On Monday, July 4, 2022 at 8:06:43 PM UTC+2, sergi o wrote:

> I am tired of spoon feeding you.

BIG SURPRISE!!! :-)

On 7/4/2022 1:08 PM, Fritz Feldhase wrote:
> On Monday, July 4, 2022 at 8:06:43 PM UTC+2, sergi o wrote:
>
>> I am tired of spoon feeding you.
>
> BIG SURPRISE!!! :-)

ok, ok,

I just bought 4 dozen NOS "feeding" spoons off Ebay for $3.00, bulk buy from some institution.

On 7/4/2022 8:31 AM, WM wrote:
> bwr fml schrieb am Montag, 4. Juli 2022 um 04:21:06 UTC+2:
>
>> Polite honest serious brief question.
>>
>> Is there any chance that you might be able to come up with the smallest
>> consistent list of axioms that forms the foundation and distinguishes
>> your math from what other people are using?
>
> I do not use different axioms but only the axioms and theorems of ZFC. The only difference is that I take them seriously. For instance I take the bijection between ℕ and ℚ seriously in that I assume that every fraction has got an index. That is not a matter of "limit in the infinite". I assume Cantor's function
> k = (m + n - 1)(m + n - 2)/2 + m
> with the resulting sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
>
> I represent the positive fractions by the well known matrix
>
> 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, ...
> ...
>
> and use the indices 1, 2, 3, ..., abbreviated by X, initially covering the first column
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...

Why ? there is no reason at all to do this.

>
> (in oder to prove by the bijection of the indices with the integer fractions n/1 that all natural numbers are applied). According to Cantos sequence they are distributed over the whole matrix step by step:
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> ...
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> ...
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> ...
>
> and so on.

so you you do your swapadoodlefestival, and arrive at your meaningless butchered matrix.

>
> The result is that the X will never cover the whole matrix although every definable fraction will get an index.

because your proceedure, and process are both wrong.

Use EQUATIONS. if you did you would find out much sooner you are kaputsky.

>
>> And get everyone to agree to preface all future posts with
>> "Assuming this list of axioms:..."?
>
>
>> If the answer is "no" then I would understand completely.
>> I was only hoping to find some way quit the endless arguments about
>> things that do not seem to really address the deeper questions.
>

<snip repeated crap>

> Regards, WM

William schrieb am Montag, 4. Juli 2022 um 19:11:13 UTC+2:
> On Monday, July 4, 2022 at 2:01:03 PM UTC-3, WM wrote:
> > William schrieb am Montag, 4. Juli 2022 um 17:20:43 UTC+2:
> > > On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
> > > > William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> > > > > On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
> > > >
> > > > > > we know that for all of them [the successors] are gone.
> > > > >
> > > > > So what? No statement about "all of them",
> > > > All of them are the set.
> > > Indeed "all of them" is the set. No statement is made about the set. Only a statement about the *elements* of the set.
> > > Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.
> > Correct. We have the elements 1, 2, 3, ...
> Indeed, each element of the set is followed by ℵ₀ successors but the *set* not is followed by ℵ₀ successors.

In the sequence of intersections
E(1) ∩ E(2) ∩ E(3) ∩ ... = { }
there are no natural numbers remaining because each natural number is deleted by one and only one endsegment according to

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

How can ℵ₀ natural numbers be deleted between every endsegment
|∩{E(k) : k ∈ ℕ}| = ℵ₀
∩{E(k) : k ∈ ℕ} = { }
and all endsegments?

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 19:22:17 UTC+2:
> On Monday, July 4, 2022 at 7:06:17 PM UTC+2, WM wrote:

> > We have the elements 1, 2, 3, ... .
> > Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.
> Who cares?

Intelligent mathematicians should do.
>
> We aren't interested in natural numbesr which are or aren't "deleted by definable endsegments".

You prove that all natnumbers are deleted by using the fact that every endsegmet deletes one natnumber. But for most natnumbers this is wrong because:
|∩{E(k) : k ∈ ℕ}| = ℵ₀
>
> All we are interested in is the provable set-theoretic fact:
>
> ∩{E(k) : k ∈ ℕ} = { } .
> > How many endsegments are required to
> get an empty intersection?
>
> The answer is: ___INFINITELY MANY ensegements.___

But no definable endsegment reduces the infinite intersection to less than an infinite intersection. However, the empty intersection is claimed to be produced by endsegments.
>
> Now the important fact concerning
>
> ∩{E(k) : k ∈ ℕ} = { }
>
> is the following:
>
> ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). In words: "For each and every natural number there's an ensegement not containing it."

Definable endsegments leave always ℵ₀ natnumbers.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.

> ∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

Definable endsegments leave always ℵ₀ natnumbers.

> ~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Definable endsegments leave always ℵ₀ natnumbers.
>
> Hence
>
> ∩{E(k) : k ∈ IN} = { }
>
> You know, these steps form a so called PROOF.
>
> Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }
>
Actually this proof proves dark endsegments.

Regards, WM

On Monday, July 4, 2022 at 5:15:58 PM UTC-3, WM wrote:

> In the sequence of intersections
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { }

This is not a sequence of intersections. It is one intersection. It is the intersection of the elements of a sequence. A sequence of intersections is

(E(1), E(1) ∩ E(2), E(1) ∩ E(2) ∩ E(3) , ... )

Note that each *element* of the sequence of intersections is infinite (trivial induction: if element n is infinite, then element n+1 is infinite), but the *intersection of the elements* of the sequence is empty (trivial induction: if n is not in the intersection, then n+1 is not in the intersection).

--
William Hughes

FromTheRafters schrieb am Montag, 4. Juli 2022 um 19:56:16 UTC+2:
> WM explained :

> > According to
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > it is a step-by-step process.
> No, it isn't.

It is for every definable endsegment.
{1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...
If it is not a step-by-step process for some other endsegments, then they must be dark.
What else could prevent steps?

Regards, WM

zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 20:01:45 UTC+2:
> måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:

> > According to
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > it is a step-by-step process.
> no it IS NOT!

It is for every definable endsegment.
{1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...
If it is not a step-by-step process for some other endsegments, then they must be dark.
What else could prevent steps?

> > > The intersection is empty because for every element n, it is not in E(n+1)
> > For almost all natural numbers every definable endsegment fails.
> "definable" is meaningless.
>
> There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE that does it.

Bingo! Only collections can accomplish what no individual can do. That is called dark numbers.

> > For almost all natural numbers every definable endsegment fails.
> >
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

> None of which is relevant for the intersection to be empty

because individuals cannot accomplish it. Only dark endsegments can empty the intersection collectively.

Regards, WM

William schrieb am Montag, 4. Juli 2022 um 23:09:15 UTC+2:
> On Monday, July 4, 2022 at 5:15:58 PM UTC-3, WM wrote:
>
> > In the sequence of intersections
> > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }
> This is not a sequence of intersections. It is one intersection. It is the intersection of the elements of a sequence. A sequence of intersections is
>
> (E(1), E(1) ∩ E(2), E(1) ∩ E(2) ∩ E(3) , ... )
>
> Note that each *element* of the sequence of intersections is infinite (trivial induction: if element n is infinite, then element n+1 is infinite), but the *intersection of the elements* of the sequence is empty (trivial induction: if n is not in the intersection, then n+1 is not in the intersection).

Of course. The facts are clear. Now the question rises: How can these elements of the sequence of intersections produce the empty the intersection if every element of the sequence of intersections obeys

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

Regards, WM

On Monday, July 4, 2022 at 11:07:44 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 19:22:17 UTC+2:
> >
> > <bla bla bla> because: |∩{E(k) : k ∈ ℕ}| = ℵ₀

No, you psyhotic asshole full of shit, "|∩{E(k) : k ∈ ℕ}| = ℵ₀" is wrong, since |∩{E(k) : k ∈ ℕ}| = 0.

Remember:

> > All we are interested in is the provable set-theoretic fact:
> >
> > ∩{E(k) : k ∈ ℕ} = { } .

i.e. |∩{E(k) : k ∈ ℕ}| = 0.

> > > How many endsegments are required to get an empty intersection?
> > >
> > This question has been answered MANY TIMES ALREADY, YOU SILLY ASSHOLE FULL OF SHIT!
> >
> > The answer is: ___INFINITELY MANY ensegements___,

you psychotic asshole full of shit!

> But

No "but", you fucking asshole.

THIS IS A/THE CORRECT ANSWER. No "but".

> > Now the important fact concerning
> >
> > ∩{E(k) : k ∈ ℕ} = { }
> >
> > is the following:
> >
> > ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). In words: "For each and every natural number there's an ensegement not containing it."
> >
> > Hence
> >
> > ∩{E(k) : k ∈ IN} = { }
> >
> > You know, these steps form a so called PROOF.
> >
> > Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }

And now fuck off, you demented and psychotic asshole full of shit!

Math is definitely not your thing.

On Monday, July 4, 2022 at 11:15:44 PM UTC+2, WM wrote:
>
> [Infinite] collections [i. e. infinite sets] can accomplish [this].

Indeed!

> That is called

set theory.

Well done, Mückenheim. Keep up the good work!

WM formulated on Monday :
> FromTheRafters schrieb am Montag, 4. Juli 2022 um 19:56:16 UTC+2:
>> WM explained :
>
>>> According to
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> it is a step-by-step process.
>> No, it isn't.
>
> It is for every definable endsegment.
> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1,
> n+2, ...} --->... If it is not a step-by-step process for some other
> endsegments, then they must be dark. What else could prevent steps?

Nothing prevents it, it is just your inability to grasp the concept of
infinity or *all* without counting to it. Nobody else I know has any
problem at all with it.

on 7/4/2022, WM supposed :
> zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 20:01:45 UTC+2:
>> måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:
>
>>> According to
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> it is a step-by-step process.
>> no it IS NOT!
>
> It is for every definable endsegment.
> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1,
> n+2, ...} --->... If it is not a step-by-step process for some other
> endsegments, then they must be dark. What else could prevent steps?
>
>>>> The intersection is empty because for every element n, it is not in E(n+1)
>>> For almost all natural numbers every definable endsegment fails.
>> "definable" is meaningless.
>>
>> There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE
>> that does it.
>
> Bingo! Only collections can accomplish what no individual can do. That is
> called dark numbers.

What is the difference between the set and its elements?

On Monday, July 4, 2022 at 6:20:56 PM UTC-3, WM wrote:
> William schrieb am Montag, 4. Juli 2022 um 23:09:15 UTC+2:
> > On Monday, July 4, 2022 at 5:15:58 PM UTC-3, WM wrote:
> >
> > > In the sequence of intersections
> > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }
> > This is not a sequence of intersections. It is one intersection. It is the intersection of the elements of a sequence. A sequence of intersections is
> >
> > (E(1), E(1) ∩ E(2), E(1) ∩ E(2) ∩ E(3) , ... )
> >
> > Note that each *element* of the sequence of intersections is infinite (trivial induction: if element n is infinite, then element n+1 is infinite), but the *intersection of the elements* of the sequence is empty (trivial induction: if n is not in the intersection, then n+1 is not in the intersection).
> Of course. The facts are clear.

Yes. Note there is no contradiction. Everything else is just a question of how intuitive things are.

> Now the question rises: How can these elements of the sequence of intersections produce

No. The elements of the sequence of intersections does not "produce" the empty set. The intersection of the elements *is* the empty set.
The question is:

How can the intersection of all the elements of the sequence be empty if

>... every element of the sequence of intersections obeys
>
> ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.

and thus only one element can be "deleted" by each element.
No problem: there are an infinite number of elements and each element removes a different element.

We only get a problem if we try to analyse this as a stepwise process. At which step does the remaining set of elements become finite?

To the strictly limited extent that this makes sense we note that there are an infinite number of steps after each element of the sequence. No matter how far we go there are still an infinite number of steps left in which the remaining infinite number of elements of |N can be "deleted".

More correctly we can say that the intersection of the elements of an infinite sequence cannot be obtained by a stepwise process
No matter how many steps you take you are still no closer to the end.

--
William Hughes

måndag 4 juli 2022 kl. 21:15:58 UTC+1 skrev WM:
> William schrieb am Montag, 4. Juli 2022 um 19:11:13 UTC+2:
> > On Monday, July 4, 2022 at 2:01:03 PM UTC-3, WM wrote:
> > > William schrieb am Montag, 4. Juli 2022 um 17:20:43 UTC+2:
> > > > On Monday, July 4, 2022 at 12:10:16 PM UTC-3, WM wrote:
> > > > > William schrieb am Montag, 4. Juli 2022 um 16:19:26 UTC+2:
> > > > > > On Monday, July 4, 2022 at 10:42:48 AM UTC-3, WM wrote:
> > > > >
> > > > > > > we know that for all of them [the successors] are gone.
> > > > > >
> > > > > > So what? No statement about "all of them",
> > > > > All of them are the set.
> > > > Indeed "all of them" is the set. No statement is made about the set.. Only a statement about the *elements* of the set.
> > > > Each *element* of set of natural numbers is followed by ℵ₀ successors does not mean the *set* , "all of them", is followed by ℵ₀ successors.
> > > Correct. We have the elements 1, 2, 3, ...
> > Indeed, each element of the set is followed by ℵ₀ successors but the *set* not is followed by ℵ₀ successors.
> In the sequence of intersections
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { }
> there are no natural numbers remaining because each natural number is deleted by one and only one endsegment according to
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> How can ℵ₀ natural numbers be deleted between every endsegment
> |∩{E(k) : k ∈ ℕ}| = ℵ₀
> ∩{E(k) : k ∈ ℕ} = { }
> and all endsegments?
>
> Regards, WM

Your first equation is wrong

måndag 4 juli 2022 kl. 22:07:44 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 19:22:17 UTC+2:
> > On Monday, July 4, 2022 at 7:06:17 PM UTC+2, WM wrote:
>
> > > We have the elements 1, 2, 3, ... .
> > > Many of them are deleted by definable endsegments. Most of them are not deleted by definable endsegments.
> > Who cares?
> Intelligent mathematicians should do.

Mathematicians are more intelligent than you. And you are not a mathematician.

> >
> > We aren't interested in natural numbesr which are or aren't "deleted by definable endsegments".
> You prove that all natnumbers are deleted by using the fact that every endsegmet deletes one natnumber. But for most natnumbers this is wrong because:
> |∩{E(k) : k ∈ ℕ}| = ℵ₀

False, it is equal to 0

> >
> > All we are interested in is the provable set-theoretic fact:
> >
> > ∩{E(k) : k ∈ ℕ} = { } .
> > > How many endsegments are required to
> > get an empty intersection?
> >
> > The answer is: ___INFINITELY MANY ensegements.___
> But no definable endsegment reduces the infinite intersection to less than an infinite intersection. However, the empty intersection is claimed to be produced by endsegments.

your "definable" is undefined and meaningless

> >
> > Now the important fact concerning
> >
> > ∩{E(k) : k ∈ ℕ} = { }
> >
> > is the following:
> >
> > ∀k ∈ ℕ: En ∈ ℕ: k !e E(n). In words: "For each and every natural number there's an ensegement not containing it."
> Definable endsegments leave always ℵ₀ natnumbers.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.
> > ∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}
> Definable endsegments leave always ℵ₀ natnumbers.
> > ~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}
> Definable endsegments leave always ℵ₀ natnumbers.
> >
> > Hence
> >
> > ∩{E(k) : k ∈ IN} = { }
> >
> > You know, these steps form a so called PROOF.
> >
> > Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }
> >
> Actually this proof proves dark endsegments.

It doesn't, you make them up

>
> Regards, WM

måndag 4 juli 2022 kl. 22:15:44 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 20:01:45 UTC+2:
> > måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:
>
> > > According to
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > it is a step-by-step process.
> > no it IS NOT!
> It is for every definable endsegment.

It is not a step by step process

> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...

That is a sequence, not a process.

> If it is not a step-by-step process for some other endsegments, then they must be dark.

False, mathematics do not do step by step processes, it doesn't even do processes because nothing takes time.

> What else could prevent steps?

The fact none is ever taken to begin with.

> > > > The intersection is empty because for every element n, it is not in E(n+1)
> > > For almost all natural numbers every definable endsegment fails.
> > "definable" is meaningless.
> >
> > There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE that does it.
> Bingo! Only collections can accomplish what no individual can do. That is called dark numbers.

Nope, It means that infinite and finite are very different. You argue finite intersection and then ask about the infinite one.

> > > For almost all natural numbers every definable endsegment fails.
> > >
> > > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .
> > None of which is relevant for the intersection to be empty
> because individuals cannot accomplish it. Only dark endsegments can empty the intersection collectively.
>
> Regards, WM

FromTheRafters schrieb am Montag, 4. Juli 2022 um 23:43:28 UTC+2:
> WM formulated on Monday :
> > FromTheRafters schrieb am Montag, 4. Juli 2022 um 19:56:16 UTC+2:
> >> WM explained :
> >
> >>> According to
> >>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> >>> it is a step-by-step process.
> >> No, it isn't.
> >
> > It is for every definable endsegment.
> > {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1,
> > n+2, ...} --->... If it is not a step-by-step process for some other
> > endsegments, then they must be dark. What else could prevent steps?
> Nothing prevents it,

Then it should go until the end. According to Cantor the end ω and the state ω + 1 behind the end are existing.

> it is just your inability to grasp the concept of
> infinity or *all* without counting to it. Nobody else I know has any
> problem at all with it.

Most are too credulous to see the problem or to ask for its explanation. It is this one:

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

Regards, WM

FromTheRafters schrieb am Montag, 4. Juli 2022 um 23:47:42 UTC+2:
> on 7/4/2022, WM supposed :
> > zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 20:01:45 UTC+2:
> >> måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:
> >
> >>> According to
> >>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> >>> it is a step-by-step process.
> >> no it IS NOT!
> >
> > It is for every definable endsegment.
> > {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1,
> > n+2, ...} --->... If it is not a step-by-step process for some other
> > endsegments, then they must be dark. What else could prevent steps?
> >
> >>>> The intersection is empty because for every element n, it is not in E(n+1)
> >>> For almost all natural numbers every definable endsegment fails.
> >> "definable" is meaningless.
> >>
> >> There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE
> >> that does it.
> >
> > Bingo! Only collections can accomplish what no individual can do. That is
> > called dark numbers.
> What is the difference between the set and its elements?

The set is one object. Its elements are 0 or 1 or many.

Regards, WM

William schrieb am Dienstag, 5. Juli 2022 um 00:03:55 UTC+2:
> On Monday, July 4, 2022 at 6:20:56 PM UTC-3, WM wrote:

> > How can the intersection of all the elements of the sequence be empty if
>
> >... every element of the sequence of intersections obeys
> >
> > ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.
> and thus only one element can be "deleted" by each element.
>
> No problem: there are an infinite number of elements and each element removes a different element.

All definable endsegments leave almost all natnumbers.
>
> We only get a problem if we try to analyse this as a stepwise process. At which step does the remaining set of elements become finite?

It is a stepwise process:
{1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...
If it is not a step-by-step process for some other endsegments, then they must be dark.
What else could prevent steps?

> To the strictly limited extent that this makes sense we note that there are an infinite number of steps after each element of the sequence. No matter how far we go there are still an infinite number of steps left in which the remaining infinite number of elements of |N can be "deleted".

No matter how far we go, the intersection is not empty because there is always a next step which must find some natnumber to delete.
>
> More correctly we can say that the intersection of the elements of an infinite sequence cannot be obtained by a stepwise process

Then ∀k ∈ ℕ: E(k+1) = E(k) \ {k} is wrong?

> No matter how many steps you take you are still no closer to the end.
>
Why then do you believe in the existence of an end?

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 08:22:41 UTC+2:
> måndag 4 juli 2022 kl. 22:07:44 UTC+1 skrev WM:
> > Definable endsegments leave always ℵ₀ natnumbers.

> > Actually this proof proves dark endsegments.
> It doesn't, you make them up

How do the ℵ₀ natnumbers of every definable endsegment geht lost when
∩{E(k) : k ∈ ℕ} = { } ?

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 08:24:14 UTC+2:
> måndag 4 juli 2022 kl. 22:15:44 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 4. Juli 2022 um 20:01:45 UTC+2:
> > > måndag 4 juli 2022 kl. 18:10:35 UTC+1 skrev WM:
> >
> > > > According to
> > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > > it is a step-by-step process.
> > > no it IS NOT!
> > It is for every definable endsegment.
> It is not a step by step process
> > {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...
> That is a sequence, not a process.

Of course, the results of a process can form a sequence. Here we see a loss of natnumbers step by step. Usually this is called a process.

> > > There needs no fucking specific endsegmetn, it is the COLLECTION AS A WHOLE that does it.
> > Bingo! Only collections can accomplish what no individual can do. That is called dark numbers.
> Nope, It means that infinite and finite are very different.

Of course. Every finite is definable. Every infinite is mostly undefinable.

> You argue finite intersection and then ask about the infinite one.

Yes. Every definable endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 23:39:44 UTC+2:
> On Monday, July 4, 2022 at 11:15:44 PM UTC+2, WM wrote:
> >
> > [Infinite] collections [i. e. infinite sets] can accomplish [this].
>
> Indeed!
>
> > That is called
>
> set theory.
>
But it is claimed, mistakenly or by blatant deceit, that every element can be defined and used individually.

Regards, WM

Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 23:28:16 UTC+2:

> > > In words: "For each and every natural number there's an ensegement not containing it."

Maybe. But for almost all natural numbers there is no individual endsegment not containing them because
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ .

Regards, WM

On Tuesday, July 5, 2022 at 4:21:58 PM UTC+2, WM wrote:

> But it is claimed [...] that every element can be defined and used individually.

1. Define

a) "can be defined"

b) "can be used individually"

2. Provide a quote for your claim.

- Who (author) claimed / where (textbook, paper, article) "that every element [in IN] can be defined and used individually"?