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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.logic
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Wed, 27 Nov 2024 11:57:06 +0100
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On 26.11.2024 20:32, Richard Damon wrote:
> WM <wolfgang.mueckenheim@tha.de> wrote:
>> On 26.11.2024 18:24, Richard Damon wrote:
>>> WM <wolfgang.mueckenheim@tha.de> wrote:
>>
>>>>> And nothing in the logic says that white hats go away.
>>>>
>>>> Nothing in logic allows that. But Cantor claims it erroneously.
>>
>>> Cantor NEVER had the white hats go away, he just showed you could match
>>> every number with a black hat
>>
>> That means all white hats must disappear under black hats. That means
>> the white and black must be exchanged until no white remains. That is
>> impossible.

> Why is there a white hat under the black hat?

That is irrelevant. The black hats must cover all numbers ℕ but cannot 
because when black and white hats are exchanged never a white hat is 
deleted.
> 
> They are two *DIFFERENT* sets, one with the number {1, 2, 3, 4, …} each
> with a white hat, and one with the numbers {10, 20, 30, 40, …} each with a
> black hat. Just because we gave 10 a white hat in the first set doesn’t
> give the 10 in the second set that hat, then are DIFFERENT sets.

We have the black hats from the second set {10, 20, 30, 40, …}. They 
shall cover all numbers The numbers divisible by 10 from the first set ℕ 
= {1, 2, 3, 4, …} have been covered at the outset.
> 
> We can exchange the white hat from the first set on 1 with the black hat on
> the second set on 10.

We can first cover all numbers 10n of the first set ℕ with black hats. 
That does not increase or decrease the set of black hats.

> We then swap hats between 1st set 2 and 2nd set 20, and so on for the first
> set n and the second set 10n, and we see that ALL the numbers in the first
> set get their black hats and all the numbers in the second set get white
> hats, and thus we have proven that this is a bijection, and the sets use
> have the same cardinality.
> 
> You are just showing you don’t understand what is happening,

Do you understand that the bijection is not disturbed in the least when 
we first cover all numbers 10n of the first set ℕ with black hats. We 
can proceed then precisely in the same way as you say, can't we? What is 
different in your opinion?

Regards, WM