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From: Moebius <invalid@example.invalid>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
 (extra-ordinary)
Date: Fri, 10 Jan 2025 03:01:10 +0100
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Am 10.01.2025 um 02:48 schrieb Moebius:
> Am 10.01.2025 um 02:45 schrieb Moebius:
>> Am 10.01.2025 um 02:19 schrieb Chris M. Thomasson:
>>> On 1/9/2025 5:15 PM, Moebius wrote:
>>>> Am 09.01.2025 um 22:12 schrieb Chris M. Thomasson:
>>>>> On 1/9/2025 8:18 AM, WM wrote:
>>>>>> On 09.01.2025 10:56, FromTheRafters wrote:
>>>>>>> WM explained :
>>>>>>
>>>>>>>> The set {1, 2, 3, ...} is smaller by one element than the set 
>>>>>>>> {0, 1, 2, 3, ...}.
>>>>>>>
>>>>>>> Both sets are equal in size
>>>>>>
>>>>>> No. Both sets appear equal (although everybody can see that they 
>>>>>> are not) when measured by an insufficient tool.
> 
> Hint: WM here meant (of course): "Both sets appear equal IN SIZE ..."

Hint@WM: The size of {1, 2, 3, ...} EQUALS the size of {0, 1, 2, 3, ...} 
when "measured" by the "tool" /equivalence/.

See: https://www.britannica.com/science/set-theory/Equivalent-sets

____________________________________________________________________

Hint: Using Zermelo's definition of the natural numbers we have 1 = {0}, 
2 = {1}, 3 = {2}, 4 = {3}, ...

And hence {1, 2, 3, 4, ...} = {{0}, {1}, {2}, {3}, ...}

If we NOW compare

                  {{0}, {1}, {2}, {3}, ...}       (= {1, 2, 3, 4, ...})
with
                  { 0 ,  1 ,  2 ,  3 , ...} ,

does ist STILL make sense to claim "everybody can see that they are not 
equal in size"?

> .
> .
> .
>