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

On 12/28/2024 11:50 AM, WM wrote:

On 28.12.2024 15:12, Jim Burns wrote:

On 12/27/2024 5:24 PM, Ross Finlayson wrote:

The, "almost all", or, "almost everywhere",
does _not_ equate to "all" or "everywhere",

>
Correct.
⎛ In mathematics, the term "almost all" means
⎜ "all but a negligible quantity".
⎜ More precisely, if X is a set,
⎜ "almost all elements of X" means
⎜ "all elements of X but those in
⎜ a negligible subset of X".
⎜ The meaning of "negligible" depends on
⎜ the mathematical context; for instance,

>
A good example is the set of FISONs.
Every FISON contains only
a negligible quantity of natural numbers.

Yes.
For each two FISONs ⟦0,j⦆, ⟦0,k⦆
#⟦0,j⦆ ≠ #(⟦0,j⦆∪⦃j⦄) and #⟦0,k⦆ ≠ #(⟦0,k⦆∪⦃k⦄)
For each two FISONs ⟦0,j⦆, ⟦0,k⦆
#⟦0,j+k⦆ ≠ #(⟦0,j+k⦆∪⦃j+k⦄)
and ⟦0,j+k⦆ is a FISON
For each FISON ⟦0,j⦆
the numbers in ⟦0,j⦆ are negligibly.many
in comparison to
the numbers in any FISON but not.in ⟦0,j⦆
( ⋃ᵢ₌᳹₀{⟦j,j+i⦆} holds more than any ⟦j,j+k⦆ or ⟦0,k⦆

A generous estimation is:
Every FISON contains
less than 1 % of all natural numbers.
There is no FISON that contains more than 1 %.

Yes.

Therefore the union of all FISONs contains less than 1 % of all natural numbers.

No.
The union of FISONS contains
100% of all members of any FISON.
Still,
each FISON contains
less than 1% of the union of FISONs.

Outside of the union of FISONs are almost all natural numbers.

No.
#⟦0,j⦆ ≠ #(⟦0,j⦆∪⦃j⦄) determines
which FISONs are unioned.
Nothing outside all of those ⟦0,j⦆
and outside of their union,
is a natural number.
What #⟦0,j⦆ ≠ #(⟦0,j⦆∪⦃j⦄) determines
is more than any ⟦0,k⦆ in FISONs outside ⟦0,j⦆,
which is almost all of them in FISONs.