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

On 1/5/2025 1:14 PM, WM wrote:

On 05.01.2025 19:03, joes wrote:

Am Sun, 05 Jan 2025 12:14:47 +0100 schrieb WM:

On 04.01.2025 21:38, Chris M. Thomasson wrote:

For me,
there are infinitely many natural numbers, period...
Do you totally disagree?

>
No.
There are actually infinitely many natural numbers.
All can be removed from ℕ, but only collectively
ℕ \ {1, 2, 3, ...} = { }.
It is impossible to remove the numbers individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

>
Well yes,
the size of N is itself
not a natural number.
Big surprise.

>
ℕ cannot be covered by FISONs,
neither by many nor by their union.
If ℕ could be covered by FISONs
then one would be sufficient.

ℕ is the set of finite.ordinals.
ℕ holds each finite ordinal.
ℕ holds only finite.ordinals.
⎛ A FISON is a set of finite.ordinals
⎝ up to that FISON's maximum (finite.ordinal) element.
A finite.ordinal is an ordinal
smaller.than fuller.by.one sets.
Lemma 1.
⎛ For sets A∪{a} ≠ A and B∪{b} ≠ B
⎜⎛ if A is smaller.than B
⎜⎝ then A∪{a} is smaller.than B∪{b}
⎝ #A < #B  ⇒  #(A∪{a}) < #(B∪{b})
Lemma 1
is true for both the darkᵂᴹ and the visibleᵂᴹ.
Consider finite.ordinal k.
Finite: ⟦0,k⦆ is smaller.than ⟦0,k⦆∪⦃k⦄
A = ⟦0,k⦆
A∪{a} = ⟦0,k⦆∪⦃k⦄
B = ⟦0,k⦆∪⦃k⦄ = ⟦0,k+1⦆
B∪{b} = (⟦0,k⦆∪⦃k⦄)∪⦃k+1⦄ = ⟦0,k+1⦆∪⦃k+1⦄
⎛ By lemma 1
⎜ if ⟦0,k⦆ is smaller.than ⟦0,k+1⦆
⎜ then ⟦0,k⦆∪⦃k⦄ is smaller.than ⟦0,k+1⦆∪⦃k+1⦄
⎜
⎜ If
⎜ k is in ℕ and
⎜ k is finite and
⎜ ⟦0,k⦆ is smaller.than ⟦0,k⦆∪⦃k⦄
⎜ then
⎜ ⟦0,k+1⦆ is smaller.than ⟦0,k+1⦆∪⦃k+1⦄ and
⎜ k+1 is finite and
⎝ k+1 is in ℕ.
k ∈ ℕ  ⇒  k+1 ∈ ℕ
is true for both the darkᵂᴹ and the visibleᵂᴹ.

If ℕ could be covered by FISONs
then one would be sufficient.

ℕ is the set of finite.ordinals.
A FISON is a set of finite.ordinals
up to that FISON's maximum (finite.ordinal) element.
If one FISON covered ℕ,
that FISON.cover would equal ℕ,
and the maximum of that FISON.cover
would be the maximum.of.all finite.ordinal.
However,
no finite.ordinal k is the maximum.of.all.
k ∈ ℕ  ⇒  k+1 ∈ ℕ
That is true for both the darkᵂᴹ and the visibleᵂᴹ.
Contradiction.
No one FISON covers ℕ.

ℕ cannot be covered by FISONs,
neither by many nor by their union.

No.
ℕ is the set of finite ordinals.
Each finite.ordinal k is in
at least one FISON: ⟦0,k⟧
Each finite.ordinal is in
the union of FISONs
The union of FISONs covers
the set ℕ of finite.ordinals

But for all we have:
Extension by 100 is insufficient.

Correct.
Which is weird, but accurate.
The source of that weird result is lemma 1.
⎛ For sets A∪{a} ≠ A and B∪{b} ≠ B
⎜⎛ if A is smaller.than B
⎜⎝ then A∪{a} is smaller.than B∪{b}
⎝ #A < #B  ⇒  #(A∪{a}) < #(B∪{b})
It would be great if you (WM) did NOT
find lemma 1 weird,
but it is what it is.