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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: Mon, 9 Dec 2024 22:50:23 +0100
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On 09.12.2024 21:07, Python wrote:
> Le 09/12/2024 à 20:40, Crank Wolfgang Mückenheim from Hochschule 
> Augsburg aka WM a écrit :
>> On 09.12.2024 20:18, Python wrote:
>>> Le 08/12/2024 à 23:34, Crank Wolfgang Mückenheim from Hochschule 
>>> Augsburg aka WM a écrit :
>>>> On 08.12.2024 19:01, Jim Burns wrote:
>>>>> On 12/8/2024 5:50 AM, WM wrote:
>>>>
>>>>>> ∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
>>>>>> What can't you understand here?
>>>>>
>>>>> {E(i):i} is the set.of.all non.empty end.segments.
>>>>>
>>>>> ⋂{E(i):i} is the intersection.of.all
>>>> non.empty end.segments.
>>>>>
>>>>> ∀n ∈ ℕ:
>>>>> {E(i):i}∪{E(n+1)} = {E(i):i}
>>>>> Each is "already" in.
>>>>
>>>> Not the empty endsegment.
>>>> ∀n ∈ ℕ: E(n) is non-empty. But not every E(n+1).
>>>
>>> You could hardly write something worse and more wrong that that.
>>>
>>> The very core property of N is that if n ∈ ℕ then n+1 ∈ ℕ.
>>
>> That is correct for definable natural numbers and even for almost all 
>> dark natural numbers.
>>
>> The very core property of analysis is that equal sequences have equal 
>> limits if they have limits at all.
> 
> E(1)∩E(2)∩...∩E(n) = E(n)
> 
> Lim E(1)∩E(2)∩...∩E(n) = {}
> Lim E(n) = {}
> 
> The are equal.

Not in a set theory where every endsegment is infinite.

An empty limit endsegment requires finite predecessors because there is 
only one way to emptiness allowed, namely this one:
∀k ∈ ℕ : E(k+1) = E(k) \ {k}.

Regards, WM