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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: The non-existence of "dark numbers"
Date: Wed, 19 Mar 2025 20:30:27 +0100
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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.
> 
>> 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.
All FISONs fail to empty ℕ when applied at once, because mathematics is 
not time-dependent.

Regards, WM