Obections to Cantor's Theory (Wikipedia article)
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From: Robert Low on 27 Jul 2005 17:22 Tony Orlow (aeo6) wrote: > Daryl McCullough said: >>That's false, no matter how many times you say it. No finite >>set can contain every (finite) natural. Why? Because every finite >>set of naturals has a largest element, and there is no largest >>finite natural. > That's false, no matter how many times you say it. And the only difference is that Daryl can give a proof that he's right. Sometimes life just isn't fair, is it?
From: Tony Orlow on 27 Jul 2005 17:23 Robert Low said: > Tony Orlow (aeo6) wrote: > > 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: > > > But nobody has disagreed with that. The point of contention is > whether the union over all n of S^n contains an infinite string; > it doesn't, but you claim it does. I claim that if it doesn't then it isn;t an infinite set, which is the case for finite n. > > When you answer the question: > > "How many elements does the set of all finite integers contain?" A finite number, as I have said. That is all that can be said on that matter. > > you will be enlightened, and all these mysteries will become > clear to you. > If I prove that there is no n in N for which the set of all strings of length less than or equal to n is infinite, then I have proven that no finite limit on the length of strings can produce an infinite set. The number of strings, 2^(n+ 1)-2, is finite for all finite n. If your strings are all finite, you have a finite number of them. -- Smiles, Tony
From: Robert Low on 27 Jul 2005 17:26 Tony Orlow (aeo6) wrote: > Robert Low said: > >>Tony Orlow (aeo6) wrote: >> > 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: >> >> >>But nobody has disagreed with that. The point of contention is >>whether the union over all n of S^n contains an infinite string; >>it doesn't, but you claim it does. > > I claim that if it doesn't then it isn;t an infinite set, which is the case for > finite n. You keep on proving something that is so, and then claiming to have proven something entirely different. As Darryl keeps saying, you seem to be unable to distinguish between for all x there exists y such that x<y and there exists y such that for all x x<y >>"How many elements does the set of all finite integers contain?" > > A finite number, as I have said. That is all that can be said on that matter. Except that you think that adding one to this finite number makes it infinite. You really need to say that, too.
From: Daryl McCullough on 27 Jul 2005 17:07 Tony Orlow (aeo6) wrote: >Daryl McCullough said: >> Tony, you are just winging it, making up your answers as you go, >> without any regard to whether they are consistent with what you've >> already said. That's the point of rigorous definitions and axioms--they >> keep the mathematician honest. He can't just make stuff up, he has to >> follow the rules---even if they are rules that the mathematician made >> up himself. >If I am just winging it, I'm doing pretty good. No, you're not. You are wandering out biting your own tail, contradicting yourself. >> Let's go through the steps, and you can say which ones you >> disagree with: >> >> 1. First, do you agree that if S is a finite set, then >> the number of elements in S must be some (finite) natural >> number? > >That's the definition of a finite set, more or less. (yeay, yeah bijections, I >know) >> >> If that's not the case, then I don't know what you mean by >> finite (and I suspect that neither do you). >> >> 2. Second, do you agree that if n is any natural number, >> and S has exactly n+1 elements, then S has a largest element? >Sure, if you can identify the elements in it. >> So, from 1 and 2, it follows that >> >> Let FN = the set of finite natural numbers. If FN is finite, >> then FN has a largest element. >I disagree with that blanket statement, as I have said many times. That's what I mean---you are just winging it, making things up as you go along. You agree that every finite set has a largest element, and you agree that FN is a finite set, but then you say that FN doesn't have a finite element. You are contradicting yourself. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 27 Jul 2005 17:30
imaginatorium(a)despammed.com said: > Tony Orlow (aeo6) wrote: > > 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. > > Er, no it isn't. At least if the following proof is supposed to be the > relevant one. > > > > 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. > > (1) at least doesn't need to be an axiom, since it follows from > elementary arithmetic. > > > > 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. > > No comment on this on, I see. > > ... > > > > 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. > > Absolutely. There is no n in N for which the set of strings of length > up to n is infinite. Right. (You've done this before, though; you > vacillate between correctly proving one thing, then restating it as > something else, which is false.) No, you mean, you missed the point before. > > If for any finite integer the strings are limited to this length, then > the set is finite. If, however, the strings are not limited to the > length of any single integer, but may have a length of any integer, > then the set is infinite. As always, you have to be careful reading the > quantification of the two parts of this sentence correctly: > > (a) For any string, there is an integer which is its length. > (b) There does not exist an integer such that none of the strings > exceeds it in length. The absence of a largest finite does not in any way invalidate this proof. There is no finite length of string that can possibly produce an infinite set of strings. If you restrict your strings to finite lengths, you cannot have an infinite set. "largest finite" has absolutlely no bearing on this fact. In order for 2^(n+1)-2 to be infinite, n must be infinite, but you do not allow n to be infinite, because you restrict your strings to finite lengths. This is beyond clear. You folks are hopeless. > > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony |