Re: abundance of irrationals!)

"r.e.s." <r.s@xxxxxxxxxxxxxxxx> wrote in message news:<xHe7e.5724$lP1.2862@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
> "*** T. Winter" <***.Winter@xxxxxx> wrote ...
> > mueckenh@xxxxxxxxxxxxxxxxx (W. Mueckenheim) writes:
>
> > > SUM 1/2^k < 1
> > > k e N
> >
> > Wrong when you use the standard definition of that infinite sum.
> > Because that is lim {n -> oo} sum {k <= n} 1/2^k. But you refuse
> > to accept that definition, so you are using your own definition.
> > Are you finally willing to reveal how *you* define that infinite sum?
>
> Since making the above claim, W.M. has steadfastly *refused* to
> state the definition he is using for SUM[k in N](1/2^k). Isn't
> it reasonable to suppose he *has* no definition, and does not
> understand what he himself has written?

I use precisely that definition which Cantor uses to construct his
antidiagonal. Take a number from each line which is indexed by a
natural number (but not by oo). Hence do not perform the limit
process. Take Sum b_k * 10^-k without letting k become actually
infinite.

If, however, a limit process is agreed upon to be automatically
performed for Cantor's antidiagonal, then it must also be assumed for
the binary sequence
0.1;
0.10; 0.11;
0.001; 0.101; 0.011; 0.111;
....

Then all the reals of (0,1) are members of the sequence, hence, are
countable and well-ordered.
Because: For any epsilon > 0 there is a number of the sequence which
is less than this epsilon apart from a given irrational number.

What is the difference?

Regards, WM
.

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