Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 07. Nov 2024, 20:06:25
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <d79e791d-d670-4a5a-bd26-fdf72bcde6bc@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13
User-Agent : Mozilla Thunderbird
On 11/7/2024 3:46 AM, WM wrote:
On 06.11.2024 21:20, Jim Burns wrote:
[...]
>
Use the intervals
I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n].
Since n and q_n can be in bijection,
these intervals are sufficient to cover all q_n.
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
Yes. Except boundaries.
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S
ℝ is the set of points.between.splits of ℚ
Consider ℝ.points x < x″ with
their foresplits F F″ and hindsplits H H″ of ℚ
F ᵉᵃᶜʰ< x ≤ᵉᵃᶜʰ H and F∪H = ℚ
F″ ᵉᵃᶜʰ< x ≤ᵉᵃᶜʰ H″ and F″∪H″ = ℚ
F ≠⊂ F″
H ⊃≠ H″
F″∩H ≠ {}
F″∩H is
the set of rationals between x and x″
F″∩H is not empty, because x < x″
Each ℝ.interval [x,x″] holds rationals.
There is an enumeration of ℚ⁺
the set of ratios of ℕ⁺ countable.to from.1
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ/jₖ
⎜ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-((iₖ+jₖ)-1)((iₖ+jₖ)-2)/2
⎜ jₖ = (iₖ+jₖ)-iₖ
⎝ (iₖ+jₖ-1)(iₖ+jₖ-2)/2+iₖ = k
⎛ If you must have ℚ instead of ℚ⁺
⎜ I can add bells and whistles to the enumeration.
⎜ It won't significantly change the argument,
⎝ but it will give you more squiggles to look at.
There is an ε.cover of ℚ⁺
-- not scrunched together, but spread across ℚ⁺ --
of closed intervals with irrational endpoints
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜ x⁽ᵋₖ = iₖ/jₖ-2ᵏ⁻¹ᐟ²⋅ε
⎝ xᵋ⁾ₖ = iₖ/jₖ+2ᵏ⁻¹ᐟ²⋅ε
Consider a point xᵒᵘᵗ not.in the ε.cover.
xᵒᵘᵗ ∉ ⋃(ε.cover)
For each interval [x,x″] such that x < xᵒᵘᵗ < x″
there are points in ℚ⁺ which are also in ⋃(ε.cover)
there are points not.in ⋃(ε.cover): xᵒᵘᵗ
xᵒᵘᵗ is in the boundary of ⋃(ε.cover)
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".
Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.
A boundary so much more than its set
is not what we think of as "boundary"
but ℚ⁺ is not what we thing of as "interval".