Sujet : Re: how
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 03. Jun 2024, 09:57:10
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <6350e436-cfa4-427d-989e-1b0d42814aae@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 6/1/2024 11:15 AM, WM wrote:
Le 31/05/2024 à 21:15, Jim Burns a écrit :
On 5/31/2024 1:15 PM, WM wrote:
Ask colleagues
(without pointing to our discussion)
whether they agree that
in the course of exchanging elements,
infinitely many elements can disappear.
Ask further whether
in the accumulation point of the sequence (1/n)
infinitely many unit fractions
can populate one and the same point.
>
Even better:
Ask whether
∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
>
No.
"No", what?
⎛ No
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y
⎛ No
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y does not imply
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
⎛ No
⎜ I (WM) don't know what it means to say
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
⎛ No
⎜ I (WM) don't dare ask my colleagues
⎜ whether ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
⎛ No
⎜ I (WM) will snip this again, un.asking
⎜ whether ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
⎛ No
⎜ something something unimagined and unrelated to
⎜ whether ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
?