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

On 11/10/2024 4:35 AM, WM wrote:

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.

The set
   {3,4,5}
does not _change_ to the set
   {6,7,8}
because
our sets do not change.
The word 'set' has many uses.
In some uses of 'set', sets change.
Break a plate, and your dinnerware set changes.
Not here.
In this use of 'set',
as in mathematics more generally,
sets do not change.
Add 3 to each of {3,4,5}.
You do not change {3,4,5}.
{3,4,5} continues to be {3,4,5}.
You get a different set, {6,7,8].
{6,7,8} has never been {3,4,5}.
The not.changing nature of our sets is a choice,
although it is such an ancient, universal, and
enormously.useful choice that it is easy
to overlook that a choice has been made.
But our sets could have been like dinnerware sets.
However,
our sets are not like dinnerware sets.
Our sets do not change.
⎛ I have argued for the correctness of that choice.
⎜ But, however good or bad my argument is,
⎜ that is the choice.
⎜ Pretending otherwise is like
⎝ pretending dogs are cats.

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

"It" refers to who or what?
Me? You? Chuck Norris? That finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲?
I will continue talking about that 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.

But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
For that finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲, because 𝗶𝘁 is finite,
it is enough for us to see that
each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true.or.not.first.false.
If we see that, we know that
each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true.
Each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true,
even if
a 𝗰𝗹𝗮𝗶𝗺 refers to one of infinitely.many.
There may be infinitely.many possible.referents,
but we know the 𝗰𝗹𝗮𝗶𝗺 is true
because
there are only finitely.many 𝗰𝗹𝗮𝗶𝗺𝘀.

It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.

A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false,
will forever remain
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false.

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.

Our sets do not change.
If matheologians would "concede" a claim which
they do not make and have never made,
would you (WM) stop
filling your students' heads with that gibberish?

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.
Our sets do not change.
? That is prohibited by geometry.
Consider geometry.
Similar triangles have
corresponding sides in the same ratio.
Consider these points, line.segments, and triangles
in the ⟨x,y⟩.plane
A = ⟨0,-1⟩
B = ⟨0,0⟩
C = ⟨x,0⟩  with 0 < x < 1
D = ⟨1,0⟩
E = ⟨1,y⟩  with points A C E collinear.
△ABC and △EDC are similar
△ABC ∼ △EDC
μA͞B = 1
μB͞C = x
μE͞D = y
μD͞C = 1-x
Similar triangles.
μA͞B/μB͞C = μE͞D/μD͞C
1/x = y/(1-x)
y = 1/x - 1
x = 1/(y+1)
To each point C = ⟨x,0⟩ in (0,1)×{0}
there corresponds
exactly one point E = ⟨1,y⟩ in {1}×(0,+∞)
and vice versa.
(0,1)×{0} is not stretched over {1}×(0,+∞)
{1}×(0,+∞) is not shrunk to (0,1)×{0}
They both _are_
And their points correspond
by line A͞C͞E through point A.
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