Re: Cantor and the binary tree



*** T. Winter wrote:
> In article <1116958479.555107.284630@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> > *** T. Winter wrote:
> > > In article <1116939502.814879.192170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
> > mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > > If we accept that, in binary digits, SUM{n = 1 ... oo} 2^-n =
> > > > 0.111... = 1
> > >
> > > You may note that that is *not* an infinite sum...
>
> Note what I said here. It is not a sum of infinitely many terms.
>
> > > > .
> > > > 0 1
> > > > 0 1 0 1
> > > > ..................
> >
> > Any path is an infinite sequence of bits which by multiplying with 2^-n
> > and summing up establishes an infinite series representing a real
> > number. Every combination of countably many bits is realized by
> > definition.
>
> Mathematics does *not* define summing up infinitely many terms. It uses
> limits in this case.

Does mathematics allow me to write
0.
1
1
1
....
?

> > Why should 0.010101... not exist in that tree? Every path is infinite
> > by definition as is 0.010101..., by definition.
>
> The path does exist, but there is no node at the end. Or if you wish
> there is a node at the and (as J.H. Conway does with his surreal numbers),
> the number of nodes is uncountable.

There can be no node at the end, because there is no end. We do not
need any node at the end in order to show that the set of nodes is
equivalent to that of paths. It is shown by the following steps. Please
point out which step is wrong.

1) Each real number of (0,1) is given by a path stretching over
infinitely many nodes (bits).
2) All nodes (bits) of the tree belong to a countable set.
3) A node can only exist within a path.
4) Any node increases the number of paths by 1 from 1 coming in, to 2
going out. 2 - 1 = 1.
5) Any node increases the number of nodes by 1.

Regards, WM

.

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