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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Sat, 14 Dec 2024 22:40:48 +0100
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On 14.12.2024 19:53, Richard Damon wrote:
> On 12/14/24 10:46 AM, WM wrote:
>> On 14.12.2024 12:06, joes wrote:
>>> Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
>>>> On 14.12.2024 09:30, Mikko wrote:
>>>>> On 2024-12-13 10:28:44 +0000, WM said:
>>>>>> On 13.12.2024 10:46, Mikko wrote:
>>>>>>
>>>>>>> Between any two intervals there is space and that space contains
>>>>>>> other intervals.
>>>>>> No. Starting from a point in the complement the cursor will hit a
>>>>>> first interval. This is true for all visible intervals.
>>>>> False. From a point that is not a part of an interval no interval is
>>>>> the nearest one because another interval is nearer.
>>>> IF ALL intervals and their endpoints are existing as invariable points
>>>> on the real line this cannot happen. In potential infinity however
>>>> between any two points new intervals come into being.
>>> They are ALREADY there.
>>
>> Therefore they cannot appear after the cursor has passed their 
>> positions. Every interval and every end of an interval would be hit by 
>> the cursor.
>> 
> Where did the cursor come from in the first place?

It starts in the complement of the intervals of measure 3 covering 
rational numbers. If the cursor is thrown by chance, the chance is 3/oo 
= 0 that it hits an interval.
> 
> And why did it pass them when you tried to place t?

It passes an interval when it moves.
> 
> This is your old problem of there not being a "next" in a dense set.

In a geometry where all points exist, all points can be passed. But the 
set of intervals is not dense. It would be dense if all rationals were 
covered.

Regards, WM