On 9/12/2022 3:57 PM, WM wrote:
> 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.

And what is that description?
The description is what I intend to provide.

Remember the strategy:
_Describe_ one of infinitely-many.
Derive new claims only by truth-preserving ways.

That's why we need the description itself,
as the seed from which we grow new claims.

The admonition "Get some seed" is not enough.
We need the seed.

>> 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.

There's a split.

What's on the other side of the split
is one question.
Can we get to the other side
is another question.

For this split, with all of ⋃𝓕 on this side,
we know that we can't get to the other side
_even without knowing what's on the other side_

We know that no step
which starts on the ⋃𝓕 side
ends on the other side

because
we know that each step
which starts on the ⋃𝓕 side
ends on the ⋃𝓕 side

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

WM <askasker48@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> 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.

Again, no swaps. Apparently now things get and lose indexes. I avoided
such hand-waving terms in my argument. Did you not appreciate the
clarity of it? And since it was all in WMaths, there could be no
confusion about dark numbers and so on.

>> > 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.

Don't keep lying about it. "Swaps in Cantor" means he wrote about
swaps. But you mean your inaccurate re-imagining of Cantor includes
swaps. Stop it. It's not intellectually honest.

--
Ben.

WM <askasker48@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> 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.

Of course I don't understand that -- I have no idea what a definable
place is. I don't know what you mean by the result of endless sequence
of swap is. All compounded by your not saying what swaps you mean.
It's something to do with k = (m + n - 1)(m + n - 2)/2 + m but you
really need to say what is swapped with what.

So I made a better argument using only WMaths. Spelled out the swaps
and I gave a reasonable definition of the result of an endless sequence
of swap. Did you understand it? If you did, why did you not say it was
right or point out where I went wrong?

>> 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.

So no, there is no progress on defining set membership, equality and
difference in WMaths. Students will just have to guess what they mean
and take your word for it that e ∈ S and S \ {e} = S is possible.

Of course the matters for my argument because it's done in WMaths, but I
assume that, had you understood it, you would have pointed out anywhere
I presumed to know what was and was not in a set.

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

Oh I did. Go back and look. You'll see that when I responded about
other topics it is because /you/ wanted a distraction. For example, you
brought up end segments again, not me. And it was you that brought up
set convergence by saying the demonstrably wrong

> There are no set limits in my book.

But when such remarks of yours lead to embarrassing conclusions, it is
/I/ who must focus to the topic of your choosing!

Right from the start, all I wanted was to help you understand how the
result you claim for Cantor can be reached using only WMaths. Of course
you will do anything and everything to distract from that because you
hate talking about WMaths.

--
Ben.

On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
> 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.

This is called a discontinuity. Yes, the card() function is not continuous, because, e.g., card(lim{n -> oo} E(n)) =/= lim{n -> oo} card(E(n)).

On 9/12/2022 7:47 PM, Gus Gassmann wrote:
> On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
>> 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.
>
> This is called a discontinuity. Yes, the card() function is not continuous, because, e.g., card(lim{n -> oo} E(n)) =/= lim{n -> oo} card(E(n)).

for WM;

Math is full of discontinuities, checkout the Gamma function for negative numbers. (yea, its not natural numbers, nor sheeps counting)

So is real life, your fridge uses a bang-bang controller, which is uber discontinuous, on/off, yet provides very accurate temperature control and long
relay/motor life. AND your car, and that LED lightbulb, and the computer screen you are looking at..., the music you listen too,...

On Monday, September 12, 2022 at 3:30:06 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:
> >
> > Actually, intersection is not a "stepwise" process
> >
> What step cannot be checked?

Consider:

∩{E(k) : k e IN}.

Where do you see a "step" here, you blithering idiot?!

EOD

On Monday, September 12, 2022 at 3:36:27 PM UTC+2, WM wrote:
> 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.

Holy shit!

The statement (sic!) ∀k ∈ ℕ: E(k+1) = E(k) \ {k} just states that the "difference" between E(k+1) and E(k) consists at most of one element (namely the element k if k e E(k)) for each and every k in IN.

No "proof" at all.

You are dumb like a brick, Mückenheim!

Now ∩{E(k) : k ∈ ℕ} = { } (the 2nd statement) is IMPLIED by ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (and E(1) = IN).

There's not even a HINT of a proof for this fact.

It's even worse with the 1st statement:

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

Here we need even more "machinery".

So let me repeat my statement:

| Look, you silly asshole full of shit: I asked for a PROOF (of something).
|
| What you've presented here are 3 (true) statements which hardly are a proof of anything.
|
| Actually the first 2 statements follow from E(1) = IN, |IN| = ℵ₀ (and some theorems concerning |.| and IN) and (*). But there's not even a HINT of a PROOF for this fact.
|
| You are completely mad, man.

EOD

On Monday, September 12, 2022 at 3:21:23 PM UTC+2, WM wrote:
> 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?
> >
> The jump is before the zero of the empty set.

No, there is no such "jump", you fucking idiot!

SINCE THERE IS NO "before the zero of the empty set". <facepalm>

Remember, your IDIOTIC / BRAINDEAD claim was:

WM> If all endsegments are infinite,

which they are

WM> then there is a jump from more than one element to 0 elements.

Nope, there is no such jump.

Hint: For each and every endsegment E there is an endsegment E' such that E > E' > {}.

=> NO JUMP.

EOD

On Tuesday, September 13, 2022 at 2:47:41 AM UTC+2, Gus Gassmann wrote:
> On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
> > 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.
> >
> This is called a discontinuity.

But it (whatever you like to call it) is CERTAINLY not "before the zero of the empty set" (won't you think so)? :-P

On Monday, 12 September 2022 at 22:59:09 UTC-3, Fritz Feldhase wrote:
> On Tuesday, September 13, 2022 at 2:47:41 AM UTC+2, Gus Gassmann wrote:
> > On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
> > > 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.
> > >
> > This is called a discontinuity.
> But it (whatever you like to call it) is CERTAINLY not "before the zero of the empty set" (won't you think so)? :-P
OK. It was late, and I only read his shit to the word "jump". More I could not tolerate. My bad.

Ben Bacarisse schrieb am Dienstag, 13. September 2022 um 01:55:20 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 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.
> Of course I don't understand that -- I have no idea what a definable
> place is.

It is precisely what you wanted to teach me. The result of all indexing. Only indexed fractions are visible. You and other short-sighted mathematicians thought that were all.

> I don't know what you mean by the result of endless sequence
> of swap is. All compounded by your not saying what swaps you mean.
> It's something to do with k = (m + n - 1)(m + n - 2)/2 + m but you
> really need to say what is swapped with what.

Every index k = 1, 2, 3, ... is at the beginning residing in the first column. Every index is taken from there to the place calculated by k = (m + n - 1)(m + n - 2)/2 + m, as Cantor taught us.
>
> So I made a better argument using only WMaths. Spelled out the swaps
> and I gave a reasonable definition of the result of an endless sequence
> of swap. Did you understand it?

Chuckle. I described it in the OP.

> If you did, why did you not say it was
> right or point out where I went wrong?

I did. I said that limits are irrelevant because it was not denied that every visible fractions is indexed in the end. The visible matrix looks like this

XXX...
XXX...
XXX...
....

> Right from the start, all I wanted was to help you understand how the
> result you claim for Cantor can be reached using only WMaths.

Okay, now you know that I had this result long ago.

> Of course
> you will do anything and everything to distract from that because you
> hate talking about WMaths.

No. But here I prefer talking about set theory.

I am looking forward to your full understanding of my argument.

Regards, WM

Gus Gassmann schrieb am Dienstag, 13. September 2022 um 02:47:41 UTC+2:
> On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
> > 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.
> This is called a discontinuity. Yes, the card() function is not continuous, because, e.g., card(lim{n -> oo} E(n)) =/= lim{n -> oo} card(E(n)).

The deletion of elements of endsegments is a continuous process or function.. There is no jump or discontinuity.

It is as continuous as the deletion of O's in the matrix

XOOO...
XOOO...
XOOO...
XOOO...
....

But here never any O is deleted. Only a fool or a dishonest person can pretend that he believes in the deletion of all O's in the limit.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 03:36:25 UTC+2:
> On Monday, September 12, 2022 at 3:30:06 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 21:18:53 UTC+2:
> > >
> > > Actually, intersection is not a "stepwise" process
> > >
> > What step cannot be checked?
> Consider:
>
> ∩{E(k) : k e IN}.
>
> Where do you see a "step" here

This is the last step - if it exists. Allother steps are here ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 03:48:44 UTC+2:
> On Monday, September 12, 2022 at 3:36:27 PM UTC+2, WM wrote:
> > 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.

> The statement (sic!) ∀k ∈ ℕ: E(k+1) = E(k) \ {k} just states that the "difference" between E(k+1) and E(k) consists at most of one element

This is a rule which proves that every loss in the function of (intersections of) endsegments is one natnumber per step. No step no loss.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 03:56:37 UTC+2:

> WM> If all endsegments are infinite,
>
> which they are
>
> WM> then there is a jump from more than one element to 0 elements.
>
> Nope, there is no such jump.

Endsegments get empty by ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
If the limit exists and is empty, then it gets empty by precisely the same process. - Or by abracadabra. - But not in my math.

Regards, WM

On 9/13/2022 8:38 AM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 13. September 2022 um 02:47:41 UTC+2:
>> On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
>>> 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.
>> This is called a discontinuity. Yes, the card() function is not continuous, because, e.g., card(lim{n -> oo} E(n)) =/= lim{n -> oo} card(E(n)).
>
> The deletion of elements of endsegments is a continuous process or function. There is no jump or discontinuity.

yet all endsegments remain the same size, ℵ₀, there is no "emptying"

>

>
> Regards, WM

On 9/13/2022 8:52 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 03:56:37 UTC+2:
>
>> WM> If all endsegments are infinite,
>>
>> which they are
>>
>> WM> then there is a jump from more than one element to 0 elements.
>>
>> Nope, there is no such jump.
>
> Endsegments get empty by ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Wrong, all endsegments remain the same size, ℵ₀, they do not get emptied.

> If the limit exists and is empty, then it gets empty by precisely the same process. - Or by abracadabra. - But not in my math.

your math is sloppy math full of obvious errors, even your books.
You are a troll and a crank who does not know math.

>
> Regards, WM

It happens that WM formulated :
> 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.

Is zero a natural number today?

Sergio formulated on Tuesday :
> On 9/13/2022 8:38 AM, WM wrote:
>> Gus Gassmann schrieb am Dienstag, 13. September 2022 um 02:47:41 UTC+2:
>>> On Monday, 12 September 2022 at 10:21:23 UTC-3, WM wrote:
>>>> 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.
>>> This is called a discontinuity. Yes, the card() function is not
>>> continuous, because, e.g., card(lim{n -> oo} E(n)) =/= lim{n -> oo}
>>> card(E(n)).
>>
>> The deletion of elements of endsegments is a continuous process or
>> function. There is no jump or discontinuity.
>
> yet all endsegments remain the same size, ℵ₀, there is no "emptying"

He seems to think that an increasing endsegment index sequence means
that the emptyset cannot be somehow 'reached' whatever that means.
That's why I thought the term 'Delusion Monotony' was so appropriate.

On Tuesday, September 13, 2022 at 3:41:38 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 03:36:25 UTC+2:
> >
> > Consider:
> >
> > ∩{E(k) : k e IN}.
> >
> > Where do you see a "step" here
> >
> This is the last step

The last step? OF WHAT PROCESS, YOU FUCKING ASSHOLE FULL OF SHIT?!

Hint: Intersection is not a "stepwise" process.

On Tuesday, September 13, 2022 at 4:44:01 PM UTC+2, FromTheRafters wrote:

> That's why I thought the term 'Delusion Monotony' was so appropriate.

It sure is!

Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 17:55:27 UTC+2:
> On Tuesday, September 13, 2022 at 3:41:38 PM UTC+2, WM wrote:

> Hint: Intersection is not a "stepwise" process.

You talk about the dark domain?
For every definable endsegment we can follow the steps:
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} .
This does not end but cannot traverse the dark domain. Only the result of the last step ω is visible again, the empty set.

Regards, WM

On 9/13/2022 1:15 PM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 13. September 2022 um 17:55:27 UTC+2:
>> On Tuesday, September 13, 2022 at 3:41:38 PM UTC+2, WM wrote:
>
>> Hint: Intersection is not a "stepwise" process.
>
> You talk about the dark *QUACK*!! domain?
> For every definable *QUACK*!! endsegment we can follow the *QUACK*!! steps:
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k} .
> This does not end but cannot *QUACK*!! traverse the dark *QUACK*!! domain. Only *QUACK!! the result of the last *QUACK*!!step *QUACK*!! ω is visible *QUACK*!! again, the empty set *QUACK*!!.
>
> Regards, WM

Exceptional Quack Density!!

10/38 = 26%

On Sunday, September 11, 2022 at 9:03:51 PM UTC+2, Fritz Feldhase wrote:
> On Sunday, September 11, 2022 at 8:48:56 PM UTC+2, WM wrote:
> >
> > there is a jump <bla>
> >
> Where exactly?

Now in a recent answer to JB you just wrote

| A jump does not exist.

YOU LYING PIECE OF SHIT!!

On Tuesday, September 13, 2022 at 3:56:37 AM UTC+2, Fritz Feldhase wrote:
> On Monday, September 12, 2022 at 3:21:23 PM UTC+2, WM wrote:
> > 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?
> > >
> > The jump is before the zero of the empty set.
> >
> No, there is no such "jump", you fucking idiot!
>
> SINCE THERE IS NO "before the zero of the empty set". <facepalm>
>
> Remember, your IDIOTIC / BRAINDEAD claim was:
>
> WM> If all endsegments are infinite,
>
> which they are
>
> WM> then there is a jump from more than one element to 0 elements.
>
> Nope, there is no such jump.
>
> Hint: For each and every endsegment E there is an endsegment E' such that E > E' > {}.
>
> => NO JUMP.

Now in a recent answer to JB you just wrote

| A jump does not exist.

YOU LYING PIECE OF SHIT!!