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

On 26.11.2024 15:50, Richard Damon wrote:

On 11/26/24 8:58 AM, WM wrote:

An actual bijection is assumed: Consider the black hats at every 10 n and white hats at all other numbers n. It is possible to shift the black hats such that every interval (0, n] is completely covered by black hats. There is no first n discernible that cannot be covered by  black hat. But the origin of each used black hat larger than n is now covered by a white hat. Without deleting all white hats it is not possible to cover all n by black hats. But deleting white hats is prohibited by logic. Exchanging can never delete one of the exchanged elements.

 But a bijection is NOT a set to itself, but between two sets.

The set of natural numbers divisible by 10 and the set of natural numbers are two different sets.

 The white hats aren't on the "OTHER" numbers, (unless you bijection is from numbers zero mod 10 to number non-zero mod 10), but ALL numbers have a white hat, and every tenth has a black hat, and the bijection shows we can pair every black hat to a white hat.

Let all numbers n have white hats and in addition the 10n have black hats.
Claim: We can cover every white hat by a black hat. But when we take a black hat from 20 to give it to 2, then 20 has a white hat, and so on. How can the white hats disappear?

 

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Which element of which infinite set did not participate in the bijection?

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That is the crucial point! The white hats remain as long as logic is valid. But their carriers cannot be found. They are dark.

 And nothing in the logic says that white hats go away.

Nothing in logic allows that. But Cantor claims it erroneously.
Regards, WM