When can power sets be limit cardinals?

Question

My original question (posted in here at the Math.SE) was:

Is it possible to create a model of ZFC, so that the cardinality of each power set is a limit cardinal (as opposed to GCH where they are always successor cardinals)?

Obviously, from Easton's theorem, this is possible for regular cardinals. I also showed that in this model strong limit cardinals must also be fixed points of the $\aleph$ function (using Shelah's upper bound from PCF), and therefore such a model would mean $2^{\aleph_\delta} \geq \aleph_{\delta^+}$. But whether we can improve Shelah's upper limit from $2^{\aleph_\delta} < \aleph_{{|\delta|}^{+4}}$ to $2^{\aleph_\delta} < \aleph_{\delta^+}$ is an open question. So for strong limit cardinals, we don't know whether such a model is possible (even for one cardinal, let alone all of them).

This leaves the singular weak limit $\delta$ case, which isn't interesting unless we assume the continuum function doesn't become constant for all $\gamma$ such that $\kappa < \gamma < \delta$. Assuming this, can we create a model of where the cardinality of the power set of singular weak limit are limit cardinals?

For a concrete example, take the following rule (which satisfies Easton's theorem, and the known limitations for singular cardinals as surveyed in this 2002 ICM paper by Gitik (MR1989201, Zbl 1030.03032)): $$2^{\aleph_\alpha} = \aleph_{\omega_{\alpha + 1}}$$

Answer 1

The answer to your question is yes. In the Foreman-Woodin model The generalized continuum hypothesis can fail everywhere, $2^\kappa$ is weakly inaccessible for all infinite cardinals $\kappa.$ To be more precise, Foreman and Woodin proved the following:

Theorem 1. Con(ZFC+there exists a supercompact cardinal and infinitely many inaccessibles above it) implies Con(ZFC+$\forall \kappa, 2^\kappa$ is weakly inaccessible).

In fact we can reduce their large cardinal assumption by replacing the supercompact cardinal with a strong cardinal. See my paper with Yair Hayut On Foreman's maximality principle.

Even more surprisingly, we can prove the following:

Theorem 2 Con(ZFC+there exists a strong cardinal) implies Con(ZFC+$\forall \kappa, 2^\kappa$ is a singular cardinal).

For the proof, see again my paper with Yair Hayut stated above.

Remark. I may mention that in general you can not talk about Shelah's bound in models of Theorems 1 or 2. The point is that in these models, the strong limit points are all fixed points of the $\aleph$-function, and Shelah's bound talks about non-fixed points of the $\aleph$-function.