Re: how

Le 11/04/2024 à 01:05, Richard Damon a écrit :

On 4/10/24 4:14 PM, WM wrote:

Le 10/04/2024 à 01:06, Richard Damon a écrit :

On 4/9/24 8:16 AM, WM wrote:

 

Nope, because ONE set is not TWO Sets.

 In the set ℚ there are as many indices n/1 as are indices n in ℕ. If indexing all fractions was possible, it was possible with indices n/1. But it isn't.

>

I didn't say "N_applied", I said N.

>
But what you can use belongs to ℕ_applied. Otherwise show a natural number that completes the bijection, i.e., which has not infinitely many pairings on front.

>
Nope, you can use ALL of the Natural Numbers.
>

You can use only a small minority because almost all remain unused:
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

 Which ones were unused by e = 2*n?

Those of {1, 2, 3, ...} with less than ℵo successors.

 

Regards, WM