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.
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2$\begingroup$ The assertion in your title could be expressed more concisely as $2^{\aleph_0}=2^{\aleph_\omega}$. $\endgroup$– bofSep 12, 2022 at 4:05
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2$\begingroup$ See en.wikipedia.org/wiki/Easton%27s_theorem . $\endgroup$– Emil JeřábekSep 12, 2022 at 6:18
1 Answer
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.
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$\begingroup$ It may also be worth noting that it holds if the continuum is a real-valued-measurable cardinal. $\endgroup$– bofSep 12, 2022 at 4:55