Timeline for How big is the proper class of all sets?
Current License: CC BY-SA 3.0
15 events
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| Oct 6, 2013 at 21:17 | vote | accept | CommunityBot | ||
| Oct 6, 2013 at 21:15 | vote | accept | CommunityBot | ||
| Oct 6, 2013 at 21:16 | |||||
| Jul 8, 2013 at 9:41 | vote | accept | CommunityBot | ||
| Jul 8, 2013 at 9:41 | |||||
| Jun 29, 2013 at 17:13 | comment | added | Asaf Karagila♦ | @TheUser: Whitehead problem is absolutely solved (every Whitehead group, of every cardinality, is free). | |
| Jun 29, 2013 at 17:10 | comment | added | The User | Well, $V=L$ was just a random famous example which is not implied by large-cardinal axioms. However, now I am interested: Could you give any example of “$V=L$ish” behaviour of non-set-theorists? One thing related to $V=L$ comes to my mind: Many non-logicians think that explicitly definable sets cannot behave badly (measurability etc.), however, this contradicts $V=L$, which allows a definition of a $\Delta^1_2$-well-order of the reals, thus it is “$V\ne L$ish” behaviour (but they do not know). But of course (hopefully?) they do not use such assumptions in their proofs. | |
| Jun 29, 2013 at 16:47 | comment | added | Asaf Karagila♦ | @TheUser: I feel that in fact most mathematicians are working in $L$, in the sense that it decides a lot of things. It is true that most of them aren't even aware of this issue, or care about this issue. But philosophically, I feel, that this is the case. You probably refer to inaccessible cardinals and universes, but remember that those are compatible with $L$. It's when a lot of non-logicians will have to assume $0^\#$ when we'll see people caring about $V=L$ as well. | |
| Jun 29, 2013 at 16:43 | comment | added | The User | @AsafKaragila Some large cardinal axioms (well, certainly not Reinhardt cardinals) are probably the most widely used non-conservative extensions of ZFC outside of mathematical logic. Nobody uses, say, V=L, as far as I know. | |
| Jun 29, 2013 at 12:47 | comment | added | Asaf Karagila♦ | Sure, you can do almost all the ordinary mathematics within the bound "There is no power set to the real numbers", too. But when you suddenly add large cardinals into the picture, it becomes your burden to give excellent justifications as to why this is a good idea. Because large cardinals rarely appear in ordinary mathematics (even if they do have influence on properties of "small sets"). Moreover there had been no appearances of Reinhardt cardinals yet. | |
| Jun 29, 2013 at 12:35 | comment | added | Asaf Karagila♦ | Ali, take it from someone who has been working on the axiom of choice related questions for the past two years. You always start with the notion that you can go to the farthest generalization, and remove a lot of unnecessary assumptions. However that is rarely the case. Much more often than not when you remove all the assumptions you are left without much structure to work with. See how Presburger arithmetic is complete, while Robinson isn't. Why is that? Because Presburger is too weak, and lacks structure for a meaningful manipulation of Godel numbers. | |
| Jun 29, 2013 at 12:34 | comment | added | user36136 | We can do anything in ordinary mathematics without foundation. In the other words the axiom of foundation is not necessary for the foundation of mathematics which is the main goal of stating the axioms of set theory. Other assumptions just simplify the life of set theorists by removing "hidden" and "odd" objects of the Cantor's "heaven" such as large cardinals and non well founded sets. | |
| Jun 29, 2013 at 12:21 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Jun 29, 2013 at 12:19 | comment | added | Asaf Karagila♦ | But usefulness of an axiomatic system comes from what you can do with it, not from what limitation it may free you from. We can do a whole lot in $\sf ZFC$ which makes it useful. And we use transfinite recursion often enough to suggest that regularity is essential. What would the consistency of a Reinhardt cardinal without regularity do you much good? | |
| Jun 29, 2013 at 12:14 | comment | added | user36136 | Dear Asaf. Thank you for your answer. But I think about some disadvantages of the axiom of foundation in set theory. For example it imposes a limitation on the tree of large cardinals by the Kunen inconsistency theorem which uses this axiom in its proof and implies that the Reinhardt cardinal does not exist. One can compare this situation between assumption "V=L" (in my notation "G=L") and the large cardinal axiom "0-sharp". So the axiom of foundation is a constructibility kind axiom which says "G=V" and limits our imagination of the largeness of the world as same as "useful" axiom "G=L"! | |
| Jun 29, 2013 at 11:45 | comment | added | Asaf Karagila♦ | I would add a remark that in modern set theory sets are considered well-founded, because there is a big advantage to this assumption: we get $\in$-induction, and ranks and whatnot; so in some sense "proper class of all sets" is taken, in the modern context, to be $V$ from the start. | |
| Jun 29, 2013 at 11:39 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |