Obections to Cantor's Theory (Wikipedia article)
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From: quasi on 28 Jul 2005 19:45 On 28 Jul 2005 12:49:34 -0700, Han.deBruijn(a)DTO.TUDelft.NL wrote: >Look what some of those mathematicians have done: > >0 = {} , 1 = {{}} , 2 = {{},{{}}} , 3 = {{},{{}},{{},{{}}}} , ... > >Let's get physical now. The empty set is nothing. Putting curly >brackets around it still makes it nothing. Thus: > >1 = {{}} = {} = 0 , 2 = {{},{{}}} = {{},{}} = {{}} = {} = 0 . > >Do I have to proceed? No. You can't create something from nothing. >There is emptyness all over the place. Let me try to use the box concept to clear up your (possibly deliberate) misunderstanding. View sets as analogous to boxes, and at least for finite sets, the number of objects inside the box is the number of elements. So the idea of representing (or defining, take your pick) 0,1,2,... by: 0 = {} , 1 = {{}} , 2 = {{},{{}}} , 3 = {{},{{}},{{},{{}}}} , ... is reasonable since: 0 is a box with no elements, 1 is a box containing a box, so it has 1element, 2 is a box containing 2 boxes, so it has 2 elements, etc. No one said you were supposed to look inside those elements to see what they contained. Even in the real world you are ignoring the fact that the boxes themselves may have value (or weight) Imagine each box is made of gold, all of equal value (if empty), and you were given a choice of the contents of 0, 1 or 2. I think you would come to your senses and choose 2. Alternatively, think of it in terms of weight. If each box has equal weight if empty, then box 0 weighs less than box 1, which weighs less than box 2, etc. If each box weighs 1 pound empty, and if you had to move one of the boxes 0, 1, 2, 3, ... 100, which would be the least work? Once again, I'm confident you would come to your senses and opt for box 0 instead of box 100. Hopefully now you see that the formal definition of 0, 1, 2, ... was designed to count the contents just 1 level down. You're only allowed to go one level down -- those are the rules. So what if the contents are just boxes -- count them, and don't look inside (don't be nosy, just count them). quasi
From: malbrain on 28 Jul 2005 16:52 MoeBlee wrote: > >From a post by Han.deBru...(a)DTO.TUDelft.NL: > > > > Let's get physical now. > > It seems that you miss that set theory and mathematics are not a > narrative of the physical universe and set theory and mathematics do > not denote with words that pick out objects and even concepts of the > physical universe in the way that everyday language or physical > sciences do. For that matter, mathematics can't be tied to a particular > theory of the physical universe, since, such theories are about > contingent states-of-affairs, This is a negation of the history of mathematics. Mathematics has a "current state-of-affairs" in relation to REALITY also. Ask any high-school or junior-college teacher of mathematics. karl m
From: Virgil on 28 Jul 2005 17:37 In article <MPG.1d52e7463e55be54989fda(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > David Kastrup said: > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > No, I understand what you mean by "countable". My issue with your > > > "uncountability of the powerset of the naturals" is that you can > > > easily form a bijection between the members of the powerset and the > > > set of infinite bit strings denoting set membership in the subsets, > > > > Yes. > > > > > and between those bit strings and whole numbers, > > > > No. > > > > > and the only reason this bijection is rejected is because of the > > > refusal to allow infinite whole numbers in the set of whole numbers, Infinite whole numbers need not be any part of the set of natural numbers. > > > despite the fact that an infinite set of whole numbers requires > > > infinite whole numbers, > > > > Whining does not make it so. It requires arbitrarily large numbers, > > but none of them need to be infinite. > I am not whining and you are not correct. TO is whining and DK is correct. Two strikes in one short sentence! > > > > > so you don't consider the infinite bitstrings to represent whole > > > numbers. It's like all the wrong choices have been made, almost on > > > purpose, in order to make some grand distinction which really isn't > > > there. > > > > Well, it's like you try to fix something that you perceive as a > > problem by making it much worse. > > > > > If you allow infinite whole numbers, as is required, > > > > Whining won't make it so. > And making statements like that doesn't make it any less so. It requires assuming infinite naturals exist and are required to prove that they exist and are required. Without that assumption, there is no need for them at all. The Peano properties do very nicely without them.
From: Virgil on 28 Jul 2005 17:43 In article <MPG.1d52e828a17493c1989fdb(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Daryl McCullough said: > > Tony Orlow (aeo6) wrote: > > > > >Daryl McCullough said: > > > > >> So, you agree that for *finite* sets, two sets have the same > > > > >> bigulosity if and only if there is a bijection between the two? > > >> But that no longer holds for infinite sets? > > >> > > >> Then how is bigulosity an improvement over cardinality? > > >Because Bigulosity takes into account the nature of the bijection > > >in order to determine a precise relative size of infinity. > > > > In other words, bigulosity is whatever you want it to be, and so > > you have a lot more flexibility. Just make it up as you go along. > > > > A = the set of natural numbers { 0, 1, 2, ... } B = the set of > > base ten numerals { "0", "1", "2", ... } C = the set of base > > two numerals { "0", "1", "10", "11", "100", ... } > > > > A and B have the same bigulosity. A and C have the same bigulosity. > > But B and C do *not* have the same bigulosity (clearly C has a > > smaller bigulosity than B). That's bigulosity for you... > > > > -- Daryl McCullough Ithaca, NY > > > > > If you consider your numerals to have N digits, then base ten has > 10^N elements and base 2 has 2^N elements So that TO is declaiming that the number of naturals is dependent on the base in which they are to be represented? Those mushrooms TO is nibbling must be really potent!
From: malbrain on 28 Jul 2005 17:45
Virgil wrote: > It requires assuming infinite naturals exist and are required to prove > that they exist and are required. Without that assumption, there is no > need for them at all. The Peano properties do very nicely without them. It refers to the axiom of infinity. karl m |