Re: how

Le 14/04/2024 à 22:39, Jim Burns a écrit :

On 4/14/2024 3:12 PM, WM wrote:

Le 13/04/2024 à 21:16, Jim Burns a écrit :

 

if  n < ω
then  2⋅n < ω

>
That is impossible because
doubling is a linear operation.

 You (WM) have decided that
ω is like all the numbers n < ω

Cantor has decided, that ω is an ordinal which can be counted and passed by counting (Hilbert).

 Whatever it might mean
to put ω and 1 on the same line,

It is Cantor's number classes. See Transfinity p. 42.

 If n is a number
different.in.size from its nearest.neighbors,
then 2⋅n is a number
different.in.size from its nearest.neighbors.
 If n is a number less than
  the least.upper.bound of numbers
  different.in.size from their nearest.neighbors,
then 2⋅n is a number less than
  the least.upper.bound of numbers
  different.in.size from their nearest.neighbors.

That is wrong if all natnumbers are present already such that no further natnumbers fits below ω.

 If  n < ω
then  2⋅n < ω

That is true if not all natnumbers are present, blocking all places for finite ordinals.
Regards, WM