Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 19 Jul 2005 17:08 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > There is no reason to expect the natural number corresponding to > >> > the subset to be a member of that subset. > >> > >> There is no such expectation. The only expectation is that _every_ > >> natural number is _either_ a member of its corresponding subset, or > >> not. > > Okay. > >> _Depending_ on that, the constructed subset will either _not_ or > >> _do_ contain the number, respectively. > > Redundant, but okay. > > Sure it is redundant. But you should be the last person to complain > about your need to get this hammered bit by bit into your skull. > > >> This constructed subset then does not correspond to _any_ natural > >> number. > > > This is where it goes wrong. > > Uh yes. That's the conclusion: that the assumption of the mapping > goes wrong, since it fails to cover the constructed subset. > > > The consructed subset corresponds to the natural number denoted by > > the binary string constructed from the right to the left, where each > > successive bit is a 1 if the successive natural is a member, and 0 > > if it is not. > > You are assuming a particular mapping, the proof holds for _any_ > mapping. So let's see where your mapping takes us. Since 2^n>n > always, the subset we get is simply N itself, at least as long as we > assume that N only contains finite numbers. However, that is an > assumption that you are not willing to make, so let is call this > particular subset of N by the name Q. > > So what natural number corresponds to Q according to your logic? I'll > take a daring guess at your muddled thoughts and suppose 111...111 or > something like that. Is 111...111 itself a member of Q? Well, that's an interesting question. yes, it is 111...111, and 111...111 is also a member of N, but those two are a little different. The first is a map of membership, and has N number of 1's. The second, is a digital number, and presumably denotes N-1. Therefore, it has log2(N) 1's, rather than N. To represent 2^N reals, you need N digits, but to represent N naturals, you only need log2(N) digits. > > > It is so basic, I cannot even quite see what erroneous assumption > > you are making. It seems like you are assuming subset number X must > > contain X as a member. If so, how do you justify this assumption? > > I don't. But I assume that for every X, subset number X must either > contain number X as a member, or doesn't. And if it does, subset Q > will not contain number X as a member, and if it doesn't, subset Q > will contain number X as a member, and so subset Q differs from every > subset X at the position of number X. This is essentially the diagonal argument in disguise. You are assuming you have the same number of numbers as digits, which is clearly a false assumption when using any base greater than 1. > > > -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 17:10 David Kastrup said: > Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > > > David Kastrup said: > >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > >> > >> > David Kastrup said: > >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> >> > >> >> > David Kastrup <dak(a)gnu.org> wrote: > >> >> > > >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: > >> >> >> > It has been known for a proof to be put forward, and fully accepted > >> >> >> > by the mathematical community, with a fatal flaw only spotted years > >> >> >> > later. > >> >> >> > >> >> >> In a concise 7 line proof? Bloody likely. > >> >> > > >> >> > I doubt it had seven lines, but I really don't know how many. > >> >> > Probably many more than seven. > >> >> > >> >> It was seven lines in my posting. You probably skipped over it. It > >> >> is a really simple and concise proof. Here it is again, for the > >> >> reading impaired, this time with a bit less text: > >> >> > >> >> Assume a complete mapping n->S(n) where S(n) is supposed to cover all > >> >> subsets of N. Now consider the set P={k| k not in S(k)}. Clearly, > >> >> for every n only one of S(n) or P contains n as an element, and so P > >> >> is different from all S(n), proving the assumption wrong. > >> > >> > I still do not get this. You have a set of naturals {0,1,2,3...}, > >> > and a set of binary numbers {0,1,10,11,100,101,110,111,....}. Surely > >> > there is a bijection between these two sets. > >> > >> Fine. > >> > >> > So, what is the problem if one interprets the binary numbers (with > >> > implied leading zeroes) as being a map of each subset, where each > >> > successive bit represents membership in thesubset by each successive > >> > natural number? > >> > >> Ok, so let's construct P. It is actually easy enough, since n<2^n, > >> and so P=N. So what number corresponds to N itself in your mapping? > >> > > An infinite string of 1's: 1111.....1111. This is (2^aleph_0)-1. > > > > Infinite whole numbers are required for an infinite set of whole > > numbers. > > And what number corresponds to the subset of N without the number > 1111.....1111 in it? > > What is this, 20 questions? It's 01111....1111. Duh! -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 17:11 The World Wide Wade said: > In article > <MPG.1d472484f809e37c989f2c(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > The idea of uncountability as being equivalent to "larger than the set of > > naturals" is unfounded. There is no reason to believe that larger sets cannot > > be enumerated. the power set of the naturals can be enumerated and bijected > > with the naturals, as I described in another post, as long as infinite > > natural > > numbers are allowed. > > Sort of like saying "if 0 = 1" is allowed. > No, more like saying "if infinite digits are allowed", which is what is required to have an infinite set of digital numbers using a finite base. That was a useless comment. I shouldn't even be responding. -- Smiles, Tony
From: Tony Orlow on 19 Jul 2005 17:12 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 -- Smiles, Tony
From: Stephen Montgomery-Smith on 19 Jul 2005 17:11
Alec McKenzie wrote: > I am not making either of those points, and I don't see why you > should think I am: > > 1. I have no reason to believe there is such an enumeration, and > I have never suggested that there is. > > 2. I have no reason to believe there really is a flaw in the > proof, and I have never suggested that there is. > > The point I am trying to make is precisely the one I wrote: > > "The fact that no flaw has yet been correctly identified does > not lead to a certainty that such a flaw cannot exist." > > I still believe this to be true. Would you deny it? I think that Hamlet said it well (V i): How absolute the knave is! we must speak by the card, or equivocation will undo us. Although I also like Mercutio's commentary (Romeo and Juliet III i): Thou! why, thou wilt quarrel with a man that hath a hair more, or a hair less, in his beard, than thou hast: thou wilt quarrel with a man for cracking nuts, having no other reason but because thou hast hazel eyes: what eye but such an eye would spy out such a quarrel? Thy head is as fun of quarrels as an egg is full of meat, and yet thy head hath been beaten as addle as an egg for quarrelling: thou hast quarrelled with a man for coughing in the street, because he hath wakened thy dog that hath lain asleep in the sun: didst thou not fall out with a tailor for wearing his new doublet before Easter? with another, for tying his new shoes with old riband? Stephen |