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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: Thu, 19 Dec 2024 22:37:11 +0100
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On 19.12.2024 19:29, Jim Burns wrote:
> On 12/19/2024 9:50 AM, WM wrote
>>> ⋂{E(1),E(2),...} is empty.
>>> Other numbers are not in the argument.
>>
>> But E(1) is full.
>> The only way to get rid of content is
>> to proceed by
>> ∀k ∈ ℕ:
>> ∩{E(1), E(2), ..., E(k+1)} =
>> ∩{E(1), E(2), ..., E(k)} \ {k},
>> i.e. to lose one number per term of
>> the function
>> f(k) = ∩{E(1), E(2), ..., E(k)}.
> 
> f(k) = E(k) = ⟦k,ℵ₀⦆
> 
> Yes,
> after infinitely.many one.number.losses,
> infinitely.many numbers are lost.

That means all numbers are lost by loss of one number per term. That 
implies finite endsegments.
> 
> However,
> each number is not.after infinitely.many numbers,
> each loss is not.after infinitely.many losses.

As long as any natural number remains, the sequence of lost numbers is 
finite (ended by the remaining number).
> 
>> The empty term
> 
> There is no darkᵂᴹ or visibleᵂᴹ empty term.

Finite intersections of endsegments preceding the empty intersection of 
all endsegments are dark.
> 
> ⎛ Assume otherwise.

For my argument I don't need empty endsegments.

>> The empty term
>> has lost all natnumbers one by one.
> 
> One.by.one and (darkᵂᴹ or visibleᵂᴹ) endlessly.

If Cantor uses all indices for his bijections the intersection of all 
endsegments must be empty.

Regards, WM