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On 12/16/24 3:30 AM, WM wrote:Seems to be so. Unless I am missing something, WM seems to suggest that Cantor Pairing does not work with all natural numbers. 0 aside for a moment. Even though it works with 0 as well, anyway... WM has a personal issue here? For him to overcome? Can he see the "light" so to speak?On 15.12.2024 21:21, joes wrote:But N isn't, so the sets [1, n] aren't what the bijection is defined on.Am Sun, 15 Dec 2024 16:25:55 +0100 schrieb WM:>On 15.12.2024 12:15, joes wrote:Those are all finite.Am Sat, 14 Dec 2024 17:00:43 +0100 schrieb WM:>Therefore we use all [1, n].??? The bijection is not finite.Then deal with all infinitely many intervals [1, n].No, we are not forbiding "detailed" analysisThose who try to forbid the detailed analysis are dishonestThat pairs the elements of D with the elements of ℕ. Alas, it canBut we aren't dealing with intervals of [1, n] but of the full set.
be proved that for every interval [1, n] the deficit of hats
amounts to at least 90 %. And beyond all n, there are no further
hats.
swindlers and tricksters and not worth to participate in scientific
discussion.
All n are finite.
And all the intervals are finite, and thus not the INFINITE set N, which is where the bijection occurs.>>Wonrg. There is no natural n that „covers N”.With and without limit.Only in the limit.The intervals [1, n] cover the full set.Why can't he? The problem is in the space of the full set, not theThe problem is that you can't GET to "beyond all n" in the pairing,If this is impossible, then also Cantor cannot use all n.
as there are always more n to get to.
finite sub sets.
All intervals do it because there is no n outside of all intervals [1, n]. My proof applies all intervals.
Thus your "proof" is just a LIE.
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