Re: 2N=E

On 02.11.2024 21:53, Jim Burns wrote:

On 11/2/2024 1:24 PM, WM wrote:

On 02.11.2024 13:20, joes wrote:

 

Why "absorbed"?
Do you think
some multiple of a power of 2 is not natural?

>
If all multiples of 2 smaller than ω are doubled,
then this doubling results in larger numbers than doubled.

 If n is finite  ∧  ω ≤ n+n
then ω is finite.

That might appear so in a set without dark numbers. It is not true when dark numbers come into play.

 ⎜ ⟦0,n⟧ is finite, which means that
⎜ each subset of ⟦0,n⟧ is two.ended.or.{}
⎜
⎜ Each subset of ⟦n,n+n⟧ is two.ended.or.{}

Yes, but the order of dark numbers cannot be determined.
Every number n < ω is finite. ω/2 is finite.

'Finite' means
each subset is two.ended.or.{}
'Infinite' means not finite.

Every interval (0, n) is finite because n is finite. But for dark numbers n this cannot be seen. The dark realm appears as infinite. It cannot be counted through.
Regards, WM