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

On 1/15/2025 1:17 PM, WM wrote:

On 15.01.2025 16:16, Jim Burns wrote:

On 1/14/2025 4:07 AM, WM wrote:

Half are new.

>
A step is never from finite to infinite.

>
∀n ∈ ℕ: 2n =/= n.

A step is never from finite to infinite.
⎛ For there to be, after finite j+k₀,
⎜ a first.infinite.sum j+(k₀+1),
⎜ there must be a step from finite j+k₀
⎜ to infinite (j+k₀)+1
⎜  (j+k₀)+1 = j+(k₀+1)
⎜
⎜ There is no step from finite to infinite.
⎜ There is no first.infinite.sum from finites
⎜ By well.order,
⎝ there is no infinite.sum from finites.
Lather, rinse, and repeat.
⎛ For there to be, after finite j×k₀,
⎜ a first.infinite.product j×(k₀+1),
⎜ there must be a sum from finite j and j×k₀
⎜ to infinite j+(j×k₀)
⎜  j+(j×k₀) = j×(k₀+1)
⎜
⎜ There is no infinite.sum from finites.
⎜ There is no first.infinite.product from finites
⎜ By well.order,
⎝ there is no infinite.product from finites.
⎛ For there to be, after finite j^k₀,
⎜ a first.infinite.power j^(k₀+1),
⎜ there must be a product from finite j and j×k₀
⎜ to infinite j×(j^k₀)
⎜  j×(j^k₀) = j^(k₀+1)
⎜
⎜ There is no infinite.product from finites.
⎜ There is no first.infinite.power from finites
⎜ By well.order,
⎝ there is no infinite.power from finites.

A step is never from finite to infinite.

>
∀n ∈ ℕ: 2n =/= n.

There is no infinite.product from finites.
∀n ∈ ℕ:  2×n ∈ ℕ