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

On 12/14/2024 5:26 AM, WM wrote:

On 14.12.2024 05:54, Jim Burns wrote:

On 12/13/2024 2:31 PM, WM wrote:

On 13.12.2024 20:00, Jim Burns wrote:

On 12/13/2024 6:25 AM, WM wrote:

If endegments were defined as
E(n) = {n+1, n+2, ...}:
>
E(0) = {1, 2, 3, ...}
E(1) = {2, 3, 4, ...}
E(2) = {3, 4, 5, ...}
...
E(ω-1} = { }.
Then this change from content to index
would even be more obvious.

One problem which
Eᑉ(ω-1) = {}
has  is that
'finite' is NOT defined the way in which
you (WM) think 'finite' should be,
which means
ω is NOT defined the way in which
you (WM) think ω should be.

>
Don't say what not is.

For sets A and B with one.to.one A.to.B
B is not.smaller.than A
[1] (see below)
For non.empty sets A and B
such that B is not.smaller.than A
emptier.by.one B\{b} is not.smaller.than
emptier.by.one A\{a}
A finite ordinal is smaller.than its successor.
An infinite ordinal is not.smaller.than its successor.
[2] (see below)
Each infinite ordinal bounds all finite ordinals.
ω is the first upper bound of finite ordinals.

Which means
⎛ if k is finite
⎝ then k+1 is finite

>
This contradicts the fact that
nothing remains but
every element can go only as a single.

⎛ if k is finite
⎝ then k+1 is finite
⎛ Assume otherwise.
⎜ Assume k is finite and k+1 is infinite.
⎜
⎜ k+1 is infinite.
⎜ ⟦0,k+1⦆ is not.smaller.than ⟦0,k+1⦆∪{k+1}
⎜ ⟦0,k+1⦆ = ⟦0,k⦆∪{k}
⎜
⎜ ⟦0,k⦆∪{k} is not.smaller.than
⎜ ⟦0,k+1⦆∪{k+1}
⎜
⎜ From [1] above, it follows that
⎜ emptier.by.one ⟦0,k⦆ is not.smaller.than
⎜ emptier.by.one ⟦0,k+1⦆
⎜ ⟦0,k+1⦆ = ⟦0,k⦆∪{k}
⎜
⎜ ⟦0,k⦆ is not.smaller.than ⟦0,k⦆∪{k}
⎜ k is infinite.
⎜
⎜ However, k is finite.
⎝ Contradiction.
[3]
Therefore,
⎛ if k is finite
⎝ then k+1 is finite

E(ω-1} = { }.

Don't say what not is.

If ω-1 exists
then
⎛ ω-1 is last.before.ω
⎜ ω is first bound of the finites
⎜ ω-1 is not any before.ω bound of the finites
⎜ ω-1 is not infinite (see [2])
⎜ ω-1 is finite
⎜ ω-1 is smaller.than (ω-1)+1, also finite (see [3])
⎜ (ω-1)+1 is smaller.than (ω-1)+2, also finite
⎜ ω bounds w-1, (ω-1)+1, (ω-1)+2
⎜ ω-1 < (ω-1)+1 < (ω-1)+2 ≤ ω
⎝ ω-1 is not last.before.ω
If ω-1 exists, ω-1 gives us the contradictions.
Therefore,
ω-1 doesn't exist.
----
[1]
For non.empty sets A and B such that
B is not.smaller.than A
emptier.by.one B\{b} is not.smaller.than
emptier.by.one A\{a}
B is not.smaller.than A
one.to.one g: A ⇉ B
Define
⎛ f(g⁻¹(b)) = g(a)
⎝ otherwise f(x) = g(x)
one.to.one f: A\{a} ⇉ B\{b}
B\{b} is not.smaller.than A\{a}
----
[2]
Each infinite ordinal bounds all finite ordinals.
⎛ Assume otherwise.
⎜ Assume ¬(finite k < infinite ψ)
⎜
⎜ From [4] below
⎜ for two (well.ordered) ordinals,
⎜ whether finite or infinite,
⎜ (k < ψ) ∨ (ψ < k)
⎜
⎜ infinite ψ < finite k
⎜ ⟦0,ψ⦆ s ⟦0,k⦆
⎜
⎜ infinite ψ
⎜ ⟦0,ψ⦆ is not.smaller.than ⟦0,ψ⦆∪{ψ}
⎜ ⟦0,ψ⦆ is not.smaller.than ⟦0,ψ⦆∪{k}
⎜ ⟦ψ,k⦆ is not.smaller.than ⟦ψ,k⦆
⎜ ⟦0,ψ⦆∪⟦ψ,k⦆ is not.smaller.than ⟦0,ψ⦆∪⟦ψ,k⦆∪{k}
⎜ ⟦0,k⦆ is not.smaller.than ⟦0,k⦆∪{k}
⎜ infinite k
⎜
⎜ However,
⎜ k is finite.
⎝ Contradiction.
Therefore,
each infinite ordinal bounds all finite ordinals.
----
[4]
For two of the (well.ordered) ordinals,
whether finite or infinite,
(k < ψ) ∨ (ψ < k)
Non.empty {k,ψ} holds a first element.
Either it's k, and k < ψ
or it's ψ, and ψ < k