Re: how

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
except maybe ω is humongous,
whatever "humongous" may happen to mean.
You (WM) probablvy decided that
because,
once, you saw something like
1, 2 ,3 , ..., ω, ω+1 ω+2, ...
Whatever it might mean
to put ω and 1 on the same line,
_what we mean_ by ω is
the least.upper.bound of numbers which are
different.in.size from nearest.neighbors.
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.
If  n < ω
then  2⋅n < ω