
Re: Matheology § 210
Posted:
Feb 9, 2013 6:09 AM


On 9 Feb., 00:01, Virgil <vir...@ligriv.com> wrote:
> > It is simple to prove it in mathematics. Every term of the sequence > > 2^n is finite. And there is no infinite level. > > Every f 2^n is exceeded by most others so there cannot be a last member, > which requires more than any finite number of finite levels.
There need not be a last member in order to see that all are finite. Or do you believe that there is an infinite natural number?
All crosssections 2^n are natural numbers. So the sequence of cross sections is a subsequence of 1, 2, 3, ... and has the same limit as the latter, namely the ordinary infinite, called oo and later, by Cantor, called omega and even later by Cantor called aleph_0. > > What term or terms does WM want to use for > "more than any finite number finite levels"?
The term is infinity, the limnit is the same (improper limit) as of the supersequence 1, 2, 3, ... of 2, 4, 8, ..., denoted by oo.
Regards, WM

