An uncountable countable set
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From: mueckenh on 7 Jul 2006 10:36 Virgil schrieb: > In article <1152182749.926627.310460(a)a14g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > > All sequences, even all single transpositions which I apply can be > > > > enumerated by natural numbers, just like the lines of Cantor's list. > > > > > > Yup, so there are infinitely many. > > > > Are yo sure? How could an enumeration by natural numbers supply > > infinity? > > If an enumeration rule exhausts the naturals, the set being enumerated > is Dedekind infinite. > > > > > There are only finite natural numbers. > > But infinitely many (an endless supply) of those finite natural numbers. That does not make their sizes infinite. It is completely irrelevant here. > > > In addition to cardinality we have to do here with ordinality. The > > > sequence > > > of transpositions you give has order type w * w. > > > > No. There is a first element and there remains a first element. > > The order type w*w indicates a well ordering, so any nonempty subset > has a first element. But this is not the order type of the set being > permuted, but of the set of permutations being applied to it. > > > > > > I have shown that the transpositions can be enumerated by natural > > > > numbers. There is no number oo. There is neither such a number in > > > > Cantor's list. All we do is enumerated. Othewise it would not be > > > > defined at all. Neither in Cantor's list. Therefore it is irrelevant > > > > whether or not something has to be "executed in order". We are in the > > > > countable domain and do not leave it. > > But transpostitions do not always commute: As right operators on a list > (a b) (b c) = (a c b) but (b c) (a b) = (a b c) > > So changing their order of execution can change their effect. Why should we change their order? > > > > > > That makes no sense, again. The transpositions have to be execute in the > > > order given. > > > > Of course. But that does not exclude that these transposition can be > > executed and finished (if Cantor's list can be finished). > > But it prohibits them from being executed out of their prescribed order. > Since the order of execution induces a well ordering of the set of > transpositions, there would have to be a first transpostion producing > any given effect. What is the first transposition on a well ordering of > the rationals producing an ordering that is dense? What is the first real number which cannot be named by a finite set of letlers? There are such numbers. Which is the first, e.g., by size, or in an assumed well-ordering of the reals? Regards, WM
From: mueckenh on 7 Jul 2006 10:38 Virgil schrieb: > In article <1152183139.346593.62140(a)m73g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > Anyone is quite free to reject any axiom set, but no one is free to > > > impose any prohibition of any axiom set on others. > > > > Unless there appears a conradiction. > > Not even then. Though if there were any contradiction WITHIN an axiom > system there would be little point in pursuing that system further. What axiom is not obeyed by my transpositions? Regards, WM
From: mueckenh on 7 Jul 2006 10:41 Virgil schrieb: > In article <1152183314.431125.225030(a)m79g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > To say n --> oo does not require that there exist any "oo". > > > > > > It is merely an abbreviation for "as n increases unboundedly beyond any > > > given natural". > > > > But always remaining a natural itself, yeah. And the number of naturals > > < n is always a finite number, so that it never does reach infinity > > too. > > But it reaches past every finite value!!! There is always a finite value past every finite value. So we are and remain sufficiently save within the domain of finit values. Regards, WM
From: mueckenh on 7 Jul 2006 10:45 Virgil schrieb: > > There is not a remote member, but each natural is finite. Therefore, > > how far we may go, there is never an infinite number of transpositions > > accumulated. And the Cantor diagonal has not an infinite number of > > digits. > > But it is provably different from each and every number in the list. > If "mueckenh" disputes this, he must find some listed number equal to > the number described by any of Cantor's diagonal rules, or at least show > that some such number must exist. By construction of the list we have to compare K = 0.111... only with numbers 1 0.1 2 0.11 3 0.111 .... n 0.111...1 .... Those are unary representations of 1,2,3,..., which we may abbreviate conveniently by n. There are (by tertium non datur) two possible alternatives: 1) K has a 1 at a position p where no n has a 1. E p A n : p =/= n 2) K has not a position p where no n has a 1. ~(E p A n : p =/= n) If (1) is true, then the position p in question cannot be enumerated by a natural number, hence it is undefined. If (2) is true, then we observe that for any position where the 1 coincides in p and n also all positions with m < n do coincide. The case you mentioned above that position n may coincide but position n-1 may not, simply cannot occur. We get from (2) A p E n : p = n and by construction of n = 0.111...1 we obtain that all 1's in K coincide with the 1's of n. This case would invalidate Cantor's diagonal proof. > And after than, Mueckenheim must show why Cantor's first proof is invalid No problem, that is easy to do. > and also why Cantor's proof that no set surjects to its power set is > invalid. That has been done already. > And then Mueckenheim can take on all those many other proofs of > the uncountability of the reals. > > Only when he has finished all of that can Mueckenheim claim that the > Cantor theorem is invalid. There are no further proofs but only some variants of three proofs mentioned. Even the second and the third are fairly similar. Regards, WM
From: Dik T. Winter on 7 Jul 2006 11:02
In article <1152282125.757193.28320(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > Therefore: As my sequence of transpositions > (1, 2) > (2, 3) > (1, 2) > (3, 4) > (2, 3) > (1, 2) > ... > determines exactly which transposition will be executed as the n-th, > everything is determined and everything is finite. > > > > And again, it maintains the well order as long as you remain in the finite > > domain. So up to every finite n, > > THERE IS NO OTHER n ! ! ! Indeed. > > you have ordered the first n elements > > of the well-ordered list, retaining a finite list. And back again to > > problem 1 with this. How do you define that when n grows without bound? > > Also without bound n is always finite, or it would not be a natural > number! Ok, good, so your sequence of transpositions will never give the standard order of the rationals (as you claimed). > > How do you define the "limit"? And if you define that, is that "limit" > > also well-ordered? Those are things you have to prove. > > There is no limit other than in Canrtor's diagonal. EVERY set initially > indexed by natural numbers will remain indexed by natural numbers. And > it will unavoidably become ordered by size too. There *is* a limit, as I wrote already: > > > But Cantor's arguing is that without epsilon. > > > > I do not know Cantor's argument exactly. I think that he implicitly > > uses that result. In my formulation it was abundantly clear that I did > > use it. > > Cantor argues: If for all natural numbers something is defined (like > the bijection n <--> 2n or the replacement of 5 by 4) then there is no > further reason for any limit process. Cantor in his paper about the > diagonal argument does not at all consider any limit process. How then does he show that the diagonal he obtains is a real number, without (implicitly) using a limit? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |