Timeline for Is "$2^{|X|} = \aleph_{|X|^+}$ for all infinite sets $X$" consistent with ZFC?
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| Sep 9 at 13:26 | comment | added | David Gao | @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. | |
| Sep 9 at 13:21 | comment | added | Asaf Karagila♦ | 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. | |
| Sep 9 at 13:17 | comment | added | Asaf Karagila♦ | @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. | |
| Sep 8 at 21:57 | comment | added | Jesse Elliott | Sorry for the ambiguity in notation. I was using $|X|^+$ to refer to the cardinal successor to $|X|$. | |
| Sep 8 at 12:41 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Sep 8 at 11:53 | comment | added | David Gao | 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 $\aleph_{\kappa+1}$. | |
| Sep 8 at 10:40 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Sep 8 at 10:40 | history | undeleted | Jesse Elliott | ||
| Sep 8 at 9:23 | history | deleted | Jesse Elliott | via Vote | |
| Sep 8 at 9:23 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Sep 8 at 9:17 | history | asked | Jesse Elliott | CC BY-SA 4.0 |