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On 23.01.2025 16:18, Jim Burns wrote:For every natural, {1, …, n} exists.On 1/23/2025 3:43 AM, WM wrote:ℕ is a superset of each FISON Each set which is a superset of eachWrong.
FISON is a superset of ℕ
Cantor's theorem does not force the existence of a necessary set.That's a longer.winded way to say that ℕ is the union of all FISONsThe union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON.
N is not a FISON."All" is not more than the repeated "each".
'All' is complete, whatever you (WM) mean by 'complete'.
From the finite.length description of a FISON,
we know that up to a FISON is not complete, is not all.
Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Since every FISON is theIs there supposed to be a universal quantifier there?
union of all its predecessors we get F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in the union of all F(n), find the first exception.WDYM believe in the union?
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