On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:
> > > So we have established the fact that an irrational number has no node > > of its own. > > No number in any infinite binary tree has any node "of its own", as > every node has two child nodes belonging to necessarily different > numbers.
That is correct, but only establishes the fact that no actually infinite path can be distinguished from all rational paths as should be possible in a Cantor-list - but is not.
> > The first bit causes that the anti- > > diagonal differs from 2^-1 of all entries and agrees with 2^-1, the > > second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of > > the list agreeing with the anti-diagonal. There is no bit that would > > leave zero entries agreeing with the diagonal, since there is no bit > > at position infinity. > > Thus there is always at least one bit of any listed entry disagreeing > with the antidiagonanl, just as the Cantor proof requires.
In a list containing every rational: Is there always, i.e., up to every digit, an infinite set of paths identical with the anti- diagonal? Yes or no?