Obections to Cantor's Theory (Wikipedia article)
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From: Randy Poe on 27 Jul 2005 12:45 Tony Orlow (aeo6) wrote: > > Why do you keep saying that? It's provably false. The set of all > > finite strings is an infinite set. It's infinite by *your* definition > > of infinite, in the sense that it is "without end". The set of all > > finite strings is the union of > > > > S_1 = the set of strings of length 1 > > S_2 = the set of strings of length 2 > > S_3 = the set of strings of length 3 > > ... > > > > The collection of subsets S_n goes on without end. > So, each of these sets is finite right, given finite S and L? > There are an infinite number of such finite sets? > Do they then go, say, from S_1 to S_oo? > And S_1 is the set of strings of length 1, and S_2 is the set of strings of > length 2, etc, so S_n is the set of strings of length n? > Okay. What length are the strings in S_oo? Who cares about S_oo? That's not one of the S_n. The question being contested is, how many strings are there in the union S_1 U S_2 U S_3 U ... where the union is over all the S_n for finite natural numbers n? You say that's a finite number. You say there's some finite L which is the size of the largest string in this collection. But if you believe there's a finite L, don't you think there's a set S_L in this collection? And isn't there a set S_L+1? - Randy
From: Robert Low on 27 Jul 2005 12:47 Tony Orlow (aeo6) wrote: > Robert Low said: >>Are there rationals in [0,1) where p and/or q have to be infinite? > Interesting question. In order to have an infinite number of them, yes, you > would need either an infinite number of numerators or an infinite number of > denominators, both of which are whole numbers. So, you would require infinite > values in either numerator or denominator, in order to achieve an infinite set. OK, then, so now I consider the ratio 314159.../100000000000, where both are 'infinite integers'. (I'm clearly playing the game of 'it looks a bit like' where I ought to be playing the game of 'I can prove'; but that's for obvious reasons.) It looks to me as if their ratio is pi, which is not rational. Anyway, I'd still like to know how many elements there are in the set of all finite integers. When you can answer that, there might be some scope for further development.
From: malbrain on 27 Jul 2005 13:09 Robert Low wrote: > Tony Orlow (aeo6) wrote: > > Robert Low said: > >>Are there rationals in [0,1) where p and/or q have to be infinite? > > Interesting question. In order to have an infinite number of them, yes, you > > would need either an infinite number of numerators or an infinite number of > > denominators, both of which are whole numbers. So, you would require infinite > > values in either numerator or denominator, in order to achieve an infinite set. > > OK, then, so now I consider the ratio 314159.../100000000000, where both > are 'infinite integers'. (I'm clearly playing the game of 'it looks > a bit like' where I ought to be playing the game of 'I can prove'; > but that's for obvious reasons.) It looks to me as if their ratio is > pi, which is not rational. If you start from the premise that the reals are countable, it makes sense that all the irrationals suddenly become rational, doesn't it? karl m
From: Tony Orlow on 27 Jul 2005 13:19 Daryl McCullough said: > Tony Orlow writes: > > >Barb, you're not saying anything new. I have heard it all before. I am not > >drawing my conclusions in any such confused way > > Uh, yes you are. Why don't you write down what you consider to be > valid axioms for working with infinite sets, and then try to write > a formal proof of your claim > > If a set S of strings is infinite, then S contains some infinite > strings. > I guess you can't understand a proof if it has natural language. Your statement is not quite right. This is what I have proven: Given an alphabet of finite size S, an infinite set of unique strings made from this alphabet must contain infinite strings. Here's my two axioms: (1) N=S^L, where N is the size of the set of ALL strings of length L, constructed from an alphabet of size S. N is the upper limit on the number of strings from a set of S symbols and a maximum length of L. (2) A^B is finite if and only if A and B are finite. Given (2), S^L is finite iff S and L are finite. Given finite S, S^L is finite iff L is finite. Given (1), N is finite iff L is finite. Conversely, N is infinite iff L is infinite. .... Well, I just got back from getting lunch, during which it occurred to me that it would be good to put this proof in inductive format, and watch you try to discredit it while hanging on to your proof that all naturals are finite. This proof relies on some assumptions. Let me know if you disagree with any of them: (1) All naturals are finite. (you already "proved" this one) (2) The number of strings of length L that can be constructed from a set of symbols of size S (set size, not symbol size, remember (sigh)), is S^L. (3) For finite A and B, both A^B and A+B are finite. Are we good to go? Okay. Proof that f(n), the number of strings in the set of all strings up to and including length n in N, on a finite alphabet of size S, is finite: 1. For n=1, we have S^1 strings of length 1, for a total of S strings less than or equal to 1 in length. f(1)=S is finite, as stated. 2. For a finite number of srings of length n or less, we can add S^(n+1) strings of length n+1, to get f(n+1),the number of strings of length n+1 or less. S is finite and n+1 (a natural number) is finite, so S^(n+1) is finite. S^(n+1) is finite and f(n) is finite, so f(n+1)=f(n)+S^(n+1) is finite. 3. Therefore, for all n in N, f(n), the number of strings up to and including n symbols, which are created from a finite alphabet, is finite. There is no n in N, in other words, which maximum length of string will allow you to have an infinite set of strings. I eagerly await any response on this, but I won't hold my breath. > A proof consists of a sequence of statements such that every statement > is *either* an axiom, or follows from previous statements by logical > deduction. > > -- > Daryl McCullough > Ithaca, NY > > -- Smiles, Tony
From: Robert Low on 27 Jul 2005 13:18
malbrain(a)yahoo.com wrote: > If you start from the premise that the reals are countable, it makes > sense that all the irrationals suddenly become rational, doesn't it? If I started from the premise that the reals were countable, it would probably make sense that up was down and black was white, too. But I don't see what that has to do with Tony's claims that an infinite set of integers must contain an infinite integer. |