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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: 2N=E
Date: Sat, 2 Nov 2024 20:17:16 -0400
Organization: i2pn2 (i2pn.org)
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On 11/2/24 5:34 PM, WM wrote:
> 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.

It *IS* true in a set without dark numbers, and every Natural Number is 
not dark, so we don't need your "dark numbers", they are just your cruch 
to handle the fact that you logic can't actually handle the full 
infinite set of Natural Numbers.

>>
>> ⎜ ⟦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.

But it can't be, becuase if w/2 was finite, and thus had a finite value 
of x. Then 2*x would be infinite, and thus there were less than 2*x 
Natural Numbers, and thus not an infinite number.

Remember, 2 times any Natural Number is a Natural Number, provable by 
mathematics.

>> '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.

And thus is not finite.

Your logic is just full of contradictions, as your system has blown 
itself, and your mind, to smithereens, leaving the dark hole behind it.

> 
> Regards, WM
>