An uncountable countable set
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From: mueckenh on 5 Jul 2006 17:15 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. > > > > >But to answer you argument of the missing limit: Cantor's diagonal has > >also this very same problem. 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. > > >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). > > 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. Regards, WM
From: mueckenh on 5 Jul 2006 17:18 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. Regards, WM
From: mueckenh on 5 Jul 2006 17:26 Dik T. Winter schrieb: > You consider aleph_0 to be a number. I think this is a problem, but I let > it pass. It is not a number, but it is said to be a number. > > > And to > > make sure that every digit is covered by your proof, you must find out > > whether "if" or "else" has to be applied. > > Why? We know that either "if" or "else" has been applied, that is sufficient. The same is with my transpositions. > > > > > Leave Cauchy out of the play. There is no justification to reach all > > > > last digits of the diagonal number by his epsilons. His specification > > > > would only show that all columns of the list up to nn have been > > > > covered. > > > > > > Apparently you do not understand what Cauchy involves. > > > > Do you know it better than you know Cantor's writings? > > > > There is the positive epsilon arbitrarily small. But every epsilon > > covers infinitely many digits of Cantors diagonal. To follow Cantor, on > > the other hand, *every* digit has the same weight as the first one. > > There is no limit-process at all. > > Apparently you do not know Cauchy well. According to Cauchy, any number > that is written decimally has a limit, and that limit is a real number. > And, no, I have not read much from Cantor, so I have no idea how I should > interprete your representation here. But what you claim above to be Cauchy's is exactly Cantor's (and before him Meray's) theory of irrational numbers. Not Cauchy's. He is that guy with the epsilon. > > > > The second sequence of transpositions defines the second "infinite > > > permutation", but that one overlaps the first "infinite permutation". > > > > No problem. For a given well-order to start with, every result is > > determined. > > When you use finitely many overlapping sequences of "infinite permutations". When I stay within the domain of finite numbers. > > > > > This single definition leads to infinitely comparisons, unless the > > > > numbers in the list are all known. > > > > > > By representing a list it is assumed that all members are known. If they > > > are not all known, it is not a list. > > > > The given well-order to start with is also known, and every result is > > determined. > > For finitely many "infinite permutations". Just like it is for a list > (which is a map from N to something, where N contains finite elements only). If the transpositions are well-ordered, they are indexed by finite numbers - up to the last one. Regards, WM
From: Virgil on 5 Jul 2006 18:06 In article <1152132958.281422.235990(a)m79g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > On the other hand, let f map R on P(N) with the only condition on f be > > that f map a non-natural to K= {n in N: n !in f(n)}, then f can quite > > easily both exist and be a bijection. > > That is silliest simplicity. Try this: K = {x in R: x notin f(x)},. Can you show that for an arbitrary function f:R --> P(N). such a set actually exists as a member of P(N), i.e., K being a subset of N? I think not! Note that to be a member of P(N), a subset of N, it must be a set of only natural numbers, and not contain any other sorts of numbers.
From: Virgil on 5 Jul 2006 18:13
In article <1152133235.172340.249300(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > What I think is that there has to be a function exactly defined before > > K exits and when you have defined that you already have that function, > > so K is not in the range of the function. The P(N) is different because > > it contains all possible K:s. > > Insn't that a bit too much naive intuition? That is hard logic. Which is possibly why "mueckenh" can't deal with it. |