Re: Replacement of Cardinality

Le 16/08/2024 à 18:48, Richard Damon a écrit :

On 8/16/24 12:21 PM, WM wrote:

Le 15/08/2024 à 18:17, joes a écrit :
 

0 is not a unit fraction, so not an end; there is no smallest.

 We can reduce the interval (0, x) c [0, 1] such that x converges to 0.
Then the number of unit fractions diminishes. Finally there is none remaining. But never, for no interval (0, x), more than one unit fraction is lost. Therefore there is only one last unit fraction.
 Regards, WM

 No, you can't because the count of the unit fractions in (0, x) is always aleph_0. When you reduce the size to get the next smaller one, the count is still aleph_0, as that is how arithmetic on infinite values work.

It works so for cranks.

 The value never converges to 0, as you can never get off aleph_0. To converge, you first need to find a value of x where the count of unit fractions in (0, x) is a finite number, but no finite number x has that property.

Then more than one diminish simultaneously. Contradiction.
This proves a first one: ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 .
Regards, WM