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

On 1/20/25 7:33 AM, WM wrote:

On 19.01.2025 14:29, FromTheRafters wrote:

WM formulated the question :

On 19.01.2025 11:42, FromTheRafters wrote:

WM presented the following explanation :

On 18.01.2025 12:03, joes wrote:

Am Fri, 17 Jan 2025 22:56:13 +0100 schrieb WM:

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Correct. If infinity is potential. set theory is wrong.

And that is why set theory doesn't talk about "potential infinity".

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Nevertheless it uses potential infinity.

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No, it doesn't.

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Use all natnumbers individually such that none remains. Fail.

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This makes no sense.

 It is impossible.

Because logic that insists on dealing with an INFINITE set one by one is illogical except for a being that is itself INFINITE and thus capable of INFINITE action.

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All "bijections" yield the same cardinality because only the potentially infinite parts of the sets are  applied.

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No, it is because these bijections show that some infinite sets' sizes can be shown to be equal even if no completed count exists.

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They appear equal because no completed count exists.

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No, they are the same size when it is shown there is at least one bijection.

 Every element of the bijection has almost all elements as successors. Therefore the bijection is none.

Nope, the logic that can't see the completion at infinity is broken.

 

All natnumbers in bijections have ℵ₀ not applied successors.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Only potential infinity is applied.

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You mean that only finite sets are involved.

 Of course Infinitely many successors prevent that their predecessors are infinitely many.
 Regards, WM