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

On 06.11.2024 15:22, Jim Burns wrote:

On 11/6/2024 5:35 AM, WM wrote:

On 05.11.2024 18:25, Jim Burns wrote:

On 11/4/2024 12:32 PM, WM wrote:

 

The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

>
⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

>
This plentiness does not change
when the intervals are translated.

 ⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.
 Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?

In fact, when ordered that way they include only intervals around the positive integers because the natural numbers already are claimed to be indexing all fractions. (Hence not more intervals are required.)

 Isn't that a plentiness which changes?

No.

The intervals are closed with irrational endpoints.

 'Exterior' seems like a good way to say
'not in contact'.

Every point outside is not an endpoint and is not in contact.

 It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.

Closed intervals with irrational endpoints prohibit any point outside. Boundary or not! Of course there are points outside. This shows that the rationals are not countable.

 Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.
 

Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.
>
There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.

>
Nice try.
But there are points outside of intervals,

 Are any of these points.outside
  ⅟2 from any interval? ⅟3? ⅟4?

If not, then the measure of the real axis is less than 3.>

If there is no point with more.than.⅟2
between it and any midpoint,

In your first configuration, there are points with more than 1/3 between it and a midpoint. If the intervals are translated, the distance may become smaller for some points but necessarily becomes larger for others. Shuffling does not increase the sum of the intervals.

There are 3/oo of all points exterior.

 Did you intend to write "interior"?

Of course.

 An exterior point is in
an open interval holding no rational.

and therefore no irrational either.

 There are no
open intervals holding no rational.

That is true but shows that not all rationals are caught in intervals because they are not countabel.

 There are no exterior points.

If and only if all rationals could be enumerated!
That has been disproved.

Therefore not all rationals are enumerated.

 Explain why.

There are rationals outside of all intervals in the infinite space outside of the intervals covering less than 3 of the infinite space.

Contradiction.

 It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.

There is no difference between outside, exterior and you "boundary points". The latter are only created by your inability to define small enough intervals.

 

Something of your theory is inconsistent.

 Your intuition is disturbed by
an almost.all boundary.

No. Your boundary is nonsense. If a point is outside of an interval, then it is irrelevant whether you can construct an open interval not covering points of the interior. It is outside.

 Disturbed intuitions and inconsistencies
are different.

Tricks relating to your inability are not acceptable.
Regards, WM