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On 05.06.2025 09:32, Mikko wrote:That may be good enough when you want to prove something that we alreadyOn 2025-06-04 17:32:28 +0000, WM said:If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has one successor less, that is infinitely many (ℵo) successors.
On 04.06.2025 08:43, Mikko wrote:No -> there.On 2025-06-03 13:17:57 +0000, WM said:Cantor shows it.I had only to show that in Cantor's set theory proofs by arithmetic induction are possible.Which you didn't show.ℵo - 1 = ℵo.That confirms my proof:So far good. But no P[n] -> P[n+1] and no induction.
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.It is relevant to the extent that you cannot learn form it how a directI need not learn it, because I did it.
proof should be presented.
The expression "subtracting them individually" should be representedThis expression is represented by
mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
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