Is the statement "$2^{|X|} = \aleph_{|X|^+}$ for all infinite sets $X$" consistent with ZFC?
For a point of reference, see Can the continuum be a singular cardinal?
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1$\begingroup$ If by $|X|^+$ you meant the successor ordinal of $|X|$, I believe this is true in Easton’s model that satisfies $2^\kappa = \aleph_{\kappa+1}$ for all regular cardinals $\kappa$, since in that model $2^\kappa$ for a singular $\kappa$ is the minimal cardinal $\lambda$ s.t. $\lambda \geq \aleph_{\eta+1}$ for all cardinal $\eta < \kappa$ and $\text{cf}(\lambda) > \kappa$, which is exactly $\aleph_{\kappa+1}$. $\endgroup$– David GaoCommented Sep 8 at 11:53
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$\begingroup$ Sorry for the ambiguity in notation. I was using $|X|^+$ to refer to the cardinal successor to $|X|$. $\endgroup$– Jesse ElliottCommented Sep 8 at 21:57
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$\begingroup$ @David: The use of $\alpha^+$ as the successor ordinal in some texts which present only the very basic introduction to set theory is somehow understandable (since $x\cup\{x\}$ is, in a good sense, a successor of $x$ as a set). But the notation is unambiguously used in set theory to mean the cardinal successor. Indeed, one of the few cases of notation meaning "pretty much just the one thing" that I can think of in set theory. $\endgroup$– Asaf Karagila ♦Commented Sep 9 at 13:17
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$\begingroup$ Relevant: mathoverflow.net/q/368387/7206 and mathoverflow.net/q/138308/7206 and mathoverflow.net/q/226887/7206 which might have something of an answer to your actual question. $\endgroup$– Asaf Karagila ♦Commented Sep 9 at 13:21
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1$\begingroup$ @AsafKaragila I know. I use $^+$ to denote cardinal successor and $+ 1$ to denote ordinal successor as well. (I even wrote $\kappa + 1$ in my comment.) I simply don’t know the answer to the question if $^+$ meant cardinal successor, but I do know the answer if it’s replaced by ordinal successor, so I only wrote a comment (instead of an answer), just in case the OP used $^+$ in a nonstandard way. That’s not the case, as OP has indicated now, but I think it’s still reasonable to keep the comment given its relevance to the question, especially the discussion below Joel’s answer. $\endgroup$– David GaoCommented Sep 9 at 13:26
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This is not a full answer, but let me mention that your hypothesis has large cardinal strength, because it implies the failure of the singular cardinals hypothesis.
Assume $2^\kappa=\aleph_{\kappa^+}$ for every infinite cardinal $\kappa$. (I am taking your $|X|^+$ to refer to the cardinal successor.) Let $\kappa$ be a singular strong limit cardinal. By your hypothesis, we have $2^\kappa=\aleph_{\kappa^+}$, but this is strictly bigger than $\kappa^+$, since successor cardinals are never $\aleph$-fixed points, and so SCH fails in every instance.
The failure of SCH has the large cardinal strength of a measurable cardinal $\kappa$ of Mitchell rank $\kappa^{++}$.
Meanwhile, if you assert your property $2^{\kappa}=\aleph_{\kappa^+}$ only for regular cardinals $\kappa$, then this is equiconsistent with ZFC by Easton's theorem.
answered Sep 8 at 12:41
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$\begingroup$ Thank you for your answer. If we assume that my statement is true just for regular cardinals, then what does Easton's model say $2^\kappa$ is for singular cardinals $\kappa$? $\endgroup$ Commented Sep 8 at 22:07
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$\begingroup$ That is what David Gao's comment is about on the main question. $\endgroup$ Commented Sep 8 at 22:34
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$\begingroup$ Sure. I thought it might reduce to a simpler description in this particular case. I'll think about it some more. $\endgroup$ Commented Sep 8 at 22:47
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1$\begingroup$ According to what he says, it seems that we will have $2^\kappa=\aleph_{\kappa+1}$ for singular infinite cardinals $\kappa$ in the relevant Easton model, since such a cardinal is always a limit cardinal, and so $2^\lambda$ for regular $\lambda<\kappa$ will be $\aleph_{\lambda^+}<\aleph_\kappa$, meaning that $2^{<\kappa}=\aleph_\kappa$ and so $2^\kappa$ is $\aleph_{\kappa+1}$ as he explains. $\endgroup$ Commented Sep 9 at 0:54