Re: how

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?

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

But none of them have less than ℵo successors, because if have less, the all have less since earlier ones have only finitely more, so NO numbers have ℵo successors, and thus you don't have the Natural Numbers.
You logic just doesn't create the actual set of the Natural Numbers, so goes BOOM when you try to use if for that.

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Regards, WM