An uncountable countable set
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From: Virgil on 5 Jul 2006 19:21 In article <1152133662.805514.156210(a)a14g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1151961303.406453.116760(a)75g2000cwc.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Consider a surjective mapping from N on the set of even positive > > > numbers & K. > > > > Only if we consider a function g and K = K(f). > > > > > Or consider a surjective mapping from R on the set of rational numbers > > > & K. > > > > Only if we consider a function g and K = K(f). > > > > > Or Consider a surjective mapping from R on the set {1, 2, 3, K}. > > > > Only if we consider a function g and K = K(f). > > > > > > > > > > But without readjusting f by g. That is forbidden in case of Cantor's > > > diagonal as well as in our case. > > > > Then "mueckenh" does not understand either 'Cantor's diagonal' or your > > case. > > > > In the Cantor "diagonal" on has a choice of uncountably many diagonals > > rules to choose from, any one of which is sufficient to prove the > > incompleteness of the list. > > Wrong. There are no uncountable sets. There are in ZFC and NBG. Uncountably many of them.
From: Virgil on 5 Jul 2006 19:25 In article <1152133786.204995.294260(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > Wrong. The scheme is a fixed matrix and each line can be applied > > > simultaneously. > > > > > > (1,2), (3,4), (5,6), ... > > > 1, (2,3), (4,5), (6,7), ... > > > (1,2), (3,4), (5,6), ... > > > 1, (2,3), (4,5), (6,7), ... > > > (1,2), (3,4), (5,6), ... > > > 1, (2,3), (4,5), (6,7), ... > > > > 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.
From: Virgil on 5 Jul 2006 19:29 In article <1152133914.661066.91950(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > 0.111.. is not in the list, then it must have more digits than can be > > > > > indexed (and hence, can exist). > > > > > > > > They are satisfactorily indexed by the infinite set of finite natural > > > > numbers, N. > > > > > > You just proved that there are infinitely digit positions which are not > > > indexed by natural numbers (*all* of which are given in the list). > > > 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. So that the actuality is that both of "mueckenh"'s alternatives are wrong simultaneoulsy.
From: Virgil on 5 Jul 2006 19:39 In article <1152134107.848676.131670(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Hartley schrieb: > > > In message <1151961927.534777.45250(a)v61g2000cwv.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de writes > > > > > >David Hartley schrieb: > > > > > > > > >> If this is defined for every element of A, we can say that the sequence > > >> has a limit, and it is f. > > >> > > >> > > >> Now, granting that your process works as you hope, you get a sequence of > > >> well-orderings of the rationals whose limit is the usual ordering. But, > > >> the corresponding sequence of bijections N -> Q does *not* have a limit. > > >> (You need only consider the images of 1, these diverge to -oo.) > > > > > >As I only consider positive rationals, these go to the smallest > > >positive rational, but that is the same problem. > > > > 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. Such bijections do exist, but are not an order isomorphisms, merely a bijections. > > > > > > > >But to answer you argument of the missing limit: Cantor's diagonal has > > >also this very same problem. Cantor's "diagonal has no problems of validity, only problems of comprehesibility. "Mueckenh", for example, hasn't a clue. > > > In case of the Cauchy-limit of an > > >irrational number, we see (a_k)*10^-k disappear, because it becomes as > > >small as desired. In the Cantor-diagonal however, there is always a > > >digit with the same weight and importance as the first one, even for k > > >diverging to oo. > > > > This is a complete non sequitur, as well as nonsense. Even if there were > > a problem with the diagonal argument, > > It is. > > > it would not alter the simple fact > > that: > > >> Your > > >> example does not "preserve the bijective mapping between N and Q". > > But it would, if actual infinity existed. It doesn't in ZFC of NBG where actual infinities actually exist. > > > > >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. > > > > 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. Except possibly in "mueckenh"'s head.
From: Virgil on 5 Jul 2006 19:43
In article <1152134296.084553.209780(a)a14g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Hartley schrieb: > . > > > > > >The bijection (one-to-one corespondence) between rationals and naturals > > >is preserved. This is if not a well-order so at least more then set > > >theory can supply for the natural order of Q. > > > > Your original claim was that your process would produce a bijection from > > the naturals to the rationals which still respected the ordering, which > > is now the usual one. This would of course be the contradiction you are > > looking for, but you have not offered any attempt to prove it. > > If infinity existed and if infinite processes could an be brought to an > end, then this result would unavoidably occur. But it does not because > it cannot because there is no actual infinity. Perhaps not in "mueckenh"'s world, but in the world of ZFC and NBG, and a number of other worlds, "actual" infinities actually do exist.. That "mueckenh"'s world is so small, is only "mueckenh"'s problem, no one else's. |