An uncountable countable set
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From: mueckenh on 6 Jul 2006 09:13 Dik T. Winter schrieb: > In article <1152134107.848676.131670(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > David Hartley schrieb: > ... > > > So you acknowledge that your process doesn't give, in the limit, a > > > bijection N -> Q+ ? > > > > Such a bijection does not exist, because there is no smallest positive > > rational. > > There is no need of such for a bijection. Consider the set of positive > numbers P and the set of negative numbers N. Clearly there is a bijection > from P to N: f(x) = -x. But N has no smallest element. A bijection is > not necessarily order preserving. You are right. But the transpositions which I defined are order preserving. Regards, WM
From: mueckenh on 6 Jul 2006 09:16 David Hartley schrieb: > > Infinite processes can be brought to an end if they have a clearly > defined limit. You haven't defined what you mean by the limits of the > orderings or mappings produced by your process. To make "this result > occur unavoidably" you need to give definitions of both limits and show > that the limit of the mappings is an order-isomorphism from N to the > limit of the orderings. Using the definitions I offered, the sequence of > orderings does have a limit, and it is the usual ordering of the > positive rationals. The limit of the associated sequence of mappings > from N would be an order-isomorphism if it existed but - no surprise - > it doesn't. The limit is clear. The question is why it is not reached. > > Unless you at least try to define these limits, and to prove the > required relationship between them, I see little point in continuing > this discussion. Here is a set of transpositions of order type omega. Therefore we stay once and for all in the finite natural numbers. Start with transpositions of the indices: (1, 2) (2, 3) (1, 2) This orders the set of the first three elements by size. Then order the set of the first 4 elements by size. (1, 2), (3, 4) (2, 3) (1, 2) And so on. After the set of the first n elements has been ordered by size order the set of the first n+1 element by size. At most n-1 transpositions are required for a set of n elements. Now, the complete set of transpositions is of order type omega. And it maintains the well-order by index while it achieves well-order by size. Regards, WM
From: mueckenh on 6 Jul 2006 09:19 Virgil schrieb: > > > Since any line and the following line both act on the same elements, > > > they cannot be applied simultaneously, and they do not "commute", but > > > must be applied sequentially. > > > > What has commutation to do with this proof? > > Absence of commutativity, which is the case with certain transpositions > and sequences of transpostions, means that they must be applied in > sequence and not simultaneously as "mueckenh"'s theory requires. They are not "applied" at all but are given in zero time. The set of transpositions is predetermined like the solution of a solvable mathematical problem. It may take a long time to find it, but it is existing as son as it is posed. Regards, WM
From: mueckenh on 6 Jul 2006 09:23 Virgil schrieb: > Cantor's "diagonal has no problems of validity, only problems of > comprehesibility. "Mueckenh", for example, hasn't a clue. Are there in fact people who can believe that anybody was unable to understand Cantor's arguing? Was it such a problem for you to comprehend it? > > > > > > > >Remember, here we have the same answer as in Cantor's diagonal: All > > > >natural numbers are finite. Though there are infinitely many, each one > > > >is finite. Hence even the smallest positive rational is in one-to-one > > > >correspondence with a finite natural number. There is no natural number > > > >oo (like in Cantor's list). > > Cantor's anti-diagonal rule does not have any oo in it. So I said. > > > > > > I thought your reference above to a "smallest positive rational" was a > > > joke, you don't really believe there is such a thing, do you? > > > > No. Therefore it cannot be indexed by 1. > > > > > The rest > > > of the statements in this last paragraph are simple truths (unless > > > you're claiming "Cantor's list" contains oo), but don't answer anything, > > > as far as I can see. > > > > They answer your question about the problem witn n --> oo. n is always > > finite. > > The symbolism "n --> oo" does not require any n's other than finite > ones, so no such "problems" exist. So I said. Regards, WM
From: mueckenh on 6 Jul 2006 09:43
Virgil schrieb: > > > Having infinitely many does not require that any one of them be > > > infinitely large. > > > > > > And each of the infinitely many naturals is only finitely large. > > > > Either the diagonal number 0.111... is not distinguished from all > > finitely large numbers of the list > > 0. > > 0.1 > > 0.11 > > 0.111 > > ... > > then Cantor's proof fails. > > > > Or 0.111... is distinguished from all finitely large numbers of the > > list > > 0.1 > > 0.11 > > 0.111 > > ... > > then the digits of 0.111... cannot all be indexed by natural numbers. > > OR, as is actually the case, the endless sequence of 1's fraction > 0.111... is distinct from every finite truncation of it AND every digit > of it CAN be indexed by a natural number. Or 10 is smaller than eleven and 10 is larger than eleven. "To be different" means for all unary representations of n An : 0.111... - n =/= 0 Every digit can be indexed by a natural number means that there is no digit, which is different from all unary representations of natural numbers. By construction of 0.111... this means further that every digit at position n and all digits at smaller positions can be indexed. This means there does not exist any digit which differs from all unary representations. not forall n: 0.111... - n =/= 0 En : 0.111... - n = 0. Or is your logic unable to transform "no sequence of 0.111... differs from all sequences 0.111...1" into "there is at least one sequence 0.111..1 which is equal to 0.111..."? Regards, WM |