Re: 2N=E

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