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Apr 8, 2019 at 11:50 review Close votes
Apr 8, 2019 at 18:45
May 16, 2010 at 19:34 answer added Timothy Chow timeline score: 57
Apr 29, 2010 at 17:27 answer added Peter LeFanu Lumsdaine timeline score: 7
Apr 28, 2010 at 11:22 vote accept Andrea Ferretti
Apr 27, 2010 at 20:58 comment added Andrea Ferretti Your objection seems right. I just mean that I want a model, that is, a set of sets such that with the usual relationship of containment satisfy the ZFC axioms. But I'm not sure it makes sense anymore...
Apr 27, 2010 at 20:22 comment added Miguel What theory could you use to prove this statement?<blockquote>if Set1 is consistent, then it cannot prove the consistence of Set2.</blockquote>What does it mean to say "Set1 is consistent"? In what level is that statement? Like there is a naive set theory Set1 used to define a formal logic Logic1 within which Set2 is a theory, is there a naive logic (Logic0?) used to reason about Set1? Is the statement "Set1 is consistent" a statement in Logic0? Is there hope of proving "If Set1 is consistent it cannot prove the consistency of Set2" within Logic0 which, being naive, isn't too powerful?
Apr 27, 2010 at 19:33 answer added kakaz timeline score: -1
Apr 27, 2010 at 11:32 answer added Charles Stewart timeline score: 4
Apr 27, 2010 at 9:17 history edited Andrea Ferretti CC BY-SA 2.5
added 11 characters in body
Apr 27, 2010 at 9:16 comment added Andrea Ferretti @unknown: I know that some logicians do not accept this point of view, but I had to choose one point of view to make the question clear, and I chose the one which is majoritary among mathematicians.
Apr 27, 2010 at 9:15 comment added Andrea Ferretti @Mike: sorry for my inaccuracies, and thank you for your amendments!
Apr 27, 2010 at 6:25 comment added Qfwfq (b.t.w., it was not me to downvote the question, which is indeed nice)
Apr 27, 2010 at 6:24 comment added Qfwfq "On the first level is the set theory mathematicians use all day. This has axioms, but is not a theory in the usual sense of logic. Indeed, to speak about logic we already need sets (to define alphabets and so on). In this naif set theory we develop logic, in particular the notions of theory and model. We call this theory Set1" I do not agree with this assumption.
Apr 27, 2010 at 2:49 comment added Mike Shulman Also, strictly speaking the existence of universes is not independent of the usual axioms of set theory, it's just not provable from them. Independent would mean that it's also not refutable from them. While no one has yet managed to refute it, the incompleteness theorem actually implies that, if universes are consistent with ZF, then we can't prove that to be the case (in ZF).
Apr 27, 2010 at 2:45 comment added Mike Shulman Minor objection: the Godel sentence in the first incompleteness theorem is not "true in PA." In fact, I don't know what that would mean. In a stronger theory than PA one can prove that it is true in the natural numbers, which are then usually considered the "intended model" of the internal PA in this stronger theory, but I don't think that should be called "true in PA."
Apr 27, 2010 at 0:11 answer added Carl Mummert timeline score: 22
Apr 26, 2010 at 23:26 comment added Qiaochu Yuan My guess is someone didn't read the question closely and assumed this was a question whose answer is "no; see the incompleteness theorem."
Apr 26, 2010 at 21:57 comment added Andrea Ferretti I just got a downvote. It would be nice to explain the reason, so that I can improve the question.
Apr 26, 2010 at 21:40 answer added Neel Krishnaswami timeline score: 7
Apr 26, 2010 at 20:36 history edited Andrea Ferretti CC BY-SA 2.5
added 455 characters in body; edited title
Apr 26, 2010 at 18:54 history asked Andrea Ferretti CC BY-SA 2.5