Obections to Cantor's Theory (Wikipedia article)
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From: Dik T. Winter on 19 Jul 2005 22:19 In article <MPG.1d4722e516a9e4df989f2b(a)newsstand.cit.cornell.edu> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: .... > The only reason to reject this bijection is > if one clings to the idea that all natural numbers are finite, which is > impossible. Back on your horse again. Tell me about the binary numbers (extended to the left with 0's) where the leftmost 1 is in a finite position. Are all those numbers finite? Are there only finitely many of them? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 19 Jul 2005 22:23 In article <MPG.1d4722e516a9e4df989f2b(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: > > >> Alec McKenzie <mckenzie(a)despammed.com> writes: > > >> > > >> > "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. > > >> > > >> Uh, what? There is nothing fuzzy about the proof. > > >> > > >> Suppose that a mapping of naturals to the subsets of naturals exists. > > >> Then consider the set of all naturals that are not member of the > > >> subset which they map to. > > >> > > >> The membership of each natural can be clearly established from the > > >> mapping, and it is clearly different from the membership of the > > >> mapping indicated by the natural. So the assumption of a complete > > >> mapping was invalid. > > >> > > >> > Yet that is just what pro-Cantorians appear to be asserting, with no > > >> > justification that I can see. > > >> > > >> Uh, where is there any room for doubt? What more justification do you > > >> need apart from a clear 7-line proof? It simply does not get better > > >> than that. > > > > > Is the above your 7-line proof? it makes no sense. > > > > If you don't get it. > > > > > 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. > > This constructed subset then > > does not correspond to _any_ natural number. > This is where it goes wrong. This is where TO's understanding fails, but the argument does not fail. > 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. Any subset can be denoted as a string of bits, and any string > of > bits can denote a natural number. > The only reason to reject this bijection is if one clings to the idea > that all natural numbers are finite, which is impossible. The existence of infinite natural numbers requires that at some point one must be able to add 1 to a finite natural and get an infinite natural. If TO can swallow this camel and still strain at the gnat of uncountability, his whole mind needs a rebuild from the ground up. > > > > > if this rests on the diagonal proof, > > > > Rather the other way round. It is more basic than the diagonal proof. > 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? > > > >
From: Virgil on 19 Jul 2005 22:31 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. The meaning of "uncountable" is a matter of definition. One may, as TO does, object to claims of its being instanciated, but it is perfectly well-defined. > There is no reason to believe that larger sets cannot > be enumerated. What does "larger than" mean, then? There is no reason to believe that larger sets cannot be enumerated. That TO does not understand the reasons does not invalidate them. > 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. But to allow infinite naturals means that one must have a finite natural so large that adding one to it gives an infinite result, since every natural, except the first, is produced by adding 1 to a previous natural.
From: Barb Knox on 19 Jul 2005 22:27 In article <ITSnetNOTcom#virgil-CEB650.19511919072005(a)comcast.dca.giganews.com>, Virgil <ITSnetNOTcom#virgil(a)COMCAST.com> wrote: >In article <MPG.1d4715811811605b989f27(a)newsstand.cit.cornell.edu>, > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > >> David Kastrup said: >> > Alec McKenzie <mckenzie(a)despammed.com> writes: >> > >> > > "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. >> > >> > Uh, what? There is nothing fuzzy about the proof. >> > >> > Suppose that a mapping of naturals to the subsets of naturals exists. >> > Then consider the set of all naturals that are not member of the >> > subset which they map to. >> > >> > The membership of each natural can be clearly established from the >> > mapping, and it is clearly different from the membership of the >> > mapping indicated by the natural. So the assumption of a complete >> > mapping was invalid. >> > >> > > Yet that is just what pro-Cantorians appear to be asserting, with no >> > > justification that I can see. >> > >> > Uh, where is there any room for doubt? What more justification do you >> > need apart from a clear 7-line proof? It simply does not get better >> > than that. >> > >> > >> Is the above your 7-line proof? it makes no sense. > >It makes sense to those who have sufficient mental capacity to >understand it. Those who either cannot or will not understand it, but >cannot fault it, are irrelevant. I like Randall Holmes' sig on that subject: "And God posted an angel with a flaming sword at the gates of Cantor's paradise, that the slow-witted and the deliberately obtuse might not glimpse the wonders therein." [snip] -- --------------------------- | BBB b \ Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | -----------------------------
From: Virgil on 19 Jul 2005 22:35
In article <MPG.1d4725fbdfa9bb24989f2d(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > There is a simple, demonstrably valid proof of Cantor's Theorem in ZF > > set theory. So you must think the proof is unsound. Which axiom of ZF > > do you believe to be false? > > > > Chris Menzel > > > > > I was asked that before, and never got around to fully analyzing the axioms > for > lack of time, but the diagonal proof suffers from the fatal flaw of assuming > that the diaginal traversal actually covers all the numbers in the list. If it misses any of them, it must miss a first one. Which is the first one misssed? If there is not a first one missed, then none are missed. |