Re: Replacement of Cardinality

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.

 The input set was the Natural Numbers and w,

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.

∀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.

 Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

Not for all dark numbers.
Regards, WM