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From: "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Fri, 27 Dec 2024 21:08:36 -0800
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On 12/18/2024 4:25 AM, Richard Damon wrote:
> On 12/17/24 4:51 PM, WM wrote:
>> On 17.12.2024 13:34, Richard Damon wrote:
>>
>>>> According to Cantor the "bijection" uses all n and nothing more.
>>>
>>> Right, but no FISON uses contains ALL n.
>>
>> But all FISONs contain/are all n.
>>
>> Regards, WM
>>
>>
> 
> ???? The FISONs are all finite, so NONE of them contain *ALL* n.

Ditto. :^)

{ { 1 }, { 1, 2 }, { 1, 2, 3 }, ... }

Infinite FISONs, they are all finite. No FISON has infinity, or else it 
would not be a FISON to being with? Fair enough?

> 
> You can have a FISON for ANY n, but a give FISON isn't for ALL n.
> 
> You don't seem to understand the difference.
> 
> You don't seem to understand that the whole can be more than the sum of 
> its parts.