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On 12/21/2024 6:34 AM, WM wrote:Well, anybody can just build "infinite-middle",On 20.12.2024 19:48, Jim Burns wrote:>On 12/19/2024 4:37 PM, WM wrote:>That means all numbers are lost by loss of>
one number per term.
>
That implies finite endsegments.
Q. What does 'finite' mean?
Consider end.segments of the finite cardinals.
>
Q. What does 'finite' mean,
'finite', whether darkᵂᴹ or visibleᵂᴹ?
>Here is a new and better definition of endsegments>
>
E(n) = {n+1, n+2, n+3, ...} with E(0) = ℕ.
>
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
means that the sequence of endsegments can decrease only by one
natnumber per step.
E(n+1) is larger.than each of
the sets for which there are smaller.by.one sets.
E(n+1) isn't any of
the sets for which there are smaller.by.one sets.
>
E(n+1) isn't smaller.by.one than E(n).
E(n+1) is emptier.by.one than E(n)
>Therefore the sequence of endsegments>
cannot become empty
Yes, because
the sequence of end.segments
can become emptier.one.by.one, but
it cannot become smaller.one.by.one.
>(i.e., not all natnumbers can be applied as indices)>
Each finite.cardinal can be applied,
which makes the sequence emptier.by.one
but does not make the sequence smaller.by.one.
>unless the empty endsegment is reached,>
Each end.segment is emptier.by.one.
No end.segment is smaller.than the first end.segment ℕ
The empty end.segment is not reached.
>
No finite.cardinal is in common with
all infinitely.many
infinite.end.segments of
finite.cardinals.
>
Nothing other.than a finite.cardinal is in
any end.segment of the finite.cardinals,
or in their intersection.
>
The intersection of all infinitely.many
infinite end.segments of finite.cardinals
is not an end.segment
but is empty.
>
Q. What does 'finite' mean?
>unless the empty endsegment is reached,>
The empty end.segment, not.existing, is not.reached.
>
The intersection.of.finitely.many is not.empty.
The intersection.of.all is empty.
>and>
before finite endsegments,
endsegments containing only 1, 2, 3, or n ∈ ℕ numbers,
have been passed.
⎛ Assume end.segment E(n) of the finite.cardinals
⎜ holds only finite.cardinal.k.many finite.cardinals.
⎜
⎜ k.sized E(n) holds
⎜ kᵗʰ.smallest, 1ˢᵗ.largest finite.cardinal E(n)[k]
⎜
⎜ If a finite.cardinal larger.than E(n)[k] exists,
⎜ it would also be in end.segment E(n) and
⎜ larger.than 1ˢᵗ.largest E(n)[k]: a contradiction.
⎜
⎜ Thus,
⎜ a finite.cardinal larger.than E(n)[k] doesn't exist.
⎜
⎜ However,
⎜⎛ for each finite.cardinal j,
⎜⎝ larger.than.j finite.cardinal j+1 exists.
⎜
⎜ Larger.than.E(n)[k] finite.cardinal E(n)[k]+1 exists.
⎝ Contradiction.
>
Therefore,
end.segment E(n) of the finite.cardinals does not hold
only finite.cardinal.many finite.cardinals.
>
There are no finite end.segments of the finite.cardinals.
>
Q. What does 'finite' mean?
>These however, if existing at all, cannot be seen.>
They are dark.
Darknessᵂᴹ and visibilityᵂᴹ don't change any of this.
There are no finite end.segments of the finite.cardinals.
>
We know it by the method of
assembling finite sequences of claims (proofs),
each claim of which is true.or.not.first.false (valid),
and holding those claims.we.know (theorems),
because
a finite sequence of claims,
each claim of which true.or.not.first.false,
holds only true claims.
>
Some claims (definitions)
we know are true because
we know how we have defined things.
>
Some claims (valid inferences)
we know are not.first.false because
we can inspect the finite sequence of claims.
>
None of _the claims_ are darkᵂᴹ,
whatever the status of _what the claims are about_
>
Darknessᵂᴹ or visibilityᵂᴹ of finite.cardinals
don't change _the claims_
>>>That means all numbers are lost by loss of>
one number per term.
>
That implies finite endsegments.
No.
Yes, each number is lost by loss of
one number per term.
However,
each end.segment is not finite.Then the last endsegment is empty.>
There is no last end.segment of the finite.cardinals.
⎛ For each finite.cardinal j,
⎝ larger.than.j finite.cardinal j+1 exists.
contradicts a last end.segment.
>
>
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