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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Sun, 29 Dec 2024 11:34:08 +0100
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On 28.12.2024 18:31, Jim Burns wrote:
> On 12/28/2024 9:03 AM, WM wrote:

>> [0, ω-1] = [0,ω⦆ = ℕ =/= [0, ω]
> 
> Yes,
> ⟦0,ω⦆  =  ℕ  ≠  ⟦0,ω⟧
> 
> However,
> ⎛ Assume k < ω
> ⎜ #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
> ⎜ #⟦0,k+1⦆ ≠ #(⟦0,k+1⦆∪⦃k+1⦄)
> ⎝ k+1 < ω

That holds for almost all natural numbers k.
It cannot hold for an actually infinite system without disappearing Bob.

For example, all endsegments obey the sequence defined by
∀k ∈ ℕ : E(k+1) = E(k) \ {k+1}.
If all natnumbers exist, then all will leave the terms of the sequence, 
but cannot other than in single steps including the dark endsegments
E(ω-3) = {ω-2, ω-1}, E(ω-2) = {ω-1}, E(ω-1) = { }.

∀k ∈ ℕ : E(k+1) = E(k) \ {k+1} does not allow another alternative.

Regards, WM