Re: 2N=E

On 29.10.2024 18:19, Jim Burns wrote:

On 10/28/2024 4:01 PM, WM wrote:

Relevant is only that
the density in the interval is halved,
the number remains,
the interval is doubled.

 For each finite ordinal n
there is a larger finite double n+n
 For each finite double n+n
there is a larger finite ordinal n+n+1
 The finite ordinal interval and
the finite double interval are the same.
 'Infinite' does not mean what you want it to mean.
 

I have no preference.
If infinity is complete, the we can double all natural numbers with the result
(0, ω)*2 = (0, ω*2). The some products are in the interval (ω, ω*2). Then your claim concerns only the elements of the potentially infinite collection of definable numbers.
If infinity is only potential, then your claim concerns all numbers. Reason: When all existing numbers are doubled then larger numbers are created but those can be natural numbers because the multiplied first set was not complete.
Note: It is impossible that all doubled numbers of an interval are elements of this interval. If someone claims this, then he is a fool.
Regards, WM