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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Fri, 8 Nov 2024 17:30:23 +0100
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On 08.11.2024 14:09, Mikko wrote:
> On 2024-11-07 13:21:42 +0000, WM said:
> 
>> On 07.11.2024 10:22, Mikko wrote:
>>> On 2024-11-06 17:55:15 +0000, WM said:
>>>
>>>> On 06.11.2024 16:04, Mikko wrote:
>>>>> On 2024-11-06 10:01:21 +0000, WM said:
>>>>
>>>>>> I leave ε = 1. No shrinking. Every point outside of the intervals 
>>>>>> is nearer to an endpoint than to the contents.
>>>>>
>>>>> This discussion started with message that clearly discussed limits 
>>>>> when
>>>>> ε approaches 0. The case ε = 1 was only about a specific unimportant
>>>>> question.
>>>>
>>>> When ε approaches 0 then the measure of the real axis is, according 
>>>> to Cantor's results, 0. That shows that his results are wrong.
>>>
>>> It is not the measure of the real axis but the set of rationals. The
>>> real axis more than just the rationals. The irrationals are also a
>>> part of the real axis.
>>
>> But not between irrational points.
> 
> Real axis contains both real and irrational numbers and nothing else.
> Between any two points of the real axis there are both rational and
> irrational points.

If Cantors enumeration of the rationals is complete, then all rationals 
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and none is 
outside. Therefore also irrational numbers cannot be there. Of course 
this is wrong. It proves that not all rational numbers are countable and 
in the sequence.

>> It is Cantor's result that all rationals are countable, hence inside 
>> my intervals.
> 
> That is but what you said above is not.
> 
>> But we can use the following estimation that should convince everyone:
>>
>> Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n 
>> and q_n can be in bijection, these intervals are sufficient to cover 
>> all q_n. That means by clever reordering them you can cover the whole 
>> positive axis except "boundaries".
> 
> Depends on the type of n.

The n are the natural numbers.
> 
>> And an even more suggestive approximation:
>> Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
> 
> Likewise.

These are the intervals sketched:
 >> J(n) = [n - 1/10, n + 1/10]
 >> --------_1_--------_2_--------_3_--------_4_--------_5_--------_...

Only a very hard believer can believe that by shuffling the intervals 
they could cover the real line infinitely often.

Regards, WM

>