Re: Replacement of Cardinality

Le 01/08/2024 à 02:09, Richard Damon a écrit :

On 7/31/24 10:27 AM, WM wrote:

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:

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But what number became ω when doubled?

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ω/2

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

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

 In other words, outside the Natural Nubmer, all of which are defined and definable.

That is simply nonsense. Do you know what an accumalation point is? Every eps interval around 0 contains unit fractions which cannot be separated from 0 by any eps. Therefore your claim is wrong.

 

>
The input set was the Natural Numbers and w,

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

 They may be "dark" but they are not Natural Numbers.

They are natural numbers.

 Natural numbers, by their definition, are reachable by a finite number of successor operations from 0.

That is the opinion of Peano and his disciples. It holds only for potetial infinity, i.e., definable numbers.

I assume completness.

 I guess you definition of "completeness" is incorrect.
 If I take the set of all cats, and the set of all doges, can there not be a gap between them?

What is the reason for the gap before omega? How large is it? Are these questions a blasphemy?

 

 

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

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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.

 Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,

That does ny formula not say. It says for all n which have successors, there is  distance between 1/n and 1/(n+1).

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

 Not for all dark numbers.

 Maybe not for dark numbers, but it does for all Natural Numbers, as that is part of their DEFINITION.

It is the definition of definable numbers. Study the accumulation point. Define (separate by an eps from 0) all unit fractions. Fail.
Regards, WM