Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 11/17/2024 2:50 AM, WM wrote:

On 16.11.2024 23:36, Jim Burns wrote:

On 11/16/2024 2:54 PM, WM wrote:

Therefore
the set of intervals cannot grow.

>
An infinite set can match a proper superset
without growing.

>
But with shrinking.

No.
An infinite set
can match some proper supersets without growing  and
can match some proper subsets without shrinking.
Sets which can't aren't infinite.

When it matches
first itself and then a proper subset,
then it has decreased.

There is no even number which,
after the evens matching the integers,
will not be an even number.
There is no even number which,
before the evens matching the integers,
was not an even number.
Even numbers do not change their evenicity.
Our sets do not change.

The set of even numbers has fewer elements
than the set of integers.

The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.
Because it is infinite.
A _finite_ set can be ordered so that
each non.empty subset holds a first and a last.
An infinite set is _not.that_
However an infinite set is ordered,
some of its non.empty subsets
don't have a first or don't have a last,
not visibly and not darkly.

Because it is infinite.

>
The interval [0, 1] is infinite because
it can be split into infinitely many subsets.
But its measure remains constant.

There is no reason   except naivety
to believe that the intervals [n - 1/10,  n + 1/10]
could cover the real line infinitely often.

There is no reason except
naivete and an almost fanatical devotion to the Pope.
https://www.youtube.com/watch?v=D5Df191WJ3o
No, wait! There is no reason except
naivete, an almost fanatical devotion to the Pope, and
⎛ k ↦ ⟨i,j⟩ ↦ k
⎜
⎜ (i+j) := ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ i := k-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = k