Re: 2N=E

On 10/30/2024 11:27 AM, WM wrote:

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.

Your preferences lead you
to treat ω is as though it is finite.

If infinity is complete,
the we can double all natural numbers with the result
(0, ω)*2 = (0, ω*2).

n ∈ [0,ω)  ⇒
∃⟨0,1,...,n-1,n⟩  ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩  ⇒
n+n ∈ [0,ω)
n+n ∈ [0,ω)  ⇒
∃⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩  ⇒
∃⟨0,1,...,n-1,n⟩  ⇒
n ∈ [0,ω)

The some products are in the interval (ω, ω*2).

ω is infinite.
If you declare ω is finite, or
if you declare claims requiring finite ω,
ω does not become only finite.
ω becomes both finite and infinite.
ω becomes not.existing.