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): $$2^{\aleph_\alpha} = \aleph_{\aleph_{\alpha + 1}}$$


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.

  • 3
    $\begingroup$ I have a simple follow up question, which you may already know the answer to - is there a model where $2^\kappa$ is a limit cardinal $\iff$ $\kappa$ is a limit cardinal. $\endgroup$
    – Alon Navon
    Dec 25, 2015 at 0:29
  • $\begingroup$ @AlonNavon Unfortunately, I don't know the answer. $\endgroup$ Dec 27, 2015 at 4:35
  • 2
    $\begingroup$ For completeness of MathOverflow cross-references, @AlonNavon's question in the comment above has been asked as a separate question here: mathoverflow.net/q/321543 $\endgroup$
    – Gro-Tsen
    Jan 23, 2019 at 16:34

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