Cantor Confusion
in [Math]
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From: Dik T. Winter on 21 May 2007 22:58 In article <1179750984.210277.71110(a)n15g2000prd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Mai, 04:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > Pardon, I meant "countably many definitions". This is implied by the > > > finity of every definition. If there were infinite definitions, then > > > there were uncountably many definitions. > > > > Yup. So the question remains: "you disagree with common mathematical > > terminology?" > > No, it is obvious is that a finite definition means a definition which > is given by a finite number of words. Infinite definitions are > nonsense, according to Cantor. Yup, you are still living some hundred years ago. Mathematics has gone forward since that time. > > But whatever, you can not apply the diagonal argument > > to "finitely defined numbers". You can supply a list of finite > > definitions, but not all of them define a number. And for the > > diagonal argument a list of numbers is needed. > > Correct. Therefore it is clear that the finitely defined numbers > cannot be put in a list, i.e., they cannot be in bijection wit N. They are in bijection with a subset of N, so see below. > My > argument should only show that there are countable sets (like the > paths in the tree) which cannot be put in a bijection with N. They can, but that requires an infinite definition. Look at the proof that if there is an injection from A to B and an injection from B to A, that there is a bijection between A and B. Schr�der-Bernstein, of about the same time as Cantor. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 21 May 2007 23:00 In article <1179751344.202204.88880(a)y18g2000prd.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 21 Mai, 05:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > > The following wm-proof certainly even in your opinion belongs to the > > > diagonal proofs considered by Cantor: > > > > > > 0) mmm... > > > 1) wmmm... > > > 2) wwmmm... > > > 3) wwwmmm... > > > 4) wwwwmmm... > > > ... .......... > > > > > > And if the list can be considered as a completed entity, then there > > > must be all natural numbers in the first column. And there must be a > > > line with all natural indexes mapped on w's, i.e., no w must be > > > missing (as would be the case if one m was present). > > > > Why? Show a proof. You are again assuming that there is a last natural > > number. > > If the diagonal is complete, then the list must also be complete, > because the diagonal cannot be longer or broader than the list. If the > list is complete, then there must be a line with only w's, because > otherwise the list is not complete. Back again to this nonsense. By the axiom of infinity the list can be completer without there being a last element. > This is also shown by considering the limit for n --> oo (line with n > w's - diagonal) = 0 What has a limit to do with it? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 22 May 2007 15:50 On 22 Mai, 03:53, Virgil <vir...(a)comcast.net> wrote: > In article <1179777652.039258.88...(a)y2g2000prf.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 20 Mai, 21:09, Virgil <vir...(a)comcast.net> wrote: > > > In article <1179668114.743217.104...(a)p47g2000hsd.googlegroups.com>, > > > > There cannot be in any set theory of repute any infinite set which is > > > not actually infinite. > > > > Whatever WM insists on as being potentially but not actually infinite, > > > it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, > > > but not a set. > > > So the union U(T(n)) of all finite trees T(n), formed in an obvious > > way, is an actually infinite binary tree AIBT with actually countable > > sets of paths and nodes and edges. > > I will presume that "AIBT" should be "CIBT". and the set of paths is not > countable, but otherwise, you are correct. AIBT = actually infinite binary tree AIBT . The set of paths in the AIBT is countable, because a countable union of countable sets is countable. > > >> (2) ==> p is nothing but a union of finite paths. > > > > Not neccessarily. That is the case only if one requires that a > > > path be no more than a set of nodes. > > > But every path has a soul in addition? > > It has a set of edges as well as a set of nodes. Neither of the edges or the nodes is increased when switching from the AIBT to the CIBT. > > > > > > > > > ================================= > > > On 20 Mai, 22:11, Virgil <vir...(a)comcast.net> wrote: > > > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > Set theory is simply biased. Consider the list > > > > > > > 0.666... > > > > > > 0.3666... > > > > > > 0.33666... > > > > > > 0.333666... > > > > > > ... > > > > > > > If the diagonal number is defined by "replace 6 by 3", then we have > > > > > > two answers none of which can be preferred by logic, but the second > > > > > > of > > > > > > which is suppressed by convention. > > > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > The following wm-proof certainly even in your opinion belongs to the > > > > diagonal proofs considered by Cantor: > > > > > 0) mmm... > > > > 1) wmmm... > > > > 2) wwmmm... > > > > 3) wwwmmm... > > > > 4) wwwwmmm... > > > > ... .......... > > > > > And if the list can be considered as a completed entity, then there > > > > must be all natural numbers in the first column. > > > > As each "column" can contain only w's or m's, there are no natural > > > numbers in any column. > > > The first column consists of natural numbers only. (and of the unnatural 0) > > They are not a part of the strings, but only arguments to the function > which defines the list. My statement is nevertheless correct. Regards, WM
From: WM on 22 May 2007 16:08 On 22 Mai, 04:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179746990.290370.268...(a)z24g2000prd.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 21 Mai, 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > for what is now called "simply normal". > > > > Please do not conclude from your lacking knowledge on that of others. > > Weakly normal is a common definition. See fort instance. > >http://eom.springer.de/N/n067560.htm > > Or do you think that "weakly normal" is not common because it is > > missing in Wikipedia? I would estimate Springer somewhat higher than > > Wikipedia. > > I do not use Wikipedia in general for mathematical knowledge. My main > online source is Mathworld. And of course the large library of my > institute. Springer Online is edited by a man from CWI, Amsterdam. > > > > > > Yes, so what? The Champerowne constants and the Copeland-Erdos constants > > > are *not* rational. Read just below your quote, where they give a > > > Champerowne constant. > > > > So what? I never said that every however normal number must be > > rational. But 0.01234567890123456789... (weakly normal to base 10) > > and 0,012012012... (weakly normal to base 3) *are* rational. > > So your statement was irrelevant as a response to my statement. Your statement was irrelevant to my book. > But from > your book: > "also eine sogenannte normale irrationale Zahl, die keine erkennbares > Muster der Ziffernfolge aufweist" Note the last sentence! > In the first place, a normal number can have a well defined sequence of > digits (as the Champerowne numbers show). Of course, but then we could compute the numerals. > And, in the second place, there > are numbers without a well defined sequence of digits that are *not* normal. And there are red cars, which are not normal cars. > (Normal here meaning normal to a particular base.) The two are not > equivalent. So even if no rule can be given for the digits of pi, that > does *not* mean that it is normal. But if pi is normal with no recognizable pattern, then no rule can be given. (If I say A ==> B then I do not imply B ==> A.) Regards, WM
From: Virgil on 22 May 2007 16:48
In article <1179863445.912400.167960(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Mai, 03:53, Virgil <vir...(a)comcast.net> wrote: > > In article <1179777652.039258.88...(a)y2g2000prf.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 20 Mai, 21:09, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1179668114.743217.104...(a)p47g2000hsd.googlegroups.com>, > > > > > > There cannot be in any set theory of repute any infinite set which is > > > > not actually infinite. > > > > > > Whatever WM insists on as being potentially but not actually infinite, > > > > it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, > > > > but not a set. > > > > > So the union U(T(n)) of all finite trees T(n), formed in an obvious > > > way, is an actually infinite binary tree AIBT with actually countable > > > sets of paths and nodes and edges. > > > > I will presume that "AIBT" should be "CIBT". and the set of paths is not > > countable, but otherwise, you are correct. > > AIBT = actually infinite binary tree AIBT . The set of paths in the > AIBT is countable, because a countable union of countable sets is > countable. WM has yet to show that the set of paths of an an AIBT is a countable union of countable sets. Others have shown directly that a CIBT does not have a merely countable set of all paths. WM is asserting a difference between an AIBT and a CIBT. As CIBTs are what occur ini mathematics, one must presume that AIBTs only occur in MathUnRealism. > > > > >> (2) ==> p is nothing but a union of finite paths. > > > > > > Not neccessarily. That is the case only if one requires that a > > > > path be no more than a set of nodes. > > > > > But every path has a soul in addition? > > > > It has a set of edges as well as a set of nodes. > > Neither of the edges or the nodes is increased when switching from the > AIBT to the CIBT. But, apparently, the set of paths is increased, since in a CIBT, as previously defined, that set of paths is not countable. In a CIBT, there is, for every subset of N, a path unique to that subset, which branches left from the level of each member of that subset of N, and right from the level of each non-member. Thus the set of paths bijects with the power set of N. And in ZF and NBG that is of cardinality greater than that of N. Apparently there are subsets of N for which an AIBT has no corresponding paths, so it is not a complete tree. > > > ================================= > > > > > On 20 Mai, 22:11, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1179663816.282116.232...(a)h2g2000hsg.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > Set theory is simply biased. Consider the list > > > > > > > > > 0.666... > > > > > > > 0.3666... > > > > > > > 0.33666... > > > > > > > 0.333666... > > > > > > > ... > > > > > > > > > If the diagonal number is defined by "replace 6 by 3", then we > > > > > > > have > > > > > > > two answers none of which can be preferred by logic, but the > > > > > > > second > > > > > > > of > > > > > > > which is suppressed by convention. > > > > > > > > But, again, that is *not* the diagonal proof of Cantor. > > > > > > > The following wm-proof certainly even in your opinion belongs to the > > > > > diagonal proofs considered by Cantor: > > > > > > > 0) mmm... > > > > > 1) wmmm... > > > > > 2) wwmmm... > > > > > 3) wwwmmm... > > > > > 4) wwwwmmm... > > > > > ... .......... > > > > > > > And if the list can be considered as a completed entity, then there > > > > > must be all natural numbers in the first column. > > > > > > As each "column" can contain only w's or m's, there are no natural > > > > numbers in any column. > > > > > The first column consists of natural numbers only. > (and of the unnatural 0) > > > > They are not a part of the strings, but only arguments to the function > > which defines the list. > > My statement is nevertheless correct. Not if constrained to string members. mmm... wmmm... wwmmm... wwwmmm... wwwwmmm... .... ........ Is essentially the same list, but without the numbers. |