Re: Incompleteness of Cantor's enumeration of the rational numbers

On 23.11.2024 06:32, Moebius wrote:

Am 23.11.2024 um 06:18 schrieb Moebius:

On 22.11.2024 16:11, joes wrote:

Am Fri, 22 Nov 2024 15:51:11 +0100 schrieb WM:

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the sets of naturals and of prime numbers [can] cover each other.
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As it should. You can give each prime an index.

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Indeed! The two formulas
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| p(1) = min P
| p(n+1) = min {p e P : p > p(n)}   (for all n e IN)
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(recursively) define the function p: IN --> P. Where IN is the set of all natural numbers and P is the set of all prime numbers.
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Hint: p(1) = 2, p(2) = 3, p(3) = 5, ...
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Actually, if p e P, then there is an (index) n e IN such that p(n) = p.
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(It's easy to prove that p: IN --> P is a bijection.)

 p(n) is the n-th prime number in the sequence of prime numbers ordered by size.

Up to every definable prime there is a bijection. But not for all naturals.
Let every unit interval on the real axis be coloured white. Cover the unit intervals of prime numbers by red hats. It is impossible to shift the red hats so that all unit intervals of the whole real axis get red hats. There are too few prime numbers.
Regards, WM