Obections to Cantor's Theory (Wikipedia article)
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From: Daryl McCullough on 20 Jul 2005 11:22 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >>> There is nothing in Peano's axioms that states explicitly that all >>> natural numbers are finite. Let's get more specific, and consider the sets S_n = the set of all natural numbers less than n. Are you claiming that there is a natural number n such that S_n is not finite? What definition of finite are you using? -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 20 Jul 2005 11:42 Virgil said: > In article <MPG.1d4733ca67be65e3989f31(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > David Kastrup said: > > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > > > > > David Kastrup said: > > > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > >> > > > >> > Now, I am not familiar, I think, with the proof concerning > > > >> > subsets of the natural numbers. Certainly a power set is a larger > > > >> > set than the set it's derived from, but that is no proof that it > > > >> > cannot be enumerated. > > > >> > > > >> Uh, not? > > > > > > > Yes, not. "Larger" is not a synonym for "uncountable" except in > > > > Cantorland, and that is a leap and an assumption. > > > > > > "Larger" is a synonymon for "can't be surjected onto from" in set > > > theory. And "uncountable" is a synonymon for "larger than the set of > > > naturals". It is not a leap or an assumption, but simply a > > > definition. > > > > > > >> > Is this the same as the proof concerning the "uncountability" of > > > >> > the reals? > > > >> > > > >> It's pretty similar. > > > > Figures. > > > >> > > > >> Assume a set X can be put into complete bijection with its powerset > > > >> P(X) such that we have a mapping x->f(x) where x is an element from X > > > >> and f(x) is an element from P(X). Now consider > > > >> Q = {x in X|x not in f(x)}. Clearly, for all x in X we have > > > >> Q unequal to f(x), since x is a member of exactly one of f(x) and Q. > > > >> So Q is missing from the bijection. > > > >> > > > >> > > > > Again with the "Clearly". You might want to refrain from using the > > > > word, and just try to be clear, without hand-waving. > > > > > > > > There is no requirement that subset number x include x as a member, > > > > > > Quite so. But there is a requirement that subset number x _either_ > > > include x as a member _or_ not include x as a member. Only one of > > > those two statements can be true. And then Q _either_ not includes x > > > as a member _or_ does include it, respectively. > > > > > > You are free to choose your mapping as you want to. But once you have > > > chosen your mapping, each subset number x _either_ includes x as a > > > member _or_ it doesn't. Whether it does, can be chosen independently > > > for every x. But once you are through, for every particular x, x will > > > be in f(x), or it won't. And depending on that, x won't be in Q, or > > > it will. > > Actually, as I think about it, given this natural mapping of the naturals to > > the subsets of naturals, subset number x will ONLY include x as a member for > > subsets 0, 1 and 2. Beyond that, subset x will NEVER contain x. So, you have > > non-empty Q for the null set, and the singletons {1} and {2}. So, what does > > that prove? > > The mapping from X to P(X) is not "natural", it is 'arbitrary', meaning > that it can be anything and cannot be assumed to have any special > properties. > > In particular, f(0) = f(1) = f(2) = X is possible, whenever {0,1,2} is > a subset of X, . > I defined the mapping and it's as natural as it gets. Each succesive bit represents membership of each successive element. It doesn't get any more natural than that.n Every unique infinite string of bits represents a unique subset of the naturals, so I don't know WHAT that last sentence means. There is a 1-1 correspondence between infinite bit strings and subsets. -- Smiles, Tony
From: Tony Orlow on 20 Jul 2005 11:44 Virgil said: > In article <MPG.1d4736d25730cfe7989f35(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > Stephen J. Herschkorn said: > > > Should a reputable encyclopedia contain an entry devoted entirely to > > > people who think the earth is flat? > > > An entry only for those who think that sun revolves the earth? > > > An entry devoted specifically to those who think that man never landed > > > on the moon? > > > To those who insist there is a smallest positive real number? > > > > > > > > 000...000.000...001 > > It is not a real number unless there is a finite number of zeroes > specified for each ellipsis. and once that is done, it is not a smallest > positive real, since that is a mythical beast. > True in ways. It is an infinitesimal unit, which can be considered as a smallest real. -- Smiles, Tony
From: Daryl McCullough on 20 Jul 2005 11:25 Tony Orlow <aeo6(a)cornell.edu> said: >> > but the diagonal proof suffers from the fatal flaw of assuming that >> > the diaginal traversal actually covers all the numbers in the list. >> > Any complete list of digital numbers of a given length, even a given >> > infinite length, is exponentially longer in members than wide in terms >> > of the digits in each member. Therefore, the diagonal traversal only >> > shows that the anti- diagonal does not exist in the first aleph_0 >> > terms. Yes, that's exactly what the diagonal argument proves. There is no list of length aleph_0 that contains all real numbers. -- Daryl McCullough Ithaca, NY
From: Tony Orlow on 20 Jul 2005 12:02
Robert Low said: > Dave Rusin wrote: > > Robert Low <mtx014(a)coventry.ac.uk> wrote: > >>Tony Orlow (aeo6) wrote: > >>>Stephen J. Herschkorn said: > >>>>To those who insist there is a smallest positive real number? > >>>000...000.000...001 > >>And how many 0's are there after that decimal point? N? log_2(N)? N-1? > >>N+1? (Whatever the hell any of those answers mean...) > > Tsk, tsk, Robert, you've not been paying attention! There are N of > > them; the "1" is in the N-th place, i.e. this decimal expansion is > > simply of the number 10^(-N). > > I think I get it now. N is the biggest number, so if you divide > 0.0...01 (N 0's) by 10, you actually get zero, not 0.00...01 > (N+1 0's). Of course, the naive amongst us might object that > this means that 10*0=0.0...01; this is just silly, because > it's making the entirely unwarranted assumption that multiplication > or real numbers is associative, and there's no reason for > that to be true of the infinitely small ones. > > > Hmm, looks like I haven't been paying attention either... > > Well, I must be smarter than you, because I've figured > it out :-) > Okay, you boys can have your fun. I had introduced terminology for nested infinities and infinitesimals in another thread, which I will try to include in web pages at some point. Until then, have fun! -- Smiles, Tony |