On 7/31/2023 6:06 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 31. Juli 2023 um 02:39:22 UTC+2:
>> However,
>> what you list are not self-contradictions.
>>
>> They are counter-your-intuition,
>> no more than that.
>
> They are self-ontradictions.
> The endsegmente keep infinitely many numbers
> but their intersections gets empty.
> Where remain the infinitely many numbers?
Yes,
their intersection is empty.
Each split F∥E ⇉ ⋃{FISON} is into
a FISON F = ⟨0…i⟩ and
an end segment E = ⟨i⁺¹…⟩
Each number n ∈ ⋃{FISON}
is in a FISON ⟨0…i⟩ and
is not-in an end segment ⟨i⁺¹…⟩
Each number n not-in an end segment ⟨i⁺¹…⟩
is not-in their intersection ⋂{end}
Each number n ∈ U{FISON}
is not-in their intersection ⋂{end}
thus
their intersection ⋂{end} is empty.
> The endsegmente keep infinitely many numbers
Yes,
each end segment is infinite.
Each split F∥E ⇉ ⋃{FISON} is into
a FISON F = ⟨0…i⟩ and
an end segment E = ⟨i⁺¹…⟩
⟨0…i⟩ is 1×1 2-ended: finite.
⟨i⁺¹…⟩ is 1×1 1-ended: infinite.
No mₓ is last
in ⋃{FISON} or in any ⟨i⁺¹…⟩
because
⟨0…mₓ⟩ is followed by ⟨0…mₓ,mₓ⁺¹⟩
and so mx is followed by mx⁺¹
in ⋃{FISON} or in any ⟨i⁺¹…⟩
>>> Inclusion-monotonic sequence of
>>> infinite sets cannot have
>>> an empty intersection.
>>>
>>> ∀x ∈ (0, 1]: NUF(x) = ℵo is wrong
>>> at least for the first unit fractions.
>
> An increase happens from 0 to ℵo.
> It cannot happen at one point,
It happens at zero
in each neighborhood N₀ of 0
there is a ℵ₀-many 1×1 1-ended ⅟n-sequence.
It is not the same sequence in
each neighborhood.
there is no ℵ₀-many 1×1 1-ended ⅟n-sequence
in each neighborhood N₀ of 0
Only 0 is in each neighborhood N₀ of 0
> according to mathematics.
> ∀n ∈ ℕ:
> 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> is mathematics, not intuition.
0 is not a unit fraction.
¬∃n ∈ ℕ⁺: ⅟n = 0
Your formula about unit fractions
does not apply.
It's too bad you (WM) never took
any physics classes.
You would have learned that
formulas can't be used randomly,
anywhere the mood takes you.
>>> Bob cannot disappear.
>
> By simplest logic:
> In lossless exchange no loss can happen.
For each cell p/q
there is a last lossless swap p/q↔n/1
p/q↔n/1 leaves an X in p/q
If Bob Not-An-X is in p/q
then lossless p/q↔n/1 is not swapped.
B⇒¬S
If Bob Not-An-X is in any cell
then not all lossless swaps are swapped.
S⇒¬B
If all lossless swaps are swapped
then Bob Not-An-X is not in any cell.
>>>> Actual-infinityᵂᴹ is not _our work_
>>>
>>> It is although you don't understand it.
>>
>> In _our_ work
>> things always equal themselves,
>> unlike with actual-infinityᵂᴹ
>
> Things always equal themselves,
Then,
each set A can match itself
A ⟷ⁱᵈ A
id(x) = x
Suppose set B contains
a proper subset C which contains
a proper subset D
B ⊃≠ C ⊃≠ D
and C can match D
C ⟷ᶠ D
Then
B contains a proper subset which
B can match.
and
no actually-infiniteᵂᴹ
not-matching-proper-subset but
proper-subset-matching-its-proper-subset
set exists.
Respectively,
C and B\C can match D and B\C
C ⟷ᶠ D
B\C ⟷ⁱᵈ B\C
and
B can match D∪(B\C), a proper subset.
B ⟷ᵍ D∪(B\C)
g(x) = f(x) if x ∈ C
g(x) = x if x ∈ B\C
> but we cannot know all of them.
Knowing all of them isn't needed
in order to know something about
each one of them.
I know that each natural number has
a unique prime factorization.
And yet, I do not know infinitely-many
natural numbers.
There is no conflict between those claims.
> We cannot know the first unit fractions.
> But they must exist
> if infinite sets exist at all.
No!
The opposite of that!
If infinite ordered sets exist,
there must exist ordered sets
without a first or without a last.
| 3. (Paul Stäckel) S can be given a total
| ordering which is well-ordered both forwards
| and backwards. That is, every non-empty
| subset of S has both a least and a greatest
| element in the subset.
|
https://en.wikipedia.org/wiki/Finite_set
You call a finite set "infinite" and
-- surprise! -- you get "contradictions".
> The latter however is questionable.
The sequence of
1×1 2-ended ⟨0…n⟩ ⁺¹ing from 0 is
1×1 1-ended
because
each ⟨0…n⟩ is followed by ⟨0…n,n⁺¹⟩
no ⟨0…n⟩ is a second end.
Question that. <shrug>