Re: Incompleteness of Cantor's enumeration of the rational numbers

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