Re: abundance of irrationals!)

In article <1115726749.843879.120570@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Virgil wrote:
>
> > > Then consider this list 2 please:
> > >
> > > .
> > > 0 1
> > > 0 1 0 1
> > > 0 1 0 1 0 1 0 1
> > > ...
> > >
> > > and read from top to bottom (like the Chinese do or like we do in
> order
> > > to read the diagonal number of a list).
> > >
> > > Then you find ALL the bits of 1/3 = 0.010101... in the positions
> > > 1,2,3,6,11,22,... The recursive definition of that sequence a_n
> (with n
> > > the line number and a the position in the line) is
> > >
> > > a_1 = 1 (Position of the first 0 after the point)
> > > a_n = 2*a_(n-1), if n is even,
> > > a_n = -1 + 2*a_(n-1), if n is odd.
> > >
> > > In this way ALL reals of the interval (0,1) are given by that list
> 2.
> > >
> > > We have, as easy can be proved:
> > > 1) The set of all bit positions of list 2 is countable.
> > > 2) The number of all bit positions is larger than the number of all
> > > reals in the interval (0,1).
>
> "Larger" means in set theory: "Not smaller than ..."

WRONG! In set theory, A larger than B means that there is an injection
from B to A but no injection from A to B.

And since the 'number' of bit positions is no more than the 'number' of
naturals and the 'number' of reals is at least as large as the 'number'
of naturals, there CANNOT be more bit positions than reals.

So that WM is doubly wrong here.
> >
> > If you think that your point (2) is so easily proved, then prove it.
> > Since you equate the bit positions to the naturals, you proof would
> have
> > to find a mappinig, N -> R, from the naturals to the reals that can
> be
> > SHOWN to be an ONTO mapping.
> > Merely claiming that it can be done isn't good enough. Even James
> Harris
> > can do that.
>
> You do not doubt the countability of my list-2-bit-positions. Therefore
> I need to show the onto-mapping from bit-positions - -> [0,1).
> This is easily seen by the fact that every real deviates from very
> other real at least at one bit-position.

That only shows that bit-positions map onto bit-positions, which is
trivial, not that they map onto all reals, which is false.
>
> Every such real an be formed by the bits available. For every pair of
> reals there must be a first bit where they differ.

So?

Actally, WM has entirely ignored the problem of dual representations.

Among many other things.

WM seems to have incredible ingenuity at creating ways to deceive
himself.
>
> Regards, WM
.

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