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

On 11/27/24 5:57 AM, WM wrote:

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.

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Nothing in logic allows that. But Cantor claims it erroneously.

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Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat

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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.

Your just showing your inconsistancy here.
You switch between a white hat being revealed from under the black hat when it is moved from 10n to n, and the hats being switched.
The first happens when you have ONE set of numbers, putting the white hats on all of them, and the black hats on every tenth, and you move the black hats on top of the white hats, and every white hat (at n) gets covered with the black hat from 10n, and none left over, so the bijection is confirmed
The second happens when you have TWO sets of numbers, one with all the natural numbers with white hats, and the second with the multiples of 10 with black hats, and we swap between them and find that we end up with ALL the natural numbers now with black hats, and all the set of multiples of 10s with white hats, so again the bijection is confirmed.
In no cases were white hats "deleted" because in both cases we still had an infinite set of numbers with white hats (just in the first, they are all under black hats). Your arguement is thus just irrelevent and shows that you are just not thinking clearly (or at all)

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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.

No, that isn't the bijection being talked about.
You can't disprove a bijection by not following it.
You don't seem to understand Cantor's claim, it isn't that every possible attempt at a bijection will succeed, it is that *IF* there is one, the sets are equal.

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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.

And thus prove that you don't understand what Cantor was talking about.
You just refuse to follow the rules, so mathematics has put you into the mathematical loony bin for it, and blown up your brain with your contradctions.

 

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.
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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
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Of course it is, since you can't complete the first part, you never get to the actual bijection.
You are just proving your stupidity, and that you have a funny-mental condition where you just don't understand what real mathematics is.