Re: Incompleteness of Cantor's enumeration of the rational numbers (not extra-ordinary, fragments)

On 12/07/2024 11:59 AM, Jim Burns wrote:

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:

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⎜ With {} NOT as an end.segment,

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all endsegments hold content.

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But no common.to.all finite.cardinals.

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Show two endsegments which
do not hold common content.

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

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

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Not actually infinitely many however.

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

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More than finitely many are finitely many,

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You (WM) define it that way,
which turns your arguments into gibberish.
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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.

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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.

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All the fite cardinals are actually infinitely many.

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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.

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A set of finite.cardinals
with a finite.cardinal upper.bound
is finite.
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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,

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

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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.

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Thank you.
That simplifies the expression of my point.
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Considering all non.empty end.segments of
all finite.cardinals:
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Each finite.cardinal indexes
one end.segment.
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Each end.segment is indexed by
one finite.cardinal.
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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.

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

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Being.true is not an activity.
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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.

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

Each content is index in a later set.

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Only if all content is lost.

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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.
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∀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.
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⎛ 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'.

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

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It doesn't _come down to_ the empty set. At all.
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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/>
>
>

>
If you're not going to be addressing the extra-ordinary
proper, then it seems what would be polite would be
to change the subject line to better reflect you
stop-short straw-man soft-ball sock-bot slush-mop.
>
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