Re: Replacement of Cardinality

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.

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

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

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Why?

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Because otherwise there was a gap below ω.

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

 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?

 

∀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, and that one has an n+2, which has an n+3 and so on continuing without bound.
If you want there to be a n that doesn't have an n+1 in the set, then you are not talking about the actual Natural Numbers, because you logic system just can't handle them.
That just proves you are using too primative of tools for the task at hand, which has been obvious for years.

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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.
Of course, if you logic is based on lying about definitions, it isn't good for must.

 Regards, WM