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

On 12/21/2024 4:58 PM, WM wrote:

On 21.12.2024 20:32, Jim Burns wrote:

E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.

>
No.
E(n+2) is smaller than E(n+1)
by one element, namely n+2.

What does 'finite' mean?
One way to answer is:
⎛ Set A is finite if A∪{x} is larger, for A∌x
⎝ Set Y is infinite if Y∪{x} isn't larger, for Y∌x
ℕ is the set of finite.cardinals.
which means
ℕ is the set of cardinals #A of
  sets A smaller.than A∪{x} for A∌x
For each finite set A,
its cardinality #A is in ℕ
For each set Y without #Y in ℕ
Y is infinite, and
Y∪{x} isn't larger than Y, for Y∌x
E(n+2) is the set of all finite.cardinals > n+2
E(n+1)  =  E(n+2)∪{n+2}
E(n+2)∪{n+2} isn't larger.than E(n+2)
⎛ because
⎜ E(n+2) is without #E(n+2) in ℕ
⎜⎛ because,
⎜⎜ for each finite.cardinal j in ℕ
⎜⎜ #E(n+2) isn't j
⎜⎜⎛ because
⎜⎜⎜ E(n+2) contains
⎝⎝⎝ a larger.than.j subset E(n+2)\E(n+j+3)
#E(n+2) isn't any of the finite.cardinals in ℕ
E(n+2) isn't any of the sets
smaller.than fuller.by.one sets.
E(n+2) is emptier.by.one than E(n+1), but
E(n+2) isn't smaller.by.one than E(n+1).