Sujet : Re: 2N=E
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 25. Oct 2024, 19:30:58
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <d781add9-cd6b-4fff-9601-111c74f4ae32@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
User-Agent : Mozilla Thunderbird
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.
…