Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 12/17/24 4:13 AM, WM wrote:

On 17.12.2024 00:52, Richard Damon wrote:

On 12/16/24 3:30 AM, WM wrote:

On 15.12.2024 21:21, joes wrote:

 

Therefore we use all [1, n].

Those are all finite.

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All n are finite.

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But N isn't, so the sets [1, n] aren't what the bijection is defined on.

 Every element is the last element of a FISON [1, n]. ℕ is the set of all FISONs. I use all FISONs. ∀n ∈ ℕ: f([1, n]) =< 1/10.
Ever heard of the effect of the universal quantifier?

But your logic can't deal with ALL Fisons.
Note, the mapping isn't in your [1, n] but in N.
Your logic that if it holds for all FISONs, it holds for N, is what shows that 0 == 1, so we see that logic is broken when it is applied to truly infinite things.

 

All intervals do it because there is no n outside of all intervals [1, n]. My proof applies all intervals.

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And all the intervals are finite, and thus not the INFINITE set N, which is where the bijection occurs.

 According to Cantor the "bijection" uses all n and nothing more.

Right, but no FISON uses contains ALL n.

 Regards, WM