Re: abundance of irrationals!)

Randy Poe said:
>
> 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
>
>
Excuse me, but the nodes of an infinitely deep binary tree can obviously be
enumerated in a linear manner, corresponding to binary integers and thus to the
naturals. If you want to complain that it would require infinite digits for
most values, that's irrelevant. it's still an enumeration of the reals.
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.
--
Smiles,

Tony
.

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