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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: Thu, 28 Nov 2024 17:09:35 +0100
Organization: A noiseless patient Spider
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On 28.11.2024 16:50, Richard Damon wrote:
> On 11/28/24 9:17 AM, WM wrote:
>> On 28.11.2024 14:01, Richard Damon wrote:
>>> On 11/28/24 7:20 AM, WM wrote:
>>>> On 27.11.2024 22:20, Richard Damon wrote:
>>>>> On 11/27/24 3:09 PM, WM wrote:
>>>>
>>>>>> It is completed! Every number 10n starts wit a black hat.
>>>>>
>>>>> Right, and it gives that hat to n, so every number gets one.
>>>>>
>>>>> And, 10n will get back a hat from 100n, so it still have one at the 
>>>>> end.
>>>>>
>>>> It seems so, but that is impossible. Up to every 10n, the interval 
>>>> 1, 2, 3, ... 10n has a covering of 1/10 only. The limit of this 
>>>> sequence is 1/10 too. And since this is true for all natural 
>>>> numbers, where should additional hats come from?
>>
>>> No, it is possilbe,
>>
>> It is impossible that the sequence 1/10, 1/10, 1/10, ... has another 
>> limit than 1/10.

> But that makes the error of a wrong category.
No.

> The properties of the infinte set is not just the properties of the 
> series of the finite sets that approach it.

Maybe. In this case it is precisely this.

Look: If for all intervals 1, 2, 3, ..., n the covering is 1/10, then 
there are no natnumbers outside of all intervals and there are no hats 
outside of all intervals. Therefore only a fool could believe that 
infinitely many black hats were supplied after all. If you wish to be a 
fool you may claim that. My students would never come down that much.

Regards, WM