Obections to Cantor's Theory (Wikipedia article)
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From: Robert Low on 19 Jul 2005 16:02 Tony Orlow (aeo6) wrote: > Robert Low said: >>Of course. But that isn't the issue. The issue is the existence >>of a bijection between either of these sets and the power set >>of the set of naturals. So, which binary number does your >>procedure associate with the set of *all* natural numbers >>divisible by 3? > Again, that infinite set is denoted by the infinite whole number > 100100100...100100100. Notice, every third bit, from right to left, is a 1. > Those are the multiples of 3. So, you're saying that you can find a bijection between N and its power set, as long as you get to say that N isn't what everybody else thinks it is. See elsewhere for the problem with this.
From: Tony Orlow on 19 Jul 2005 16:04 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. -- Smiles, Tony
From: David Kastrup on 19 Jul 2005 16:10 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? > 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. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Alec McKenzie on 19 Jul 2005 16:14 Stephen Montgomery-Smith <stephen(a)math.missouri.edu> wrote: > Alec McKenzie wrote: > > "Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote: > > > > > >>Can anti-Cantorians identify correctly a flaw in the proof that there > >>exists no enumeration of the subsets of the natural numbers? > > > > > > In my view the answer to that question a definite "No, they > > can't". > > > > However, the fact that no flaw has yet been correctly identified > > does not lead to a certainty that such a flaw cannot exist. Yet > > that is just what pro-Cantorians appear to be asserting, with no > > justification that I can see. > > As best I can see from your other posts, you are making one of two points: > > 1. There is such an enumeration, because set thoery is inconsistent. > Yes, I cannot be sure that cannot happen, but it would not invalidate > the proof (because the theorem would be both true and not true > simultaneously). If that is your problem, I think your issue is with > proof by contradiction, not with Cantor's argument us such. > > 2. There really is a flaw in the proof, but mathematicians have somehow > simply not seen it. While one cannot totally discount this possibility, > the chances of this being the case is so extraordinarily small that for > all practical purposes it is just not the case. We are talking > probabilities like that of a Monkey sitting at a typewriter and dashing > off a Shakespeare play. In principle, yes it can happen, in reality, > you should worry more about UFO's abducting you. 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? -- Alec McKenzie mckenzie(a)despammed.com
From: David Kastrup on 19 Jul 2005 16:15
Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > Robert Low said: >> Tony Orlow (aeo6) wrote: >> > David Kastrup said: >> >> > 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. >> >> Of course. But that isn't the issue. The issue is the existence of >> a bijection between either of these sets and the power set of the >> set of naturals. So, which binary number does your procedure >> associate with the set of *all* natural numbers divisible by 3? >> > Again, that infinite set is denoted by the infinite whole number > 100100100...100100100. Notice, every third bit, from right to left, is a 1. > Those are the multiples of 3. Is this number a member of the set of all natural numbers divisable by 3? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |