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Subject: Re: Replacement of Cardinality
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From: WM <wolfgang.mueckenheim@tha.de>
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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?
> 
>>>> ∀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.

Regards, WM