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

On 12/11/2024 9:48 AM, WM wrote:

On 11.12.2024 05:06, Jim Burns wrote:

Which, for end.segments of final.cardinals,
they are not the same.
>
#⋂{E(i):i} = #{} = 0
>
⋂{#E(i):i} = ⋂{ℵ₀:i} = ℵ₀

>
There is no use for # but only for
⋂{E(i):i} = {}.
>
The problem is that
every definable endsegment is infinite.
But if all natural numbers are applied
in a Cantor bijection,
then they all must leave the endsegments
by the only possible process:
E(k+1) = E(k) \ {k}

Each "leaves" by
not.being.in.common.with.all.end.segments.
∀k ∈ ℕ:
k ∉ E(k)\{k} = E(k+1) ⊇ ⋂{E(i):i} ∌ k
⋂{E(i):i} = {}.
----
A finite.cardinal is Original Cardinal.
⎛ When your sheep head out to the field to graze,
⎜ as each sheep passes, put a pebble in your pocket.
⎜ When they head in at end.of.day,
⎜ as each sheep passes, take a pebble out.
⎜ When your pocket is empty, all your sheep are in.
⎝ That's Original Cardinal.
Define a finite.cardinal as one of
the (well.ordered) ordinals which
are larger.than one.element.emptier
and smaller.than one.element.fuller.
⎛ A larger set cannot map one.to.one
⎝ into a smaller set.
The new cardinal which is grown.to or shrunk.to
is also finite, which means that
the new can also grow by one and
shrink by one, unless it's zero.
⎛ For all sets, even for dark.finite.cardinals,
⎜ if S is
⎜ not.smaller.than one.element.fuller S∪{c}
⎜ then S is
⎜ not.larger.than one.element.emptier S\{d}
⎜
⎜ Assume
⎜ S is not smaller than S∪{c}
⎜ one.to.one g: S∪{c} ⇉ S
⎜
⎜ Define
⎜⎛ one.to.one f: S ⇉ S\{d}
⎜⎜ f(g⁻¹(d)) = g(c)
⎜⎝ otherwise f(x) = g(x)
⎜
⎝ S is not.larger.than S\{d}
ℕ is the set of all finite.cardinals.
If dark.finite.cardinals can grow by one,
then they are in ℕ too, but otherwise not.
ℕ\{0} is one.element.emptier than ℕ
However,
ℕ\(0) is not.smaller.than ℕ
⎛ Assume otherwise.
⎜ Assume finite.cardinal 𝔊 exists such that
⎜ ⟦0,𝔊⦆ ⇉ ℕ
⎜ ⟦0,𝔊+1⦆ ⇉| ℕ
⎜
⎜ However,
⎜ 𝔊+1 ∈ ℕ
⎜ ⟦0,𝔊+1⦆ ⊆ ℕ
⎜ ⟦0,𝔊+1⦆ ⇉ ℕ
⎝ Contradiction.
Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ
That is the reason that
the end.segments of ℕ
stay the same larger.than.any.finite size
as they lose each element of ℕ.