Re: how

Le 31/05/2024 à 21:15, Jim Burns a écrit :

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

You should try to think logically. Beyond 0 there is no unit fraction. Hence before 0 there is a last one.

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.

Yes.

>
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:

No.
Regards, WM