fom
Posts:
1,969
Registered:
12/4/12


Re: Matheology § 233
Posted:
Mar 31, 2013 9:34 AM


On 3/31/2013 3:31 AM, WM wrote: > On 31 Mrz., 02:56, Virgil <vir...@ligriv.com> wrote: >> In article >> <2bc13fff5cbb43dda06a218c68c99...@m9g2000vbc.googlegroups.com>, >> >> >> >> >> >> WM <mueck...@rz.fhaugsburg.de> wrote: >>> On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote: >> >>>>>> 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? >> >>>> The set of paths in any Complete Infinite Binary Tree which agree with >>>> any particular path up to its nth node is equinumerous with the set of >>>> all paths in the entire tree i.e., is uncountably infinite. >> >>> This was the question: In a list containing every rational: Is there >>> always, i.e., up to every digit, an infinite set of paths identical >>> with the antidiagonal? Yes or no? >> >> Lists and trees are different. And antidiagonals derive from lists, not >> trees. >> The entries in list are well ordered. >> The entries in a Complete Infinite Binary Tree are densely ordered. > > The entries in form of nodes are well ordered. Anything else is not > added to the list. > >> Those order types are incompatible. > > This was the question: In a list containing every rational: Is there > always, i.e., up to every digit, an infinite set of paths (rational > numbers) identical with the antidiagonal? Yes or no?
As noted before, WM's questions often make no sense.
As explained before:
> But note that the question also demonstrates WM's > complete lack of understanding of the diagonal > argument. > > He has been told time and time again that it is > an argument scheme which only has application > under certain assumptions. > > He chooses to believe otherwise for the agenda > of his fanaticism. > > Suppose one is given a countable listing of > the rationals (with the appropriate restriction > on double representation) according to the > infinite listing of an expansion. > > Suppose one performs a diagonalization on > that listing. > > Is the resultant a rational number? No. > > What may be concluded? That the rational numbers > do not exhaust the capacity of the algorithm > to generate representations if that algorithm > is to generate a representation for every > rational number. > > Although the burden of proof lies with WM > concerning the nature of the diagonal, it > is a simple matter to understand if one > uses a Baire space representation instead. > > In the Baire space, rationals are in correspondence > with eventually constant sequences. Since, > by construction, a list of Baire space rationals > would exhaust all of the eventually constant > sequences, the resultant of a diagonal argument > could not have an eventually constant sequence > unless the original premise had been false.

