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

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.

 I mean 'exterior' in the topological sense.
 For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

The intervals are closed with irrational endpoints.

 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, and they are closer to interval ends than to the interior, independent of the configuration of the intervals. Note that only 3/oo of the points are inside.

 An exterior point which is not
a positive distance from any interval
is not an exterior point.

Positive is what you can define, but there is much more in smaller distance. Remember the infinitely many unit fractions within every eps > 0 that you can define.

 Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

There are 3/oo of all points exterior.

 Instead, there are boundary points.
  For each x not.in the intervals,
  each open set Oₓ which holds x
  holds points in the intervals and
  points not.in the intervals.
x is a boundary point.

The intervals are closed

The rationals are dense

 Yes.
Each multi.point interval [x,x′] holds
rationals.
 

but the intervals are not.

 No.

Therefore not all rationals are enumerated.

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜  iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝  jₖ = k-iₖ
proves that
the rationals are countable.

Contradiction. Something of your theory is inconsistent.
Regards, WM