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

On 27.11.2024 15:16, Richard Damon wrote:

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

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.

It is irrelevant, but assume the hats being switched. Black hats for numbers 10n, white hats for all others.

The second happens when you have TWO sets of numbers,

We have one set, namely ℕ to be covered with black hats. The natural numbers 10n are already covered. Now the black hats have to be moved to cover all natural numbers.

In no cases were white hats "deleted" because in both cases we still had an infinite set of numbers with white hats

But Cantor claims that all can be replaced by black hats.

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.

This bijection comes afterwards.

 You can't disprove a bijection by not following it.

You can follow it now. The number of black hats remains the same.

 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.

Now show that there is one.

 

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?

Of course it is, since you can't complete the first part.

It is completed! Every number 10n starts wit a black hat.
Regards, WM