Timeline for Can the set of endomorphisms of $(\mathbb R,+)$ have cardinality strictly between $\frak c$ and $\frak{c^c}$?
Current License: CC BY-SA 4.0
10 events
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| Sep 30, 2019 at 12:47 | comment | added | Asaf Karagila♦ | @Tim: It's unfortunate that some books define it in one way, and other books in the other way. Since the majority of books focus on ZFC, rather than ZF, then you'd think it doesn't really matter. But then things like this happen. | |
| Sep 30, 2019 at 12:17 | comment | added | Tim Campion | @AsafKaragila You just doubled my confusion by giving two incompatible definitions of “cardinal” in the same comment! :) Since only the first one applies to the continuum I suppose I have my answer for the purposes of this question, but my confusion over the choiceless meaning of “cardinal” will just have to persist I suppose... | |
| Sep 30, 2019 at 10:33 | comment | added | Todd Trimble | @AsafKaragila In fact it didn't occur to me to read it any other way. | |
| Sep 30, 2019 at 9:15 | comment | added | Asaf Karagila♦ | @Todd: Actually it's the only sensible way to read this question, since if you talk about cardinals as well-ordered sets only, then either $\Bbb R$ can be well-ordered, in which case we fallback to the AC situation, or it's not a cardinal to begin with, in which case the question is meaningless. Since the question is not quite meaningless, reading it as "cardinals are initial ordinals" makes no sense. :-) | |
| Sep 30, 2019 at 9:14 | comment | added | Asaf Karagila♦ | @Tim: Cardinals, without choice, refer to (and should refer to) the general notion of cardinality. But since ordinals are just so damn useful, cardinals are defined as initial ordinals for well-ordered sets, and Scott equivalence classes otherwise (i.e., take the class-sized equivalence class modulo bijections, and cut it on the least rank it's not empty). | |
| Sep 29, 2019 at 21:55 | comment | added | Todd Trimble | @TimCampion Tim, yes, I assume we are talking about the large poset that is the posetal collapse of the preorder defined by $X \leq Y$ iff there exists an injection from $X$ to $Y$ (by Cantor-Schroeder-Bernstein, $X \leq Y$ and $Y \leq X$ iff $X$ and $Y$ are isomorphic). The question is certainly sensible under this assumption. | |
| Sep 29, 2019 at 20:33 | comment | added | Tim Campion | I hate to interject with an uninformed comment, but what does cardinality mean without choice? Are we talking about sets modulo existence of a bijection? Or something about maps to or from cardinals? | |
| Sep 29, 2019 at 18:00 | comment | added | Asaf Karagila♦ | Martin Sleziak pointed out, correctly, on MSE that choice is unnecessary for the cardinal arithmetic at the end of the first bullet point. | |
| Sep 29, 2019 at 12:40 | history | edited | user35370 | CC BY-SA 4.0 |
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| Sep 29, 2019 at 0:33 | history | asked | user35370 | CC BY-SA 4.0 |