Racist mathematics proves dark numbers.

[Ganzhinterseher]

The intersection over a set {E(1), E(2), E(3), … E(n)}
of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:

∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:

∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

However, it is impossible to find a natural number common to all E(n) satisfying (*).

This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.

The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.

Regards, WM

[Ben Bacarisse]

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

> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

This is true statement about infinitely many finite intersections.

> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

This is a false statement about a single infinite intersection.

(It's a neat trick to try to hide the difference between a bound and a
free variable, but with luck your students should see through it.)

--
Ben.

[Ganzhinterseher]

Ben Bacarisse schrieb am Donnerstag, 25. Februar 2021 um 12:37:42 UTC+1:
> Ganzhinterseher <claus.v...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > The intersection over a set {E(1), E(2), E(3), … E(n)}
> > of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
> >
> > ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
> This is true statement about infinitely many finite intersections

which, when appearing together, will do the same as when appearing as last elements.

Why should they change their effect?

Regards, WM

[Sergio]

On 2/25/2021 4:43 AM, Ganzhinterseher wrote:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>
> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
>
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>
> However, it is impossible to find a natural number common to all E(n) satisfying (*).

wrong, E(N)

>
> This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.

you have NO MONTONY at all. (google is your fiend)

>
> The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.

since you cannot identify, you cannot prove.

>
> Regards, WM
>

[Sergio]

On 2/25/2021 5:59 AM, Ganzhinterseher wrote:
> Ben Bacarisse schrieb am Donnerstag, 25. Februar 2021 um 12:37:42 UTC+1:

>> Ganzhinterseher <claus.v...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>> The intersection over a set {E(1), E(2), E(3), … E(n)}
>>> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>>>
>>> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>> This is true statement about infinitely many finite intersections.
>>> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>> This is a false statement about a single infinite intersection.
>

> It is about infinitely many endegments. You can imagine every endsegment of (*) as a filter which, in all environments, i.e., independent of the presence of other endsegments, removes up to n-1 natural numbers from an incoming set and lets pass an infinite set of natural numbers.

>>
>> (It's a neat trick to try to hide the difference between a bound and a
>> free variable
>

> The applied endsegments are defined by (*). All of them can be applied in (**).

>
>> , but with luck your students should see through it.)
>

> Only anti-racist mathematicians will accept your argument.


so racist mathematicians will not accept his argument, and agree with you.

> Here the matter are not variables but endsegments which in all environments let pass an infinite set.
>
> Regards, WM
>

[Ben Bacarisse]

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

> Ben Bacarisse schrieb am Donnerstag, 25. Februar 2021 um 12:37:42 UTC+1:
>> Ganzhinterseher <claus.v...@gmail.com> writes:

>> > The intersection over a set {E(1), E(2), E(3), … E(n)}
>> > of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>> >
>> > ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>> This is true statement about infinitely many finite intersections.
>> > ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>> This is a false statement about a single infinite intersection.
>

> It is about infinitely many endegments.

Not in doubt. Do you accept that it is a false statement about a single
intersection involving an infinite collection of end segments?

--
Ben.

[Ganzhinterseher]

Of course I don't because you are wrong.

You can imagine every endsegment of (*) as a filter which, in all environments, i.e., independent of the presence of other endsegments, removes up to n-1 natural numbers from an incoming set and lets pass up to an infinite set of natural numbers. Why should any of them change its effect?

Regards, WM

[Dan Christensen]

On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>
> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*).

Poorly stated. I'm guessing you mean that, for all n in N, we have E(n) = {n, n+1, n+2, ... } and ∩{E(1), E(2), E(3), … E(n)} = E(n).

> The intersection however does not depend on positions...

We have: E: N --> Pow(N).

> - a set has no order anyway.

Irrelevant even if it were true.

> Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
>

Wrong. While the finite intersection ∩{E(1), E(2), E(3), … E(n)} = E(n) is infinite for all n in N, we have ∩{E(1), E(2), E(3), … } = { } since there is no element common to every end-segment. It's just basic high-school math, Mucke. I guess that's why you still don't get it. Oh, well...

More absurd quotes from Wolfgang Muckenheim (WM):

“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16

“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20

“1/9 has no decimal representation.”
-- sci.math, 2015/09/22

"0.999... is not 1."
-- sci.logic 2015/11/25

“Axioms are rubbish!”
-- sci.math, 2014/11/19

“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05

“No set is countable, not even |N.”
-- sci.logic, 2015/08/05

“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05


Slipping ever more deeply into madness...

“There is no actually infinite set |N.”
-- sci.math, 2015/10/26

“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26

“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28

“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30

"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09

“The set of numbers will get empty after all have numbers been used.”
-- sci.math, 2016/08/24

“I need no set theory.”
-- sci.math, 2016/09/01

A special word of caution to students: Do not attempt to use WM's “system” (MuckeMath) in any course work in any high school, college or university on the planet. You will fail miserably. MuckeMath is certainly no shortcut to success in mathematics.

Using WM's “axioms” for the natural numbers, he cannot even prove that 1=/=2. His goofy system is truly a dead-end.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Sergio]

I see the problem,

WM does not use "variables", algebra is not a language he speaks.

Instead he presents this overly obtuse show, rather than state the
obvious there are oo numbers after a given k in an infinite series of
natural numbers.

And WMs failure of not accepting the fact that his "dark numbers", are
simply "unknown numbers", strongly indicates he has not and cannot use
variables, therefore his Math knowledge is limited to below that of
Algebra.


>
> Regards, WM
>

[Ganzhinterseher]

Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 15:39:42 UTC+1:
> On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:

> > - a set has no order anyway.
> Irrelevant even if it were true.

Relevant and true.

> > Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
> >
> Wrong. While the finite intersection ∩{E(1), E(2), E(3), … E(n)} = E(n) is infinite for all n in N, we have ∩{E(1), E(2), E(3), … } = { } since there is no element common to every end-segment.

That is one argument. The other is this:

∩{E(n) | E(n) is satisfying ∩{E(1), E(2), E(3), … E(n)} =/= { }} =/= { }

Why should an endsegment E(n), leaving not an empty set when standing at the end, leave an empty set when not standing at the end. Note that only such endsegments are intersected.

Regards, WM

[Dan Christensen]

Exactly so! He also doesn't seem to understand what the intersection of two or more sets is. He is singularly unqualified to teach math at any level.

>
> >
> > Regards, WM
> >

[Dan Christensen]

On Thursday, February 25, 2021 at 12:08:21 PM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 15:39:42 UTC+1:
> > On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:
>
> > > - a set has no order anyway.
> > Irrelevant even if it were true.
> Relevant and true.

Wrong again, Mucke.

> > > Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
> > >
> > Wrong. While the finite intersection ∩{E(1), E(2), E(3), … E(n)} = E(n) is infinite for all n in N, we have ∩{E(1), E(2), E(3), … } = { } since there is no element common to every end-segment.
> That is one argument. The other is this:
>
> ∩{E(n) | E(n) is satisfying ∩{E(1), E(2), E(3), … E(n)} =/= { }} =/= { }
>

We don't disagree on that. It is when you take the intersection of ALL end-segments that your argument goes off the rails, Mucke. An element x is the intersection of 2 or more sets if and only if x is an element of EACH of those sets. Why can't you understand that?

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com

[Dan Christensen]

On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote: ...

RACIST mathematics??? Sorry, that's just too weird. I have to ask, what do you mean by that? I hope it is a mistranslation. Or a typo.

Dan

[Sergio]

math is not "racist", it is "White Supremacist",
according to affore mentioned LA racist bigoted teachers nutcases,

[Mostowski Collapse]

How many dark numbers are there? billions and billions?

Donald Trump Says Billions And Billions And Billions
https://www.youtube.com/watch?v=u_aLESDql1U

LoL

[Dan Christensen]

Doesn't make sense in the context of mathematical foundations. Is WM simply grasping at straws?

Dan

[Sergio]

bet its translation error. Spaten Lager w Bratwurst an Brezen ist gut an
einem kalten Tag

[Python]

Crank Wolfgang Mueckenheilm, aka Ganzhinterseher wrote:
> ... the conventional racist mathematics which does not ...

Given the fact that you've posted numerous actual racists
statements on this place and other, Crank Wolfgang Mueckenheim,
from Hoschschule Augsburg, this post of yours is utterly
outrageous.

You are racist, you've proven it. You are also a fraud.
"conventional" mathematicians rejecting your sophistries
is only a question of decency.

[Sergio]

Trump is BacK !!

The nation rewards Trump for getting the vaccine out to the nation in
only 6 months! by giving him the prediency, replacing feeble Joe Biden
who was found left in a closet for a week.

Biden has already had too much BAT SOUP, he has bat wings for eyes (cant
see his eyes on TV, cuze they are GLAZED OVER)

[Ben Bacarisse]

Ganzhinterseher <claus.v...@gmail.com> writes:

> Ben Bacarisse schrieb am Donnerstag, 25. Februar 2021 um 15:34:13 UTC+1:
>> Ganzhinterseher <claus.v...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>> > Ben Bacarisse schrieb am Donnerstag, 25. Februar 2021 um 12:37:42 UTC+1:
>> >> Ganzhinterseher <claus.v...@gmail.com> writes:
>> >> > The intersection over a set {E(1), E(2), E(3), … E(n)}
>> >> > of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>> >> >
>> >> > ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>> >> This is true statement about infinitely many finite intersections.
>> >> > ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>> >> This is a false statement about a single infinite intersection.
>> >
>> > It is about infinitely many endegments.
>> Not in doubt. Do you accept that it is a false statement about a single
>> intersection involving an infinite collection of end segments?
>
> Of course I don't because you are wrong.

Which part don't you accept? (a) that (**) is a statement about a
single intersection involving an infinite collection of end segments, or
(b) that (**) is false? You can of course be doubly wrong by not
accepting either.

--
Ben.

[Ganzhinterseher]

Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 18:19:02 UTC+1:
> On Thursday, February 25, 2021 at 12:08:21 PM UTC-5, Ganzhinterseher wrote:
> > Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 15:39:42 UTC+1:
> > > On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:
> >
> > > > - a set has no order anyway.
> > > Irrelevant even if it were true.
> > Relevant and true.
> Wrong

Unlike a sequence, a tuple or a permutation of a set, the order in which the elements of a set are listed in roster notation is irrelevant, so {6, 11} is the same set as {11, 6}, and {2, 6, 4}, {4, 2, 6}, {4, 6, 2}, {6, 2, 4} or {6, 4, 2} all represent the same set. (Wikipedia)

> > > > Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
> > > >
> > > Wrong. While the finite intersection ∩{E(1), E(2), E(3), … E(n)} = E(n) is infinite for all n in N, we have ∩{E(1), E(2), E(3), … } = { } since there is no element common to every end-segment.
> > That is one argument. The other is this:
> >

> > ∩{E(n) | E(n) is satisfying ∩{E(1), E(2), E(3), … E(n)} =/= { }} =/= { } (**)

> >
> We don't disagree on that.

You did this morning.

> It is when you take the intersection of ALL end-segments

Try to increase your text comprehension. Read the OP again. I take only the intersection of all endsegments E(n) which are defined by

∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

> An element x is the intersection of 2 or more sets if and only if x is an element of EACH of those sets.

Alas the intersection (**) cannot be empty. An endsegment E(n), leaving not an empty set when standing at the end, will not leave an empty set when not standing at the end. Note that only such endsegments are intersected.

Regards, WM

[Ganzhinterseher]

Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 18:57:17 UTC+1:
> On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote: ...
>
> RACIST mathematics??? Sorry, that's just too weird. I have to ask, what do you mean by that? I hope it is a mistranslation. Or a typo.

In US schools correct mathematics is denounced as racist. T-shirts with imprint 2 + 2 = 4 are denounced as racist.

When I argue that an endsegment E(n), leaving not an empty set when standing at the end, cannot leave an empty set when not standing at the end, this will certainly be condemned as racist.

Regards, WM

[Ganzhinterseher]

Mostowski Collapse schrieb am Donnerstag, 25. Februar 2021 um 19:51:19 UTC+1:
> How many dark numbers are there?

aleph_0.

You can imagine every endsegment E(n) of

∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

as a filter which, in all environments, i.e., independent of the presence of other endsegments, removes up to n-1 natural numbers from an incoming set and lets pass up to an infinite set of natural numbers.

The idea that this action is independent of the position of the endsegment and no other opinion is admitted in mathematics shows racism.

∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

but since no natnumbers can be found in the intersection there must exist dark natnumbers.

Regards, WM

[Python]

You are a fraud and a depicable person, Crank Wolfgang Mueckenheim, from
Hochschule Augsburg.

On top of that, you are INSANE.

[Chris M. Thomasson]

On 2/25/2021 2:43 AM, Ganzhinterseher wrote:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>

> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>

> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:

>
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>

> However, it is impossible to find a natural number common to all E(n) satisfying (*).
>
> This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.
>
> The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.

Racist math.... Imvvho, that means the teachers right? Thinking that the
color of a students skin has anything to do with his/her ability to
learn math. Wow!

[Chris M. Thomasson]

Are you sober right now?

[Ganzhinterseher]

(**) is an intersection over all endsegments E(n) which satisfy (*), i.e., which, when apearing at the end, leave an infinite set in the intersection. The number of endsegmnts is potentially infinite.

> or
> (b) that (**) is false?

(**) is true because we can imagine every endsegment as a filter which, in all environments, i.e., independent of the presence of other endsegments, removes up to n-1 natural numbers from an incoming set and lets pass up to an infinite set of natural numbers. Therefore the intersection of these endsegments cannot be empty. That is an irrefutable fact.

But it is impossible to observe a natural number in the intersection. Therefore either ZFC is inconsistent or there are dark natnumbers.

Regards, WM

[Ganzhinterseher]

Chris M. Thomasson schrieb am Donnerstag, 25. Februar 2021 um 23:02:01 UTC+1:


> Racist math.... Imvvho, that means the teachers right? Thinking that the
> color of a students skin has anything to do with his/her ability to
> learn math. Wow!

https://nypost.com/2021/02/12/adding-wokeness-oregon-promotes-teacher-program-to-subtract-racism-in-mathematics/

Regards, WM

[Sergio]

you call them your little buddies, your "Dark Numbers", the subservient
background numbers that fail at everything, and you say you have proved
they are that way for a reason...

>
> Regards, WM
>

[Sergio]

we should all chip in and get him a "2 + 2 = 4" T shirt on one side, and
"Ask Me About Dark Numbers" on the other side

[Python]

Crank Wolfgang Mueckenheim, aka Ganzhinterseher wrote:

Nothing there supports your silly and disgusting rant, Crank Wolfgang
Mueckenheim, from Hochschule Augsburg.

Of course, being a racist yourself, and you've proven being so in
numerous posts here, you're only looking for adding confusion and
promote your fraud and evade responsibility for your abusing behaviour.

As a teacher in a CS engineering school, I recently covered numeration
systems across ages up to computing and several other related topics.

I covered in particular the Indian-arabic origin of the usual decimal
system, how something similar was used by the Maya civilization as well
as in China. How and why Al-jabr and Al-Kwarismi are the roots of the
words algebra and algorithm. How different was the Greek and Chinese
style of mathematical arguments and how the current international style
emerged from the foundation crisis in the XIXth Century.

See, Crank Wolfgang Mueckenheim, contrarily to you, I'm not teaching
fraud, I'm not teaching fallacies, I'm not abusing students and at
the same time, I'm fighting racism, not spreading it on Usenet as YOU
do.

[Chris M. Thomasson]

That might sell! :^)

[Sergio]

On 2/25/2021 3:34 PM, Ganzhinterseher wrote:
> Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 18:19:02 UTC+1:
>> On Thursday, February 25, 2021 at 12:08:21 PM UTC-5, Ganzhinterseher wrote:
>>> Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 15:39:42 UTC+1:
>>>> On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:
>>>
>>>>> - a set has no order anyway.
>>>> Irrelevant even if it were true.
>>> Relevant and true.
>> Wrong
>
> Unlike a sequence, a tuple or a permutation of a set, the order in which the elements of a set are listed in roster notation is irrelevant, so {6, 11} is the same set as {11, 6}, and {2, 6, 4}, {4, 2, 6}, {4, 6, 2}, {6, 2, 4} or {6, 4, 2} all represent the same set. (Wikipedia)

you had to look that up in the wiki ??



<snip crap>

[Ben Bacarisse]

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

> (**) is an intersection over all endsegments E(n) which satisfy (*),

> i.e., which, when apearing...

Ah. no. "Which" nothing. Every end segment satisfies (*). ∩{E(n) |
E(n) is satisfying (*)} is just ∩{ E(n) | n ∈ N } which, as you know, is
empty.

> (**) is false because we can imagine every endsegment...

but in another reply you say

> (**) is true because we can imagine every endsegment...

Have you decided yet?

--
Ben.

[Dan Christensen]

On Thursday, February 25, 2021 at 4:35:03 PM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 18:19:02 UTC+1:
> > On Thursday, February 25, 2021 at 12:08:21 PM UTC-5, Ganzhinterseher wrote:
> > > Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 15:39:42 UTC+1:
> > > > On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote:
> > >
> > > > > - a set has no order anyway.
> > > > Irrelevant even if it were true.
> > > Relevant and true.
> > Wrong

> Unlike a sequence, a tuple or a permutation of a set, the order in which the elements of a set are listed in roster notation is irrelevant, so {6, 11} is the same set as {11, 6}, and {2, 6, 4}, {4, 2, 6}, {4, 6, 2}, {6, 2, 4} or {6, 4, 2} all represent the same set. (Wikipedia)

So what????

> > > > > Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
> > > > >
> > > > Wrong. While the finite intersection ∩{E(1), E(2), E(3), … E(n)} = E(n) is infinite for all n in N, we have ∩{E(1), E(2), E(3), … } = { } since there is no element common to every end-segment.
> > > That is one argument. The other is this:
> > >
> > > ∩{E(n) | E(n) is satisfying ∩{E(1), E(2), E(3), … E(n)} =/= { }} =/= { } (**)
> > >
> > We don't disagree on that.
> You did this morning.

You are wrong again, Mucke.

> > It is when you take the intersection of ALL end-segments
> Try to increase your text comprehension. Read the OP again. I take only the intersection of all endsegments E(n) which are defined by
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

But that is true for EVERY end-segment. As we have already agreed, ∩{E(1), E(2), E(3), … E(n)} = E(n) = {n, n+1, n+2, ...}

> > An element x is the intersection of 2 or more sets if and only if x is an element of EACH of those sets.

> Alas the intersection (**) cannot be empty. An endsegment E(n), leaving not an empty set when standing at the end, will not leave an empty set when not standing at the end. Note that only such endsegments are intersected.

Pure gibberish! If you understood even basic high-school math, you would know that ∩{E(1), E(2), E(3), … } = { }.

[Dan Christensen]

On Thursday, February 25, 2021 at 4:39:10 PM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Donnerstag, 25. Februar 2021 um 18:57:17 UTC+1:
> > On Thursday, February 25, 2021 at 5:43:22 AM UTC-5, Ganzhinterseher wrote: ...
> >
> > RACIST mathematics??? Sorry, that's just too weird. I have to ask, what do you mean by that? I hope it is a mistranslation. Or a typo.
> In US schools correct mathematics is denounced as racist. T-shirts with imprint 2 + 2 = 4 are denounced as racist.
>

I see that your grasp of social issues is no better than your grasp of mathematics, Mucke.

Dan

[Sergio]

On 2/25/2021 3:55 PM, Ganzhinterseher wrote:

> (**) is an intersection over all endsegments E(n) which satisfy (*), i.e., which, when apearing at the end, leave an infinite set in the intersection. The number of endsegmnts is potentially infinite.

no, the # of your "endsegments" is infinite.

no need to do intersections, or E(n) or (*) etc. nor endsegments

just take the ordered set of natural numbers to infinity, stop at k, etc
etc...

what you call dark numbers are simply unknown numbers from k to oo.

>
>> or
>> (b) that (**) is false?
>

> (**) is false because we can imagine every endsegment as a filter which, in all environments, i.e., independent of the presence of other endsegments, removes up to n-1 natural numbers from an incoming set and lets pass up to an infinite set of natural numbers. Therefore the intersection of these endsegments cannot be empty. That is an irrefutable fact.

and that does not prove anything at all, there is stuff out there, an
infinite amount of naturel numbers

>
> But it is impossible to observe a natural number in the intersection. Therefore either ZFC is inconsistent or there are dark natnumbers.

wrong, just look at k+1 or 2*k etc.

>
> Regards, WM
>
>

[Ganzhinterseher]

No. But this is at a level that might be comprehensible by Christensen.

Regards, WM

[Ganzhinterseher]

Sergio schrieb am Freitag, 26. Februar 2021 um 04:58:57 UTC+1:

> what you call dark numbers are simply unknown numbers from k to oo.

aleph_0 of them are unknowable. For all knowable we have

∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:

∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

But the numbers needed to fill the right side cannot be known.

Regards, WM

[Ganzhinterseher]

Even if so, my sentence is correct. Your no is wrong.

> ∩{E(n) | E(n) is satisfying (*)} is just ∩{ E(n) | n ∈ N } which, as you know, is
> empty.

That is wrong. Fact is that every endsegment has an effect that is independent of its position. Do you doubt that?

>
> > (**) is false because we can imagine every endsegment...

I wrote this in error in a hurry and deleted it.

Regards, WM

[Sergio]

so you agree:

In the ordered set of natural numbers to infinity, stop at k,

"dark numbers" are simply "unknown numbers" from k to oo.

like lawyers using Latin phrases, your use of set notation provides
flash, but no math.

[Ganzhinterseher]

Sergio schrieb am Freitag, 26. Februar 2021 um 15:11:02 UTC+1:
> On 2/26/2021 4:37 AM, Ganzhinterseher wrote:
> > Sergio schrieb am Freitag, 26. Februar 2021 um 04:58:57 UTC+1:
> >
> >> what you call dark numbers are simply unknown numbers from k to oo.
> >
> > aleph_0 of them are unknowable. For all knowable we have
> >
> > ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
> >
> > Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
> >
> > ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
> >
> > But the numbers needed to fill the right side cannot be known.

> so you agree:
>
> In the ordered set of natural numbers to infinity, stop at k,

No. If k is in the set, then also k+1, 2k, k^k etc aqre in the set. Fior every k. So there is no stop. Try to grasp it.

> "dark numbers" are simply "unknown numbers" from k to oo.

None of the numbers mentioned above are in

∩{E(n) | E(n) is satisfying (*)}.

But this set cannot be empty by logical reasoning (since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position). Therefore there must be dark numbers or no actual infinity.

Regards, WM

[Sergio]

On 2/26/2021 12:05 PM, Ganzhinterseher wrote:
> Sergio schrieb am Freitag, 26. Februar 2021 um 15:11:02 UTC+1:
>> On 2/26/2021 4:37 AM, Ganzhinterseher wrote:
>>> Sergio schrieb am Freitag, 26. Februar 2021 um 04:58:57 UTC+1:
>>>
>>>> what you call dark numbers are simply unknown numbers from k to oo.
>>>
>>> aleph_0 of them are unknowable. For all knowable we have
>>>
>>> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>>>
>>> Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:
>>>
>>> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>>>
>>> But the numbers needed to fill the right side cannot be known.
>
>> so you agree:
>>
>> In the ordered set of natural numbers to infinity, stop at k,

In the ordered set of natural numbers to infinity, stop at k,
"dark numbers" are simply "unknown numbers" from k to oo.

that is fact.

>
> No. If k is in the set, then also k+1,

no, as we stopped at k,

> 2k, k^k etc aqre in the set. Fior every k. So there is no stop. Try to grasp it.

you are simply replacing k, with bigger k's (recursion)

if you assume that, then there are no dark numbers at all,
all the way out to oo.

you have shown that there are no dark numbers by above, your reasoning.

>
>> "dark numbers" are simply "unknown numbers" from k to oo.
>
> None of the numbers mentioned above are in
>
> ∩{E(n) | E(n) is satisfying (*)}.

except when n < k

>
> But this set cannot be empty by logical reasoning (since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position). Therefore there must be dark numbers or no actual infinity.

by dark numbers, you mean unknown numbers.

your above statement still just boils down to:

"In the ordered set of natural numbers to infinity, stop at any k, "dark

numbers" are simply "unknown numbers" from k to oo."

where you use n, I use k


>
> Regards, WM
>

[mitchr...@gmail.com]

There is the absence of quantity that can't be seen as anything but dark...
Zero comes first with fundamental nonzero beyond it.

Mitchell Raemsch

[Mostowski Collapse]

So you call yourself Racist, since you proves dark numbers?

LoL

Ganzhinterseher schrieb am Donnerstag, 25. Februar 2021 um 11:43:22 UTC+1:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>

> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>

> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:

>
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>

> However, it is impossible to find a natural number common to all E(n) satisfying (*).
>
> This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.
>
> The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.
>

> Regards, WM

[Chris M. Thomasson]

On 2/26/2021 1:22 PM, Mostowski Collapse wrote:
> So you call yourself Racist, since you proves dark numbers?

It sure seems that way!

[Ben Bacarisse]

Your intersection (**) omitted no end segments. It is empty. Since you
like functions and analysis, you could use the product of the indicator
(AKA characteristic) functions to prove that, given

e_n(i) = [i >= n] (using Iverson's bracket notation),

the product

Product_{k ∈ N} e_k = z where z(i) = 0.

Can you prove that? Do you need a hand? Maybe one of your students
could do it for you?

>> ∩{E(n) | E(n) is satisfying (*)} is just ∩{ E(n) | n ∈ N } which, as you know, is
>> empty.
>
> That is wrong. Fact is that every endsegment has an effect that is
> independent of its position.

Of course. This is an intersection of a set of sets. The intersection
is provably empty because every element of N is missing from at least
one of the intersected sets. Whatever you think the "position" of an
end segment is, it will have no bearing on the result.

--
Ben.

[FromTheRafters]

Chris M. Thomasson presented the following explanation :

> On 2/26/2021 1:22 PM, Mostowski Collapse wrote:
>> So you call yourself Racist, since you proves dark numbers?
>
> It sure seems that way!

Are some dark numbers darker than others, or do they all look the same?

[Sergio]

Intellectual discussion Julian Bond and Garrett Morris on SNL, proves
that their are darker and lighter "dark numbers". (cant play the video
anymore in woke wacko america)

https://snltranscripts.jt.org/76/76rblackperspective.phtml

[Mostowski Collapse]

Time for reverse racism. Declare the natural numbers also dark.

[Sergio]

they are very dark, even black, when you print them,

1,2,3,4,5,6,...see?

WM's *dark numbers turn BLACK* when he incantinatiates them!

[FromTheRafters]

Sergio presented the following explanation :

> they are very dark, even black, when you print them,
>
> 1,2,3,4,5,6,...see?
>
> WM's *dark numbers turn BLACK* when he incantinatiates them!

1,2,3,4, ,6,7,...

Number five, please step forward. You are NOT passing for black, nice
try though.

[Ganzhinterseher]

Mostowski Collapse schrieb am Freitag, 26. Februar 2021 um 22:22:19 UTC+1:
> So you call yourself Racist, since you proves dark numbers?

ODE would call me so.

Regards, WM

[Ganzhinterseher]

Ben Bacarisse schrieb am Freitag, 26. Februar 2021 um 23:02:00 UTC+1:

> Ganzhinterseher <claus.v...@gmail.com> writes:

> Since you
> like functions and analysis, you could use the product of the indicator
> (AKA characteristic) functions to prove that, given

No problem. But I prefer to prove also the contrary: Every endsegment acts like a filter which, independent of its position and of the presence of other endsegments, removes up to n - 1 natural numbers from an incoming set and lets pass the rest.

Note two important features:
---every endsegment
and
--- the intersection is independent of the position since sets are not ordered anyhow.

> Can you prove that?

You proof is easy for definable elemnts. But you seem to be reluctant of considering the other side. To much afraid ro confirm a contradiction?

> Of course. This is an intersection of a set of sets. The intersection
> is provably empty because every element of N is missing from at least
> one of the intersected sets.

The intersection cannot be empty by the definition of the sequence:

E(n+1) = E(n) \ {n}

unless finite sets appear.

> Whatever you think the "position" of an
> end segment is, it will have no bearing on the result.

Correct. Therefore the endsegments which appear at the end in
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
will not change their effect when appearing not at the end.

And if you have a set of red elements, then there will not be a green element, even if there are infinitely many red elements. Same with endsegments leaving infinitely many natnumbers.

Are you unable to see this? Small wonder that ZFC is free of contradictions.

Regards, WM

[Python]

Crank Wolfgang Mueckenheim, aka Ganzhinterseher wrote:

Ordinary Differential Equation? Why that, Crank Wolfgang Mueckenheim,
from Hoschschule Augsburg?

by the way, about racism, here a two of your numerous racists statements
posted here:

Crank Wolfgang Mueckenheim, aka Ganzhinterseher wrote:

> O sorry, I forgot. Of course *all* races are the most clever ones. In
> fact there are no races. All people look alike and all have same
> colour and same bodies and same intellectual abilities and same
> mathematical skills and very soon also males will be able to
> bear childern and the we will be in the state of eternal rapture.

https://groups.google.com/g/sci.math/c/MTDeITKEyqQ/m/LgCclbNVBgAJ

Crank Wolfgang Mueckenheim, aka Ganzhinterseher wrote:

> In my opinion the Jews are the most clever race. The reason is perhaps
> that they have been mistreated over two millenia wherever they lived
> (not only in Germany). Only the best could survive in reasonable
> conditions. Contrary to their close Semitic relatives who have
> not been persecuted. An impressive example of Darwin's teachings.

https://groups.google.com/g/sci.math/c/MTDeITKEyqQ/m/XxQrSoFTBgAJ

[Chris M. Thomasson]

LOL! :^D

[Mostowski Collapse]

Yeah these Eugenics conferences must have been fun:

Following the opening addresses of the Congress, delivered in the
Auditorium of the Museum on the evening of the twentv-second, a
reception was held under the auspices of the Committee on Reception
and Entertainment in the Hall of the Age of Man. The inspiring and
beautiful surroundings, the exhibition of anthropological material,
and the mural decorations of prehistoric mammals

Program:

Saturday, September 24. 2-30 p.m. to 5 p.m.
Dr. Henry A. Cotton, Medical Director, New Jersey State Hospital, Trenton, N.J.
"Inheritance in Mental Disorders."
Dr. Abraham Myerson, 483, Beacon Street, Boston, Mass. "Inheritance of MentaI.
Diseases."
[...]
Dr. Helen Dean King, Wistar Institute of Anatomy and Biology, Philadelphia, Pa.
"Is Inbreeding Injurious?"
Dr. Howard J. Banker, Eugenics Record Office, Carnegie Institution of Washington,
Long Island, N.Y. "An Ideal Family History."
Etc...

http://belmont.bme.umich.edu/wp-content/uploads/sites/377/2018/02/2-The-Second-International-Congress-of-Eugenics.pdf

Will we soon see a new one at Augsburg Crank institute?

[Mostowski Collapse]

So how does WMs program of cleaning mathematics from infinity progress?
Is there already an armada of students, that brings the new knowlegde into
the world. Especially that there are dark numbers?

This is also fun:

6. Toward Civic Biology
The prime duty of the good citizen of the right type is to leave his or her
blood behind in the world, and we have no business permitting the
perpetuation of the wrong type.
Theodore Roosevelt

We know enough about agriculture so that the agricultural production
of the country could be doubled if the knowledge were
applied. We know enough about disease so that if the knowledge
were utilized, infectious and contagious diseases would be substantially
destroyed in the United States within a score of years; we know
enough about eugenics so that if the knowledge were applied, the
defective class would disappear within a generation.1

https://www.facinghistory.org/sites/default/files/publications/Race_Membership.pdf

[Sergio]

Number five should be Black Listed ;)

[Sergio]

FYI, potential Dark number five was outted for trying to pass as black,
but got caught, five is now Black Listed.

see => 1,2,3,4, ,6,7,...

[the black numbers above were incantinatiated by WM, which were dark
numbers at the time to him, as he was not thinking about them, or he
dropped the list they were on, but he restarted counting nearby German
rocks, and WMs count started at 1, "One is the first, before the second,
but only one before the second, 3 is right out!" (i digress) but the
numbers go from dark to black as shown above. Finally, potential
Progress in Dark Numbers!]

[zelos...@gmail.com]

>∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

Thsi is false.

>This is a contradiction

This isn't a contradiction because you're the one doing shit wrong.

>at least according to the conventional racist mathematics which does not tolerate contradictions

How is mathematics racist?

Have you gone full SJW crazy too?

[Ganzhinterseher]

zelos...@gmail.com schrieb am Montag, 1. März 2021 um 07:36:01 UTC+1:
> >∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
> Thsi is false.

No. Every endsegment acts like a filter which removes from an incoming set up to n-1 natnumbers and let's pass up to aleph_0 natnumbers. The position of the endsegment in the intersection does not play a role. Therefore every endsegment that is in

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

at last position and let's pass an infinite set of natnumbers cannot change its effect when appearing at not last position. Since in

∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

only such endsegments are applied, the result cannot be empty.

Regards, WM

[zelos...@gmail.com]

>No. Every endsegment acts like a filter which removes from an incoming set up to n-1 natnumbers and let's pass up to aleph_0 natnumbers. The position of the endsegment in the intersection does not play a role. Therefore every endsegment that is in

It is false because your

E(n) is satisfying (*)

means all endsegments and the intersection of all endsegments is empty.

>only such endsegments are applied, the result cannot be empty

It is empty because (*) applies to ALL endsegments and ALL endsegments intersected is empty.

You have not shown any element that is in there, but I can show you that NO element can be in it.

[Ganzhinterseher]

Then we have a contradiction, because every endsegment appearimg at last position in

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

will not yield an empty intersection when appearing at any position in

∩{ E(1), E(2), E(3),... }.

Regards, WM

[zelos...@gmail.com]

We got no contradiction because the first ones are all finite, the last one is infinite.

2 different cases with 2 very different starting points. The fact they yield different end results is not a contradiction.

[Ganzhinterseher]

zelos...@gmail.com schrieb am Montag, 1. März 2021 um 14:28:31 UTC+1:
> måndag 1 mars 2021 kl. 14:23:39 UTC+1 skrev Ganzhinterseher:
> > zelos...@gmail.com schrieb am Montag, 1. März 2021 um 13:07:51 UTC+1:
> > > >No. Every endsegment acts like a filter which removes from an incoming set up to n-1 natnumbers and let's pass up to aleph_0 natnumbers. The position of the endsegment in the intersection does not play a role. Therefore every endsegment that is in
> > > It is false because your
> > >
> > > E(n) is satisfying (*)
> > >
> > > means all endsegments and the intersection of all endsegments is empty.
> > > >only such endsegments are applied, the result cannot be empty
> > > It is empty because (*) applies to ALL endsegments and ALL endsegments intersected is empty.
> > >
> > > You have not shown any element that is in there, but I can show you that NO element can be in it.
> > Then we have a contradiction, because every endsegment appearimg at last position in
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> > will not yield an empty intersection when appearing at any position in
> >
> > ∩{ E(1), E(2), E(3),... }.

> We got no contradiction because the first ones are all finite, the last one is infinite.

But all infinitely many endsegments appearing in { E(1), E(2), E(3),... } are appearing in (*) already, showing that they do not yield empty intersections. And the intersection can become empty only by the effect of these endsegments.

Regards, WM

[zelos...@gmail.com]

but all intersections in (*) is FINITE

but (**) is INFINITE

That is teh difference and thati s why it becomes empty. Everything else is IRRELEVANT.

[Dan Christensen]

Wrong again, Mucke. From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }. Since there is no number common to EVERY end segment, this intersection will be empty. There is no contradiction.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Ganzhinterseher]

zelos...@gmail.com schrieb am Montag, 1. März 2021 um 14:43:05 UTC+1:

> but all intersections in (*) is FINITE

and definable

>
> but (**) is INFINITE
>
> That is teh difference and thati s why it becomes empty. Everything else is IRRELEVANT.

Relevant is that all intersections of infinitely many endsegments contain dark endsegments. Definable are only those which do not yield an empty intersection.

Proof: Try to define more endsegments than are in

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

Fail. But these do not yield empty intersections because the effect does not depend on their position. And their effect is provably not resulting in an empty intersection.

Regards, WM

[Ganzhinterseher]

Dan Christensen schrieb am Montag, 1. März 2021 um 14:59:27 UTC+1:
> On Monday, March 1, 2021 at 8:23:39 AM UTC-5, Ganzhinterseher wrote:

> > Then we have a contradiction, because every endsegment appearimg at last position in
> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> >
> > will not yield an empty intersection when appearing at any position in
> >
> > ∩{ E(1), E(2), E(3),... }.
> >

> From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }.

And, what is more important, no further endsegment will be there. Hence, the intersection can only be empty, if the endsegments which do not produce an empty intersection when being at last position will produce an empty intersection when being at other position.

> Since there is no number common to EVERY end segment, this intersection will be empty.

Either the effect of an endsegment depends on its position or there is no empty intersection and hence a contradiciton.

Regards, WM

[Ben Bacarisse]

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

q

> Ben Bacarisse schrieb am Freitag, 26. Februar 2021 um 23:02:00 UTC+1:
>> Ganzhinterseher <claus.v...@gmail.com> writes:
>
>> Since you
>> like functions and analysis, you could use the product of the indicator
>> (AKA characteristic) functions to prove that, given
>
> No problem.

Great. So you accept, now, that your (**) is false? Are you going to
correct the record?

>> Can you prove that?
>
> You proof is easy for definable elemnts. But you seem to be reluctant
> of considering the other side. To much afraid ro confirm a
> contradiction?

The direct proof, by induction, that no elements of N are in the
intersection is trivial. If you think you can prove the converse, go
ahead, but waffle will not hack it.

--
Ben.

[Python]

Crank Wolfang Mueckenheim, aka Ganzhinterseher wrote:
> zelos...@gmail.com schrieb am Montag, 1. März 2021 um 13:07:51 UTC+1:
>>> No. Every endsegment acts like a filter which removes from an incoming set up to n-1 natnumbers and let's pass up to aleph_0 natnumbers. The position of the endsegment in the intersection does not play a role. Therefore every endsegment that is in
>> It is false because your
>>
>> E(n) is satisfying (*)
>>
>> means all endsegments and the intersection of all endsegments is empty.
>>> only such endsegments are applied, the result cannot be empty
>> It is empty because (*) applies to ALL endsegments and ALL endsegments intersected is empty.
>>
>> You have not shown any element that is in there, but I can show you that NO element can be in it.
>
> Then we have a contradiction, because every endsegment appearimg at last position in
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)

There is no "position" involved here, what follow "∩" is a set,
of sets, Crank Wolfang Mueckenheim, from Hochschule Augsburg.

> will not yield an empty intersection when appearing at any position in
>
> ∩{ E(1), E(2), E(3),... }.

There is no "position" involved here, what follow "∩" is a set,
of sets, Crank Wolfang Mueckenheim, from Hochschule Augsburg.

Moreover, even if you would make sense of "position" by considering
sequences of sets, or ordered set of sets for that matters, it
would still be wrong as the value of both expressions would stay
independent of the order chosen.

This is a new (quite silly) kind of sophistry of yours: invoke a
property (position) that is not even involved in order to "prove"
a point, while even if this property were involved it wouldn't
change anything, and the "proof" would still be wrong.

It may be a test of docility for the students you are abusing
(so an outrageous shame), because you actually ask them to repeat
such nonsense, or a new symptom of your mental illness which is
worse and worse every single day Crank Wolfgang Mueckenheim.

Anyway, this is definitely a good argument to fire you from you
teaching position at Hochschule Augsburg, Crank Wolfgang Mueckenheim.

[Dan Christensen]

On Monday, March 1, 2021 at 10:32:48 AM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Montag, 1. März 2021 um 14:59:27 UTC+1:
> > On Monday, March 1, 2021 at 8:23:39 AM UTC-5, Ganzhinterseher wrote:
>
> > > Then we have a contradiction, because every endsegment appearimg at last position in
> > >
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> > >
> > > will not yield an empty intersection when appearing at any position in
> > >
> > > ∩{ E(1), E(2), E(3),... }.
> > >
> > From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }.

> And, what is more important, no further endsegment will be there. Hence, the intersection can only be empty, if the endsegments which do not produce an empty intersection when being at last position will produce an empty intersection when being at other position.

There is no "position" of an end-segment. Where are you getting this nonsense?

> > Since there is no number common to EVERY end segment, this intersection will be empty.

> Either the effect of an endsegment depends on its position or there is no empty intersection and hence a contradiciton.
>

Wrong again, Mucke. There is no "effect" or "position" of an end-segment. The intersection of all end-segments is empty. Just because you don't understand this does NOT mean there is contradiction.

Learn some basic algebra, Mucke. Again, read and work through the examples at https://www.mathsisfun.com/algebra/introduction.html DO YOUR HOMEWORK!

[Sergio]

correction,

Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Potenzial

[Sergio]

monotonic exclusion not using monotony Ants

[zelos...@gmail.com]

>and definable

Which means all natural numbers/endsegments are definable.

>Relevant is that all intersections of infinitely many endsegments contain dark endsegments.

Incorrect, there is nothign "dark" in it, they are all definable as you say it.

>Try to define more endsegments than are in

This does not in anyway show what you want, a pure non-sequitor as always.

[Ganzhinterseher]

Python schrieb am Montag, 1. März 2021 um 18:18:37 UTC+1:

> > Then we have a contradiction, because every endsegment appearing at last position in

> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> There is no "position" involved here, what follow "∩" is a set,
> of sets,

Correct. That is the argument that I apply:

> > will not yield an empty intersection when appearing at any position in
> >
> > ∩{ E(1), E(2), E(3),... }.
> There is no "position" involved here, what follow "∩" is a set,
> of sets,

Correct.

>
> Moreover, even if you would make sense of "position" by considering
> sequences of sets,

Obviously you cannot understand written text. I stated that the position does not interfere.That is my argument.

> or ordered set of sets for that matters, it
> would still be wrong as the value of both expressions would stay
> independent of the order chosen.

So it is. Can you try to increase your readings skills?

Regards, WM

[Ganzhinterseher]

Ben Bacarisse schrieb am Montag, 1. März 2021 um 17:21:10 UTC+1:
> Ganzhinterseher <claus.v...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> q
> > Ben Bacarisse schrieb am Freitag, 26. Februar 2021 um 23:02:00 UTC+1:
> >> Ganzhinterseher <claus.v...@gmail.com> writes:
> >
> >> Since you
> >> like functions and analysis, you could use the product of the indicator
> >> (AKA characteristic) functions to prove that, given
> >
> > No problem.
> Great. So you accept, now, that your (**) is false?

You can't distinguish ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { } ?

> Are you going to correct the record?

Chuckle, I'll try to correct your understanding by the simplest example possible:

You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction. For the subtraction of *undefined* unit fractions however this is possible: "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction.

The same is true for ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { }.

> > You proof is easy for definable elemnts. But you seem to be reluctant
> > of considering the other side. To much afraid ro confirm a
> > contradiction?
> The direct proof, by induction, that no elements of N are in the
> intersection is trivial.

Is my proof with unit fractions too difficult for you?

Regards, WM

[Python]

Crank Wofgang Mueckenheim, aka Ganzhinterseher wrote:
> Python schrieb am Montag, 1. März 2021 um 18:18:37 UTC+1:
>
>>> Then we have a contradiction, because every endsegment appearing at last position in
>>>
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
>> There is no "position" involved here, what follow "∩" is a set,
>> of sets,
>
> Correct. That is the argument that I apply:

So involving "position" is pointless...

>>> will not yield an empty intersection when appearing at any position in
>>>
>>> ∩{ E(1), E(2), E(3),... }.
>> There is no "position" involved here, what follow "∩" is a set,
>> of sets,
>
> Correct.

So involving "position" is pointless...

>> Moreover, even if you would make sense of "position" by considering
>> sequences of sets,
>
> Obviously you cannot understand written text. I stated that the position does not interfere.That is my argument.

And position do not interfere, if considered, in Set Theory. You
have no argument.

>> or ordered set of sets for that matters, it
>> would still be wrong as the value of both expressions would stay
>> independent of the order chosen.
>
> So it is. Can you try to increase your readings skills?

You are grotesque, Crank Wolfgang Mueckenheim, from Hochschule
Augsburg.

You admit that "position" is no applicable in the arguments where
you use "position" as a key point, but you pretend that your
arguments stand...

You admit that if "position" were applicable, then it wouldn't
affect ANY of the results (in Set Theory) of the expressions
involved, but you still pretend that your argument that Set
Theory is wrong because the results should be the same, stands.

You should try to increase your mental skills, they are very,
very low, Crank Wofgang Mueckenheim, from Hochschule Augsburg.

This is actually so silly on your part that I wonder if it is
merely a docility test for your victims (aka students): you
are checking that you have enough authority to make them repeat
complete nonsense in exams. This is outrageously criminal.

[Ganzhinterseher]

Dan Christensen schrieb am Montag, 1. März 2021 um 18:18:37 UTC+1:
> On Monday, March 1, 2021 at 10:32:48 AM UTC-5, Ganzhinterseher wrote:
> > Dan Christensen schrieb am Montag, 1. März 2021 um 14:59:27 UTC+1:
> > > On Monday, March 1, 2021 at 8:23:39 AM UTC-5, Ganzhinterseher wrote:
> >
> > > > Then we have a contradiction, because every endsegment appearimg at last position in
> > > >
> > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> > > >
> > > > will not yield an empty intersection when appearing at any position in
> > > >
> > > > ∩{ E(1), E(2), E(3),... }.
> > > >
> > > From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }.
>
> > And, what is more important, no further endsegment will be there. Hence, the intersection can only be empty, if the endsegments which do not produce an empty intersection when being at last position will produce an empty intersection when being at other position.
> There is no "position" of an end-segment.

There is the last position in the written formula for E(k). But it is irrelevant for the result.

> > > Since there is no number common to EVERY end segment, this intersection will be empty.
>
> > Either the effect of an endsegment depends on its position or there is no empty intersection and hence a contradiciton.
> >

> There is no "effect" or "position" of an end-segment.

Just that is my argument. Are your problems with logic also extending to English language?

> The intersection of all end-segments is empty.

Of course. But the intersection of *defined* endsegments is not empty. Can you grasp the difference? Here is a very simple example:

You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction. For the subtraction of *undefined* unit fractions however this is possible: "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction.

Regards, WM

[Ganzhinterseher]

zelos...@gmail.com schrieb am Dienstag, 2. März 2021 um 07:24:36 UTC+1:
> >and definable
>
> Which means all natural numbers/endsegments are definable.

You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction. For the subtraction of *undefined* unit fractions however this is possible: "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction. So there are unit fractions which cannot be identified individually.

Regards, WM

[Ganzhinterseher]

Python schrieb am Dienstag, 2. März 2021 um 11:18:41 UTC+1:


> >>> ∩{ E(1), E(2), E(3),... }.
> >> There is no "position" involved here, what follow "∩" is a set,
> >> of sets,
> >
> > Correct.
> So involving "position" is pointless...

Fine that you finally have grasped it.

> You admit that "position" is no applicable in the arguments where
> you use "position" as a key point,

I mention position in order to show that the effect of endsegment E(k) is independent of its position.

> You admit that if "position" were applicable, then it wouldn't
> affect ANY of the results (in Set Theory) of the expressions
> involved,

The effect of endsegment E(n) is the same in ∩{E(1), E(2), ..., E(n)} as in ∩{E(k) | k ∈ ℕ }.

> This is outrageously criminal.

No, it is set theory.

Regards, WM

[Alan Mackenzie]

Ben Bacarisse <ben.u...@bsb.me.uk> wrote:
> Ganzhinterseher <claus.v...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)

>> Ben Bacarisse schrieb am Freitag, 26. Februar 2021 um 23:02:00 UTC+1:
>>> Ganzhinterseher <claus.v...@gmail.com> writes:

>>> Since you like functions and analysis, you could use the product of
>>> the indicator (AKA characteristic) functions to prove that, given

>> No problem.

> Great. So you accept, now, that your (**) is false? Are you going to
> correct the record?

>>> Can you prove that?

>> You proof is easy for definable elemnts. But you seem to be reluctant
>> of considering the other side. To much afraid ro confirm a
>> contradiction?

> The direct proof, by induction, that no elements of N are in the
> intersection is trivial. If you think you can prove the converse, go
> ahead, but waffle will not hack it.

Ben, don't forget that if you wrestle with a pig, you both get dirty,
but the pig enjoys it.

> --
> Ben.

--
Alan Mackenzie (Nuremberg, Germany).

[Ben Bacarisse]

Alan Mackenzie <a...@muc.de> writes:

> Ben Bacarisse <ben.u...@bsb.me.uk> wrote:
>> Ganzhinterseher <claus.v...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)

<stuff>

> Ben, don't forget that if you wrestle with a pig, you both get dirty,
> but the pig enjoys it.

Are you assuming I don't? My main purpose (apart from a bit of fun such
as telling him his students could prove things he can't) is to put the
name of his course and both spellings of his name on posts. As a
result, there is a tiny chance that one of host students will find his
Usenet nonsense.

--
Ben.

[Sergio]

On 3/2/2021 4:19 AM, Ganzhinterseher wrote:
> Dan Christensen schrieb am Montag, 1. März 2021 um 18:18:37 UTC+1:
>> On Monday, March 1, 2021 at 10:32:48 AM UTC-5, Ganzhinterseher wrote:
>>> Dan Christensen schrieb am Montag, 1. März 2021 um 14:59:27 UTC+1:
>>>> On Monday, March 1, 2021 at 8:23:39 AM UTC-5, Ganzhinterseher wrote:
>>>
>>>>> Then we have a contradiction, because every endsegment appearimg at last position in
>>>>>
>>>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
>>>>>
>>>>> will not yield an empty intersection when appearing at any position in
>>>>>
>>>>> ∩{ E(1), E(2), E(3),... }.
>>>>>
>>>> From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }.
>>
>>> And, what is more important, no further endsegment will be there. Hence, the intersection can only be empty, if the endsegments which do not produce an empty intersection when being at last position will produce an empty intersection when being at other position.
>> There is no "position" of an end-segment.
>
> There is the last position in the written formula for E(k). But it is irrelevant for the result.

You can't have it both ways.
position or not position, that is the question.

>
>>>> Since there is no number common to EVERY end segment, this intersection will be empty.
>>
>>> Either the effect of an endsegment depends on its position or there is no empty intersection and hence a contradiciton.
>>>
>> There is no "effect" or "position" of an end-segment.
>
> Just that is my argument. Are your problems with logic also extending to English language?
>
>> The intersection of all end-segments is empty.
>
> Of course. But the intersection of *defined* endsegments is not empty. Can you grasp the difference? Here is a very simple example:

nonsense
Your "defined" by strangers, in a room, a rapping an a tapping, while
flashing without recording anything, having a flipping NUMBER party,
intentionally killing off Dark Numbers by EXPOSING them to the LIGHT, is
not part of "set theory", but part of "silly theory".

[Sergio]

so, you are having a problem with math nomenclature.

"You cannot define a unit fraction 1/n such that the intervall [1/n, 1]
subtracted from [0, 1] leaves no unit fraction."

so pick an n, call it k.

the interval [1/k, 1] subtracted from [0,1] = [0, 1/k]

EZ Cheesy.

now, limit as k => oo of [0, 1/k] = [0,0] = [0]

I do not see any unit fractions in there, just a 0

note: I did not have to resort to your strange "defined", or
"identified" by 3rd parties before use theorm.

The bigger question is why do any of that 1/n removal stuff in a real
interval ? it is always full of numbers.

"no unit fractions left" is meaningless, it is just intentional
obfuscation.

instead of fractions, invert it all, and go back to natural numbers, and
prove your case there.

counting numbers

1,2,3,4,5.....,k,...,oo

[Chris M. Thomasson]

On 2/25/2021 2:43 AM, Ganzhinterseher wrote:
> The intersection over a set {E(1), E(2), E(3), … E(n)}
> of endsegments E(k) = (k, k+1, k+2, ...) of ℕ is not empty:
>
> ∩{E(1), E(2), E(3), … E(n)} =/= { } (*)
>
> There are infinitely many endsegments E(n) which give infinite intersections when they appear at last position in (*). The intersection however does not depend on positions - a set has no order anyway. Since the position is irrelevant, the intersection of all of these endsegments is infinite and not empty also when they appear at not last position:

>
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>

> However, it is impossible to find a natural number common to all E(n) satisfying (*).
>
> This is a contradiction, at least according to the conventional racist mathematics which does not tolerate contradictions. But knows about exclusion monotony.
>
> The only remedy is this: Although you cannot find any natural number in (**) this does not prove an empty intersection because all findable natural numbers have infinitely many dark, i.e., not findable successors.
>

HYPER TROLL!

https://youtu.be/Ipbj_DF3Zt8

[Ben Bacarisse]

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

> Ben Bacarisse schrieb am Montag, 1. März 2021 um 17:21:10 UTC+1:
>> Ganzhinterseher <claus.v...@gmail.com> writes:
>> > Ben Bacarisse schrieb am Freitag, 26. Februar 2021 um 23:02:00 UTC+1:
>> >> Ganzhinterseher <claus.v...@gmail.com> writes:
>> >
>> >> Since you
>> >> like functions and analysis, you could use the product of the indicator
>> >> (AKA characteristic) functions to prove that, given
>> >
>> > No problem.
>> Great. So you accept, now, that your (**) is false?
>
> You can't distinguish ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { } ?

I can see why you want to change the subject, but this was your (**):

∩{E(n) | E(n) is satisfying (*)} =/= { } (**)

and you went on to say that "(**) is false".

>> Are you going to correct the record?
>
> Chuckle, I'll try to correct your understanding by the simplest
> example possible:

I'd rather you just corrected what you started the thread with. That
would be the honest thing to do.

--
Ben.

[Dan Christensen]

On Tuesday, March 2, 2021 at 5:19:10 AM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Montag, 1. März 2021 um 18:18:37 UTC+1:
> > On Monday, March 1, 2021 at 10:32:48 AM UTC-5, Ganzhinterseher wrote:
> > > Dan Christensen schrieb am Montag, 1. März 2021 um 14:59:27 UTC+1:
> > > > On Monday, March 1, 2021 at 8:23:39 AM UTC-5, Ganzhinterseher wrote:
> > >
> > > > > Then we have a contradiction, because every endsegment appearimg at last position in
> > > > >
> > > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { } (*)
> > > > >
> > > > > will not yield an empty intersection when appearing at any position in
> > > > >
> > > > > ∩{ E(1), E(2), E(3),... }.
> > > > >
> > > > From your definition, we have ∩{E(1), E(2), ..., E(k)} = E(k) for all k ∈ ℕ. EVERY such end-segment will be an element of { E(1), E(2), E(3),... }.
> >
> > > And, what is more important, no further endsegment will be there. Hence, the intersection can only be empty, if the endsegments which do not produce an empty intersection when being at last position will produce an empty intersection when being at other position.
> > There is no "position" of an end-segment.

> There is the last position in the written formula for E(k).

What is the "position" of the end-segment {3, 4, 5, 6, ... }?

> But it is irrelevant for the result.

This is true of ALL your claims here, Mucke. Why should this be any different?

> > > > Since there is no number common to EVERY end segment, this intersection will be empty.
> >
> > > Either the effect of an endsegment depends on its position or there is no empty intersection and hence a contradiciton.
> > >
> > There is no "effect" or "position" of an end-segment.

> Just that is my argument.

Yes. {3, 4, 5, 6, ... } for example is an end-segment. It has no defined "effect" or "position."

> > The intersection of all end-segments is empty.

> Of course.

So, you ARE capable of learning from your mistakes. VERY good!

> But the intersection of *defined* endsegments is not empty.

I spoke too soon. You have apparently learned nothing, Mucke. Very sad and pathetic, but not surprising in your case.

You yourself have DEFINED all end-segments as follows: S is an end-segment iff there exists n in N such that S = {x in N : x >= n }

> Can you grasp the difference?

Do YOU grasp the difference between what is and is not defined to be an end-segment? The above definition tells you not only what is an end-segment, but also what is NOT an end-segment. By this definition, the set { 3, 5, 7, 9, ... }, for example, is NOT defined to be an end-segment. Can you see why? Yes, every element of S is a natural number x >= 3, but what is missing?

> Here is a very simple example:
> You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction.

True. The well-defined unit fractions 1/(n+1), 1/(n+2), 1/(n+3), ... are all in [0, 1] \ [1/n, 1]. So what? It's just basic high-school math. Hmmm... It seems you STILL haven't been doing your homework, have you, Mucke?

> For the subtraction of *undefined* unit fractions however this is possible.

No need to conjure up any of your mysterious "undefined" or "dark" numbers. This result does not depend on the existence of any such nonsensical constructs.

> "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction.

So, what???

That link again: https://www.mathsisfun.com/algebra/introduction.html

DO YOUR HOMEWORK, MUCKE!

[zelos...@gmail.com]

>You can't distinguish ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { } ?

Given N_def = N, they are both empty.

>You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction

This does not in anyway demonstrate "dark numbers" or "undefinable fractions" or "undefinable numbers".

[zelos...@gmail.com]

A non-sequitor, the fact there isn't a specific one does not mean they all aren't "identified individually"

[Ganzhinterseher]

Ben Bacarisse schrieb am Mittwoch, 3. März 2021 um 02:36:01 UTC+1:
> Ganzhinterseher <claus.v...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)

> >> Great. So you accept, now, that your (**) is false?
> >
> > You can't distinguish ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { } ?
> I can see why you want to change the subject, but this was your (**):
> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
> and you went on to say that "(**) is false".

No. "E(n) is satisfying (*)" means E(n) is definable:

∩{E(1), E(2), E(3), … E(n)} =/= { } (*)

Otherwise it could not appear there.

> >> Are you going to correct the record?
> >
> > Chuckle, I'll try to correct your understanding by the simplest
> > example possible:
> I'd rather you just corrected what you started the thread with. That
> would be the honest thing to do.

Read what I wrote for newbies. You need it.

You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction. For the subtraction of *undefined* unit fractions however this is possible: "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction.

The same is true for ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { }.

∩{E(k) | k ∈ ℕ_def } = ∩{E(n) | E(n) is satisfying (*)}

This is different from ∩{E(k) | k ∈ ℕ} over "all" natural numbers.

Regards, WM

[Ganzhinterseher]

Dan Christensen schrieb am Mittwoch, 3. März 2021 um 06:02:00 UTC+1:

> Do YOU grasp the difference between what is and is not defined to be an end-segment?

Every defined endsegment can appear at the last position of

∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) =/= { }

and will not yield an empty intersection even if appearing at not last position. All those and only those endsegments are gathered in

∩{E(k) | k ∈ ℕ_def } =/= { } .

> > Here is a very simple example:
> > You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction.
> True. The well-defined unit fractions 1/(n+1), 1/(n+2), 1/(n+3), ... are all in [0, 1] \ [1/n, 1].

So what? It's just basic high-school math.

But most matheologians are unable to understand that this proves the existence of unit fractions which cannot be addressed individually. Because each individually addressed 1/n has infinitely many successors 1/(n+1), 1/(n+2), 1/(n+3), ... most of which cannot be addressed individually. Never less will remain. If addressing them collectively however, all can be removed. Then there are no further successors. This is the difference between visible and dark endsegments.

> > For the subtraction of *undefined* unit fractions however this is possible.
>
> No need to conjure up any of your mysterious "undefined" or "dark" numbers. This result does not depend on the existence of any such nonsensical constructs.

It shows the existence of unit fractions which exist because they can be removed but cannot be removed individually.

> > "All unit fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no unit fraction.
> So, what???

Try to think whenever you'll have the chance.

Regards, WM

[Ganzhinterseher]

zelos...@gmail.com schrieb am Mittwoch, 3. März 2021 um 07:33:44 UTC+1:

> >You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction
> This does not in anyway demonstrate "dark numbers" or "undefinable fractions" or "undefinable numbers".

It demonstrates numbers that cannot be removed individually but can be removed collectively. Don't you agree?

Regards, WM

[zelos...@gmail.com]

If by "individually" you mean "step by step method like an algorithm" them you are still wrong because anyone can be removed then by it but it takes time. And yes at no point along the step chain is all gone.

Neither of which has anythign to do with your dark numbers.

[Chris M. Thomasson]

Not exactly sure what you mean. [x, y], say y is one, and x is trying to
get there... Well, its already there with 1/1 = 1 so we have [1, 1]. Are
you trying to go, say 1/4 where we have

[0, 1] <- start here
[.25, 1]
[.5, 1]
[.75, 1]
[1, 1]

?

I am sorry for missing what you are getting at.

[Dan Christensen]

On Wednesday, March 3, 2021 at 6:01:37 AM UTC-5, Ganzhinterseher wrote:
> Dan Christensen schrieb am Mittwoch, 3. März 2021 um 06:02:00 UTC+1:
>
> > Do YOU grasp the difference between what is and is not defined to be an end-segment?
> Every defined endsegment can appear at the last position of
>
> ∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) =/= { }
>

Or the "first": ∀k ∈ ℕ_def: ∩{E(k), E(k-1), ..., E(1)} = E(k)

"First" or "last" has no meaning in this context. It seems you are deliberately trying to create confusion by introducing meaningless jargon.

> and will not yield an empty intersection even if appearing at not last position. All those and only those endsegments are gathered in
> ∩{E(k) | k ∈ ℕ_def } =/= { } .

More of your meaningless jargon! We have ∩{E(k) | k ∈ ℕ } = { } Just basic high-school math, Mucke. You really should learn some. See https://www.mathsisfun.com/algebra/introduction.html to get you started.

> > > Here is a very simple example:
> > > You cannot define a unit fraction 1/n such that the intervall [1/n, 1] subtracted from [0, 1] leaves no unit fraction.
> > True. The well-defined unit fractions 1/(n+1), 1/(n+2), 1/(n+3), ... are all in [0, 1] \ [1/n, 1].
>
> So what? It's just basic high-school math.

You're learning! You see, there is no need for your mysterious, "dark" or "undefined" numbers.

> But most matheologians are unable to understand that this proves the existence of unit fractions which cannot be addressed individually.

Still a crank at heart, eh, Mucke? When will you learn?

> Because each individually addressed 1/n has infinitely many successors 1/(n+1), 1/(n+2), 1/(n+3), ... most of which cannot be addressed individually.

"Individually addressed???" "Successors???" More of your meaningless jargon! When will you learn?

There are infinitely many, well-defined unit fractions less than 1/n for any n in N+ if that's what you mean. Basic ELEMENTARY-school math, Mucke!

> Never less will remain. If addressing them collectively however, all can be removed. Then there are no further successors. This is the difference between visible and dark endsegments.

Pure gibberish! Again, you need nothing more than basic high school to prove that ∩{E(k) | k ∈ ℕ } = { } since there is no number common to every end-segment.

DO YOUR HOMEWORK, MUCKE!!! https://www.mathsisfun.com/algebra/introduction.html

[Ben Bacarisse]

Ganzhinterseher <claus.v...@gmail.com> writes:

> Ben Bacarisse schrieb am Mittwoch, 3. März 2021 um 02:36:01 UTC+1:
>> Ganzhinterseher <claus.v...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>
>> >> Great. So you accept, now, that your (**) is false?
>> >
>> > You can't distinguish ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { } ?
>> I can see why you want to change the subject, but this was your (**):
>> ∩{E(n) | E(n) is satisfying (*)} =/= { } (**)
>> and you went on to say that "(**) is false".
>
> No.

I am sure you did. Where you mistaken to say that (**) is false?

> You cannot define a unit fraction 1/n such that the intervall [1/n, 1]
> subtracted from [0, 1] leaves no unit fraction. For the subtraction of
> *undefined* unit fractions however this is possible: "All unit
> fractions" or {1/1, 1/2, 1/3, ...}, subtracted from [0, 1] leaves no
> unit fraction.
>
> The same is true for ∩{E(k) | k ∈ ℕ } = { } and ∩{E(k) | k ∈ ℕ_def } =/= { }.

Great. We both agree that ∩{E(k) | k ∈ ℕ } = { }. Why did you spend so
long refusing to accept this simple fact?

--
Ben.