Re: how

On 5/31/2024 1:15 PM, WM wrote:

Le 30/05/2024 à 22:24, Jim Burns a écrit :

On 5/30/2024 3:44 AM, WM wrote:

NUF(x) = 1 between x = 0 and x = 1/10​^10​^10​^100000.

>
No, NUF(x) ≠ 1

>
That would imply that
not between all unit fractions distances existed.

No.
x > ⅟n > ⅟(n+1) > 0

Therefore it is wrong.
>

| Assume you are correct.
| Assume NUF(x) = 1  for
| 0 < x < 1/10​^10​^10​^100000.
|
| unit fraction ⅟n exists: 0 < ⅟n < x
| no unit fraction exists between 0 and ⅟n
|
| However,
| unit fraction ⅟(n+1) exists: 0 < ⅟(n+1) < ⅟n
| Contradiction.

>
Yes, if all n had successors,
there would be a contradiction.

For each n countable.to from 0
n+1 is countable.to from n
n+1 is countable.to (through n) from 0
Each countable.to n has a countable.to successor.
There is a contradiction.

if your chain contains only correct conclusions,
then your start must be wrong.

>
Either one of EAX is wrong
⎛ E. The empty set ∅ exists.
⎜ A. For sets x and y, adjunct x⨭y = x∪{y} exists.
⎝ X. Two equi.membered sets are equal sets.
>
or you are wrong.

>
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