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

On 24.11.2024 18:22, Richard Damon wrote:

On 11/24/24 7:26 AM, WM wrote:

On 24.11.2024 13:12, Richard Damon wrote:

On 11/24/24 5:31 AM, WM wrote:

On 24.11.2024 03:22, Richard Damon wrote:

On 11/23/24 4:11 PM, WM wrote:

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Cover the unit intervals of prime numbers by red hats. Then shift the red hats so that all unit intervals of the positive real axis get red hats.
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And you can, as
the red hat on the number 2, can be moved to the number 1
the red hat on the number 3, can be moved to the number 2
the red hat on the number 5, can be moved to the number 3

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A very naive recipe.

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But it works.

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It fails in every step to cover the interval (0, n] with hats taken from this interval.

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But that isn't the requirement.
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The requirement it to map from ALL Prime Natural Numbers to ALL Natural Numbers
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Yes, for every n that belongs to a tiny initial segment.

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No, for EVERY n.
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Show one that it doesn't work for!

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The complete covering fails in every interval (0, n] with hats taken from this interval.

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Which isn't the interval in question.
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Your funny-mental fallacy is that you think an infinite set can be thought of as just some finite set allowed to keep growning until it reaches infinity,
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That is just the wrong model.
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so all the numbers get covered.

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No.

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WHich one doesn't.

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Almost all. The reason is simple mathematics. For every interval (0, n] the relative covering is 1/10, independent of how the hats are shifted. This cannot be remedied in the infinite limit because outside of all finite intervals (0, n] there are no further hats available.

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But finite sets aren't infinite sets, and don't act the same as them.
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All finite sets are the infinite set.

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You can not just use finite mathematics on infinite sets.

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But I can use the analytical limit of the constant sequence.
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Nope, since the finite sets are not the same as the infinite set, the property you are looking at just doesn't exist in the infinite set.

The finite sets contains all hats because all natural numbers and all 10n are in finite sets. No hat is outside.>

Limit theory only works if the limit actually exists

If limits exist at all, then the limit of the sequence 1/10, 1/10, 1/10, ... does exist.

 You can get things that APPEAR to reach a limit, but actually don't.

But if infinite sets do exist, then the set ℕ does exist, and all its elements are members of finite intervals (0, n].
Regards, WM