Timeline for answer to Why worry about the axiom of choice? by Dr Strangechoice
Current License: CC BY-SA 2.5
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14 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2021 at 2:11 | comment | added | tooAnnoying | @MartinVäth Thank you for the excellent clarifying comment. You are absolutely right. | |
| Sep 26, 2020 at 18:44 | comment | added | Martin Väth | As already observed in a comment here, this argument shows only that the definition of "more" (in the sense of existence of an injection and lack of a bijection) does not make much sense without choice. Something which is not really surprising given that many sets have no cardinality (in the sense of cardinal numbers) without choice. If looked at it this way, the argument only boils down to the fact that $\mathbb R$ is such a set if all sets are Lebesgue measurable. | |
| Mar 17, 2020 at 19:20 | comment | added | Mike Rosoft | Or, as I like to visualize this: Imagine Cantor's Hotel (a competitor of Hilbert's Hotel) where rooms are indexed by real numbers. Of course the hotel has infinitely many floors, and infinitely many rooms on each floor: every floor contains precisely the rooms whose numbers differ from each other by a rational number. And in absence of axiom of choice, there can be more floors than rooms! On the other hand, you can be certain that the hotel has no less floors than rooms. | |
| Aug 17, 2019 at 20:24 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Mar 22, 2018 at 17:05 | comment | added | silvascientist | @TorstenSchoeneberg But then why is this his only post ever in the history of MO? | |
| Jan 30, 2014 at 23:43 | comment | added | Torsten Schoeneberg | @Niemi: He changed it when he became a logician. Used to be Merkwürdigauswahl. | |
| Oct 10, 2013 at 16:19 | comment | added | Niemi | Did you set up the account just to post this? | |
| Jul 2, 2010 at 15:57 | comment | added | Péter Komjáth | The statement that either there is a nonmeasurable set or R has a surjection to a larger set, is Sierpinski: Sur une proposition qui ..., Fundamenta Math, 34(1947), 157-162. It is also in my book (with Vilmos Totik) Problems and Theorems in Classical Set Theory, Springer, 2006, page 127. | |
| Apr 29, 2010 at 14:34 | comment | added | François G. Dorais | @Anton: This is not at all obvious, it's a theorem of Jean Raisonnier from the 1980's. | |
| Apr 29, 2010 at 5:41 | comment | added | Anton Geraschenko | "If every set of reals is Lebesgue measurable then $\omega_1 \not\le 2^\omega$." I feel like I'm missing something that's supposed to be clear ... could somebody explain why this line is true? | |
| Apr 29, 2010 at 5:14 | comment | added | Joel David Hamkins | Very nice ! | |
| Apr 29, 2010 at 5:07 | comment | added | Dr Strangechoice | If every set of reals is Lebesgue measurable then $\omega_1 \nleq 2^{\omega}$, but then you can partition $2^{\omega}$, or rather $\mathcal{P}(\omega\times\omega)$, into $\omega_1+2^{\omega} > 2^{\omega}$ pieces by putting two wellorderings of $\omega$ in the same piece iff they have the same order type, and all non-wellorderings into singleton pieces. | |
| Apr 29, 2010 at 4:57 | comment | added | Dan Piponi | Do you have a good citation for that? | |
| Apr 29, 2010 at 4:32 | history | answered | Dr Strangechoice | CC BY-SA 2.5 |