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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Thu, 12 Dec 2024 07:26:07 -0500
Organization: i2pn2 (i2pn.org)
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On 12/12/24 4:12 AM, WM wrote:
> On 12.12.2024 01:32, Richard Damon wrote:
>> On 12/11/24 9:32 AM, WM wrote:
>>> On 11.12.2024 03:04, Richard Damon wrote:
>>>> On 12/10/24 12:30 PM, WM wrote:
>>>>> On 10.12.2024 13:17, Richard Damon wrote:
>>>>>> On 12/10/24 3:50 AM, WM wrote:
>>>>>
>>>>>>> Two sequences that are identical term by term cannot have 
>>>>>>> different limits. 0^x and x^0 are different term by term.
>>>>>>
>>>>>> Which isn't the part I am talking of, it is that just because each 
>>>>>> step of a sequence has a value, doesn't mean the thing that is at 
>>>>>> that limit, has the same value.
>>>>>
>>>>> Of course not. But if each step of two sequences has the same 
>>>>> value, then the limits are the same too. This is the case for
>>>>>   (E(1)∩E(2)∩...∩E(n)) and (E(n)).
>>>
>>>> But the limit of the sequence isn't necessary what is at the "end" 
>>>> of the sequence.
>>>
>>> The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
> 
>> None of which are an infinite sets, so trying to take a "limit" of 
>> combining them is just improper.
> 
> Most endsegments are infinite. But if Cantor can apply all natural 
> numbers as indices for his sequences, then all must leave the sequence 
> of endsegments. Then the sequence (E(k)) must end up empty. And there 
> must be a continuous staircase from E(k) to the empty set.
> 
> Regards, WM
> 

Note, "inifinite" isn't a Natural Number, or a Real Number, so NO 
segement, specified by values, can have an "infinte endsegment".

You seem to have a definitional problem.

Of course, since you logic says that infinite sets act just like finite 
seqments, and thus no set is actually infinite, as well as that 0 is 
equal to 1 since they are both limits going to the same target, such a 
confusion is only natural.