Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

Am 10.01.2025 um 03:01 schrieb Moebius:

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 :

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The set {1, 2, 3, ...} is smaller by one element than the set {0, 1, 2, 3, ...}.

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Both sets are equal in size

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No. Both sets appear equal (although everybody can see that they are not) when measured by an insufficient tool.

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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
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 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"?

Again, referring to the sucessor operation s, we have
            {1, 2, 3, 4, ...} = {s0, s1, s2, s3, ...} .
If we NOW compare
                    {s0, s1, s2, s3, ...}       (= {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"?

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