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

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:
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According to Cantor the "bijection" uses all n and nothing more.

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Right, but no FISON uses contains ALL n.

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But all FISONs contain/are all n.
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Regards, WM
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 ???? 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.