Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

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:

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It is completed! Every number 10n starts wit a black hat.

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Right, and it gives that hat to n, so every number gets one.
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And, 10n will get back a hat from 100n, so it still have one at the end.
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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?

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No, it is possilbe,

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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