Timeline for Is there a notion of "predicative given the real numbers"?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 15, 2014 at 4:27 | history | edited | Keshav Srinivasan | CC BY-SA 3.0 |
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| Sep 30, 2013 at 2:17 | comment | added | Keshav Srinivasan | @AndrejBauer For an example of something that is impredicative even if you treat the set of real numbers as a completed totality, see the definition of inductive number at the top of my question. It's the standard definition of natural numbers in the second-order theory of real numbers, but it involves quantification over all hereditary properties, including the very property being defined, which is problematic if you're a predicativist concerning arbitary subsets of R, even if you view R itself as a completed totality. | |
| Sep 30, 2013 at 2:09 | comment | added | Keshav Srinivasan | @AndrejBauer It's OK to take suprema, because we're viewing the real numbers as a completed totality, so we're not defining a brand new real number, we're just identifying a real number that already exists. It's just like the phrase "tallest person in the room", which is impredicative, but if you view the set of people in the room as a completed totality, then you're not defining a new person, you're just identifying someone who's already in the room. | |
| Sep 29, 2013 at 15:45 | comment | added | Andrej Bauer | So if I understand you correctly, it should be ok to take suprema because we're going to take suprema of sets which were constructed predicatively (although we take the totality of all real numbers as a given in the beginning)? | |
| Sep 29, 2013 at 12:23 | comment | added | Keshav Srinivasan | @AndrejBauer It's similar to the case of arithmetic: there are people like Edward Nelson, who doesn't view the set of natural numbers as a completed totality, but only as a potential infinity, so he rejects the use of mathematical induction for formulas with quantification over the natural numbers as being impredicative. (So he don't even believe in exponentiation!) But when we talk about predicative second-order arithmetic we mean "predicative given the natural numbers", i.e. we view the natural numbers as a completed totality, but we think sets of natural numbers In a predicative fashion. | |
| Sep 29, 2013 at 12:21 | comment | added | Keshav Srinivasan | @AndrejBauer Yes, you could call the least upper bound axiom impredicative, but we're not talking about predicativity as such, but rather "predicativity given the real numbers", i.e. we're assuming that the real numbers already exist as a complete totality, and then we're trying to construct sets of real numbers in a predicative fashion. | |
| Sep 29, 2013 at 8:42 | comment | added | Andrej Bauer | Dedekind completeness produces a real number from a set of reals. Isn't it impredicative in your sense, since the real so produced can be an element of the set? | |
| Sep 29, 2013 at 4:05 | history | asked | Keshav Srinivasan | CC BY-SA 3.0 |