Re: Replacement of Cardinality

On 8/1/24 8:02 AM, WM wrote:

Le 31/07/2024 à 18:20, joes a écrit :

Am Wed, 31 Jul 2024 14:27:06 +0000 schrieb WM:

Le 31/07/2024 à 03:28, Richard Damon a écrit :

On 7/30/24 1:37 PM, WM wrote:

Le 30/07/2024 à 03:18, Richard Damon a écrit :

On 7/29/24 9:11 AM, WM wrote:

>

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

That is a bit imprecise. Even though you keep on talking about
consecutive infinities, you can't compare natural and "dark" numbers.

 Dark natnumbers are larger than defined natnumbers. Even dar natnumbers can be compared by size. ω/10^10 < ω/10 < ω/2, < ω-1.
 

ω/10^10 and ω/10 are dark natural numbers.
>

If all natural numbers exist, then ω-1 exists.

Why?

Because otherwise there was a gap below ω.

But you combined two different sets, so why can't there be a gap?

I assume completness.

Completeness of N? No number n reaches omega.

 What is immediately before ω? Is it a blasphemy to ask such questions?

It has no predicessor, just like in the Natural Numbers 0 has nothing before it.
You can expand your number system to define number there, which seems to be what you "dark numbers" are, numbers bigger than all the finite Natural Numbers, but smaller than w.

>

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every unit
fraction 1/n, there exists another unit fraction smaller than itself.

No. My formula says ∀n ∈ ℕ.

That is not a contradiction.

 It is not a contradiction to my formula if some n has no n+1.

It is a violation of the DEFINITION of the Natural Numbers.
Now, if you admit that you system doesn't actaully HAVE the actaual Natura Numbers, but some other set built by some other methd, go ahead and try to actually define that set and see what you can do with it.
Just don't claim that anyone else needs to use your inferior number system.

 Regards, WM