In article <e0b8cbfa-70a9-4be3-a08f-117e0af7fc9b@q12g2000yqj.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 21 Jun., 22:17, Virgil <Vir...@home.esc> wrote: > > > > > > > But the set of all diagonals constructed according to the scheme given > > > above is countable but cannot be listed. > > > > Whatever do yu mean by "the set of all diagonals"? > > If you mean one "diagonal" for each possible list, that set of diagonals > > is not countable. > > If you mean the set of all "diagonals" for a single list, that is not > > countable either, since each of uncountably many permutations of the > > list produces a different "diagonal". > > Take a list of all rationals. Construct, accordingto a given > substitution rule the antidiagonal. Add it at first position to the > list. Construct, according to the given rule the antidiagonal, add > it ... and so on.
You are saying given a list, create its anti-diagonal and prepend it to that list. Repeat with the new list ad infinitum. > > The number of antidiagonal is countable. Nevertheless it cannot be > listed.
I see no problem in listing the set of such anti-diagonals. They can even be listed in the order in which each anti-diagonal is to be created.
Note that the list of anti-diagonals so far created is finite so long as the number of iterations is finite, and only becomes infinite when the process achieves infinitely many iterations, and thus it is countable.
Which such anti-diagonals does WM claim remain unlisted? > > > > > > > > > > Cantor has shown that the set of all reals cannot biject with the > > > > naturals, ergo, the set of all reals is NOT of countable cardinality. > > > > > The same proof shows that some countable sets are of not countable > > > cardinality. > > > > I have yet to see such a "proof" > > That is impossible, because you will not recognize it.
Since WM's such proofs always take place in an environment incompatible with standard set theories such as FOL+ZFC, they are not binding on such theories. > > But for the lurkers: > The complete infinite binary tree contains, by definition, all real > numbers between 0 and 1 as infinite paths, i. e., as infinite > sequences { 0, 1 }^N of bits. > > > 0, > / \ > 0 1 > / \ / \ > 0 1 0 1 > / > 0 ... > > > The set { K_k | k in N } of nodes K_k of the tree is countable. > > > K_0 > / \ > K_1 K_2 > / \ / \ > K_3 K_4 K_5 K_6 > / > K_7 ... > > The tree is constructed by extending the configurations B_j as > explained below: > > _________________ > B_0 = > > K_0 > _________________ > B_1 = > > K_0 > / > K_1 > _________________ > B_2 = > > K_0 > / \ > K_1 K_2 > _________________ > B_3 = > > K_0 > / \ > K_1 K_2 > / > K_3 > _________________ > > B_4 = > > K_0 > / \ > K_1 K_2 > / \ > K_3 K_4 > _________________ > ... > _________________ > B_j = > > K_0 > / \ > K_1 K_2 > / \ > K_3 K_4 ... > ... > ... K_j > _________________ > ... > _________________ > > The complete infinite binary tree (including all those infinite paths > which consist of nodes and edges only) is constructed by a countable > number of steps. In no step more than one infinite paths is extended. > Hence there are not more than countably many infinite paths.
I have no idea what sort of definition of "countability" of infinite sets that WM is using, but it is necessarily nonstandard, and thus is irrelevant here if it does not apply precisely to those sets which may be put into bijection with the set of naturals and no others.
> Nevertheless these paths cannot be enumerated.
Thus, by any STANDARD definition of "countable", the set of them is NOT a countable set.
By definition, a set which cannot be ennumerated (or listed) is NOT countable by the definition which defines countability of infinite sets as having a bijection with the set of naturals (also called a listing, or an ennumeration).
