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

On 12/15/24 6:08 AM, WM wrote:

On 15.12.2024 11:50, Mikko wrote:

On 2024-12-14 08:42:37 +0000, WM said:
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On 14.12.2024 09:30, Mikko wrote:

On 2024-12-13 10:28:44 +0000, WM said:
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On 13.12.2024 10:46, Mikko wrote:
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Between any two intervals there is space and that space contains other
intervals.

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No. Starting from a point in the complement the cursor will hit a first interval. This is true for all visible intervals.

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False. From a point that is not a part of an interval no interval is the
nearest one because another interval is nearer.

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IF ALL intervals and their endpoints are existing as invariable points on the real line this cannot happen.

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It can. Your { [q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n] | n = 1, 2, 3, ... }
is one such set.

 That is nothing but an unfounded claim. In actual infinity of set theory all intervals and their endpoints are existing as invariable points from the beginning of the cursor's motion.
 Regards, WM
 

So? The problem is the cursor can't move without immediately hitting segments, none of which are "next" because they are dense.