Re: abundance of irrationals!)


Russell wrote:

> No. The obvious linear enumeration of *nodes* is not an
> enumeration of the reals. For that, you need to enumerate
> the different paths. That's not merely a rearrangement of
> the enumeration of nodes; if you think it is, then please
> tell us how you associate each node with one and only one
> path. Can't be done.

Tell us how you associate Cantor's antidiagonal with one and only one
line, e.g., where it ends. Cant't be done.

But the nodes can be shown to be not less than the paths. It is not
possible to say which node belongs to some single path. But in
mathematics we know of indirect proves. This runs as follows:
For EVERY real number a_1, a_2, a_3, ..., a_i, ... there is a line n
and a node in line number n, where it deviates from all other real
numbers which have the same bit-sequence a_1, a_2, a_3, ..., a_i, ...
for 1 =< i < n.

There the paths separate. The number of nodes in that line give the
number of different paths up to that line. It is just one more than the
nodes in all forgoing lines. Thsi holds for every natural number n. If
you doubt that, then tell me which number is the first exception.
>
> > Enumeration is irrelevant to the level of infinity, except insofar
as
> we must
> > rearrange the quantities in order to enumerate them linearly. Don't
> look in
> > your bag of axioms for proof of this. It's been conveniently
> excluded.
>
> Yes, it would be rather inconvenient to make that an axiom,
> since it's inconsistent with the other axioms. I think you've
> gotten stuck thinking about rearrangements that turn out not
> to exist in fact.

There is no re-arragement at all. If one believe'2 that all bits of
real numbers can be enumerated by natural numbers. But if you insist
that 1 = 2 is not a contradiction, then get happy with your opinion.

Regards, WM

.

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