WM explained on 7/5/2022 :
> 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

The cardinality of your endsegments is always Aleph_null so anything
can be stated for the thing preceding the boolean or.

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

What are you trying to say here?

WM presented the following explanation :
> 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.

What happened to your claim that a set is nothing but its elements?

On Tuesday, July 5, 2022 at 4:25:52 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 23:28:16 UTC+2:
> >
> > "For each and every natural number there's an ensegement not containing it."
> >
> Maybe.

No, not "maybe". The following is a THEOREM in ZFC (aka "set theory"):

| An e IN: Em e IN: n !e E(m). (*)

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

> But

HOLY SHIT!!! There is no "but", you silly idiot.

(*) IS IMPLIED BY your favorite formula

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Now from (*) we get:

∀k ∈ ℕ: k !e ∩{E(k) : k ∈ IN}

by definition of ∩. [ Hint: x e ∩{E(k) : k ∈ IN} iff x e E(k) for ALL k in IN. In words: x is in ∩{E(k) : k ∈ IN} iff x is in each and every endsegment. ]

Hence

~Ek ∈ ℕ: k e ∩{E(k) : k ∈ IN}

Hence

∩{E(k) : k ∈ IN} = { }

(since ∩{E(k) : k ∈ IN} c ℕ).

qed

You know, these steps form a so called PROOF.

Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }

On Tuesday, July 5, 2022 at 4:16:54 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 5. Juli 2022 um 08:24:14 UTC+2:
> >
> > You argue finite intersection and then ask about the infinite one.
> >
> Yes.

Huh?!

> Every [...] endsegment contains ℵo natnumbers. They can only get lost by dark numbers as a whole.

Well, "[WM's] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

FromTheRafters schrieb am Dienstag, 5. Juli 2022 um 16:34:36 UTC+2:
> WM presented the following explanation :

> >> What is the difference between the set and its elements?
> >
> > The set is one object. Its elements are 0 or 1 or many.
> What happened to your claim that a set is nothing but its elements?

It is correct. Like a couple is nothing but the partners.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 16:30:25 UTC+2:
> 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"

https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 212
>
> 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"?

The two, for their time and even today, strongest and profoundest arguments of St Thomas Aquinatus S. Th. I, q. 7, a. 4 [...] become invalid as soon as a principle of individuation, intention, and ordination of actually infinite numbers and sets has been found;" [G. Cantor, letter to A. Schmid (26 Mar 1887)]

Mit jeder wohlgeordneten fertigen Menge M von Alefs ist aber nach dem Bildungsgesetz der Alefs ein bestimmtes Alef gegeben, welches der Größe nach auf alle Individuen von M nächstfolgt.

FOL is restricted to quantification over individuals, the elements of the domain, often a set, but not over predicates or functions

for example, with respect to the, according to my firm conviction, actually infinite number of created individuals, in the universe as already on our earth and, most probably, even in each extended part of the space, however small it may be.

then this epitome certainly consists of infinitely many individuals; therefore we can ask about the cardinality of this set, and I am able to strictly prove that this cardinality is the first one, i.e., not a higher one." [G. Cantor, letter to W. Wundt (16 Oct 1883)] Cantor never delivered a proof of this thesis.

And finally generally accepted ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Regards, WM

On Tuesday, July 5, 2022 at 11:15:55 AM UTC-3, WM wrote:

> > 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?

The "end" is the state where all steps are done. However, there is no last step. You cannot obtain the intersection, the state at which all steps are done, from a stepwise process with no last step. There are other methods to obtain the state at which all steps are done that do work

--
William Hughes

Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 16:40:07 UTC+2:
> On Tuesday, July 5, 2022 at 4:25:52 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 23:28:16 UTC+2:
> > >
> > > "For each and every natural number there's an ensegement not containing it."
> > >
> > Maybe.
> No, not "maybe". The following is a THEOREM in ZFC (aka "set theory"):
>
> | An e IN: Em e IN: n !e E(m). (*)
> "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 . (*)

> You know, these steps form a so called PROOF.

"So called" is the key word. You cannot fail to understand the importance of (*). You have sen it often, never thought about its consequences, but now you should understand.
>
> Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }

They do not. But we can assume that ∩{E(k) : k ∈ ℕ} = { } is correct. Then we can ask how this cn become true.

Regards, WM

On Tuesday, July 5, 2022 at 5:03:06 PM UTC+2, WM wrote:
> FromTheRafters schrieb am Dienstag, 5. Juli 2022 um 16:34:36 UTC+2:
> > WM presented the following explanation :
> > >> What is the difference between the set and its elements?
> > >
> > > The set is one object. Its elements are 0 or 1 or many.
> > >
> > What happened to your claim that a set is nothing but its elements?
> >
> It is correct. Like <bla bla bla>

No, it's NOT correct, since the empty set is not nothing, you silly idiot.

Moreover the set just containing the element 1, i. e. {1}, is NOT identical with 1.

Wie gesagt, Du hast wirklich schwer einen an der Klatsche, Mann.

On Tuesday, July 5, 2022 at 5:14:47 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 16:30:25 UTC+2:
> > 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"
> >
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 212
> >
> > 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"?
> >
> The two, for their time and even today, strongest and profoundest arguments of St Thomas Aquinatus S. Th. I, q. 7, a. 4 [...] become invalid as soon as a principle of individuation, intention, and ordination of actually infinite numbers and sets has been found;" [G. Cantor, letter to A. Schmid (26 Mar 1887)]
>
> Mit jeder wohlgeordneten fertigen Menge M von Alefs ist aber nach dem Bildungsgesetz der Alefs ein bestimmtes Alef gegeben, welches der Größe nach auf alle Individuen von M nächstfolgt.
>
> FOL is restricted to quantification over individuals, the elements of the domain, often a set, but not over predicates or functions
>
> for example, with respect to the, according to my firm conviction, actually infinite number of created individuals, in the universe as already on our earth and, most probably, even in each extended part of the space, however small it may be.
>
> then this epitome certainly consists of infinitely many individuals; therefore we can ask about the cardinality of this set, and I am able to strictly prove that this cardinality is the first one, i.e., not a higher one." [G. Cantor, letter to W. Wundt (16 Oct 1883)] Cantor never delivered a proof of this thesis.
>
> And <bla>

You are very ill, man.

William schrieb am Dienstag, 5. Juli 2022 um 17:18:39 UTC+2:
> On Tuesday, July 5, 2022 at 11:15:55 AM UTC-3, WM wrote:
>
>
>
> > > 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?
> The "end" is the state where all steps are done. However, there is no last step.

If all steps are defined and there is no last step, then there is no end. An end without preceding last step cannot exist other than by dark elements.

> You cannot obtain the intersection, the state at which all steps are done, from a stepwise process with no last step.

Then you cannot enumerate all fractions either. Then this is only delusion or deceit.

> There are other methods to obtain the state at which all steps are done that do work.

Yes, they do work by dark numbers: No last one is discernible but all are there. For visible numbers this is not possible. They have an order which forces one of them to be the last.

Regards, WM

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 number does not "forces one of them to be the last".

Actually, there is no last/biggest natural number. (Still there are infinitely many of them.)

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".

Actually, there is no last/biggest natural number. (Still there are infinitely many of them.)

On Tuesday, July 5, 2022 at 12:30:10 PM UTC-3, WM wrote:
efined" has no meaning here

> [if] there is no last step, then there is no end.

Nope. There can be a state at which all steps are done, even if there is no last step. However you cannot get to this state by using a stepwise process. There are methods by which you can obtain the state. These methods do not depend on the existence of "dark elements".

--
William Hughes

On 7/5/2022 10:30 AM, WM wrote:
> William schrieb am Dienstag, 5. Juli 2022 um 17:18:39 UTC+2:
>> On Tuesday, July 5, 2022 at 11:15:55 AM UTC-3, WM wrote:
>>
>>
>>
>>>> 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?
>> The "end" is the state where all steps are done. However, there is no last step.
>
> If all steps are "defined" *putting on clown shoes*
>
>> You cannot obtain the intersection, the state at which all steps are done, from a stepwise process with no last step.
>
> Then you cannot enumerate all fractions either.

CLOWN shoes are fully one, the BS flows out. the fractions were enumerated long ago, Cantor Proved it, you cannot un-prove it.

>
>> There are other methods to obtain the state at which all steps are done that do work.
>
>

those are not Dark Clown Shoes either, anytime you use one of your Fake Math terms they light up!

https://d.newsweek.com/en/full/1898188/clown-shoes.jpg

> Regards, WM

On 7/5/2022 10:22 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 16:40:07 UTC+2:
>> On Tuesday, July 5, 2022 at 4:25:52 PM UTC+2, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 4. Juli 2022 um 23:28:16 UTC+2:
>>>>
>>>> "For each and every natural number there's an ensegement not containing it."
>>>>
>>> Maybe.
>> No, not "maybe". The following is a THEOREM in ZFC (aka "set theory"):
>>
>> | An e IN: Em e IN: n !e E(m). (*)
>> "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 . (*)

no! 17 is not in E(18) and this has nothing to do with intersection.

>
>
>> You know, these steps form a so called PROOF.
>

>>
>> Actually, they PROVE the THEOREM: ∩{E(k) : k ∈ IN} = { }
>
> They do not.

Yes, it has been proven many times to you. ∩{E(k) : k ∈ IN} = { }

>
> Regards, WM

On 7/5/2022 9:16 AM, WM wrote:
> 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} = { } ?

it is a thing you personally cannot understand, your training in math is substandard.

>
> Regards, WM

On 7/5/2022 9:21 AM, WM wrote:
> 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

we get more FUD from the Deceiver...

Claimed Ants
Mistakenly Ants
Blatant Ants
Deceit Ants
Defined as being a flipping Ants
*Used Ants* (who wants a Used Ant ?)

On 7/5/2022 10:14 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 5. Juli 2022 um 16:30:25 UTC+2:
>> 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"
>
> https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 212

WM's troll book with the beeps, honks, raps taps, flashing lights and RAINBOWS that "define" a number. but there is no requirement to remember any of it.

it is not a reference.

>>
>> 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"?
>
> The two, for their time and even today, strongest and profoundest arguments of St Thomas Aquinatus S. Th. I, q. 7, a. 4 [...] become invalid as soon as a principle of individuation, intention, and ordination of actually infinite numbers and sets has been found;" [G. Cantor, letter to A. Schmid (26 Mar 1887)]
>
> Mit jeder wohlgeordneten fertigen Menge M von Alefs ist aber nach dem Bildungsgesetz der Alefs ein bestimmtes Alef gegeben, welches der Größe nach auf alle Individuen von M nächstfolgt.
>
> FOL is restricted to quantification over individuals, the elements of the domain, often a set, but not over predicates or functions
>
> for example, with respect to the, according to my firm conviction, actually infinite number of created individuals, in the universe as already on our earth and, most probably, even in each extended part of the space, however small it may be.
>
> then this epitome certainly consists of infinitely many individuals; therefore we can ask about the cardinality of this set, and I am able to strictly prove that this cardinality is the first one, i.e., not a higher one." [G. Cantor, letter to W. Wundt (16 Oct 1883)] Cantor never delivered a proof of this thesis.
>
> And finally generally accepted ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> Regards, WM

On 7/5/2022 10:25 AM, Fritz Feldhase wrote:
> On Tuesday, July 5, 2022 at 5:03:06 PM UTC+2, WM wrote:
>> FromTheRafters schrieb am Dienstag, 5. Juli 2022 um 16:34:36 UTC+2:
>>> WM presented the following explanation :
>>>>> What is the difference between the set and its elements?
>>>>
>>>> The set is one object. Its elements are 0 or 1 or many.
>>>>
>>> What happened to your claim that a set is nothing but its elements?
>>>
>> It is correct. Like <bla bla bla>
>
> No, it's NOT correct, since the empty set is not nothing, you silly idiot.
>
> Moreover the set just containing the element 1, i. e. {1}, is NOT identical with 1.
>
> Wie gesagt, Du hast wirklich schwer einen an der Klatsche, Mann.

WM gets two spoons.

tisdag 5 juli 2022 kl. 15:16:07 UTC+1 skrev 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

There is no such thing as "definable", it is meaningless.

It is empty because no natural number is in all ENDSEGMENTS

How hard is this to get you fucking retard?

tisdag 5 juli 2022 kl. 15:16:54 UTC+1 skrev 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.

It is not a process.

We are not losing anything, we have a sequence of endsegments. It is not a fucking process because things are not happening in mathematics, they just ARE.

> > > > 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.

"definable" is meaningless so stop this stupidity.

> > 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.

Skip the "definable", it is true for ALL endsegments
There are no dark
The intersection is empty (nothing is lost), because n is not in E(n+1)
That is it, how is this too difficult for you?

I have students at age 18 that understand this shit better than you!

>
> Regards, WM

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.
>
> Actually, there is no last/biggest natural number. (Still there are infinitely many of them.)

That proves the existence of dark numbers. They can satisfy

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Visible numbers cannot satisf this because

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

Regards, WM

William schrieb am Dienstag, 5. Juli 2022 um 17:50:30 UTC+2:
> On Tuesday, July 5, 2022 at 12:30:10 PM UTC-3, WM wrote:
> efined" has no meaning here
>
> > [if] there is no last step, then there is no end.
>
> Nope. There can be a state at which all steps are done, even if there is no last step. > However you cannot get to this state by using a stepwise process.

That can only be done for all visible numbers.

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

What methods?

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

For visible numbers it is not acceptable because of

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

Regards, WM

WM wrote:

>> Nope. There can be a state at which all steps are done, even if there
>> is no last step. > However you cannot get to this state by using a
>> stepwise process.
>
> That can only be done for all visible numbers.

sure, but watch the cocaine zelenske, scared like shit. His own actors
will probably hang him, this sondre bitch. Think, how many young men died
for his oligarchs, sondre bitches, enslaving the people of the territory,
along so many years.

Zelensky's 'crazy eyes' during his meeting with Swedish PM
https://www.bitchute.com/video/UouIoGm5TDNI/

UH-OH France24 is letting it slip that Lysychansk residents are HAPPY that
Russians have come https://www.bitchute.com/video/JR3IPSPFR8S0/