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

On 12/15/2024 2:16 PM, WM wrote:

On 15.12.2024 19:53, Jim Burns wrote:

On 12/15/2024 7:00 AM, WM wrote:

Unless you have changed whatᵂᴹ you (WM) mean,
an actuallyᵂᴹ infinite set is smaller.than
a fuller.by.one set, but
it contains a potentiallyᵂᴹ infinite subset, meaning
a subset not.smaller.than a fuller.by.one set. >
Unless you have changed whatᵂᴹ you (WM) mean,
to completeᵂᴹ a potentiallyᵂᴹ infinite set means
to insert an epilogue (presumably darkᵂᴹ) so that
set+epilogue is actuallyᵂᴹ infinite.

>
This epilogue is required to empty ℕ
by |ℕ \ {1, 2, 3, ...}| = 0.
All definable numbers fail:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Each darkᵂᴹ or visibleᵂᴹ epilogue 𝔻 such that
∀d ∈ 𝔻: g(d) = d
fails to completeᵂᴹ ℕ_def.
⎛ ℕ_def
⎜⎛ ℕ_def ⊆ A  ⇐  A ∋ 0 ∧ ∀a ∈ A: A ∋ n+1
⎜⎝ ℕ_def ∋ 0 ∧ ∀n ∈ ℕ_def: ℕ_def ∋ n+1
⎜ does not change,
⎜ maybe is called.by.you not.fixedᵂᴹ anyway,
⎜ and is potentiallyᵂᴹ infinite:
⎜⎛ ∀n ∈ ℕ_def: f(n) = n+1
⎜⎜ f(Bob) = 0
⎜⎝ one.to one f: ℕ_def∪{Bob} ⇉ ℕ_def
⎜
⎜ ℕ_def is not.smaller.than ℕ_def∪{Bob}
⎝ ℕ_def is potentiallyᵂᴹ infinite.
Epilogue 𝔻 does not completeᵂᴹ ℕ_def:
⎛ ∀d ∈ 𝔻: g(d) = d
⎜ ∀n ∈ ℕ_def: g(n) = n+1
⎜ g(Bob) = 0
⎝ one.to one g: ℕ_def∪𝔻∪{Bob} ⇉ ℕ_def∪𝔻
ℕ_def∪𝔻 is not.smaller.than ℕ_def∪𝔻∪{Bob}
ℕ_def∪𝔻 is potentiallyᵂᴹ infinite.
𝔻 does not completeᵂᴹ ℕ_def
A potentiallyᵂᴹ infinite set derives
its Bob.disappearing property from being
larger.than each set without that property.
Inserting epilogue 𝔻 makes a new set
also larger.than each set without that property,
also potentiallyᵂᴹ infinite,
and not completedᵂᴹ.

(1) E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content.

>
the set of common finite.ordinals is empty.

>
Fine.

Each finite.cardinal
leaves after that.many steps,
with further steps to follow,
more steps than any other finite.cardinal.

By the law
(2) ∀k ∈ ℕ :
∩{E(1),E(2),...,E(k+1)} =
∩{E(1),E(2),...,E(k)}\{k}
the sequence gets empty one by one.

>

The limit set {}
⎛ holds all common finite.ordinals.
⎝ isn't in the sequence.

>
If all natural numbers are
individually available for indexing
then they are available for
individually leaving the intersection.

Yes.
Each finite.cardinal
indexes its own end.segment,
and leaves at the next end.segment.