Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

On 4/26/2024 10:37 AM, WM wrote:

Le 26/04/2024 à 01:11, Jim Burns a écrit :

On 4/25/2024 4:03 PM, WM wrote:

If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.

>
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n

>
That is true for definable numbers
but not for the last numbers before ω.

If any number below n canNOT be counted to from 0
then n itself canNOT be counted to from 0
Thus,
each number which CAN be counted to from 0
is not above
any number which canNOT be counted to from 0
By definition,
ω is between
numbers which CAN be counted to from 0 and
numbers which canNOT be counted to from 0
Imagine being someone who denies that definition of ω
Because the following isn't a claim about ω
you (the denier) should still admit:
if n can be counted to from 0
then n⋅2 can be counted to from n
then n⋅2 can be counted to from 0 (through n)
If ω exists as defined,
then doubling never crosses ω
(from CAN to canNOT)
Even if ω doesn't exist as defined,
then doubling never crosses
_where ω would be if ω existed_
(from CAN to canNOT)
ω is NOT a simply.humongous.instance of
the numbers 0 1 2 3 ...
ω marks a boundary between domains with
different descriptions (CAN and canNOT).
Imagine being someone who denies that
ω marks that boundary.
With or without the marker,
the domains (CAN and canNOT) remain
the domains (CAN and canNOT).