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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: The set of necessary FISONs
Date: Sun, 2 Feb 2025 17:41:05 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <31a3690769165467190929addc8f1308fc8d6d66@i2pn2.org>
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Am Sun, 02 Feb 2025 12:06:26 +0100 schrieb WM:
> On 01.02.2025 19:42, joes wrote:
>> Am Sat, 01 Feb 2025 14:40:12 +0100 schrieb WM:
>>> On 01.02.2025 14:16, joes wrote:
>>>> Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
> 
>>>>> When n can be deleted, then n+1 can be deleted. Nothing remains.
>>>> It works this way: If n can be removed, n AND n+1 can be removed.
>>>> But it doesn’t work this way: „If n can be left out, all n can be.”
>>> Which one cannot be removed?
>> Wrong question. You cannot remove infinitely many. „Infinitely many”
>> is not a natural number.
> 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]
And you want the wrong P(ω) to hold, but you cannot remove infinitely
many segments. This is different from all P(n) which together say
you can remove any finite *number of* segments (not the segments
themselves). Do you get that? There is no infinite segment, but
there *are* infinitely many.

-- 
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.