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On 22.01.2025 13:10, Richard Damon wrote:But There is no single UF(n) that equals N, because you can ony get there from the union of an infinite set of FISONs.On 1/22/25 5:34 AM, WM wrote:Can't you read?On 22.01.2025 00:41, Richard Damon wrote:>On 1/21/25 7:44 AM, WM wrote:>On 21.01.2025 13:17, Richard Damon wrote:On 1/21/25 6:45 AM, WM wrote:>All finite initial segments of natural numbers, FISONs F(n) = {1, 2, 3, ..., n} as well as their union are less than the set ℕ of natural numbers.
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Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according to Cantor's Theorem B, every non-empty set of different numbers of the first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a first element, is empty.Which is a proof of ANY, not ALL together,>
It is a proof of not any. The proof that not all together are necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.which doesn't prove your claim about the Natural Numbers.>
It proves what I said: not all are required.
And no one said you needed to take the Union of ALL the FISONs, just ALL of an infinite set of FISONs.
Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according to Cantor's Theorem B, every non-empty set of different numbers of the first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a first element, is empty.
Regards, WM
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