In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> 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 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?
WM is the only one who equates the finiteness of all members with the finiteness of the set of all members. Consicer the real inteval [0,1].
Does the finiteness of the members of that set establish the finiteness of the set itself? NO!
But for any ordered set there must be provably a last member to prove the set to be finite. > > All cross-sections 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 it is called is irrelevant once its existence is established, and it has been. > > > > 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.
Actually, proper grammar, at least in English, requires that the term be "infinite" not "infinity" --