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

followups to sci.math sci.logic
On 11/3/2024 3:38 AM, WM wrote:

Apply Cantor's enumeration of
the rational numbers q_n, n = 1, 2, 3, ... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].

= [xᵋₙ,xᵋₙ′]
xᵋₙ = qₙ - 2¹ᐟ²⁻ⁿ⋅ε
xᵋₙ′ = qₙ + 2¹ᐟ²⁻ⁿ⋅ε
μ[xᵋₙ,xᵋₙ′] = μ(xᵋₙ,xᵋₙ′) = 2³ᐟ²⁻ⁿ⋅ε

Let ε --> 0.
Then all intervals together have
a measure m < 2ε*sqrt(2) --> 0.

Yes.
For each ε > 0
⎛ there is an open ε.cover {(xᵋₙ,xᵋₙ′)ᴿ: n∈ℕ} of ℚ
⎜ ℚ ⊆ int.⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}
⎜ and
⎜ 0  ≤  μ(ℚ)  ≤
⎜ μ(⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ})  ≤¹
⎜ ∑ₙ2³ᐟ²⁻ⁿ⋅ε  =  2³ᐟ²⋅ε
⎜
⎜ ¹ '≤' not '=' because the intervals (xᵋₙ,xᵋₙ′)ᴿ
⎝ are in.line, not in.sequence, and they overlap.
0  ≤  μ(ℚ)  ≤
  glb.{μ(⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}): 0 < ε ∈ ℝ}  ≤
  glb.{2³ᐟ²⋅ε: 0 < ε ∈ ℝ}  =  0
0  ≤  μ(ℚ)  ≤  0

By construction there are
no rational numbers outside of the intervals. Further there are never
two irrational numbers without
a rational number between them.
>
This however would be the case

Maybe "This however would NOT be the case"
was intended?

This however would [not(?)] be the case
if an irrational number existed between
two intervals with irrational ends.

No.
It is still the case.
I admit that I find this difficult to picture,
but the mathematics is clear.
The intervals (xᵋₙ,xᵋₙ′)ᴿ of an ε.cover
are not strung out like a string of pearls.
They overlap, with each interval containing
a0.many smaller intervals.
Their total measure is < 2³ᐟ²⋅ε but they're smeared
like an infinitely.thin coat of butter on toast.
They're more cloud.like than necklace.like.
If you are imagining one.point "limit" intervals,
consider that
for each ε > 0
ℚ is in the interior of ⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}
and the upper bound of that set's measure
is not problematic.
"In the limit",
ℚ is not in the interior, and,
if we naively say μ(ℚ) = ℵ₀⋅0
what, exactly, is that? 0? ℵ₀?
The naive method only trades questions
for other questions.

(Even the existence of neighbouring intervals
is problematic.)

There aren't any neighboring intervals.
Any two intervals have intervals between them.

Therefore
there is nothing between the intervals,
and the complete real axis has measure 0.
>
This result is wrong
but implied by the premise that
Cantor's enumeration is complete.