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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Sun, 15 Dec 2024 12:51:21 +0200
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On 2024-12-14 21:40:48 +0000, WM said:

> 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.

Yes but none of them can be passed before passing other opoints.

-- 
Mikko