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

On 12/7/2024 6:09 AM, WM wrote:

On 06.12.2024 19:17, Jim Burns wrote:

On 12/6/2024 3:19 AM, WM wrote:

On 05.12.2024 23:20, Jim Burns wrote:

⎜ With {} NOT as an end.segment,

>
all endsegments hold content.

>
But no common.to.all finite.cardinals.

>
Show two endsegments which
do not hold common content.

I will, after you
show me a more.than.finitely.many two.

⎜ there STILL are
⎜ more.than.finite.many end.segments,

>
Not actually infinitely many however.

>
More.than.finitely.many are enough to
break the rules we devise for finitely.many.

>
More than finitely many are finitely many,

You (WM) define it that way,
which turns your arguments into gibberish.
Each finite.cardinality
cannot be more.than.finitely.many
⎛ 1 is.less.than 2, and 2 is finite.
⎜ 1 cannot be more.than.finitely.many.
⎜
⎜ 2 is.less.than 3, and 3 is finite.
⎜ 2 cannot be more.than.finitely.many.
⎜
⎜ 3 is.less.than 4, and 4 is finite.
⎜ 3 cannot be more.than.finitely.many.
⎜
⎜ etc.
⎜
⎜ If n is a finite.cardinal, then
⎜ n is.less.than n+1, and n+1 is finite.
⎝ n cannot be more.than.finitely.many.

More than finitely many are finitely many,
unless they are actually infinitely many.
Therefore they are no enough.

Call it 'potential'.
Put whipped cream on top.
Put sprinkles on the whipped cream.
None of that changes that
each finite.cardinal is followed by
a finite.cardinal,
which breaks the two.ended.subset rule,
which allows the more.than.any.finite set with Bob
to fit in a proper Bob.free subset.

For each finite.cardinal,
up.to.that.cardinal are finitely.many.
A rule for finitely.many holds.
>
All.the.finite.cardinals are more.than.finitely.many.
A rule for more.than.finitely.many holds.

>
All the fite cardinals are actually infinitely many.

Put whipped cream on top.
Put sprinkles on the whipped cream.

That is impossible as long as
an upper bound rests in the contentents of endsegments.

A set of finite.cardinals
with a finite.cardinal upper.bound
is finite.
A set of finite cardinals
without a finite.cardinal upper.bound
breaks the two.ended.subset for finite sets,
and may well break other rules for
other bounded.by.a.finite sets.

If all endsegments have content,
then not all natnumbers are indices,

>
That seems to be based on the idea that
no finite.cardinal is both index and content.

>
By an unfortunate definition (made by myself)
there is always one cardinal content and index:
E(2) = {2, 3, 4, ...}.
But that is not really a problem.

Thank you.
That simplifies the expression of my point.
Considering all non.empty end.segments of
all finite.cardinals:
Each finite.cardinal indexes
one end.segment.
Each end.segment is indexed by
one finite.cardinal.
Each finite cardinal is content of
finitely.many end.segments,
fewer than all
more.than.any.finite end.segments.
Each end.segment has as content
more.than.finitely.many finite.cardinals,
each finite.cardinal of which
is in fewer.than.all end.segments.

Elsewhere, considering one set, that's true.
No element is both
index(minimum) and content(non.minimum).
>
However,
here, we're considering all the end segments.
Each content is index in a later set.
Each non.zero index is content in an earlier set.

>
"All at once" is
the seductive attempt of tricksters.

Being.true is not an activity.
A claim which is true of
each of more.than.any.finite
is true all.at.once of
each of more.than.any.finite.

All that happens in a sequence
can be investigated at every desired step.

At each step,
investigating each up.to.that.step is not
investigating each step.
Our sets do not change.

Each content is index in a later set.

>
Only if all content is lost.

The intersection of
more.than.finitely.many
more.than.finite end.segments of
the finite.cardinals
holds only common elements,
and there are no common elements.
It is the empty set.
----
∀j,k ∈ ⟦0,ℵ₀⦆:  j+k ∈ ⟦0,ℵ₀⦆
Addition is closed
in the finite.cardinals.
∀j ∈ ⟦0,ℵ₀⦆:
|⟦0,ℵ₀⦆| ≥ |⟦0,j⟧| = j+1 > j
iow
|⟦0,ℵ₀⦆| >ᵉᵃᶜʰ ⟦0,ℵ₀⦆
The set of finite.cardinals holds
more.than.any.finite.cardinal.many.
|{⟦i,ℵ₀⦆:i∈⟦0,ℵ₀⦆}| = |⟦0,ℵ₀⦆|
thus
|{⟦i,ℵ₀⦆:i∈⟦0,ℵ₀⦆}| >ᵉᵃᶜʰ ⟦0,ℵ₀⦆
The set of end.segments holds
more.than.any.finite.cardinal.many.
∀j,k ∈ ⟦0,ℵ₀⦆:
|⟦k,ℵ₀⦆| ≥ |⟦k,k+j⟧| = j+1 > j
iow
|⟦k,ℵ₀⦆| >ᵉᵃᶜʰ ⟦0,ℵ₀⦆
Each end.segment of finite.cardinals holds
more.than.any.finite.many.
∀j,k ∈ ⟦0,ℵ₀⦆:
k ∈ ⟦j,ℵ₀⦆  ⇔  j ≤ k
∀k ∈ ⟦0,ℵ₀⦆:
∃j ∈ ⟦0,ℵ₀⦆:
¬(j ≤ k) ∧ ¬(k ∈ ⟦j,ℵ₀⦆)
iow
∀k ∈ ⟦0,ℵ₀⦆:
k ∉ ⋂{⟦i,ℵ₀⦆:i∈⟦0,ℵ₀⦆}
iow
⋂{⟦i,ℵ₀⦆:i∈⟦0,ℵ₀⦆} = {}
Their intersection is empty.
⎛ Addition is closed in the finite.cardinals.
⎜
⎜⎛ The set of finite.cardinals
⎜⎜ The set of end.segments
⎜⎝ Each end.segment
⎜ holds more.than.any.finite.cardinal.many.
⎜
⎜ The set of common finite.cardinals
⎜ in more.than.any.finite.many end.segments
⎝ is empty.

Round up the usual suspects
and label them 'definable'.

>
∀k ∈ ℕ : E(k+1) = E(k) \ {k}
cannot come down to the empty set

It doesn't _come down to_ the empty set. At all.
ℵ₀ is not an excessively.large.but.finite.cardinal.
ℵ₀ is a larger.than.any.finite.cardinal.cardinal.

∀k ∈ ℕ : E(k+1) = E(k) \ {k}
cannot come down to the empty set
in definable numbers.
No other way however is accessible.

That explains why
"Chuck Norris counted to infinity. Twice!"
is a joke. It is an impossible brag.
Excessively.large.but.finite is countable.to, in principle.
Larger.than.any.finite is not countable.to, not even darkly.
<rimshot/>