On Thursday, April 14, 2022 at 1:25:26 PM UTC+2, WM wrote:

> [The terms of] inclusion monotonic [sequences of] *infinite* sets cannot have an empty intersection.

Nonsense, idiot.

The sequence of intervals (0, 1/n] (n e lN) is a counter example.

Hint: There are infinitely many elements in each and every interval. But there is NO element in the intersection of all intervals, the intersection of all intervalls is empty.

You are dumb like shit, man.

On Thursday, April 14, 2022 at 1:30:41 PM UTC+2, WM wrote:
> William schrieb am Mittwoch, 13. April 2022 um 23:41:11 UTC+2:
> > On Wednesday, April 13, 2022 at 5:22:05 PM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 13. April 2022 um 21:07:30 UTC+2:
> >
> > >You can only notice the absence of natnumbers in the intersection
> > And thus, be definition of the intersection, the intersection in empty.
> >
> But that is contrary to mathematics.

Nonsense, you silly crank.

> The minimum is infinite for infinite endsegments.

No, you silly crank. The set of all endsegments does not contain a minimal element.

> It is <bla>

There is no "it", idiot!

horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
> On Wednesday, 13 April 2022 at 17:17:47 UTC-3, WM wrote:
> > FromTheRafters schrieb am Mittwoch, 13. April 2022 um 21:02:51 UTC+2:
> > > It happens that WM formulated :
> > > > Both can be named inclusion-monotonic, one ascending, the other descending.
> > > Descending from infinity?
> > No. The FISONs are ascending by number of elements: 1, 2, 3, ...
> > The endsegments are descending by number of elements: |ℕ|, |ℕ| - 1, |ℕ| - 2, ...
> Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|..

Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
>
> But the two statements above make it clear how your demented mind works: You think that the sequence of sizes of FISONs goes something like
> 1, 2, 3, ..., |ℕ| - 2, |ℕ| - 1, |ℕ|.

No, that is your mind. The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.

> So then, tell us: which is the first infinite FISON?

There is none.

> Likewise: What would a finite end segment look like? Name one!

They are dark. That's why transfinity needs dark numbers.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:12:52 UTC+2:
> On Wednesday, April 13, 2022 at 7:45:03 PM UTC+2, WM wrote:

> > Here is no shift but a correct statement. If you wish to attack it, find a counter example.
> The sequence of intervals [-1/n, 1/n] (n e lN) is a counter example

No. All definable intervals are infinite with infinite intersection.
>
> Hint: There are at least 100 elements in each and every interval.

No, only in all definable intervals.

> But there is only ONE element in the intersection of all intervals, namely 0.

Of all, yes. But we are looking for infinite intervals with an empty intersection.

Regards, WM

On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
> > On Wednesday, 13 April 2022 at 17:17:47 UTC-3, WM wrote:
> > > FromTheRafters schrieb am Mittwoch, 13. April 2022 um 21:02:51 UTC+2:
> > > > It happens that WM formulated :
> > > > > Both can be named inclusion-monotonic, one ascending, the other descending.
> > > > Descending from infinity?
> > > No. The FISONs are ascending by number of elements: 1, 2, 3, ...
> > > The endsegments are descending by number of elements: |ℕ|, |ℕ| - 1, |ℕ| - 2, ...
> > Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
> Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.

Congratulations on this earth-shattering insight. Yes, |ℕ| is a constant, an *INFINITE* constant. It does not follow the same arithmetic rules as the natural numbers. (Why should it?) In the words of Fritz Feldhase: Shut up, you silly crank.

On Thursday, April 14, 2022 at 2:08:41 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:12:52 UTC+2:
> > On Wednesday, April 13, 2022 at 7:45:03 PM UTC+2, WM wrote:
> > >
> > > Here is no shift but a correct statement. If you wish to attack it, find a counter example.
> >
> > The sequence of intervals [-1/n, 1/n] (n e lN) is a counter example
> >
> No.

Yes, you psychotic asshole full of shit!

> > Hint: There are at least 100 elements in each and every interval.
> >
> No

Yes. Why are lying all the time, Mückenheim?

> > But there is only ONE element in the intersection of all intervals, namely 0.
> >
> Of all, yes. But we are looking for infinite intervals with an empty intersection.

No, you demented crank.

Your claim was: "There are at least 100 elements per endsegment. They are in the intersection."

And: "Correct in this case of inclusion monotony. [...] Here is no shift but a correct statement. If you wish to attack it, find a counter example."

Now the sequence of intervals [-1/n, 1/n] (n e lN) is a counter example, you silly crank.

Hint: There are at least 100 elements in each and every interval. But there is only ONE element in the intersection of all intervals, namely 0.

You are dumb like shit, man.

Crank Wolfgang Müeckenheim, aka WM wrote:
....
> *Infinite* inclusion monotonic sets cannot have an empty intersection. The idea that the intersection of *infinite* endsegments could be empty because infinitely many sets are involved is wrong. Infinitely many sets {1} would not give an empty intersection either.

So now you're using \exists x ~P(x) as an argument against P(a).
Brilliant!

Hint: that it exists non-empty infinite intersections is not argument
for all infinite intersection to be non-empty.

You're such a kook, crank Wolfgang Müeckenheim, from Hochschule
Augsburg.

On Thursday, April 14, 2022 at 2:31:31 PM UTC+2, Python wrote:
> Crank Wolfgang Müeckenheim, aka WM wrote:
>
> The idea that the intersection of *infinite* endsegments could be empty because infinitely many sets are involved is wrong. Infinitely many sets {1} would not give an empty intersection either.

> So now you're using \exists x ~P(x) as an argument against P(a).

Or rather: ~P(b) as an argument against P(a)

> Brilliant!

indeed!

1 + 2 = 3 is wrong in Mückmath because 1 + 2 =/= 1.

Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
> On Thursday, April 14, 2022 at 1:30:41 PM UTC+2, WM wrote:
..
> > The minimum is infinite for infinite endsegments.
> The set of all endsegments does not contain a minimal element.

The set of infinite endsegments contains elements. That is important. Without any set being empty the intersection cannot be empty.

Regards, WM

horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 14:30:29 UTC+2:
> On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
> > > On Wednesday, 13 April 2022 at 17:17:47 UTC-3, WM wrote:
> > > > FromTheRafters schrieb am Mittwoch, 13. April 2022 um 21:02:51 UTC+2:
> > > > > It happens that WM formulated :
> > > > > > Both can be named inclusion-monotonic, one ascending, the other descending.
> > > > > Descending from infinity?
> > > > No. The FISONs are ascending by number of elements: 1, 2, 3, ...
> > > > The endsegments are descending by number of elements: |ℕ|, |ℕ| - 1, |ℕ| - 2, ...
> > > Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.

|ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.

> > Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
> Yes, |ℕ| is a constant, an *INFINITE* constant. It does not follow the same arithmetic rules as the natural numbers. (Why should it?)

Because ℕ consists of elements none of which can be subtracted without damaging ℕ.

Regards, WM

On 4/14/2022 8:46 AM, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
>> On Thursday, April 14, 2022 at 1:30:41 PM UTC+2, WM wrote:
> .
>>> The minimum is infinite for infinite endsegments.
>> The set of all endsegments does not contain a minimal element.
>
> The set of infinite endsegments contains elements.

No, it does not. We have proved this many times to you in this forum.

> That is important.

What is important is to realize that you, WM, cannot follow a simple Math Proof.

You get a F, for Fail, in Math.

>Without any set being empty the intersection cannot be empty.

You are a Liar, Fraud, Troll, Crank.

>
> Regards, WM

WM was thinking very hard :
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
>> On Wednesday, 13 April 2022 at 17:17:47 UTC-3, WM wrote:
>>> FromTheRafters schrieb am Mittwoch, 13. April 2022 um 21:02:51 UTC+2:
>>>> It happens that WM formulated :
>>>>> Both can be named inclusion-monotonic, one ascending, the other
>>>>> descending.
>>>> Descending from infinity?
>>> No. The FISONs are ascending by number of elements: 1, 2, 3, ...
>>> The endsegments are descending by number of elements: |ℕ|, |ℕ| - 1, |ℕ| -
>>> 2, ...
>> Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
>
> Is ℕ a constant or not?

It is a special symbol for a specific set, the set of natural numbers.

> If ℕ is a constant, then |ℕ| is a constant too.

That is a symbol representing the cardinality of a specific set, in
this case the set of natural numbers.

>> But the two statements above make it clear how your demented mind works: You
>> think that the sequence of sizes of FISONs goes something like 1, 2, 3,
>> ..., |ℕ| - 2, |ℕ| - 1, |ℕ|.
>
> No, that is your mind. The sequence of FISONs goes 1, 2, 3, ... . Above, all
> is dark. That's why |ℕ| cannot be expressed by natural numbers, neither can
> |ℕ| - 1. But they are different since |ℕ| - 1 < |ℕ|.

By the Axiom of Because I Said So?

Muckymath's core idea.

WM was thinking very hard :
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 14:30:29 UTC+2:
>> On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36
>>> UTC+2:
>>>> On Wednesday, 13 April 2022 at 17:17:47 UTC-3, WM wrote:
>>>>> FromTheRafters schrieb am Mittwoch, 13. April 2022 um 21:02:51 UTC+2:
>>>>>> It happens that WM formulated :
>>>>>>> Both can be named inclusion-monotonic, one ascending, the other
>>>>>>> descending.
>>>>>> Descending from infinity?
>>>>> No. The FISONs are ascending by number of elements: 1, 2, 3, ...
>>>>> The endsegments are descending by number of elements: |ℕ|, |ℕ| - 1, |ℕ| -
>>>>> 2, ...
>>>> Crackpottery. |ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
>
>> ℕ| - 1 = aleph_0 - 1 = aleph_0 = |ℕ|.
> Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.
>
>>> Is ℕ a constant or not? If ℕ is a constant, then |ℕ| is a constant too.
>> Yes, |ℕ| is a constant, an *INFINITE* constant. It does not follow the same
>> arithmetic rules as the natural numbers. (Why should it?)
>
> Because ℕ consists of elements none of which can be subtracted without
> damaging ℕ.

You can actually, it's only cosmetic damage though, not structural,
just relabel the elements and you are good to go. Like the worn out
labels on the keys on a computer keyboard, the keys still do the same
thing, you just need new labels.

WM submitted this idea :
> Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
>> On Thursday, April 14, 2022 at 1:30:41 PM UTC+2, WM wrote: .
>>> The minimum is infinite for infinite endsegments.
>> The set of all endsegments does not contain a minimal element.
>
> The set of infinite endsegments contains elements.

The word "infinite" is redundant here, but yes -- non-empty sets do
contain elements. We'll give you some time to submit your paper for
review, but hurry -- you don't want to be scooped.

On Thursday, April 14, 2022 at 8:25:26 AM UTC-3, WM wrote:
> Nope. (*) proves something about every *element* of the set ℕ_p. It shows nothing about the set ℕ_p.
> The set is (the union of) all its elements. Nothing more!

Piffle. And nothing less. If a set S is such that all its elements have property P, S may or may not have property P

> In particular ℕ_p has cardinality ℵ₀
> > and
> > |∩{E(k) : k ∈ ℕ_p}| = 0.
> *Infinite* inclusion monotonic sets cannot have an empty intersection.

False, It is trivial to see that for R, an inclusion monotonic set of sets, the intersection is the minimum. A finite set, R, must have a minimum. If S is infinite, it may not have a minimum. In particular if R is an infinite set of endsegments it does not have a minimum.
--
William Hughes

On Thursday, 14 April 2022 at 10:46:39 UTC-3, WM wrote:
> The set of infinite endsegments contains elements.

The elements of the "set of infinite end segments" are *END SEGMENTS*. Any questions?

On Thursday, 14 April 2022 at 10:51:12 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 14:30:29 UTC+2:
[...]
> > Yes, |ℕ| is a constant, an *INFINITE* constant. It does not follow the same arithmetic rules as the natural numbers. (Why should it?)
> Because ℕ consists of elements none of which can be subtracted without damaging ℕ.

Utter nonsense. "Damaging ℕ" has absolutely nothing to do with arithmetic rules for |ℕ|. You truly are a *FUCKING* *LUNATIC*!

On Thursday, April 14, 2022 at 8:30:41 AM UTC-3, WM wrote:
> Nope, this is trivially false for a set with no mininum.
> The minimum is infinite for infinite endsegments.

Piffle. An infinite set of endsegments cannot have a minimum.

--
William Hughes

On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
[...]

> The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark.

Above what??? Above 3? In the line above? After the three dots? If you can't even get such a simple sentence right, why do you think you have enough brains to even go to the bathroom by yourself?

On Thursday, 14 April 2022 at 10:51:12 UTC-3, WM wrote:
[...]
> Yes aleph_0 means actually infinite. But that does not mean |ℕ| - 1 = |ℕ|.

That is *EXACTLY* what it means. (Aside from the fact that aleph_0 is *NOT* the same as "actually infinite". You can't even get that right!)

WM explained on 4/14/2022 :
> William schrieb am Mittwoch, 13. April 2022 um 23:10:24 UTC+2:
>> On Wednesday, April 13, 2022 at 5:15:01 PM UTC-3, WM wrote:
>>
>>> Wrong as you can see by (*).
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ (*)
>
>> Nope. (*) proves something about every *element* of the set ℕ_p. It shows
>> nothing about the set ℕ_p.
>
> The set is (the union of) all its elements. Nothing more!

The set of odd natural numbers less than nine.

|{1,3,5,7}| = 4

Oh no!! Four is not an odd natural number less than nine, somebody must
*do* something about this travesty.

Muckymath to the rescue, by the powers of the Axiom of Because I Said
So I declare zero to be an odd natural number, sometimes.

On Thursday, 14 April 2022 at 11:45:12 UTC-3, Gus Gassmann wrote:
> On Thursday, 14 April 2022 at 09:05:20 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Donnerstag, 14. April 2022 um 12:35:36 UTC+2:
> [...]
> > The sequence of FISONs goes 1, 2, 3, ... . Above, all is dark.
> Above what??? Above 3? In the line above? After the three dots? If you can't even get such a simple sentence right, why do you think you have enough brains to even go to the bathroom by yourself?

Forgot to point out that "1, 2, 3, ... " may be many things, but it is *NOT* a sequence of FISONS.

On Thursday, April 14, 2022 at 10:46:39 AM UTC-3, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 14. April 2022 um 13:52:35 UTC+2:
> > The set of all endsegments does not contain a minimal element.
> The set of infinite endsegments contains elements.

Correct . However, it does not contain a minimal element. This is sufficient to show that the intersection is empty. One does not need the set to contain an empty element

--
William Hughes

On 4/14/2022 9:48 AM, FromTheRafters wrote:
> WM explained on 4/14/2022 :
>> William schrieb am Mittwoch, 13. April 2022 um 23:10:24 UTC+2:
>>> On Wednesday, April 13, 2022 at 5:15:01 PM UTC-3, WM wrote:
>>>> Wrong as you can see by (*).
>>
>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ (*)
>>
>>> Nope. (*) proves something about every *element* of the set ℕ_p. It shows nothing about the set ℕ_p.
>>
>> The set is (the union of) all its elements. Nothing more!
>
> The set of odd natural numbers less than nine.
>
> |{1,3,5,7}| = 4
>
> Oh no!! Four is not an odd natural number less than nine, somebody must *do* something about this travesty.

I fix;

|{1,3,5,7}| = (7+5)/(3*1)

>
> Muckymath to the rescue, by the powers of the Axiom of Because I Said So I declare zero to be an odd natural number, sometimes.

[oh no, my fix wont work in Muckymath, as it is "algebra"]

On Thursday, April 14, 2022 at 10:51:12 AM UTC-3, WM wrote:
> ℕ consists of elements none of which can be subtracted without damaging ℕ.

True. So what? The "damaged N" has the same cardinality as ℕ.

--
William Hughes