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

On 12/29/2024 8:15 PM, Ross Finlayson wrote:

On 12/29/2024 01:41 PM, Jim Burns wrote:

On 12/29/2024 3:54 PM, Ross Finlayson wrote:

Oh, you had that [0/d, 1/d, 2/d, ... oo/d]
was a thing,

>
No.
What I had was that
⋃ᵢ₌᳹₀[0/i,1/i,2/i,...,i/i]
is a thing,
but the thing which it is
is not an interval of real numbers.
>
That's still my position.
>

So, yeah, it conflicts with yourself,
yet, that's what you said.

>
No,
it isn't [what I said].

>
It's also an equi-partitioning.
>
Here of course it IS a continuous domain
or model of real numbers, ran(EF),
because being a distribution of the
natural integers at uniform random.
>
Well, go ahead then and come up with
some reason why it's not an equi-partitioning
as it fulfills otherwise being a CDF.

⋃ᵢ₌᳹₀[0/i,1/i,2/i,...,i/i] = {q∈ℚ:0≤q≤1} = [0,1]ꟴ
I suppose that one could stretch 'equi.partitioning'
into {[q,q]⊆[0,1]ꟴ}
which covers [0,1]ꟴ in
intervals each of the same (zero) length.
⋃{[q,q]⊆[0,1]ꟴ} = [0,1]ꟴ
However,
I doubt that you are referring to {[q,q]⊆[0,1]ꟴ}
Anyway, no points of ℝ\ℚ are in any of those intervals.
Which is why [0,1]ꟴ is not.connected
(you say 'not.continuous')
[0,1]ꟴ is the union of disjoint open sets.
For example, [0,⅟π)ꟴ∪(⅟π,1]ꟴ = [0,1]ꟴ

Figuring you just can't accept it, ....

"What I tell you three times is true."
-- The Bellman
----
⋃ᵢ₌᳹₀[0/i,1/i,2/i,...,i/i]
is my _least.exclusive_ guess at
what you (RF) mean when you write
d→∞ n→d n/d
I read n→d n/d as the set ⟨n/d⟩ₙͩ₌₀
I read d→∞ ⟨n/d⟩ₙͩ₌₀ as a set.limit.
I haven't yet seen how you (RF) define that limit,
other than defining it to give you the answer you want.
The least.exclusive reading excludes all
points not.in almost.all sets in the sequence.
That works out to be ⋃₁᳹⟨n/d⟩ₙͩ₌₀ = [0,1]ꟴ
[0,1]ꟴ ∌ ⅟π
[0,1]ꟴ is not.connected ('not.continuous')