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

On 1/2/2025 6:31 AM, WM wrote:

On 31.12.2024 21:26, Jim Burns wrote:

On 12/31/2024 1:20 PM, WM wrote:

Every union of FISONs which
stay below a certain threshold
stays below that threshold.
Note: Every FISON stays below 1 % of ℕ.

>
For each finite ⟦0,j⦆, there are
more.than.#⟦0,j⦆.many finite ⟦0,k⦆

>
So it is for all definable natural numbers.

So it is for each finite.ordinal.

But all natural numbers are finite,

Each finite.ordinal is finite.

and nevertheless there are
natural numbers which have only few successors

There is no finite.ordinal with
fewer finite.ordinal successors,
because
there are more.than.any.finite.many,
and therefore
its set of successors is
emptier.by.one but not smaller.by.one,
a property of only finite sets.
A darkᵂᴹ epilogue appended to the finites
does not successors for finites fewer.
Each finite is infinitely.far from the post.finites.

natural numbers which have only few successors
because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .

... the non.finite.ordinal natural.numbers.
We define natural.numbers to be
only finite.ordinals, of which
there are more.than.any.finite.many.

because when they are removed nothing remains
ℕ \ {1, 2, 3, ...} = { } .

When all the natural.numbers which are
not.in almost.all end.segments are removed,
none of them remain.

The set {⟦0,i⦆:#⟦0,i⦆<#⟦0,i+1⦆} of
all the finite ⟦0,k⦆
is not a finite set.

>
It is not a set
but a potentially infinite collection.

If you want to know how Bob disappears,
consider that #{⟦0,i⦆:#⟦0,i⦆<#⟦0,i+1⦆}
is not any finite.cardinal.