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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: The set of necessary FISONs
Date: Sun, 2 Feb 2025 14:21:21 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sun, 02 Feb 2025 12:25:49 +0100 schrieb WM:
> On 01.02.2025 20:21, Jim Burns wrote:
>> On 2/1/2025 7:56 AM, WM wrote:
> 
>>> There is the assumption that a set with U(F(n)) = ℕ exists.
It’s not an assumption. Hint: F(n+1) = F(n) u n+1 and then
U({F(n) for all n e N}).

>>> Without changing the union we can remove every element by induction.
>>> No element remains. The set does not exist.
>> 
>> Each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)} can grow¹ to another
>> initial segment F(k+1)
>> which is also finiteᵒᵘʳ, and is larger than F(k),
>> and is not larger than ⋃{F(n)}
>> {F(n}} holds each finiteᵒᵘʳ initial segment F(k) ⋃{F(n)} is larger than
>> each F(k).
> 
> But all F(n) can be discarded without changing the union.
The union of a nonempty set is not empty.

> F(1) can be discarded. If F(n) can be discarded, then F(n+1) can be
> discarded.
> Note: Mathematical induction is a method for proving that a statement
> P(n) is true for every natural number n that is, that the infinitely
> many cases P(0),P(1),P(2),P(3),... all hold. [Wikipedia]
But P(ω) does not hold.

> Therefore if U(F(n)) = ℕ, then  { } = ℕ
There are no naturals with infinite segments.

-- 
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.