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From: joes <noreply@example.org>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Thu, 12 Dec 2024 14:23:48 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:
> 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}.
The sequence is endless, has no end, is infinite.

>> 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.
It makes no sense not being able to „apply” numbers. Clearly Cantor does.
The sequence IS continuous. It’s just that you misconceive of the 
limit as reachable.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.