If $2^{\aleph_0}$ is weakly inaccessible, can every cardinal $\kappa$ in the interval $[\aleph_0,2^{\aleph_0})$ satisfy $2^\kappa = 2^{\aleph_0}$?

Question

Question. Is the following consistent with ZFC?

1. $2^{\aleph_0}$ is weakly inaccessible
2. Every cardinal $\kappa$ in the interval $[\aleph_0,2^{\aleph_0})$ satisfies $2^\kappa = 2^{\aleph_0}$.

Note that 2 implies that $2^{\aleph_0}$ is regular, so really, condition 1 can be weakened to saying that $2^{\aleph_0}$ is a weak limit cardinal.

(I am also interested in the strengthening of this statement where all instances of $\aleph_0$ are replaced by a variable $\nu$, which gets universally quantified over the regular cardinals.)

Answer 1

Yes, start with a model of $\sf GCH$ with $\mu$ inaccessible, and add $\mu$ Cohen reals. Standard arguments show that the continuum function is constant below $\mu$. This works equally well with Cohen reals being replaced by Cohen subsets of some regular $\nu$.

Now suppose that you start with a proper class of inaccessible cardinals (and $\sf GCH$), Easton's theorem tells us that the function $\nu\mapsto\min\{\mu\mid\nu<\mu\text{ inaccessible}\}$ satisfies the requirements of being a continuum function for regular cardinals. So there is an extension with the same cardinals, such that every regular cardinal satisfies that its power set is the next inaccessible.