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

On 12/8/2024 5:34 PM, WM wrote:

On 08.12.2024 19:01, Jim Burns wrote:

On 12/8/2024 5:50 AM, WM wrote:

∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't you understand here?

>
{E(i):i} is the set.of.all
non.empty end.segments.

of all the finite.cardinals.

⋂{E(i):i} is the intersection.of.all

non.empty end.segments.

of all the finite.cardinals.
Yes.
Only the non.empty,
because,
you and I are currently testing the claim that
⎛ there cannot be
⎜ an empty intersection and
⎝ no empty end.segment among the intersected.
To include an empty end.segment
would leave that claim untested.

∀n ∈ ℕ:
{E(i):i}∪{E(n+1)} = {E(i):i}
Each is "already" in.

>
Not the empty endsegment.
∀n ∈ ℕ: E(n) is non-empty.

Yes.
∀n ∈ ℕ:  n ∈ E(n) ≠ {}

But not every E(n+1).

No.
E(n) and E(n+1) are end.segments
of all the finite.cardinals.
For each E(n), there is E(n+1) ≠ {}

∀n ∈ ℕ:
(⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}

>
No. E(n+1) = { } is possible.

No.
If E(n) is a non.empty end.segment
  of all the finite.cardinals,
then E(n+1) is a non.empty end.segment
  of all the finite.cardinals.
Finite ⟦0,n⟧ and not.finite ⟦0,n+1⟧ is impossible.
⎛ Assume otherwise.
⎜ Assume finite ⟦0,n⟧ and not.finite ⟦0,n+1⟧
⎜
⎜⎛ finite ⟦0,n⟧
⎜⎜ no one.to.one from ⟦0,n⟧ to ⟦1,n⟧ exists
⎜⎝ ⟦0,n⟧ ⇉| ⟦1,n⟧
⎜
⎜⎛ not.finite ⟦0,n+1⟧
⎜⎜ g: ⟦0,n+1⟧ ⇉ ⟦1,n+1⟧: one.to.one  exists
⎜⎝ ⟦0,n+1⟧ ⇉ ⟦1,n+1⟧
⎜
⎜ However,
⎜ f: ⟦0,n⟧ ⇉ ⟦1,n⟧: one.to.one  exists,
⎜ defined as
⎜⎛ f(g⁻¹(n+1)) = g(n+1)
⎜⎝ otherwise f(k) = g(k)
⎜ ⟦0,n⟧ ⇉ ⟦1,n⟧
⎝ Contradiction.
Therefore,
finite ⟦0,n⟧ and not.finite ⟦0,n+1⟧ is impossible.
If E(n) is a non.empty end.segment
  of all the finite.cardinals,
then
⎛ n ∈ E(n)
⎜ n+1 ∈ E(n)
⎜ E(n+1) = E(n)\{n} exists
⎜ E(n+1) is a non.empty end.segment
⎝  of all the finite.cardinals.
For ℕ = ⟦0,ℵ₀⦆
∀n ∈ ℕ: E(n+1) ≠ {}

∀n ∈ ℕ:
⎛ E(1)∩E(2)∩...∩E(n) = E(n)
⎜ E(n)∩E(n+1) ≠ E(n)

>
E(n)∩E(n+1) = E(n+1)

E(n+1) ≠ E(n)

⎜ (⋂{E(i):i})∩E(n+1) = ⋂{E(i):i}

>
No.
⋂{E(i):i} is the intersection of non-empty endsegments
and as such non-empty.

⋂{E(i):i} is the intersection of ALL non-empty end.segments
and, as such,
its intersection with any non.empty end.segment
results in the same set.
On the other hand,
E(n) intersected with E(n+1)
results in a different set.
Thus, E(n) isn't ⋂{E(i):i}
For ℕ = ⟦0,ℵ₀⦆
∀n ∈ ℕ: E(n) ≠ ⋂{E(i):i}

⎝ E(n) ≠ ⋂{E(i):i}

>
Not for all n.

Not for finite n = n+1
Those n don't exist.

⋂{E(i):i} ∉ {E(i):i}
The intersection.of.all non.empty end.segments
isn't any non.empty end.segment.