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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Sat, 2 Nov 2024 22:34:44 +0100
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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