Timeline for "classes" with no cardinality; "classes" with no equality notion
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2010 at 8:10 | vote | accept | Sasha | ||
| Nov 25, 2010 at 8:09 | history | edited | Sasha | CC BY-SA 2.5 |
added 888 characters in body
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| Nov 25, 2010 at 5:43 | answer | added | Mike Shulman | timeline score: 3 | |
| Nov 25, 2010 at 3:54 | comment | added | Peter LeFanu Lumsdaine | Extracting a small point which got rather lost in my long answer: I’m not sure why you say that the collection of “beasts” shouldn’t have a cardinality. Its cardinality would have to be large (too large to live in the universe whose vector spaces we started looking at) but it seems reasonable (to me) to say it should exist. | |
| Nov 25, 2010 at 0:35 | comment | added | Andrés E. Caicedo | There is also this question: mathoverflow.net/questions/44303/cardinality-of-proper-classes/… | |
| Nov 24, 2010 at 20:11 | answer | added | Peter LeFanu Lumsdaine | timeline score: 5 | |
| Nov 24, 2010 at 19:34 | comment | added | Peter LeFanu Lumsdaine | @Andreas: if I understand the definition right, all beasts have the same domain: the class/category of vector spaces? And I’d agree that it is reasonable to have equality on beasts: they’re a 0-categorical object. | |
| Nov 24, 2010 at 19:23 | comment | added | Peter LeFanu Lumsdaine | @Stefan: I think Sasha exactly means equal, not isomorphic; and that’s why we shouldn’t want to be able to talk about whether they’re equal. | |
| Nov 24, 2010 at 17:15 | comment | added | Andreas Blass | Why should we be unable to verify whether two vector spaces are equal yet able to say whether two "beasts" are equal? The beasts look to me at least as complicated as the vector spaces. Did you perhaps mean only that we should be able to say whether two beasts with the same domain are equal? Also, I assume that "able to verify" and "can say" are intended to refer to the same concept, that this concept is neither the existence of an algorithmic procedure nor the existence of a set-theoretic expression but somewhere in between. It would certainly help if you clarified what you mean here. | |
| Nov 24, 2010 at 16:42 | comment | added | Stefan Geschke | @Ricky Demer: I was slightly confused by your statement, but now I see: There is a bijection between the universe $V$ and the class of all vector spaces over a fixed field. If global choice holds, then all proper classes, including $V$ and the class of all vector spaces over a fixed field, bijectively map to the ordinals. | |
| Nov 24, 2010 at 16:37 | comment | added | Stefan Geschke | @Ricky Demer: Yes, but global choice does not follow from ZFC. So usually, we don't assign cardinality to proper classes. @Sasha: I am not sure I understand 2). Why should we not want to be able to verify whether two vector spaces are equal? And do you really mean equal or do you mean isomorphic? | |
| Nov 24, 2010 at 10:13 | comment | added | user5810 | This class has a cardinality. If Global Choice, then its cardinality is $\operatorname{Ord}$, otherwise its cardinality is $V$. | |
| Nov 24, 2010 at 7:59 | history | asked | Sasha | CC BY-SA 2.5 |