Timeline for Equality of two circular sets
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2010 at 13:55 | comment | added | Peter LeFanu Lumsdaine | There is some relevant discussion of non-well-founded sets and the relevant axioms of foundation, anti-foundation etc at another MO question: mathoverflow.net/questions/33282/can-we-have-aa | |
| Sep 8, 2010 at 12:10 | history | edited | berater | CC BY-SA 2.5 |
deleted 24 characters in body
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| Sep 8, 2010 at 11:59 | vote | accept | berater | ||
| Sep 8, 2010 at 11:45 | comment | added | Benoît Kloeckner | @berater: you could as well check that $I=J$. It would reduce to checking $J=I$ again, which makes as much sense than the converse. | |
| Sep 8, 2010 at 11:30 | comment | added | Pietro Majer | Note that x={x} is explicitly forbidden by the ZF axiom of regularity (en.wikipedia.org/wiki/Axiom_of_regularity). But if decide that you are talking of another binary relation, that you still denote $\in$, but it is not the usual "is an element of" of ZF set theory, and X={X} may happen, then you also can have many such "self-singletons", why not. In particular, deciding if 2 sets are equal can't be decided iterating the procedure of checking if their elements are equal, like in your example. | |
| Sep 8, 2010 at 11:24 | comment | added | berater | Right, and to verify this I have to perform the same operation again and I will end up with the same problem. That makes sense. | |
| Sep 8, 2010 at 11:07 | comment | added | user5810 | You're assuming I≠J to get that I is not an element of J. | |
| Sep 8, 2010 at 11:06 | answer | added | Robin Chapman | timeline score: 5 | |
| Sep 8, 2010 at 11:00 | history | asked | berater | CC BY-SA 2.5 |