On 9/11/2022 4:54 PM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 11. September 2022 um 21:48:50 UTC+2:
>> On 9/11/2022 12:15 PM, WM wrote:

>>> Therefore the empty set is not reached
>>> in a definable way.
>>
>> No 'del-min' sequence exists from ⋃𝓕 to ∅
>
> You can't see it. It is dark:

Go ahead and try to include your dark numbers.
No 'del-min' sequence exists from ⋃𝓕⊕𝔻 to ∅

Consider the split between ⋃𝓕 and 𝔻
No 'del-min' step jumps over that split,
so no 'del-min' sequence can extend into 𝔻

No 'del-min' step jumps over that split,
because each 'del-min' step
which starts on the ⋃𝓕 side
ends on the ⋃𝓕 side.

Ultimately,
we know that because
𝐹ₙ⊕⟨n+1⟩ is a FISON too.

You can't get there from here.
https://www.urbandictionary.com/define.php?term=you%20can%27t%20get%20there%20from%20here

> ∀k ∈ ℕ:
∀k ∈ ⋃𝓕:
> ∩{E(1), E(2), ..., E(k)} =
> E(k) /\ |E(k)| = ℵ₀
> E(k) /\ |E(k)| = ℵ₀
>
> ∩{E(k) : k ∈ ℕ} = { }
∩{E(k) : k ∈ ⋃𝓕} = { }
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
∀k ∈ ⋃𝓕: E(k+1) = E(k) \ {k}

E(k) is on the ⋃𝓕 side
E(k+1) is on the ⋃𝓕 side.
They are not what you need.

>> Dark numbers do not change that.
>
> Otherwise there is a jump.

A jump would need to land on the other side.
No 'del-min' step does that.

On Monday, September 12, 2022 at 12:43:01 AM UTC+2, Fritz Feldhase wrote:
> On Sunday, September 11, 2022 at 11:23:30 PM UTC+2, Sergio wrote:
> > On 9/11/2022 3:52 PM, WM wrote:
> > >
> > > ∩{E(k) : k ∈ ℕ} = { }
> > >
> > Intersection of Endsegment k is empty (wrong)
> >
> Nope.
>
> ∩{E(k) : k ∈ ℕ} = { }
>
> is absolutely correct. (But you reading was/is wrong.)
>
> ... hint:
>
> ∩{E(1), E(2), E(3), ...} = ∩{E(k) : k ∈ ℕ}
>
> since
>
> {E(1), E(2), E(3), ...} = {E(k) : k ∈ ℕ}.

Maybe you have seen things like this:

{n^2 : n e IN} = {1, 4, 9, ...}.

Note that the "n" in "{n^2 : n e IN}" is bound (and hence not free).

WM <monteuffl@t-online.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:13:16 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>
>> > Ben Bacarisse schrieb am Freitag, 9. September 2022 um 23:06:26 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> >> Unendlichen" at Hochschule Augsburg.)
>> >> > It is what we have been discussing in extensio. It is nonsense to
>> >> > claim the indexing of all fractions if most will be remaining not
>> >> > indexed forever.
>> >> My argument was entirely within WMaths. It used a potentially infinite
>> >> bijection between potentially infinite sets to define a potentially
>> >> infinite sequence of potentially infinite functions.
>> >
>> > There is no problem.
>> You accept my argument that, in WMaths, "the result" of the swaps
>> (suitably defined) is no Os in any finite position?
>
> At the end there is in every case no definable O at any position. In
> my Mathematics, there are no undefinable O's.

You find it hard to give a straight answer. You said "no problem" but I
think you have a problem with my argument.

> Here we discuss Cantor's theory.

You can discuss Cantor with other people, but I posted to show you how
to get the result you found surprising using only WMaths. If, as seems
likely, you don't want to discuss it, that's fine. Just say you are
done.

>> > But I have been analyzing Cantor's completed infinity with a complete
>> > matrix of fractions.
>> But you don't want to analyse the consequences of the swaps in potential
>> infinity.
>
> Why not?

Because you keep talking about things that are not in WMaths:

> IN EVERY CASE there are no O visible at the end of the
> process. I never refused this. It is the proof of dark numbers.

WMaths has no dark numbers. It's the WMaths result you are avoiding.

>> And you keep not answering this question:

Why leave this text if you won't answer? Here's the question again:

In WMaths you have not yet been able to define set membership, equality
and difference (you say the definitions in your textbook are
simplifications for pedagogic reasons). Any progress on that?

I can see that it's obviously embarrassing that you don't know the
definitions of basic set operations in potential infinity, but you
should not keep cutting the question. If there is no progress, just say
so.

>> The limit is as per your book. Are there dark objects in WMaths?
>
> Look, the limit of the sequence 1/n is zero.

You don't know what I was talking about do you? You are all at sea.
Here is the limit (as per you book) that you once denied:

Given e_n(i) = [i >= n], then lim_{n->oo} e_n = z where z(i) = 0.

Do you even know what this says (in WMaths)?

And you cut this:

||| There are no set limits in my book.

|| Wrong. Functions are sets (see page 18). Some sequences of functions
|| converge to a limit function. Examples are given (see page 202).
||
|| Why would say something so obviously and demonstrably wrong?

but in another post (can you hold off posting until you've decided what
you want to say?) you said:

| Yes, but there is a difference between analysis where the terms can be
| interpreted as sets and manipulations like indexing. It seems
| important to fix this difference. I have not pondered about it because
| I was discussing always only the latter.

Your "yes" suggests you agree you were wrong, but then a whole lots of
words come along that make it unclear. Your book includes examples of
limits of convergent, non-constant sequences of sets.

--
Ben.

WM <monteuffl@t-online.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:22:53 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>> > Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:06:41 UTC+2:
>>
>> >> What positions
>> >> have Os (you asked about positions originally) "in the limit" depends on
>> >> the pattern of swaps
>> >
>> > The pattern is given by Cantor.
>>
>> You neither cited nor quoted Cantor saying anything about swaps. I
>> think you made that bit up yourself, but I would love to read how Cantor
>> defines the swaps and the result of an endless sequence of swaps.
>
> Assigning an index to a fraction is expressed by assigning an X to an
> O in my matrix. That is simple.

But you then talked about swaps.

> The only difference to Cantor's procedure is that I first assign every
> X to a unit-fraction. You must forbid this in order to save Cantor.

I'm saving WMaths. Cantor can look after himself.

Anyway, no swaps in Cantor. Got it. And now you don't want to talk
about swaps in WMaths, just the map between the first column and the
rest of the fractions? That's just a different, more messy formula.
Why do you think it's interesting?

>> > Limits are irrelevant when we ask for indexing with natural numbers,
>> > i.e. in finite steps before the limit.
>>
>> You asked about "the result" of an endless sequence of swaps. That can
>> not be revealed by the indicator function I_n for any n in N. Something
>> else is needed to make sense of the phrase "the result".
>
> Cantor's statemet that it is possible is taken as the basis.

But it was you that made up all the nonsense about swaps and the result
of an endless sequence of them. The map between the first column and
the rest of the grid is just a messy closed-form formula (in WMaths).

>> I proposed the
>> limit of the sequence of indicator function. And, behold, with some
>> patters of swaps, the limit is the function I(i) = 1, meaning no Os in
>> any position, just as you claim is the case in set theory!
>
> The result is in fact that no O's can be found in any position.

Yes, in WMaths, as I showed. But it seems you don't care about the
swaps anymore.

--
Ben.

Ben Bacarisse schrieb am Montag, 12. September 2022 um 01:56:56 UTC+2:
> WM <mont...@t-online.de> writes:
> >> You accept my argument that, in WMaths, "the result" of the swaps
> >> (suitably defined) is no Os in any finite position?
> >
> > At the end there is in every case no definable O at any position. In
> > my Mathematics, there are no undefinable O's.

> You can discuss Cantor with other people, but I posted to show you how
> to get the result you found surprising using only WMaths. If, as seems
> likely, you don't want to discuss it, that's fine. Just say you are
> done.

It seems that you have not yet comprehended the essence of my proof.
Here it is again in few words:
The result of all swaps of X and O in the matrix
XOOO...
XOOO...
XOOO...
XOOO...
....
according to Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m is this:

There is no O at any definable place of the matrix. No O can be found at any position m/n. This result is the same in clasical mathematics and in Cantor's set theory:

Surprising is only that the O's must be somewhere in the matrix because by swapping X and O none can disappear. I said this already here:

> > IN EVERY CASE there are no O visible at the end of the
> > process. I never refused this. It is the proof of dark numbers.
>
> WMaths has no dark numbers.

There are no completed infinite sets. But the dark numbers appear if we assume completed infinite sets.
>
> In WMaths you have not yet been able to define set membership, equality
> and difference (you say the definitions in your textbook are
> simplifications for pedagogic reasons). Any progress on that?

In potential infinity there are no fixed sets. But everything important for mathematics is fixed enough to be available. For instance all natural numbers required in mathematics are available. All positions finally occupied by X's are or can be made available.

Further distracting text deleted. Please focus on the fact that never an O can disappear.

Regards, WM

On 9/11/2022 9:10 AM, WM wrote:
> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:13:16 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> The limit is as per your book. Are there dark objects in WMaths?
>
> I do not talk about the possible contents of the gaps around every definable point on the real axis. There is no next point to any definable point. Therefore either there is nothing or something dark.

you are stupid.

take point A and point B, in between them is point (A+B)/2

>
>>> There are no set limits in my book.
>>
>> Wrong. Functions are sets (see page 18).
>
> Yes, but there is a difference between analysis where the terms can be interpreted as sets and manipulations like indexing.

Wrong. Indexing is not manipulation.

>
> Regards, WM

On 9/11/2022 9:10 AM, WM wrote:
> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:13:16 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> The limit is as per your book. Are there dark objects in WMaths?
>
> I do not talk about the possible contents of the gaps around every definable point on the real axis. There is no next point to any definable point. Therefore either there is nothing or something dark.

you are stupid.

take point A and point B, in between them is point (A+B)/2

>
>>> There are no set limits in my book.
>>
>> Wrong. Functions are sets (see page 18).
>
> Yes, but there is a difference between analysis where the terms can be interpreted as sets and manipulations like indexing.

Wrong. Indexing is not manipulation.

>
> Regards, WM

On Monday, September 12, 2022 at 2:50:49 PM UTC+2, WM wrote:

> ... the essence of my proof.

The essence of your "proof" is idiocy.

Ben Bacarisse schrieb am Montag, 12. September 2022 um 02:08:13 UTC+2:
> WM <mont...@t-online.de> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)

> >> >> What positions
> >> >> have Os (you asked about positions originally) "in the limit" depends on
> >> >> the pattern of swaps
> >> >
> >> > The pattern is given by Cantor.
> >>
> >> You neither cited nor quoted Cantor saying anything about swaps. I
> >> think you made that bit up yourself, but I would love to read how Cantor
> >> defines the swaps and the result of an endless sequence of swaps.
> >
> > Assigning an index to a fraction is expressed by assigning an X to an
> > O in my matrix. That is simple.
>
> But you then talked about swaps.

Since the fractions of the first column where the X's originally are residing must also get indices, these fractions have to be marked by O's after losing their index until they get another index.

> > The only difference to Cantor's procedure is that I first assign every
> > X to a unit-fraction. You must forbid this in order to save Cantor.
>
> I'm saving WMaths. Cantor can look after himself.

Cantor cannot because he is dead. But I believe that he would understand an confess like he did once in his real life.
> Anyway, no swaps in Cantor.

Swaps in Cantor if the indices are first assignet to the fraction of the first column.

> And now you don't want to talk
> about swaps in WMaths, just the map between the first column and the
> rest of the fractions?

No! The map is between the fractions of the first column and all fractions including the first column.

> Why do you think it's interesting?

Because it proves a new kind of objects of mathematics, never heard of before.

> > Cantor's statemet that it is possible is taken as the basis.
> But it was you that made up all the nonsense about swaps and the result
> of an endless sequence of them. The map between the first column and
> the rest of the grid is

Please try to understand what you are criticizing first!

> > The result is in fact that no O's can be found in any position.
> Yes, in WMaths, as I showed.

No, in every kind of mathematics, as I proved. Only because they cannot be found these numbers are dark.

> But it seems you don't care about the
> swaps anymore.

It seems that you should take a break and first understand what we are discussing.
I claimed from the beginning that all definable places are occupied by X's. See the OP: "In fact by application of (*) all O's are removed from all visible or definable matrix positions." Therefore your limit argument was irrelevant. But I proved in addition, that the O's are present nevertheless.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:03:51 UTC+2:
> On Sunday, September 11, 2022 at 8:48:56 PM UTC+2, WM wrote:
>
> > there is a jump
>
> Where exactly?

All infinite intersections
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
have constant cardinal number because ℵ₀ - 1 = ℵ₀ .
The jump is before the zero of the empty set.

Regards, WM

FromTheRafters schrieb am Sonntag, 11. September 2022 um 21:12:01 UTC+2:
> WM pretended :
> > Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 18:37:32 UTC+2:
> >> On Sunday, September 11, 2022 at 6:15:58 PM UTC+2, WM wrote:
> >
> >>>> The "deletions" do not lead to the intersection of all end segments (which
> >>>> = ∅)
> >>>>
> >>> No?
> >> No.
> >
> > Then we differ in significant precondition.
> > My opinion about all deletions is this
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > ∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
> There is no decrease in
> cardinality. The notion of a set being contained (as a proper or
> improper subset) within another set does not necessarily mean that the
> containing set is larger in cardinality despite it being so
> accomodating.

Right. But the intersection of all endsegments is empty and therefore has cardinality 0. However it does not get empty by magic spell. Between all intersections
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
and the intersection ∩{E(k) : k ∈ ℕ} = { } there is a difference of ℵ₀ elements in contradiction with ∀k ∈ ℕ: E(k+1) = E(k) \ {k} which rules every loss of elements.

Regards, WM

FromTheRafters schrieb am Sonntag, 11. September 2022 um 21:13:31 UTC+2:
> WM formulated the question :
> > FromTheRafters schrieb am Sonntag, 11. September 2022 um 20:58:25 UTC+2:
> >> WM submitted this idea :
> >
> >>> Or, put it yet another way, you must not check his method if you want to
> >>> believe in it.
> >> I just checked again,
> >
> > You failed. The set of O's cannot be removed.
> >
> >> BTW, I didn't check
> >> step by step.
> >
> > Not necessary. We know that never, in no step, any O will vanish.
> You misunderstand me, I check *his* method not yours.

You did not. You cannot. Never anybody could before I introduced the only method for valid checking: First apply all indices to the integer fractions.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:

> Actually, intersection is not a "stepwise" process

∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.

What step kcannot be checked?

Regards, WM

Fritz Feldhase schrieb am Montag, 12. September 2022 um 00:57:04 UTC+2:
> On Sunday, September 11, 2022 at 10:52:19 PM UTC+2, WM wrote:

> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> > ∩{E(k) : k ∈ ℕ} = { }
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> I asked for a PROOF (of something).
>
> What you've presented here are 3 (true) statements which hardly are a proof of anything.

They prove that never more than one natnumber can be lost.

Regards. WM

Jim Burns schrieb am Montag, 12. September 2022 um 01:00:23 UTC+2:
> On 9/11/2022 4:54 PM, WM wrote:
>
> > Otherwise there is a jump.
> A jump would need to land on the other side.
> No 'del-min' step does that.

The intersection of all endsegments is empty because all natnumbers have been lost. This can only happen one by one.

Regards, WM

WM drivels:

> XOOO...
> XOOO...
> XOOO...
> XOOO...

Boah! What a genius!

On 9/12/2022 8:28 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 11. September 2022 um 21:13:31 UTC+2:
>> WM formulated the question :
>>> FromTheRafters schrieb am Sonntag, 11. September 2022 um 20:58:25 UTC+2:
>>>> WM submitted this idea :
>>>
>>>>> Or, put it yet another way, you must not check his method if you want to
>>>>> believe in it.
>>>> I just checked again,
>>>
>>> You failed. The set of O's cannot be removed.
>>>
>>>> BTW, I didn't check
>>>> step by step.
>>>
>>> Not necessary. We know that never, in no step, any O will vanish.
>> You misunderstand me, I check *his* method not yours.
>
> You did not. You cannot. Never anybody could before I introduced the only method for valid checking: First apply all indices to the integer fractions.

*which is a mistake*.
There is no need to do that,
it does not provide any additional clarity,
you butcher the matrix of rationals
You cover up your mess with vague swapparoos

Just use a separate set of natural numbers, like Cantor did, and enumerate the fractions.

>
> Regards, WM
>
>

On 9/12/2022 8:04 AM, WM wrote:
> Ben Bacarisse schrieb am Montag, 12. September 2022 um 02:08:13 UTC+2:
>> WM <mont...@t-online.de> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>
>>>>>> What positions
>>>>>> have Os (you asked about positions originally) "in the limit" depends on
>>>>>> the pattern of swaps
>>>>>
>>>>> The pattern is given by Cantor.
>>>>
>>>> You neither cited nor quoted Cantor saying anything about swaps. I
>>>> think you made that bit up yourself, but I would love to read how Cantor
>>>> defines the swaps and the result of an endless sequence of swaps.
>>>
>>> Assigning an index to a fraction is expressed by assigning an X to an
>>> O in my matrix. That is simple.
>>
>> But you then talked about swaps.
>
> Since the fractions of the first column where the X's originally are residing must also get indices, these fractions have to be marked by O's after losing their index until they get another index.

so first column gets double swapped ?

>
>>> The only difference to Cantor's procedure is that I first assign every
>>> X to a unit-fraction. You must forbid this in order to save Cantor.
>>
>> I'm saving WMaths. Cantor can look after himself.
>
> Cantor cannot because he is dead. But I believe that he would understand an confess like he did once in his real life.
>
>> Anyway, no swaps in Cantor.
>
> Swaps in Cantor if the indices are first assignet to the fraction of the first column.

Wrong, Cantor did not do that, you did.

>
>> And now you don't want to talk
>> about swaps in WMaths, just the map between the first column and the
>> rest of the fractions?
>
> No! The map is between the fractions of the first column and all fractions including the first column.

so it is not one to one, but reusing the first column over and over again... No wonder it doesnt work.

>
>> Why do you think it's interesting?
>
> Because it proves a new kind of objects of mathematics, never heard of before.

not prove, but a spoof, dark numbers ?

>
>>> Cantor's statemet that it is possible is taken as the basis.
>> But it was you that made up all the nonsense about swaps and the result
>> of an endless sequence of them. The map between the first column and
>> the rest of the grid is
>
> Please try to understand what you are criticizing first!

then provide the algorithm for the swaps. So far it does not make any sense.

>
>>> The result is in fact that no O's can be found in any position.
>> Yes, in WMaths, as I showed.
>
> No, in every kind of mathematics, as I proved. Only because they cannot be found these numbers are dark.

you have not proved anything. it is unfounded imagination.

>
>> But it seems you don't care about the
>> swaps anymore.
>

>
> Regards, WM

On 9/12/2022 8:26 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 11. September 2022 um 21:12:01 UTC+2:
>> WM pretended :
>>> Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 18:37:32 UTC+2:
>>>> On Sunday, September 11, 2022 at 6:15:58 PM UTC+2, WM wrote:
>>>
>>>>>> The "deletions" do not lead to the intersection of all end segments (which
>>>>>> = ∅)
>>>>>>
>>>>> No?
>>>> No.
>>>
>>> Then we differ in significant precondition.
>>> My opinion about all deletions is this
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> ∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
>> There is no decrease in
>> cardinality. The notion of a set being contained (as a proper or
>> improper subset) within another set does not necessarily mean that the
>> containing set is larger in cardinality despite it being so
>> accomodating.
>
> Right. But the intersection of all endsegments is empty and therefore has cardinality 0. However it does not get empty by magic spell. Between all intersections
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

that is not all intersections, you stopped at k

> and the intersection ∩{E(k) : k ∈ ℕ} = { } there is a difference of ℵ₀ elements in contradiction with ∀k ∈ ℕ: E(k+1) = E(k) \ {k} which rules every loss of elements.

google "intersection"

>
> Regards, WM

On 9/12/2022 8:38 AM, WM wrote:
> Jim Burns schrieb am Montag, 12. September 2022 um 01:00:23 UTC+2:
>> On 9/11/2022 4:54 PM, WM wrote:
>>
>>> Otherwise there is a jump.
>> A jump would need to land on the other side.
>> No 'del-min' step does that.
>
> The intersection of all endsegments is empty because all natnumbers have been lost.

Wrong. It is because the intersection was taken.

This can only happen one by one.
>
> Regards, WM

WM pretended :
> Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:
>
>> Actually, intersection is not a "stepwise" process
>
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.
>
> What step kcannot be checked?

The zeroeth one, and the 'last' one?

On 9/12/2022 9:38 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 12. September 2022 um 01:00:23 UTC+2:
>> On 9/11/2022 4:54 PM, WM wrote:

>>> Otherwise there is a jump.
>>
>> A jump would need to land on the other side.
>> No 'del-min' step does that.
>
> The intersection of all endsegments is empty
> because all natnumbers have been lost.
> This can only happen one by one.

What does it mean to for an ordered set
to "happen"?

In order for an ordered set to "happen",
each split of precursor-sets into
before and after
must include a way to get from
before to after.

In order for a state to "happen",
there must be a last set before each split
and a first set after that split:
_Each split must have a jump, before to after_

But we put a restriction on those jumps.
The first after equals the last before with
its minimum element removed.

That is how we describe a set
"happening" _one-by-one_
with the minimum removed across a jump
but also with each split having its jump.

Some sets "happen" one-by-one.
Some sets do not "happen" one-by-one.

Each (infinite) end segment 𝐸 of ⋃𝓕
"happens" one-by-one.

For each split of the precursor-sets of 𝐸,
there is a last-before, a first-after,
and first-after equals 'del-min' last-before.

The empty set ∅ does not "happen" one-by-one.

We can order the (infinite) end segments
and ∅ and anything else you like,
and still ∅ does not "happen" one-by-one.

_Not all splits_ of ⋃𝓕 and its end segments
and anything else _have a jump_
One split,
between the (infinite) end segments and
whatever is after the (infinite) end segments,
cannot have a jump.

No one knows what is after the (infinite)
end segments, but it doesn't matter.
A jump across that spit would need to
start on an (infinite) end segment and
end on something else, something not
an (infinite) end segment.
We know that can't be, because
each 'del-min' step
which starts on an (infinite) end segment
ends on an (infinite) end segment.

Go ahead and append your dark end segments.
We know that there is no jump to them.
We know, _not_ because we know what your
dark things are, but because we know where
all the lit jumps land.

> The intersection of all endsegments is empty
> because all natnumbers have been lost.
> This can only happen one by one.

∅ does not "happen" one-by-one.

Also,
∅ is the intersection of all (infinite)
end segments.

The requirement that an intersection "happen"
one-by-one is superfluous. An intersection
is the set of elements which are in all
the intersected sets.

///

FromTheRafters schrieb am Montag, 12. September 2022 um 17:18:09 UTC+2:
> WM pretended :
> > Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:
> >
> >> Actually, intersection is not a "stepwise" process
> >
> > ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.
> >
> > What step cannot be checked?
> The zeroeth one, and the 'last' one?

The dark ones cannot be checked, that is true. My question unmasked the opinion that there are no dark ones (and of course no last ones).

Regards, WM

Jim Burns schrieb am Montag, 12. September 2022 um 19:12:41 UTC+2:
> On 9/12/2022 9:38 AM, WM wrote:
> > Jim Burns schrieb am Montag,
> > 12. September 2022 um 01:00:23 UTC+2:
> >> On 9/11/2022 4:54 PM, WM wrote:
>
> >>> Otherwise there is a jump.
> >>
> >> A jump would need to land on the other side.
> >> No 'del-min' step does that.
> >
> > The intersection of all endsegments is empty
> > because all natnumbers have been lost.
> > This can only happen one by one.
> What does it mean to for an ordered set
> to "happen"?

It describes the changes in the course of going through the ordered set.
>
> In order for an ordered set to "happen",
> each split of precursor-sets into
> before and after
> must include a way to get from
> before to after.

This way must exist, if the result "after all" does exist.
>
> In order for a state to "happen",
> there must be a last set before each split
> and a first set after that split:
> _Each split must have a jump, before to after_

If all finite intersections are infinite and the final intersection is empty, then something must cause this change.
>
> But we put a restriction on those jumps.
> The first after equals the last before with
> its minimum element removed.

All finite intersections are infinite and the final intersection is empty. What causes this change?

> The requirement that an intersection "happen"
> one-by-one is superfluous.

For all definable endsegments this requirement can be checked and confirmed
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

> An intersection
> is the set of elements which are in all
> the intersected sets.

The jump shows the existence of a new kind of objects of mathematics. Therefore checking the intersections of definable endsegments is worthwhile.

Regards, WM

On 9/12/2022 2:50 PM, WM wrote:
> FromTheRafters schrieb am Montag, 12. September 2022 um 17:18:09 UTC+2:
>> WM pretended :
>>> Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:
>>>
>>>> Actually, intersection is not a "stepwise" process
>>>
>>> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.
>>>
>>> What step cannot be checked?
>> The zeroeth one, and the 'last' one?
>
> The dark ones cannot be checked, that is true. My question unmasked the opinion that there are no dark ones (and of course no last ones).
>
> Regards, WM

Dark Unmasked Ants
Dark Ones that cannot be checked but are really Ants
Ants that say, "that is true"
Of course no last Ants
Unmasked Opinion Ants
there are no dark ones segregated Ants
Zeroeth Ants
Unchecked and Unmasked opinionated true dark Ants