Obections to Cantor's Theory (Wikipedia article)
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From: Tony Orlow on 19 Jul 2005 15:15 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. 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? So, 0 represents the null set, 1 represents the set including the first element, 10 includes just the second, 11 includes the second and third, etc.. You seem to be complaining that the fourth subset, with a numerical value of 3, only includes the first and second elements? I mean, your statement that begins with "Clearly" is not at all clear to me. Any binary number which has a 1 in the rightmost digit contains the first element, so the first element is a member of an infinite number of such subsets. What is the assumption which is supposedly wrong? > > So now it is 4 lines. And one does not need more than that. > > >> And that's what you call "with no justification that I can see". > > > > No, it is not -- what I called "with no justification that I can > > see" was something else: > > > > It was the assertion that no flaw having been found in a proof leads > > to a certainty that such a flaw cannot exist. I still see no > > justification for that. > > Fine, so you think that a four-liner that has been out and tested for > hundreds of years by thousands of competent mathematicians provides no > justification for some statement. Just over 100 years, and we've learned a lot since then. This isn't religion. > > Just what _would_ constitute justification in your book? > > -- Smiles, Tony
From: Chan-Ho Suh on 19 Jul 2005 15:14 In article <mckenzie-E70336.17334819072005(a)news.aaisp.net.uk>, Alec McKenzie <mckenzie(a)despammed.com> wrote: > David Kastrup <dak(a)gnu.org> wrote: > > > 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. > > I see you have misunderstood what I said. You seemed to be > denying that the accepted proof I mentioned (that turned out to > have a flaw) was only seven lines, and I was merely saying that > I didn't know how many lines it had. Aha, so you're saying you had a particular proof in mind? Well, let's hear which one it is.
From: Chan-Ho Suh on 19 Jul 2005 15:17 In article <87psterjcf.fsf(a)phiwumbda.org>, Jesse F. Hughes <jesse(a)phiwumbda.org> wrote: > David C. Ullrich <ullrich(a)math.okstate.edu> writes: > > > Ok, here's another question. Suppose that we want to > > prove that A implies B. Suppose that we have an > > completely flawless proof that A implies C, and > > a completely flawless proof that C implies B. > > Does the union of those two proofs constitute > > a flawless proof that A implies B? > > Am I the only one suffering from flashbacks of Achilles and the > Tortoise? > > No. :-)
From: Robert Low on 19 Jul 2005 15:24 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?
From: David Kastrup on 19 Jul 2005 15:28
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? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum |