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

On 11.01.2025 11:25, joes wrote:

Am Sat, 11 Jan 2025 11:04:56 +0100 schrieb WM:

On 11.01.2025 01:28, Richard Damon wrote:

On 1/10/25 4:48 PM, WM wrote:

On 10.01.2025 21:08, Jim Burns wrote:
>

Where OUR infinityⁿᵒᵗᐧᵂᴹ "doesn't work",
it's you who's saying it doesn't work,

You are inconsistent. You claim that all natural numbers are an
invariable set. But when all elements are doubled then your set grows,
showing it is not invariable. That is nonsense.

But the set doesn't grow.
Which element is in the doubled set that wasn't there in the first
place?

The number of elements remains constant. All odd numbers of ℕ are
deleted. That implies that new even numbers are added.

Nope. This is all just your conception of Aleph_0 as finite.

Not finite but complete.

It does
not behave like that. All countable sets are bijective to each other.

It appears so. But it is wrong. Construct a bijection between natural numbers and even natural numbers. f(n) = 2n. Then double the even numbers by multiplication. Many are not in the bijection.

 

"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

This just means it is a bijection to N, which has an order.
 

If Cantor has constructed a sequence containing all even numbers of the
original set ℕ, then the doubled even numbers are missing.

What? Doubled even numbers are also even numbers.

But not natural numbers, since *all* natural numbers have been doubled.
When we mark all natural numbers with an astrisk, then none remains. When wie double them, then none remains. The odd numbers leave. All even numbers remain. But the same number of of even numbers is larger than ω.
1, 2, 3, 4, 5, ..., ω becomes 2, 4, 6, 8, 10, ..., ω, ω+2, ω+4, ..., ω2.
Regards, WM