Re: [math-fun] Re: favorite theorem
Jim Propp wrote: << Come to think of it, I was hasty in asserting that the axiom of infinity can't be proved. It seems to me that Cantor's construction of the counting numbers (where 0 is defined as the empty set, 1 is defined as {0}, 2 is defined as {0,1}, 3 is defined as {0,1,2}, etc.) deserves to be called a constructive proof of the existence of an infinite set. It certainly doesn't FEEL like circular reasoning, even if it's unclear how one would formalize it. (Someone who knows more about Bolzano, Dedekind et al. should chime in here.)
1. Which reasoning doesn't feel like circular reasoning? 2. Without the Axiom of Infinity, I don't see how the existence of the Cantorian natural numbers implies there's a set whose members they comprise. And I doubt one can prove even the existence of each such number without the A.I. though you can clearly prove that all numbers up to any N exist. 3. I'm not fond of the Axiom of Infinity, either, and it would be very nice if it could be inferred from more natural axioms. --Dan
On Wednesday 03 May 2006 05:16, dasimov@earthlink.net wrote:
3. I'm not fond of the Axiom of Infinity, either, and it would be very nice if it could be inferred from more natural axioms.
Then you want a different version of set theory. How about NF? The only axioms you need are extensionality and stratified comprehension. (Stratified comprehension is the axiom scheme that gives you every instance of { x : P(x) } in which every variable in P can be assigned an integer "level" so that when "x in y" occurs level(y) = level(x)+1 and where "x = y" occurs level(y) = level(x). Equivalently, take formulae from an old-fashioned "type theory" and erase the types.) In NF, there's a universal set and all the Boolean algebra operations work; an axiom of infinity isn't too hard to prove. You might expect it to fall foul of a Cantorian/Russellian diagonal argument, but it doesn't seem to. (Straightforward attempts founder on the fact that the formula you'd use to define the "diagonalized" set isn't stratified.) So far as anyone knows, NF is consistent. It's also finitely axiomatizable, if you happen to care. On the other hand, the axiom of choice is provably false in NF, which most people find a bit odd. -- g
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dasimov＠earthlink.net -
Gareth McCaughan