Re: how

On 4/12/24 9:45 AM, WM wrote:

Le 12/04/2024 à 03:16, Richard Damon a écrit :

On 4/11/24 8:07 AM, WM wrote:

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:

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Nope, because ONE set is not TWO Sets.

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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.

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

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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.

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

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Which ones were unused by e = 2*n?

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

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But none of them have less than ℵo successors,

 None of them which can be used as individuals have less than ℵo successors.

NONE of them, as ALL can be used individually.
There are no "Natural Numbers" with less than ℵo successors.
Any number you might try to create with less than ℵo successors isn't a "Natural Number", as if fails to follow the definition of them.

 

because if have less, the all have less since earlier ones have only finitely more,

 No, there are infinitely many natural numbers. Only between the usable numbers there a finite difference can be calculated, because the infinitely many follow upon them and are dark.

You don't seem to understand that ALL Natural Numbers are usable individually.
Your "Dark" numbers just don't exist.
The problem is you logic can't handle the infinite set of the Natural Numbers and just blows up you logic when misapplied to it.

 Regards, WM