Re: 2N=E

On 10/25/2024 12:59 PM, WM wrote:

On 25.10.2024 18:54, Jim Burns wrote:

On 10/24/2024 10:29 AM, WM wrote:

The set, when existing completely,
covers an interval, namely (0, ω).
When its density is halved
while the number of elements is constant,
then its extension is doubled.

>
No.
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ is finite.

>
The extension is doubled in this finite set here

Irrelevant.
⟨0,1,...,n-1,n⟩ = ⟦0,n⟧
⟨0,1,...,n-1,n,n+1,...,n+n-1,n+n⟩ = ⟦0,n+n⟧

The set,[...] namely (0, ω).

⟦0,n⟧ ≠ ⦅0,w⦆
⟦0,n+n⟧ ≠ ⦅0,w⦆

as well as in my infinite set.

No.
⟦0,n⟧ ⊆ ⟦0,w⦆  ⇔
⟦0,n+n⟧ ⊆ ⟦0,w⦆
'Infinite' does not mean
what you (WM) want it to mean.