7
$\begingroup$

It is well known that the continuum cannot be $\aleph_\omega$. Instead, it can obtain any value in the aleph sequence up to that. My question is if it is consistent that all the powersets of the alephs below that, are equal to the continuum.

$\endgroup$
2

1 Answer 1

10
$\begingroup$

The answer is yes. Starting for simplicity with a model of $\mathsf{ZFC+GCH}$, if we force to add $\aleph_{\omega+1}$-many Cohen reals we will get (by the usual "nice names" analysis) $2^{\aleph_0}=2^{\aleph_\omega}=\aleph_{\omega+1}$; since $2^{\aleph_0}\le2^{\aleph_n}\le 2^{\aleph_\omega}$ for each $n\in\omega$, this gives the situation you're interested in.

More generally, if $M$ is a countable model of $\mathsf{ZFC}$ and $\alpha$ is an $M$-ordinal there is a c.c.c. (so cardinal-and-cofinality-preserving) forcing extension $N$ of $M$ satisfying $2^{\aleph_0}=2^{\aleph_\alpha}$.

It's also worth noting that the situation you're interested in is provable from $\mathsf{ZFC+MA+2^{\aleph_0}>\aleph_{\omega}}$, the point being that $\mathsf{MA}$ gives $2^\kappa=2^{\aleph_0}$ for every $\kappa<2^{\aleph_0}$. Of course, it takes effort (= iterated forcing) to show that that theory is consistent in the first place.

$\endgroup$
1
  • $\begingroup$ It may also be worth noting that it holds if the continuum is a real-valued-measurable cardinal. $\endgroup$
    – bof
    Sep 12, 2022 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.