My original question (posted in <https://math.stackexchange.com/questions/1584430/can-all-power-sets-be-limit-cardinals>) 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 in <http://www.math.tau.ac.il/~gitik/icm5.pdf>):
$$2^{\aleph\_\alpha} = \aleph\_{\aleph\_{\alpha + 1}}$$



When can Power Sets be Limit Cardinals?
The answer to your question is yes. In the Foreman-Woodin model [The generalized continuum hypothesis can fail everywhere](http://www.jstor.org/stable/2944324?origin=crossref), $2^\kappa$ 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](http://math.ipm.ac.ir/~golshani/Papers/On%20Foreman's%20maximality%20principle.pdf).


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.


