Re: The set of necessary FISONs

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.

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Which is a proof of ANY, not ALL together,

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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.

But this doesn't say that the infinite doesn't exist, and that we can't make the Natural Numbers from a union of an infinite set of FISONs.

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 FISONs,
by not having a first element, is empty.

 And, because FISONs are finite, no less than an infinite number of them should be expected to be needed.

Infinitely many fail like infinitely many traiangles would fail.

 This doesn't mean we need ALL of them, just an infinite number of them.

Contradicted by Cantor's theorem.
Regards, WM