In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 7 Apr., 12:48, William Hughes <wpihug...@gmail.com> wrote: > > On Apr 7, 12:29 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 6 Apr., 23:17, William Hughes <wpihug...@gmail.com> wrote: > > > > > > On Apr 6, 11:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 6 Apr., 22:30, William Hughes <wpihug...@gmail.com> wrote: > > > > > > > > My claim is: > > > > > > > > Let D be the set of all lines, and let > > > > > > E be any one finite subset of D. > > > > > > Then D\E is not empty. > > > > > > > Your claim is true. > > > > > > I also claim: Let D be the set of all > > > > lines, then if you remove any one finite > > > > subset of D there is something left. > > > > Do you accept this claim? > > Yes or no.- > > Of course. Yes. But what has it to do with the fact that either D is > more than all its FISs
Does WM claim that D is NOT equal to the union of the set of all its FISs?
> (then all lines of our list can be removed > without removing |N and Cantor's proof fails), or not, then Cantor's > proof holds and my counter proof holds too, since then all FISOns > include the union over all FISONs.
Wm does not have a "counterproof" to the CDA (Cantor's diaganal argument). What he claims to be a counterproof is an irrelevancy.
This is the "theorem" that WM claims disproves the CDA (Cantor Diagoanal Argument):
WM's Theorem: Consider a list that contains a complete sequence (q_k) of all rational numbers q_k. The first n digits of the anti-diagonal d are d_1, d_2, d_3, ..., d_n. It can be shown *for every n* that the Cantor-list beyond line n contains infinitely many rational numbers q_k thathave the same sequence of first n digits as the anti-diagonal d.
In the first place, unless those rationals are restricted to non-negative proper fractions, each expansion will have to allow for both a sign and a radix point, as well as digits.
And it is extrememly unlikely that, even with sucht restrictions, the "anti-diagonal" will be a rational. And since it only deals with rationlas, and neither reals not binary sequnces, it can say nothing about either sequnces of reals or sequences of binary sequnces.
And finally, the diagonal as specified by WM is with probabl;ility 1 going to be irratinal, not rational.
And generally, it is easily seen to be the claim of a mathemtaical incompetent. > > But in no case Cantor's argument remains valid.
In no case has WM shown that Cantor's argument is other than valid.