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On 11/11/2024 3:41 AM, WM wrote:
For that claim you need an infinite set of claims.Yes, I claim that.>But it will never complete>
an infinite set of claims.
We do not need an infinite ๐๐ฒ๐พ๐๐ฒ๐ป๐ฐ๐ฒ of ๐ฐ๐น๐ฎ๐ถ๐บ๐ completed.
We do not want an infinite ๐๐ฒ๐พ๐๐ฒ๐ป๐ฐ๐ฒ of ๐ฐ๐น๐ฎ๐ถ๐บ๐ completed.
But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?
This is one ๐ฐ๐น๐ฎ๐ถ๐บ:It is wrong.
โ All fractions are in bijection with
โ all natural numbers.
No, I claim that intervals can be translated. (The set of intervals remains constant in size and multitude.) For every finite subset this is possible.My intervals I(n) = [n - 1/10, n + 1/10]Your ๐ฐ๐น๐ฎ๐ถ๐บ๐ start with "Sets change".
must be translated to all the midpoints
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5,
2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
if you want to contradict my claim.
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