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.
The number of antidiagonal is countable. Nevertheless it cannot be listed. > > > > > > 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.
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:
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. Nevertheless these paths cannot be enumerated.
