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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Sat, 23 Nov 2024 22:44:36 +0100
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On 23.11.2024 22:35, joes wrote:
> Am Sat, 23 Nov 2024 21:45:06 +0100 schrieb WM:
>> On 23.11.2024 21:18, Jim Burns wrote:
>>> On 11/23/2024 5:30 AM, WM wrote:
>>>> On 22.11.2024 22:50, Jim Burns wrote:
>>>
>>>>> ℙ covers ℕ, and ℕ covers ℙ
>>>>
>>>> Let every unit interval on the infinite real axis be coloured white.
>>>> Cover the unit intervals of prime numbers by red hats.
>>>> It is impossible to shift the red hats
>>>
>>> Yes, because we are finite beings,
>>> and there are infinitely.many red hats.
>>
>> No, the reason is that every shift removes the hat from its place and
>> requires an other hat, taken from wherever, but with certainty leaving
>> an uncovered interval. That does never change.
> No. That other place where we take the hats from, larger primes, are of
> course covered by even larger primes. We don't stop at some arbitrary
> finite number, but continue forever.
> 
>>>> 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.
>>> No. Assume that that is so.
>> That need not be assumed but that is obviously so for every part of the
>> real axis.
> Not true. The first n prime numbers are obviously in bijection with the
> numbers 1 to n.

But the prime numbers come from a larger interval than (0, n]. For every 
interval (0, n] there are more hats required than are present. The 
density of prime numbers in (0, n} is about n/logn.

Could by simple reordering this formula be falsified, then number theory 
could be chucked away.
> 
>>> Assume that there are enough red hats for the first 𝔊 numbers but
>>> not enough for the 𝔊+1ᵗʰ
>> That is a mistake. If there are enough hats for G natnumbers, then there
>> are also enough for G^G^G natnumbers. Alas they leave G^G^G unit
>> intervals without hats. That is the catch!
> Right, that is your mistake. There are still hats left over.

But fewer than are required. Because n > n/logn.

Regards, WM