Re: abundance of irrationals!)


mueck...@xxxxxxxxxxxxxxxxx wrote:
> Randy Poe wrote:
> > A number x being in your list means that there is m
> > such that the m-th number in your list is x.
>
> No, these numbers exist in the same way as Cantor's antidiaonal.

Then it's not a list of real numbers.

First, Cantor's antidiagonal exists in the sense of NOT
being on the list. So you have constructed a list by
which all real numbers have the property of not being
on the list?

There is one and only one meaning of "a list of real
numbers". That is that you have an association between
every natural number m and every real number a_m on your
list.

If you respond to this remark with "no", as you did:
> > A number x being in your list means that there is m
> > such that the m-th number in your list is x.

then x is not on your list. Because that is the meaning
of "on a list". There is no other.

If you want to come up with some other construction
(some sort of infinitely deep binary tree, according
to comments from others), that's fine. But it's not
a list, and it doesn't advance your argument that
a list can contain all real numbers in (0,1).

- Randy

.