Timeline for Is the isomorphism class of a fixed cardinality a set?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| May 5, 2014 at 18:00 | comment | added | David Fernandez-Breton | @user18921: Ok., maybe "illuminating" was not the word I should have used, but nevertheless I like the argument (even if unnecessarily complicated) because I like surreal numbers :) | |
| Mar 29, 2014 at 8:44 | comment | added | goblin GONE | @DavidFernandezBreton, lol that made me laugh, bitime. (You're are being sarcastic, right?) | |
| Oct 4, 2011 at 12:21 | vote | accept | ashpool | ||
| Jun 27, 2011 at 18:42 | comment | added | David Fernandez-Breton | Still, the mention to surreal numbers makes it more illuminating... | |
| Aug 1, 2010 at 14:31 | comment | added | Stefan Geschke | This sound very complicated to me. Why mention surreal numbers? Given a nonempty set $X$, consider the class of all sets of the form $X_y=\{(x,y):x\in X\}$, where $y$ is any set. Without AC there is a bijection between $X$ and $X_y$, by mapping each $x\in X$ to $(x,y)$. Clearly the collection of all $X_y$ is a proper class. | |
| Aug 1, 2010 at 14:02 | history | answered | Anon | CC BY-SA 2.5 |