:\lambda<\kappa\} = \kappa$. $\endgroup$
    – Sam Roberts
    Commented May 14, 2014 at 6:21