How badly can GCH fail?

Question

I'm interested in "how many" cardinals we can cram between $\kappa$ and $2^\kappa$ before ZFC says "nope, too many" as measured by order type.

So my question is (and i don't know if this is well-posed), given a cardinal number $\kappa$, what is the least ordinal $\alpha$ such that ZFC proves that $\mathrm{ord}[\kappa, 2^\kappa)<\alpha$? The half open interval is intended to mean the set of all cardinal numbers between $\kappa$ and $2^\kappa$, including $\kappa$ but not including $2^\kappa$, and $\mathrm{ord}$ is meant to return the order type of this interval.

Answer 1

Let $M$ be a model of $\sf ZFC+GCH$, and let $F$ be a definable class function on the regular cardinals such that

1. $\kappa<F(\kappa)$,
2. $\operatorname{cf}(\kappa)<\operatorname{cf}(F(\kappa))$,
3. $\kappa<\lambda\implies F(\kappa)\leq F(\lambda)$.

Then there exists a model $N$ of $\sf ZFC$ such that $M\subseteq N$, and they have the same ordinals and cardinals, and in $N$ it is true that for every regular $\kappa$, $2^\kappa=F(\kappa)$.

This means that the continuum function is limited only by its most basic properties, and nothing more.

So while $\sf ZFC$ proves that $2^\kappa$ is bounded by some ordinal it is completely impossible to prove which ordinal.

Answer 2

As many as you want. There is no upper bound. The cofinality of $2^\kappa$ must exceed $\kappa$,of course. See Easton's theorem for the details regarding regular cardinals.