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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Wed, 19 Mar 2025 20:15:23 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Wed, 19 Mar 2025 20:30:27 +0100 schrieb WM:
> On 19.03.2025 19:00, Jim Burns wrote:
>> On 3/19/2025 11:55 AM, WM wrote:
> 
>>> Obviously the subtraction of the set of all FISONs which as single
>>> FISONs cannot empty Cantor's ℕ cannot empty Cantor's ℕ.
>> All the finite initial segments of ⋂𝒫ⁱⁿᵈ(ℕ)
>> empty ⋂𝒫ⁱⁿᵈ(ℕ) ⋂𝒫ⁱⁿᵈ(ℕ)\⋃{F} = {}
> Of course the inductive subsets of ℕ are eqiuvalent.
> I use Cantor's ℕ which is larger than all its inductive subsets.
It can't be larger. It contains nothing apart from the inductive set.

>>> Obviously
>> Please prove that it is obvious.
> 
> For Cantor's ℕ we have:
> |ℕ \ {1}| = ℵo |ℕ \ {1, 2, 3, ..., n}| = ℵo ==>
> |ℕ \ {1, 2, 3, ..., n+1}| = ℵo
> 
> All FISONs fail to empty ℕ when applied one after the other.
Awful wording. 

> All FISONs fail to empty ℕ when applied at once, because mathematics is
> not time-dependent.
Only you haven't "applied" all of them. That would be
N \ {1, 2, 3, ...}, which, unsurprisingly, = {}.


-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.