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

On 1/3/2025 3:39 AM, WM wrote:

On 02.01.2025 19:19, Jim Burns wrote:

On 1/2/2025 6:31 AM, WM wrote:

On 31.12.2024 21:26, Jim Burns wrote:

On 12/31/2024 1:20 PM, WM wrote:

Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.

>
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆

>
So it is for all definable natural numbers.

>
So it is for each finite.ordinal.

>
That is impossible,

No.
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
because...
For each ordinal j = ⟦0,j⦆, there is
fuller.by.one ordinal j+1 = ⟦0,j+1⦆ = ⟦0,j⦆∪⦃j⦄ and
fuller.by.two ordinal j+2 = ⟦0,j+2⦆ = ⟦0,j+1⦆∪⦃j+1⦄
For each finite.ordinal j = ⟦0,j⦆
fuller.by.one ⟦0,j+1⦆ = ⟦0,j⦆∪⦃j⦄ is larger than
finite ⟦0,j⦆
by definition of 'finite'.
For each finite.ordinal j = ⟦0,j⦆
fuller.by.two ⟦0,j+2⦆ = ⟦0,j+1⦆∪⦃j+1⦄ is larger than
fuller.by.one ⟦0,j+1⦆
by lemma 1.
[1]
For each finite.ordinal j = ⟦0,j⦆
⎛ by definition of 'finite', j+1 = ⟦0,j+1⦆ is finite
⎜ each k ∈ ⟦0,j+1⦆, ⟦0,k⦆ ⊆ ⟦0,j+1⦆ is finite
⎜ #⟦0,j⦆ < #⟦0,j+1⦆
⎝ more.than.#⟦0,j⦆.many ⟦0,k⦆ are finite.
Therefore,
for each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆
----
[1]
For each finite.ordinal j = ⟦0,j⦆
fuller.by.two ⟦0,j+2⦆ is larger than
fuller.by.one ⟦0,j+1⦆
⎛ Assume otherwise.
⎜ Assume finite ⟦0,j⦆ and
⎜ ⟦0,j+1⦆ is larger than ⟦0,j⦆ but
⎜ ⟦0,j+2⦆ is not larger than ⟦0,j+1⦆
⎜
⎜ ⟦0,j+2⦆ is not larger than ⟦0,j+1⦆
⎜
⎜ exists g one.to.one: ⟦0,j+2⦆ ⇉ ⟦0,j+1⦆
⎜⎛ g(j+2) = g(j+2)   [!]
⎜⎜ g(g⁻¹(j+1) = j+1  [!]
⎜⎝ otherwise g(i) = g(i)
⎜
⎜ define f one.to.one: ⟦0,j+2⦆ ⇉ ⟦0,j+1⦆
⎜⎛ f(j+2) = j+1          [!]
⎜⎜ f(g⁻¹(j+1)) = g(j+2)  [!]
⎜⎝ otherwise f(i) = g(i)
⎜
⎜ Swapping values conserves one.to.one.ness.
⎜
⎜ ⟦0,j+2⦆ = ⟦0,j+1⦆∪⦃j+1⦄
⎜ ⟦0,j+1⦆ = ⟦0,j⦆∪⦃j⦄
⎜
⎜ exists f one.to.one: ⟦0,j+1⦆ ⇉ ⟦0,j⦆
⎜⎛ f(g⁻¹(j+1)) = g(j+2)
⎜⎝ otherwise f(i) = g(i)
⎜
⎜ ⟦0,j+1⦆ is not larger than ⟦0,j⦆
⎜
⎜ However,
⎜ ⟦0,j+1⦆ is larger than ⟦0,j⦆
⎝ Contradiction.
Therefore,
for each finite.ordinal j = ⟦0,j⦆
fuller.by.two [0,j+2} is larger than
fuller.by.one [0,j+1)

That is impossible, because
all finite ordinals can be subtracted from ℕ without infinitely many remaining.

If
⎛ g(j+2) = g(j+2)   [!]
⎜ g(g⁻¹(j+1) = j+1  [!]
⎝ otherwise g(i) = g(i)
is one.to.one,
then
⎛ f(j+2) = j+1          [!]
⎜ f(g⁻¹(j+1)) = g(j+2)  [!]
⎝ otherwise f(i) = g(i)
is one.to.one.

If you don't understand that,
it is useless to go on.