> Date: Fri, 5 Jun 2009 08:33:47 +0200 > From: tor...@dsv.su.se > To: everything-list@googlegroups.com > Subject: Re: The seven step-Mathematical preliminaries > > > Brian Tenneson skrev: >> >> >> On Thu, Jun 4, 2009 at 8:27 AM, Torgny Tholerus <tor...@dsv.su.se >> <mailto:tor...@dsv.su.se>> wrote: >> >> >> Brian Tenneson skrev: >> > >> > >> > Torgny Tholerus wrote: >> >> It is impossible to create a set where the successor of every >> element is >> >> inside the set, there must always be an element where the >> successor of >> >> that element is outside the set. >> >> >> > I disagree. Can you prove this? >> > Once again, I think the debate ultimately is about whether or not to >> > adopt the axiom of infinity. >> > I think everyone can agree without that axiom, you cannot "build" or >> > "construct" an infinite set. >> > There's nothing right or wrong with adopting any axioms. What >> results >> > is either interesting or not, relevant or not. >> >> How do you handle the Russell paradox with the set of all sets >> that does >> not contain itself? Does that set contain itself or not? >> >> >> If we're talking about ZFC set theory, then the axiom of foundation >> prohibits sets from being elements of themselves. >> I think we agree that in ZFC, there is no set of all sets. > > But there is a set of all sets. You can construct it by taking all > sets, and from them doing a new set, the set of all sets. But note, > this set will not contain itself, because that set did not exist before. > >> >> >> >> >> My answer is that that set does not contain itself, because no set can >> contain itself. So the set of all sets that does not contain >> itself, is >> the same as the set of all sets. And that set does not contain >> itself. >> This set is a set, but it does not contain itself. It is exactly the >> same with the natural numbers, *BIGGEST+1 is a natural number, but it >> does not belong to the set of all natural numbers. *The set of >> all sets >> is a set, but it does not belong to the set of all sets. >> >> How can BIGGEST+1 be a natural number but not belong to the set of all >> natural numbers? > > One way to represent natural number as sets is: > > 0 = {} > 1 = {0} = {{}} > 2 = {0, 1} = 1 union {1} = {{}, {{}}} > 3 = {0, 1, 2} = 2 union {2} = ... > . . . > n+1 = {0, 1, 2, ..., n} = n union {n} > . . . > > Here you can then define that a is less then b if and only if a belongs > to b. > > With this notation you get the set N of all natural numbers as {0, 1, 2, > ...}. But the remarkable thing is that N is exactly the same as > BIGGEST+1. BIGGEST+1 is a set with the same structure as all the other > natural numbers, so it is then a natural number. But BIGGEST+1 is not a > member of N, the set of all natural numbers. Here you're just contradicting yourself. If you say BIGGEST+1 "is then a natural number", that just proves that the set N was not in fact the set "of all natural numbers". The alternative would be to say BIGGEST+1 is *not* a natural number, but then you need to provide a definition of "natural number" that would explain why this is the case. > The biggest advantage is that everything is finite, and you can then > really know that the mathematical theory you get is consistent, it does > not contain any contradictions. Even if you define "natural number" in such a way that there are only a finite number of them (which you haven't actually done, you've just asserted it without providing any specific definition), you still could have an infinite number of *propositions* about them if you allow each proposition to contain an unlimited number of AND and OR operators. For example, even if I say that the only natural numbers are 1,2,3, I can still make arbitrarily long propositions like ((3>1) AND (2>1)) OR (3>1)) AND ((2>3) OR (3>1)) AND ((2>3) OR ((1>3) OR ((2>1) OR ((1>3) OR (3>1))))). Of course a non-finitist would be able to prove that these infinite number of propositions are consistent, but I don't know if an ultrafinitist would (likewise a non-finitist can accept a proof that something like the Peano axioms are consistent based on an understanding of their application to a model dealing with rows of dots, even if the Peano axioms cannot be used to formally prove their own consistency). Jesse --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---

- Re: The seven step-Mathematical preliminaries Brian Tenneson
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- Re: The seven step-Mathematical preliminaries Quentin Anciaux
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- Re: The seven step-Mathematical preliminaries Brian Tenneson
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- Re: The seven step-Mathematical preliminaries Brian Tenneson
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- Re: The seven step-Mathematical preliminaries Brian Tenneson
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- RE: The seven step-Mathematical preliminaries Jesse Mazer
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- RE: The seven step-Mathematical preliminaries Jesse Mazer
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- RE: The seven step-Mathematical preliminaries Jesse Mazer
- Re: The seven step-Mathematical preliminaries Brent Meeker
- Re: The seven step-Mathematical preliminaries John Mikes
- Re: The seven step-Mathematical preliminaries Torgny Tholerus
- Re: The seven step-Mathematical preliminaries Quentin Anciaux
- Re: The seven step-Mathematical preliminaries Brent Meeker
- Re: The seven step-Mathematical preliminaries Quentin Anciaux