Re: The set of necessary FISONs

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.

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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.