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

Le 26/04/2024 à 20:53, Jim Burns a écrit :

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)

Hence not all natural numbers exist as defined for visible numbers.
Bob does not disappear.
There is a smallest unit fraction.
The intersection of infinite endsegments is not empty.
(0, ω)*2 = (0, ω*2)
There are dark numbers deviating from what you know as natural numbers.
Regards, WM