Re: Replacement of Cardinality

On 8/16/2024 12:19 PM, WM wrote:

Le 15/08/2024 à 18:16, Jim Burns a écrit :

⎜ Assume NUF(x) = 0 and x > 0

⎜ lower.bound ½⋅β and not.lower.bound ½⋅β
⎝ Contradiction.

We assume that NUF(0) = 0 and
many unit fractions are within (0, 1].
We can reduce the interval to (0, x) c [0, 1].
Let x converge to 0.
Then the number of unit fractions diminishes.

For some sets,
sets for which non.{} subsets always have both ends,
removing one element diminishes them
For other sets,
sets for which one non.{} subset has 0 or 1 end,
removing one element does not diminish them.
The set ℕᵈᵉᶠ def of diminishable ordinals is
one of the other sets.
It is not
a set for which non.{} subsets always have both ends.
ℕᵈᵉᶠ has a lower.end, but
no element of ℕᵈᵉᶠ is its upper.end,
because
for each diminishable k
diminishable k+1 disproves by counter.example
that k is the upper.end of ℕᵈᵉᶠ
ℕᵈᵉᶠ is one of the other sets.
Removing one element does not diminish ℕᵈᵉᶠ

Finally there is none remaining.
But never, for no interval (0, x),
more than one unit fraction is lost.
Therefore there is only one last unit fraction.

... and ½⋅β both is and is not a lower.bound to
the visible unit.fractions,
which apparently you (WM) accept.