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From: David R Tribble on 1 Nov 2005 15:20 David R Tribble said: >> You've offered plenty of statements about "unidentifiable largest >> naturals" and "set ranges" and "unit infinities", but you've never >> actually pinpointed how you get from the finites to the infinites. > Tony Orlow wrote: > Here's another, shorter one: > > 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. Yes, that's correct. For any finite n in N, f(n) is finite. But we already knew that. > There is no n in N, in other words, which maximum length of string will > allow you to have an infinite set of strings. Non sequiter. Yes, there is no particular n in N that gives an infinite string length (because no n in N is infinite). But that does not prove anything about the finiteness of N. In fact, it proves that N is infinite, since there is no largest n in N. The only way you could conclude that N is finite is to assume that there is some last, largest n in N, which would give you a maximum finite word size for your finite bitstrings. But you explicitly state that there is no such n in N. So you must conclude that N is not finite, but in fact is infinite. But you don't get that, do you?
From: David R Tribble on 1 Nov 2005 18:07 Albrecht Storz said: >> You argue: there is no infinite natural number since the peano axioms >> don't allow an infinite natural number. >> That's right. I agree with you. > Tony Orlow wrote: > Alas, Albrecht, it is not true. The Peano axioms start with 0 (a finite natural) and then define the successor operation. It is trivial to show that together these define an infinite number of finite naturals. (This has been shown many many times here already.) Tony believes that the successor operation somehow produces infinite naturals, at some "unidentifiable" point. His logic is that since each successor operation is equivalent to adding 1 to the previous natural, and since this is done an infinite number of times, this is all equivalent to adding an infinite number of 1s to 0, thus producing an infinite natural. The flaw here is that applying the successor operation an infinite number of times is sufficient, by itself, to produce an infinite number of finite naturals. You never "reach" an infinite natural, but you don't need to in order to get an infinite number of finite naturals. This makes perfect sense because the only way to create a natural number is to apply the successor operation to a previously defined finite natural, and this always (always!) defines another finite natural. Furthermore, the fact that, for any given natural k you can define k+1 as the next successor natural, makes it obvious that there is a never-ending supply of finite naturals, i.e., there are an infinite number of them. There cannot be an end to them, "unidentifiable" or otherwise. All of this appears to be beyond Tony's comprehension, though.
From: David Kastrup on 1 Nov 2005 19:19 "David R Tribble" <david(a)tribble.com> writes: > Albrecht Storz said: >>> You argue: there is no infinite natural number since the peano axioms >>> don't allow an infinite natural number. >>> That's right. I agree with you. >> > > Tony Orlow wrote: >> Alas, Albrecht, it is not true. > > The Peano axioms start with 0 (a finite natural) and then define the > successor operation. It is trivial to show that together these > define an infinite number of finite naturals. (This has been shown > many many times here already.) > > Tony believes that the successor operation somehow produces infinite > naturals, at some "unidentifiable" point. His logic is that since > each successor operation is equivalent to adding 1 to the previous > natural, and since this is done an infinite number of times, I don't see anybody doing this an "infinite number of times". I don't even see anybody doing it once. The Peano axioms state that each element _has_ a successor, not that this successor needs to be generated or has been generated in some manner. > this is all equivalent to adding an infinite number of 1s to 0, thus > producing an infinite natural. > > The flaw here is that applying the successor operation an infinite > number of times is sufficient, by itself, to produce an infinite > number of finite naturals. There is no such thing as "applying an infinite number of times". If it were, you would reach a particular state after that "operation", and of course, there would be no reason why you could not continue forming successors once you reach that state. It would also be inconsistent to claim that you can apply an operation aleph_0 times, but can't apply it aleph_1 times. The successor operation _can't_ be applied "an infinite number of times". There is no limit to the number of finite times you can apply it, but that does not mean that you can apply it an infinite number of times. That's just sloppy thinking. The set of naturals is an infinite set, but not because it is "generated by an infinite number of steps", but because of the properties from its definition. > You never "reach" an infinite natural, but you don't need to in > order to get an infinite number of finite naturals. You don't get to an infinite number of finite naturals by any process. You get there by _definition_. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: David R Tribble on 1 Nov 2005 18:36 Tony Orlow says... >> No element is ever identified, in any infinite bijection, >> as mapping to the last element of an endless set. > Daryl McCullough wrote: > Well, obviously if a set doesn't have a largest element, > then nothing maps to the largest element. > > Unfortunately, a power set *does* have a largest element, > namely the entire set. [...] Technically, this is incorrect. Since all the infinite subsets of P(*N) are the same size (have the same cardinality), there is no "largest" subset. Actually, that's not quite right either. All of the infinite subsets in P(N) (which are also subsets of P(*N)) are the same size, but there are infinite subsets in P(*N) (i.e., some members of P(*N) that are not also members of P(N)) that are uncountably infinite, so they are even larger subsets. This makes sense since card(P(N)) = 2^Aleph_0, but card(P(*N)) = 2^2^Aleph_0. It is these larger subsets that Tony's mapping fails to map to, among others. But nevertheless you make a good point about the lack of a largest element in *N.
From: David R Tribble on 1 Nov 2005 18:46
David R Tribble writes: >> Tony believes that the successor operation somehow produces infinite >> naturals, at some "unidentifiable" point. His logic is that since >> each successor operation is equivalent to adding 1 to the previous >> natural, and since this is done an infinite number of times, [...] > David Kastrup wrote: > I don't see anybody doing this an "infinite number of times". I don't > even see anybody doing it once. The Peano axioms state that each > element _has_ a successor, not that this successor needs to be > generated or has been generated in some manner. > > There is no such thing as "applying an infinite number of times". If > it were, you would reach a particular state after that "operation", > and of course, there would be no reason why you could not continue > forming successors once you reach that state. It would also be > inconsistent to claim that you can apply an operation aleph_0 times, > but can't apply it aleph_1 times. > > The successor operation _can't_ be applied "an infinite number of > times". There is no limit to the number of finite times you can apply > it, but that does not mean that you can apply it an infinite number of > times. > > That's just sloppy thinking. The set of naturals is an infinite set, > but not because it is "generated by an infinite number of steps", but > because of the properties from its definition. > > You don't get to an infinite number of finite naturals by any > process. You get there by _definition_. I didn't mean to imply that that is the way I think. I'm usually pretty careful not to use such misleading terminology. But I was trying to explain what I believe is Tony's logic behind his irrational beliefs. Albrecht (and others) also falls into this pit when he defines his "bijections" and correspondences between infinite sets. But I agree that the inability to see sets as "all there at once" is a key component of irrational spewage. |