An uncountable countable set
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From: mueckenh on 27 Jun 2006 16:46 Virgil schrieb: > > Consider one more time the bijective mapping from {1, a} on P({1}) = > > {{}, {1}}. a is a symbol but not a number. There are two bijections > > possible. The set M of all numbers which are non-generators cannot be > > mapped by a number although M is in the image of both the possible > > mappings. > > f: 1 --> {1} and a --> { } with M_f = { }, > > g: 1 --> { } and a --> {1} with M_g = {1}. > > {x in {1,a}: x not in f(x)} = {a}, which is not in P({1}) > {x in {1,a}: x not in g(x)} = {1,a}, which is not in P({1}) > > So your examples are both cooked. The set M of all *numbers* is asked for. Regards, WM
From: mueckenh on 27 Jun 2006 16:50 Virgil schrieb: > It is not "Cantor's list" but ANY list. > You are challenged to find a list which lists all reals, then Cantor > presented a rule for showing any list you provide is incomplete. I have already shown that any list of all or even some rationals and its diagonal number is uncountable. There is no need to consider all reals. > > And each time I implement the diagonal number K(f) into the list f > > (insert it before the first ordinary list number, for instance), you > > will construct another diagonal number K(g) and say, look here, this > > diagonal is not contained in your g. > > As that K(g) was not in M(f) , that is irrelevant. That the diagonal of Cantor's list is not in the list is irrelevant. Regards, WM
From: mueckenh on 27 Jun 2006 16:52 Virgil schrieb: > > But, independent of what were Cantor's actual thoughts, if infinity > > would exist, we could order the rationals by magnitude and maintain the > > well-order. > > Actually, that would require that no infinity exist, or at least that > the set of rational be finite, so that the rationals in their natural > order be well-ordered, as is every finite ordered set. > > That an infinite set of rationals in their usual order are not, and > cannot be, well ordered is trivial: > > In order for a set to be well ordered, every non-empty subset > must have a smallest member > > The non-empty set of positive rationals has no first element, > since for every positive rational, 1/2 of it is strictly between > it and zero. This shows that set theory is self contradictory, because the following mapping does not destroy well-order but establishes order by magnitude. For n = 1 to oo: (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, q_2n) --> (q_2n, q_2n-1) For n = 1 to oo: (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) And again: For n = 1 to oo: (q_2n-1, q_2n) --> (q_2n-1, q_2n) if q_2n-1 < q_2n, else: (q_2n-1, q_2n) --> (q_2n, q_2n-1) For n = 1 to oo: (q_2n, q_2n+1) --> (q_2n, q_2n+1) if q_2n < q_2n+1, else: (q_2n, q_2n+1) --> (q_2n+1, q_2n) .... repeat aleph_0 times. Of course we can also define all mappings in a closed loop, such that they are executed in zero time. Regards, WM
From: mueckenh on 27 Jun 2006 16:54 Virgil schrieb: > If you can describe the number as you have done, you have made it > exist, even if you cannot necessarily compare it for size with another > number. If numbers cannot be compared for size, they do not deserve the name "number". Regards, WM
From: mueckenh on 27 Jun 2006 16:56
Virgil schrieb: > In article <1151319403.823915.220910(a)c74g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb:> > > > > Virgil. Such proofs must be wrong in general, and even the allmighty > > > > could change this fact. > > > > Should read: could not change this fact. > > > > Regards, WM > > Your lack of faith in an Almighty to be almighty, is self-contradictory. I recently proved that there is no Almighty (stone, force). Regards, WM |