Re: Incompleteness of Cantor's enumeration of the rational numbers

On 11/7/2024 12:59 PM, WM wrote:

On 07.11.2024 16:29, Jim Burns wrote:

⎛ 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

>
Do you think you need the boundary in my last example?
>
When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.

No irrational is not in contact with
the union of intervals.
That's different from
⎛ no irrational is not covered by
⎝ the union of intervals.
The first is true, the second is false.
The first is true because,
for each irrational x,
each interval of which x is in its interior
holds rationals, and
rationals are points in the union of intervals.
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
#x#ₖ is the -k.th digit of
the decimal representation of real number x
d is a Cantorian anti.diagonal of
the Cantorian rational.list ⟨iₖ/jₖ⟩
#d#ₖ = (#iₖ/jₖ#ₖ+5) mod 10
The closest that d and iₖ/jₖ could be
would be if, miraculously, all prior digits matched.
Even with that and the worst case for following digits,
|d-iₖ/jₖ| ≥ 4×10⁻ᵏ
Consider this ε.cover of
closed intervals with irrational endpoints.
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜  x⁽ᵋₖ = iₖ/jₖ-2¹ᐟ²⋅10⁻ᵏ
⎝  xᵋ⁾ₖ = iₖ/jₖ+2¹ᐟ²⋅10⁻ᵏ
The infinite sum of measures = 2³ᐟ²/9
d is _not in contact with_ each interval.
However,
each interval of which d is in its interior
holds rationals, which are points in ⋃(ε.cover)
Thus, d is _in contact with_ ⋃(ε.cover)
but not with any of its intervals.

Do you believe this???

Don't you believe this???