On Tuesday, July 5, 2022 at 5:18:06 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:

> visible numbers.

"Visible" One of your many many many many ways of saying "can be written down".

> > There are methods by which you can obtain the state.
> What methods?

Induction for one. (1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.) Or contradiction (Assume the intersection is not empty. Contradiction. Hence the intersection is empty)

> > These methods do not depend on > the existence of "dark elements".
> >
> Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?

Yes. But this triviality just means that a stepwise process will not work. It does not mean that every method will not work.

--
William Hughes

William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
> On Tuesday, July 5, 2022 at 5:18:06 PM UTC-3, WM wrote:
> > William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:
> > visible numbers.
>
> "Visible" One of your many many many many ways of saying "can be written down".

Has a FISON.

> > > There are methods by which you can obtain the state.
> > What methods?
> Induction for one. (1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.)

The intersection is empty. But every definable endsegment (every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:

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

Do you accept that?

> Or contradiction (Assume the intersection is not empty. Contradiction. Hence the intersection is empty)

The intersection is empty. But all endsegments indexed by visible numbers (= numbers having FISONs) contain almost all natural numbers. (ℵ₀ is the complement of every |FISON|.)

> > > These methods do not depend on the existence of "dark elements".
> > >
> > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
> Yes. But this triviality just means that a stepwise process will not work..

No, it means that this stepwise process works until the empty set is reached. Otherwise you should not accept it.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
> tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
> > Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.
> Skip the "definable", it is true for ALL endsegments

Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong. Then ℵ₀ numbers contained in every endsegment would get lost at the end. That is in contradiction with the action of endsegments.

> There are no dark
> The intersection is empty (nothing is lost), because n is not in E(n+1)

How can you address this silly argument???
It is not true for almost all natnumbers
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .

> That is it, how is this too difficult for you?

You have agreed already that many things can happen only collectively. That is but another name for dark elements.

Regards, WM

On Tuesday, July 5, 2022 at 10:14:28 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 17:43:33 UTC+2:
> > On Tuesday, July 5, 2022 at 5:30:10 PM UTC+2, WM wrote:
> > >
> > > For [natural] numbers this is not possible. They have an order which forces one of them to be the last.
> > >
> > No, the order of the natural numbers does not "force one of them to be the last".
> >
> ω forces all natural numbers to be less.

Huh?! So ω is the last natural number in your delusional Mückenmath?

That's... FASCINATING, Mückenheim!

Did you already published this groundbreaking result?

On Tuesday, July 5, 2022 at 11:12:35 PM UTC+2, WM wrote:
> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
> >
> > Induction[:] 1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.
> >
> The intersection is empty.

Uffff... So you get it now?

> But every [...] endsegment [...] contains almost all natural numbers

Yes, of course. Actually, each and every Endsegment E has a minimal element min(E), and E = {min(E), min(E)+1, min(E)+2, ...}.

> The intersection is empty.

YES, WE KNOW THAT, DUMBO!

> But all endsegments indexed by [natural] numbers (=numbers having FISONs) contain almost all natural numbers.

YES, WE KNOW THAT, DUMBO!

Hint: Each and every Endsegment E has a minimal element min(E), and E = {min(E), min(E)+1, min(E)+2, ...}.

> > > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ? (*)
> > >
> > Yes. But this triviality just means that a stepwise process will not work.
> >
> [...] it means that this stepwise process

does not lead to an empty (or even finite) endsegment (assuming E(1) = IN).

In other words, there is no n e IN such that E(n) is finite.

Proof by induction: <trivial>

On Tuesday, July 5, 2022 at 11:18:32 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
> >
> > The intersection is empty [...], because n is not in E(n+1) [for each and every natural number n].
> >
> How can you address this silly argument???

This remark leaves me speechless. :-O

> <nonsense deleted>

On Tuesday, July 5, 2022 at 6:12:35 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
> > > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
> > Yes. But this triviality just means that a stepwise process will not work.
> No, it means that this stepwise process works until the empty set is reached.

Nope. To work a method has to produce the state where all steps are done . The stepwise process only produces states that are not the state where all steps are done. The stepwise process does not "work until the empty set is reached". The stepwise process never works
--
William Hughes

On 7/5/2022 4:12 PM, WM wrote:
> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
>> On Tuesday, July 5, 2022 at 5:18:06 PM UTC-3, WM wrote:
>>> William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:
>>> visible numbers.
>>
>> "Visible" One of your many many many many ways of saying "can be written down".
>
> Has a FISON.

"Visible Ants" are already on the ANT LIST

>
>>>> There are methods by which you can obtain the state.
>>> What methods?
>> Induction for one. (1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.)
>
> The intersection is empty. But every definable endsegment (every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀.

∩{E(1), E(2), ..., E(k)} = E(k)

that is all the above says

>
> Do you accept that?
>
>> Or contradiction (Assume the intersection is not empty. Contradiction. Hence the intersection is empty)

your MISTAKE is believing the intersection of all endsegments is an endsegment, it is not.

>
> The intersection is empty. But all endsegments indexed by visible numbers (= numbers having FISONs) contain almost all natural numbers. (ℵ₀ is the complement of every |FISON|.)
>
>>>> These methods do not depend on the existence of "dark elements".
>>>>
>>> Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
>> Yes. But this triviality just means that a stepwise process will not work.
>
> No, it means that this stepwise process works until the empty set is reached. Otherwise you should not accept it.

Wrong. You have a progression of endsegments there is no way of reaching the empty set, as it is not an endsegment.

Fools parade.

>
> Regards, WM

On 7/5/2022 4:18 PM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
>> tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
>
>>> Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.
>> Skip the "definable", it is true for ALL endsegments
>
> Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong.

wrong and intentionally misleading.

>
>
>> There are no dark
>> The intersection is empty (nothing is lost), because n is not in E(n+1)
>
> How can you address this silly argument???

n is not in E(n+1)

> It is not true for almost all natnumbers

it is true for all natural numbers. You just dont want it to be true.

> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
>
>> That is it, how is this too difficult for you?
>
> You have agreed already that many things can happen only collectively. That is but another name for dark elements.

no, your BS surrounds you.

>
> Regards, WM

tisdag 5 juli 2022 kl. 22:12:35 UTC+1 skrev WM:
> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
> > On Tuesday, July 5, 2022 at 5:18:06 PM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:
> > > visible numbers.
> >
> > "Visible" One of your many many many many ways of saying "can be written down".
> Has a FISON.

which is ALL THE NATURAL NUMBERS!

> > > > There are methods by which you can obtain the state.
> > > What methods?
> > Induction for one. (1 is not in the intersection. If n is not in the intersection then n+1 is not in the intersection. Hence the intersection is empty.)
> The intersection is empty. But every definable endsegment

Which is all endsegments, "definable" is meaningless

>(every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:

Which is again all

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

Applies to all

> Do you accept that?
> > Or contradiction (Assume the intersection is not empty. Contradiction. Hence the intersection is empty)
> The intersection is empty. But all endsegments indexed by visible numbers (= numbers having FISONs)

Which is sitll all natural numbers

>contain almost all natural numbers. (ℵ₀ is the complement of every |FISON|.)

> > > > These methods do not depend on the existence of "dark elements".
> > > >
> > > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
> > Yes. But this triviality just means that a stepwise process will not work.
> No, it means that this stepwise process works until the empty set is reached. Otherwise you should not accept it.

No, it does not! And it is not a fucking stepwise process! it is not even a process!

>
> Regards, WM

tisdag 5 juli 2022 kl. 22:18:32 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
> > tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
>
> > > Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.
> > Skip the "definable", it is true for ALL endsegments
> Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong..

Nope

>Then ℵ₀ numbers contained in every endsegment would get lost at the end. That is in contradiction with the action of endsegments.

nothing is "Lost" because there is no process going on. There is no contradiction

> > There are no dark
> > The intersection is empty (nothing is lost), because n is not in E(n+1)
> How can you address this silly argument???

You are the one making retarded arguments.

> It is not true for almost all natnumbers

It is true for ALL natural numbers that n is not in E(n+1)

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

That is true for ALL of them, both statements are true.

> > That is it, how is this too difficult for you?
> You have agreed already that many things can happen only collectively. That is but another name for dark elements.

Nope, that means nothing of the sort you retard

>
> Regards, WM

Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 23:27:59 UTC+2:
> On Tuesday, July 5, 2022 at 10:14:28 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 17:43:33 UTC+2:
> > > On Tuesday, July 5, 2022 at 5:30:10 PM UTC+2, WM wrote:
> > > >
> > > > For [natural] numbers this is not possible. They have an order which forces one of them to be the last.
> > > >
> > > No, the order of the natural numbers does not "force one of them to be the last".
> > >
> > ω forces all natural numbers to be less.

> That's... FASCINATING, Mückenheim!
>
> Did you already published this groundbreaking result?

You should try to learn the basics of the theory that you try to defend.

daß ω die kleinste transfinite Ordnungs-Zahl, d. h. die kleinste festbestimmte Zahl ist, welche größer als alle endlichen Zahlen ν [Cantor, p. 395]

Regards, WM

Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 23:36:58 UTC+2:
> On Tuesday, July 5, 2022 at 11:12:35 PM UTC+2, WM wrote:
> > William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:

> > > > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ? (*)
> > > >
> > > Yes. But this triviality just means that a stepwise process will not work.
> > >
> > [...] it means that this stepwise process
>
> does not lead to an empty (or even finite) endsegment (assuming E(1) = IN).
>
Then you do not accept (*). But that is the basic of endsegments. It follows from their definition and holds for all endsegments.

Regards, WM

William schrieb am Mittwoch, 6. Juli 2022 um 02:22:29 UTC+2:
> On Tuesday, July 5, 2022 at 6:12:35 PM UTC-3, WM wrote:
> > William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
>
> > > > Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
> > > Yes. But this triviality just means that a stepwise process will not work.
> > No, it means that this stepwise process works until the empty set is reached.
> Nope. To work a method has to produce the state where all steps are done .. The stepwise process only produces states that are not the state where all steps are done.

That proves the existence of not-steppable states. I call them dark numbers..

> The stepwise process does not "work until the empty set is reached".

It works however for all definable elemets. You cannot find any which it does not work for. This proves the existence of elements which cannot be stepped through.

> The stepwise process never works

It works for all definable numbers. But it does not work for dark numbers. That's why Cantor's enumerations must necessarily fail. A good example is here:

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
....

1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
.... ... ... ...

By steps the desired result

1/1, __, __, __, ...
1/2, __, __, __, ...
1/3, __, __, __, ...
2/2, __, __, __, ...
....

will never be reached.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:26:00 UTC+2:
> tisdag 5 juli 2022 kl. 22:12:35 UTC+1 skrev WM:

> >(every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:
> Which is again all
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀.
> Applies to all

Then ℵ₀ natnumbers are in all endsegments. Therefore not every natnumber has an endsegment where it appears last and is then deleted.

Regards, WM

onsdag 6 juli 2022 kl. 14:42:21 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:26:00 UTC+2:
> > tisdag 5 juli 2022 kl. 22:12:35 UTC+1 skrev WM:
>
> > >(every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:
> > Which is again all
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀.
> > Applies to all
> Then ℵ₀ natnumbers are in all endsegments.

Correct, we have neversaid anything else

>Therefore not every natnumber has an endsegment where it appears last and is then deleted.

Nothing is deleted, the intersection is empty because n is not in E(n+1)

Even a 15 year old can get this

>
> Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:27:22 UTC+2:
> tisdag 5 juli 2022 kl. 22:18:32 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
> > > tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
> >
> > > > Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.
> > > Skip the "definable", it is true for ALL endsegments
> > Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong.
> Nope

Only one element can be lost per endsegment. According to you infiitely many must be lost between all intersections of finitely many endsegments and the intersection of infinitely many.

> >Then ℵ₀ numbers contained in every endsegment would get lost at the end. That is in contradiction with the action of endsegments.
> nothing is "Lost" because there is no process going on.

This is a process:
{1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...

> > > There are no dark
> > > The intersection is empty (nothing is lost), because n is not in E(n+1)
> > How can you address this silly argument???
> You are the one making retarded arguments.
> > It is not true for almost all natnumbers
> It is true for ALL natural numbers that n is not in E(n+1)
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
> That is true for ALL of them, both statements are true.

Impossible. ℵ₀ natnumbers are in all endsegments.

> > You have agreed already that many things can happen only collectively. That is but another name for dark elements.
> Nope, that means nothing of the sort

It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.

Regards, WM

On Wednesday, July 6, 2022 at 3:31:31 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 23:27:59 UTC+2:
> > On Tuesday, July 5, 2022 at 10:14:28 PM UTC+2, WM wrote:
> > > Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 17:43:33 UTC+2:
> > > > On Tuesday, July 5, 2022 at 5:30:10 PM UTC+2, WM wrote:
> > > > >
> > > > > [The natural numbers] have an order which forces one of them to be the last.
> > > > >
> > > > No, the order of the natural numbers does not "force one of them to be the last".
> > > >
> > > ω forces all natural numbers to be less.
> > >
> > Huh?! So ω is the last natural number in your delusional Mückenmath?

No answer?

> > That's... FASCINATING, Mückenheim!

I meant you discovery that ω is a natural numbers [actually the last one].

> > Did you already published this groundbreaking result?
> >
> You should try to learn the basics of the theory that you try to defend.
>
> "daß ω die kleinste transfinite Ordnungs-Zahl, d. h. die kleinste festbestimmte Zahl ist, welche größer als alle endlichen Zahlen ν" [Cantor, p. 395].

Where did Cantor state that there is a last natural number [endliche Zahl]?

And where did he state that the order of the natural numbers "forces one of them to be the last"?

Is there a last natural number in your Wahnsystem, Mückenheim? Yes?

Is it ω? Really?

Hint: I did't ask for "less" but "last" - you know the difference, dumbo?

On 7/6/2022 8:33 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 23:36:58 UTC+2:
>> On Tuesday, July 5, 2022 at 11:12:35 PM UTC+2, WM wrote:
>>> William schrieb am Dienstag, 5. Juli 2022 um 22:39:29 UTC+2:
>
>>>>> Do you accept ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ? (*)
>>>>>
>>>> Yes. But this triviality just means that a stepwise process will not work.
>>>>
>>> [...] it means that this stepwise process
>>
>> does not lead to an empty (or even finite) endsegment (assuming E(1) = IN).
>>
> Then you do not accept (*).

that equation only relates two adjacent endsegments.

> But that is the basic of endsegments.

Wrong.

This is the basic of endsegments E(k) = {k,k+1,...}

> It follows from their definition and holds for all endsegments.

can you provide your definition of Endsegment ?

>
> Regards, WM

On Wednesday, July 6, 2022 at 10:38:20 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 6. Juli 2022 um 02:22:29 UTC+2:

> > The stepwise process does not "work until the empty set is reached".
> It works however for all definable elemets.

Nope. At each step it fails to produce the final state..

>You cannot find any which it does not work for.

Since it does not produce the final state , there is no step at which it works

>This proves the existence of elements which cannot be stepped through.
> > The stepwise process never works
> It works for all definable numbers. But it does not work for dark numbers. That's why Cantor's enumerations must necessarily fail. A good example is here:
<snip>
>
> By steps the desired result
<snip>
> will never be reached.

Thus the stepwise process never works.

--
William Hughes

On 7/6/2022 8:42 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:26:00 UTC+2:
>> tisdag 5 juli 2022 kl. 22:12:35 UTC+1 skrev WM:
>
>>> (every endsegment that you can choose, every endsegment that closes a finite set of endsegments) contains almost all natural numbers:
>> Which is again all
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀.
>> Applies to all
>
> Then ℵ₀ natnumbers are in all endsegments. Therefore not every natnumber has an endsegment where it appears last and is then deleted.
>
> Regards, WM

1. sets are fixed, there are no deletions.

2. there are no "last" numbers in endsegments

On 7/6/2022 8:48 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:27:22 UTC+2:
>> tisdag 5 juli 2022 kl. 22:18:32 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
>>>> tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
>>>
>>>>> Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.
>>>> Skip the "definable", it is true for ALL endsegments
>>> Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong.
>> Nope
>
> Only one element can be lost per endsegment. According to you infiitely many must be lost between all intersections of finitely many endsegments and the intersection of infinitely many.

Bla..

>
>>> Then ℵ₀ numbers contained in every endsegment would get lost at the end. That is in contradiction with the action of endsegments.
>> nothing is "Lost" because there is no process going on.
>
> This is a process:
> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...

state it using Math and Equations please.

>
>>>> There are no dark
>>>> The intersection is empty (nothing is lost), because n is not in E(n+1)
>>> How can you address this silly argument???
>> You are the one making retarded arguments.
>>> It is not true for almost all natnumbers
>> It is true for ALL natural numbers that n is not in E(n+1)
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .
>> That is true for ALL of them, both statements are true.
>
> Impossible. ℵ₀ natnumbers are in all endsegments.

It is true for ALL natural numbers that n is not in E(n+1)

∩{E(1), E(2), ..., E(k)} = E(k) trivial.

>
>>> You have agreed already that many things can happen only collectively. That is but another name for dark elements.
>> Nope, that means nothing of the sort
>
> It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.

sure, and you cannot prove any of it.

>
> Regards, WM

On Wednesday, July 6, 2022 at 3:48:13 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:27:22 UTC+2:
>
> It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.

This is not a definition, but BS.

Hint: Things? What things? Events? Mathematical objects/numbers? Can events be dark? Can numbers happen?

See: https://en.m.wikipedia.org/wiki/Equivocation
and: https://en.m.wikipedia.org/wiki/Category_mistake

On 7/6/2022 9:45 AM, Fritz Feldhase wrote:
> On Wednesday, July 6, 2022 at 3:48:13 PM UTC+2, WM wrote:
>> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:27:22 UTC+2:
>>
>> It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.
>
> This is not a definition, but BS.
>
> Hint: Things? What things? Events? Mathematical objects/numbers? Can events be dark? Can numbers happen?
>
> See: https://en.m.wikipedia.org/wiki/Equivocation
> and: https://en.m.wikipedia.org/wiki/Category_mistake

Fallacy of four terms

Equivocation in a syllogism (a chain of reasoning) produces a fallacy of four terms (quaternio terminorum). Below are some examples:

Since only man [human] is rational.
And no woman is a man [male].
Therefore, no woman is rational.[1]
The first instance of "man" implies the entire human species, while the second implies just those who are male.

A feather is light [not heavy].
What is light [bright] cannot be dark.
Therefore, a feather cannot be dark.
In the above example, distinct meanings of the word "light" are implied in contexts of the first and second statements.

All jackasses [male donkey] have long ears.
Carl is a jackass [annoying person].
Therefore, Carl has long ears.
Here, the equivocation is the metaphorical use of "jackass" to imply a simple-minded or obnoxious person instead of a male donkey.

Motte-and-bailey fallacy

Main article: Motte-and-bailey fallacy
Equivocation can also be used to conflate two positions which share similarities, one modest and easy to defend and one much more controversial. The
arguer advances the controversial position, but when challenged, they insist that they are only advancing the more modest position.

After serious thinking WM wrote :
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 08:27:22 UTC+2:
>> tisdag 5 juli 2022 kl. 22:18:32 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 19:41:08 UTC+2:
>>>> tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev WM:
>>>>> Yes. Every definable endsegment contains ℵo natnumbers. They can only get
>>>>> lost by dark numbers as a whole.
>>>> Skip the "definable", it is true for ALL endsegments
>>> Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} would be wrong.
>> Nope
>
> Only one element can be lost per endsegment. According to you infiitely many
> must be lost between all intersections of finitely many endsegments and the
> intersection of infinitely many.
>
>>> Then ℵ₀ numbers contained in every endsegment would get lost at the end.
>>> That is in contradiction with the action of endsegments.
>> nothing is "Lost" because there is no process going on.
>
> This is a process:
> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1,
> n+2, ...} --->...

The process of shifting labels?