Date: Feb 7, 2013 2:06 PM
Subject: Re: Matheology § 222 Back to the roots

On 7 Feb., 19:46, William Hughes <> wrote:
> Gosh, you are really running away
> from the fact that induction can
> show d is not in the list.

Induction can show that *your* d does not exist. Therefore it cannot
be in any list.

But if you neverthelss assume that your d exists, then among *your*
finite terms a_n = (a_n1, ..., a_nn) can (and must) also exist
infinite finite terms. Something that probably will easily be
deparadoxided by the subuncountability of supercountable

Regards, WM