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

On 11/27/24 3:09 PM, WM wrote:

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.

Npo, it shows that you 'logic' just isn't consistant, and you keep on trying to switch to improper similes to try to justify your erroneous conclusions.

 

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.

Right, from the OTHER set of numbers which contains the values that are multipe of 10.

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.

Right, every natural number can be mapped to a number that is 10 times it to get the black hat that was on that other number.

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.

It is the bijection you are talking about.

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

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

Right, there are Aleph_0 black hats, just like there are Aleph_0 Natural numbers to put them on.

>
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.

Every number n gets its hat from 10n.
That is the bijection we have been talking about.
One side has EVERY natural number in it, and all of them are covered,
The other side has EVERY multiple of 10 in it, and all of them are paired to.

>

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.

Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the end.

 Regards, WM