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

On 10.11.2024 00:27, Jim Burns wrote:

On 11/9/2024 6:45 AM, WM wrote:

Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

 Our sets do not change.
 The set
   {[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
   ⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
   {[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
   ⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩

It cannot do so because the reality of the rationals is much larger than the reality of the naturals.

 ----
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.
  In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
  has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.

But it will never complete an infinite set of claims. It will forever remain in the status nascendi. Therefore irrelevant for actual or completed infinity.
So yes, you can shift the intervals to midpoints of every finite initial segment of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ...
But you must remove them from larger natural numbers. That will never change.

In the second case, with the changing sets,
who knows?
Perhaps something else could be done,
but not that.

Certainly not. The intervals can neither grow in size nor in multitude.

Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
 They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i

But they cannot be completely transformed into each other. That is prohibited by geometry. It is possible for every finite initial segment of the above sequence, but not possible to replace all the given intervals to cover all rational midpoints.
Regards, WM