Sujet : Re: Replacement of Cardinality
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.logic sci.mathDate : 14. Aug 2024, 18:35:17
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <53b0d334-6cb7-4b96-b98e-3c1e2d75709f@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
On 8/14/2024 8:14 AM, WM wrote:
Le 13/08/2024 à 18:26, Jim Burns a écrit :
On 8/13/2024 10:27 AM, WM wrote:
Le 12/08/2024 à 20:34, Jim Burns a écrit :
On 8/12/2024 9:47 AM, WM wrote:
Before the bound there is the end,
perhaps dark.
>
0 is
greatest.lower.bound β of visibleᵂᴹ unit.fractions
>
Yes.
But the smallest unit fractions comes before.
No visibleᵂᴹ.or.darkᵂᴹ point ε > 0 is the end of
both visibleᵂᴹ and darkᵂᴹ unit.fractions.
Every ε > 0 is visible, because it is chosen.
No visibleᵂᴹ point ε > 0 is the end of
both visibleᵂᴹ and darkᵂᴹ unit.fractions.
No darkᵂᴹ point ε > 0 is the end of
both visibleᵂᴹ and darkᵂᴹ unit.fractions.
There is no lower.end of
both visibleᵂᴹ and darkᵂᴹ unit.fractions.
But ℵo unit fractions occupy an interval y > 0
to settle there.
No.
0.many unit.fractions settle.
⎛ Related:
⎝ The intersection of infinite.end.segments is empty.
For each unit fraction ⅟k
there is an initial sub.segment of (0,1]
which ⅟k is not.in.
There are ℵ₀ unit.fractions in the ⅟k.free segment,
but not ⅟k
Not every point of y has ℵo smaller unit fractions.
No.
∀ᴿy > 0: ∀k ∈ ℕ₁: 0 < ⅟⌊k+⅟y⌋ < y
Every chosen ε > 0 is larger than this interval y.
No.
¬∃ᴿy > 0: ¬∀k ∈ ℕ₁: 0 < ⅟⌊k+⅟y⌋ < y