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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: The set of necessary FISONs
Date: Wed, 22 Jan 2025 11:34:09 +0100
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On 22.01.2025 00:41, Richard Damon wrote:
> On 1/21/25 7:44 AM, WM wrote:
>> On 21.01.2025 13:17, Richard Damon wrote:
>>> On 1/21/25 6:45 AM, WM wrote:
>>>> All finite initial segments of natural numbers, FISONs F(n) = {1, 2, 
>>>> 3, ..., n} as well as their union are less than the set ℕ of natural 
>>>> numbers.
>>>>
>>>> Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is 
>>>> the first necessary FISON? There is none! All can be dropped. But 
>>>> according to Cantor's Theorem B, every non-empty set of different 
>>>> numbers of the first and the second number class has a smallest 
>>>> number, a minimum. This proves that the set of indices n of 
>>>> necessary F(n), by not having a first element, is empty.
>>
>>> Which is a proof of ANY, not ALL together,
>>
>> It is a proof of not any. The proof that not all together are 
>> necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.

> which doesn't prove your claim about the Natural Numbers.

It proves what I said: not all are required.

> But this doesn't say that the infinite doesn't exist, and that we can't 
> make the Natural Numbers from a union of an infinite set of FISONs.

According to Cantor's Theorem B, every non-empty set of different 
numbers of the first and the second number class has a smallest number, 
a minimum. This proves that the set of indices n of necessary FISONs,
by not having a first element, is empty.
> 
> And, because FISONs are finite, no less than an infinite number of them 
> should be expected to be needed.

Infinitely many fail like infinitely many traiangles would fail.
> 
> This doesn't mean we need ALL of them, just an infinite number of them.

Contradicted by Cantor's theorem.

Regards, WM