onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev 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.

Nothing is lost, there is no process.

>According to you infiitely many must be lost between all intersections of finitely many endsegments and the intersection of infinitely many.

Nothing is lost, the intersection is empty because n is not in E(n+1), simple as that.

> > >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, ...} --->...

That is a sequence, not a process.
A process takes time, things do not take time in mathematics.

> > > > 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.

Correct, but the intersection is empty. No contradiction, no dark, nothing.

> > > 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.

You don't have a definition of dark elements. THat definition of "dark" is meaningless and pointless

>
> Regards, WM

On Wednesday, July 6, 2022 at 7:20:39 PM UTC+2, zelos...@gmail.com wrote:

> A process takes time, things do not take time in mathematics.

NOT TRUE!!!1!!

Writing/reading a proof takes time!!! :-)

Thinking too!

On Wednesday, July 6, 2022 at 4:14:13 PM UTC+2, Fritz Feldhase wrote:
> 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 didn't ask for "less" but for "last" - you know the difference, dumbo?

And now for something completely different!

You [WM] wrote:

| "ω forces all natural numbers to be less."

That's a rather interesting formulation.

We might as well claim:

"The natural numbers force ω to be larger [than all natural numbers]."

Actually, I'd prefer the latter formulation.

On the other hand, your OWN CLAIM was:

| [The natural numbers] have an order which forces one of them to be the last.

That's NONSENSE.

onsdag 6 juli 2022 kl. 18:40:13 UTC+1 skrev Fritz Feldhase:
> On Wednesday, July 6, 2022 at 7:20:39 PM UTC+2, zelos...@gmail.com wrote:
>
> > A process takes time, things do not take time in mathematics.
> NOT TRUE!!!1!!
>
> Writing/reading a proof takes time!!! :-)
>
> Thinking too!

And those are processes so again, my point is made.

What is within the mathematical universe does not. They all simply exist already.

On Wednesday, July 6, 2022 at 7:53:57 PM UTC+2, zelos...@gmail.com wrote:
> onsdag 6 juli 2022 kl. 18:40:13 UTC+1 skrev Fritz Feldhase:
> > On Wednesday, July 6, 2022 at 7:20:39 PM UTC+2, zelos...@gmail.com wrote:
> > >
> > > A process takes time, things do not take time in mathematics.
> > >
> > NOT TRUE!!!1!!
> >
> > Writing/reading a proof takes time!!! :-)
> >
> > Thinking too!
> >
> And those are processes so again, my point is made.
>
> What is within the mathematical universe does not. They all simply exist already.

Of course. At least when considered from a classical point of view. (->classical mathematics)

Not sure how "intuitionism" and/or "constructivism" consider these things.

On Wednesday, July 6, 2022 at 3:11:14 PM UTC-3, Fritz Feldhase wrote:
> On Wednesday, July 6, 2022 at 7:53:57 PM UTC+2, zelos...@gmail.com wrote:
>
> > What is within the mathematical universe does not. They all simply exist already.
> Of course. At least when considered from a classical point of view. (->classical mathematics)
>
> Not sure how "intuitionism" and/or "constructivism" consider these things..

Irrelevant. WM wishes to show that taking the standard axioms of set theory leads to a contradiction. To do this it is not enough to show the standard axioms are in conflict with you theory. You must take the axioms and show a contradiction (not a result that you do note like), According to the standard axioms sets simply exist.

--
William Hughes

On Wednesday, July 6, 2022 at 8:23:05 PM UTC+2, William wrote:

> WM wishes to show that taking the standard axioms of set theory leads to a contradiction.

Does he?

Note that in one of his "works" he claims that the following is a _correct_ formulaton of the AoI (in the contect of ZFC):

ES({ } e S & (X e S => (X u {X}) e S)).

He's not wiling to rectify this erroneous formulation.

So I'd tend to doubt your claim.

On the other hand

> To do this it is not enough to show the standard axioms are in conflict with your theory. You must take the axioms and show a contradiction (not a result that you do note like).

Indeed!

> According to the standard axioms sets simply exist.

Right. According to the standard axioms! :-P

zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 19:20:39 UTC+2:
> onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev WM:

> > It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.
> You don't have a definition of dark elements.

Things that cannot happen individually but can only happen collectively are dark.

Regards, WM

William schrieb am Mittwoch, 6. Juli 2022 um 20:23:05 UTC+2:
> WM wishes to show that taking the standard axioms of set theory leads to a contradiction.

I did.

If almost all natural numbers are in every endsegment, then not every natural number is lost by one endsegment.

Regards, WM

On Wednesday, July 6, 2022 at 4:44:50 PM UTC-3, WM wrote:
>1. almost all natural numbers are in every endsegment,

Correct.

>2. not every natural number is lost by one endsegment.

Also correct. However 1 does not contradict 2. You can conclude from 1 and 2 that a stepwise process cannot "work", that is there is no step at which the intersection is obtained.

--
William Hughes

After serious thinking WM wrote :
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 19:20:39 UTC+2:
>> onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev WM:
>
>>> It is by my definition of dark elements: Things that cannot happen
>>> individually but can only happen collectively are dark.
>> You don't have a definition of dark elements.
>
> Things that cannot happen individually but can only happen collectively are
> dark.

Whatever! None of that means that Peano didn't adequately define the
natural numbers or that Cantor was wrong about countable infinity
versus the continuum. Where will the muckymath goalposts be tomorrow?

On 7/6/2022 3:36 PM, WM wrote:

> Things that cannot happen individually
> but can only happen collectively are dark.

You haven't accounted for
an individual claim about the contents of
a collection of
individuals.

Consider, for example,
| ⟨0,...,k⟩ is a collection in which,
| for each split,
| there exists a step i,i+1 which
| starts before the split and
| ends after the split.

One claim.
Infinitely-many referents.
Each referent is an individual.

What is it?

onsdag 6 juli 2022 kl. 20:44:50 UTC+1 skrev WM:
> William schrieb am Mittwoch, 6. Juli 2022 um 20:23:05 UTC+2:
> > WM wishes to show that taking the standard axioms of set theory leads to a contradiction.
> I did.

You never have done it, that is why you're a crank.

>
> If almost all natural numbers are in every endsegment, then not every natural number is lost by one endsegment.

Not a contradiction as no "losing" is happening

>
> Regards, WM

onsdag 6 juli 2022 kl. 20:36:15 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 19:20:39 UTC+2:
> > onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev WM:
>
> > > It is by my definition of dark elements: Things that cannot happen individually but can only happen collectively are dark.
> > You don't have a definition of dark elements.
>
> Things that cannot happen individually but can only happen collectively are dark.
> Regards, WM

That is not a fucking definition in mathematics

William schrieb am Mittwoch, 6. Juli 2022 um 22:06:29 UTC+2:
> On Wednesday, July 6, 2022 at 4:44:50 PM UTC-3, WM wrote:
> >1. almost all natural numbers are in every endsegment,
>
> Correct.
>
> >2. not every natural number is lost by one endsegment.
>
> Also correct. However 1 does not contradict 2.

No, 1 proves 2. If all almost are there then almost none is lost.

> You can conclude from 1 and 2 that a stepwise process cannot "work",

I do! A stepwise process works for definable numbers only:
{1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...

> that is there is no step at which the intersection is obtained.

The question is why is that so? For all definable numbers we can do the steps.
Hint: Not stepwise definable numbers I call dark numbers. But the name is not important. A rose would smell as sweet by any other name.

Regards, WM

On Thursday, July 7, 2022 at 1:22:22 PM UTC+2, WM wrote:
> William schrieb am Mittwoch, 6. Juli 2022 um 22:06:29 UTC+2:
> >
> > there is no step at which the intersection [ ∩{E(1), E(2), E(3), ...} ] is obtained.
> >
> The question is why is that so?

Because, no term in the sequence

(∩{E(1)}, ∩{E(1), E(2)}, ∩{E(1), E(2), E(3)}, ...) = (∩{E(1), E(2), E(3), ..., E(k)})_(k e IN)
equals ∩{E(1), E(2), E(3), ...}, dumbo.That's why.

Hint:

∩{E(1), E(2), E(3), ...} = { }.

On the other hand, ∩{E(1), E(2), E(3), ..., E(k)} = E(k) for all k e IN.

In addition with

| E(1) = IN
| E(n+1) = E(n) \ {n} (for all n e IN)

we get (by induction) that

Ak e IN: ∩{E(1), E(2), E(3), ..., E(k)} =/= { }.

Hence for all k e IN: ∩{E(1), E(2), E(3), ..., E(k)} =/= ∩{E(1), E(2), E(3), ...}.

In other words, ∩{E(1), E(2), E(3), ...} is not a term in (∩{E(1), E(2), E(3), ..., E(k)})_(k e IN).

qed

Why are you so bad in math, Mückenheim?

FromTheRafters schrieb am Mittwoch, 6. Juli 2022 um 23:34:49 UTC+2:
> After serious thinking WM wrote :
> > zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 19:20:39 UTC+2:
> >> onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev WM:
> >
> >>> It is by my definition of dark elements: Things that cannot happen
> >>> individually but can only happen collectively are dark.
> >> You don't have a definition of dark elements.
> >
> > Things that cannot happen individually but can only happen collectively are
> > dark.
> Whatever! None of that means that Peano didn't adequately define the
> natural numbers or that Cantor was wrong about countable infinity
> versus the continuum.

It means that almost all natural numbers cannot be reached by Peano's steps. Numbers that cannot be reached by steps are dark and cannot be used for counting.

Regards, WM

torsdag 7 juli 2022 kl. 14:25:41 UTC+1 skrev WM:
> FromTheRafters schrieb am Mittwoch, 6. Juli 2022 um 23:34:49 UTC+2:
> > After serious thinking WM wrote :
> > > zelos...@gmail.com schrieb am Mittwoch, 6. Juli 2022 um 19:20:39 UTC+2:
> > >> onsdag 6 juli 2022 kl. 14:48:13 UTC+1 skrev WM:
> > >
> > >>> It is by my definition of dark elements: Things that cannot happen
> > >>> individually but can only happen collectively are dark.
> > >> You don't have a definition of dark elements.
> > >
> > > Things that cannot happen individually but can only happen collectively are
> > > dark.
> > Whatever! None of that means that Peano didn't adequately define the
> > natural numbers or that Cantor was wrong about countable infinity
> > versus the continuum.
> It means that almost all natural numbers cannot be reached by Peano's steps. Numbers that cannot be reached by steps are dark and cannot be used for counting.
>
> Regards, WM
There are no "reaching", no "steps"
but all natural numbers is of the form S^n(0)
ALL OF THEM

Jim Burns schrieb am Donnerstag, 7. Juli 2022 um 02:51:08 UTC+2:
> On 7/6/2022 3:36 PM, WM wrote:
>
> > Things that cannot happen individually
> > but can only happen collectively are dark.
> You haven't accounted for
> an individual claim about the contents of
> a collection of
> individuals.

An infinite collection or set is definable, for instance ℕ. But its individuals cannot all be applied individually, for instance for counting.
>
> Consider, for example,
> | ⟨0,...,k⟩ is a collection in which,
> | for each split,
> | there exists a step i,i+1 which
> | starts before the split and
> | ends after the split.
>
> One claim.
> Infinitely-many referents.
> Each referent is an individual.

Not actually infinitely many referents. After actually infinitely many referents you would run out of the set or sequence because ω is the first infinite ordinal.
>
> What is it?

It is a potentially infinite collection.

Regards, WM

On 7/7/2022 6:22 AM, WM wrote:
> William schrieb am Mittwoch, 6. Juli 2022 um 22:06:29 UTC+2:
>> On Wednesday, July 6, 2022 at 4:44:50 PM UTC-3, WM wrote:
>>> 1. almost all natural numbers are in every endsegment,
>>
>> Correct.
>>
>>> 2. not every natural number is lost by one endsegment.
>>
>> Also correct. However 1 does not contradict 2.
>
> No, 1 proves 2.

Liar. 1 has nothing to do with "lost" numbers of 2

> If all almost are there then almost none is lost.

too vague.

>
>> You can conclude from 1 and 2 that a stepwise process cannot "work",
>
> I do! A stepwise process works for definable numbers only:
> {1, 2, 3, ...} --> {2, 3, 4, ...} --> {3, 4, 5, ...} --> ... --> {n, n+1, n+2, ...} --->...

swapparoofest again, no explanation either

>
>> that is there is no step at which the intersection is obtained.
>
> The question is why is that so? For all *definable*

Clown shoes are on!

zelos...@gmail.com schrieb am Donnerstag, 7. Juli 2022 um 15:29:39 UTC+2:

> There are no "reaching", no "steps"

Wrong. By n --> n+1, we reach 2 from 1, 3 from 2 and so on for all definable numbers. But we never leave the finite realm and never enter the infinite..

> but all natural numbers is of the form S^n(0)
> ALL OF THEM

Then there are no dark numbers. Then there is no actual infinity.
Between Peano's collection (reachable by step, and ω there are dark numbers not reachable by steps.

Regards, WM

On Thursday, July 7, 2022 at 8:22:22 AM UTC-3, WM wrote:
> > there is no step at which the intersection is obtained.
> The question is why is that so?

Because no matter how many steps you do there are still steps (an infinite number if fact) that have not been done.

--
William Hughes

On Thursday, July 7, 2022 at 3:25:41 PM UTC+2, WM wrote:

> almost all natural numbers cannot be reached by Peano's steps.

Each and every natural number can "be reached by Peano's steps", actually a repetition of the application of the succesor operation.

Proof by induction: 1 can "be reached" by a repetition of the application of the succesor operation (in this special case, the successor operation is applied 0 times). If the natural number n can "be reached" by a repetition of the application of the succesor operation, then n+1 can "be reached" by a repetition of the application of the succesor operation, since n+1 = s(n). Hence each and every natural number can "be reached" by a repetition of the application of the succesor operation. qed

> Numbers that cannot be reached by steps are dark and cannot be used for counting.

Indeed! Our proof SHOWS: There are no "dark" natural numbers. Each and every natural number can be used for counting. That's why the natural numbers are also called "counting numbers", you know.

See: https://www.basic-math-explained.com/counting-numbers.html

Remeber that Chuck Norris counted to infinity - twice!

On Thursday, July 7, 2022 at 3:32:08 PM UTC+2, WM wrote:

> An infinite collection or set is definable, for instance ℕ. But its individuals cannot all be applied individually, for instance for counting.

Huh?!

The natural numbers 1, 2, 3, ... cannot be "applied [i. e. used] individually, for instance for counting" in your Wahnsystem, Mückenheim?

Maybe you didn't learn it in kindergarden?

See: https://www.basic-math-explained.com/counting-numbers.html

Remember that Chuck Norris counted to infinity - twice!

[ You think that he didn't use natural numbers to perform this feat? ]

On Thursday, July 7, 2022 at 3:39:19 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 7. Juli 2022 um 15:29:39 UTC+2:
> >
> > all natural numbers is of the form S^n(0)
> > ALL OF THEM
> >
> Then there are no dark [natural] numbers.

*lol* That's what we told you ALL ALONG!

See my other post where I PROVED that fact.

> Between <bla>

Shut up, you silly crank.