Sujet : Re: how
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : sci.mathDate : 11. Apr 2024, 00:05:45
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <uv7609$n6lm$2@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Mozilla Thunderbird
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.
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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.
Which ones were unused by e = 2*n?
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You "Complete" the bijection by showing the infinite sets map one to one by the formula of the bijection.
I show that Cantor's bijection fails.
Nope, you show your misunderstanding of Cantor's Bijection fails.
Regards, WM
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