horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:47:29 UTC+2:
> On Wednesday, 13 April 2022 at 10:41:36 UTC-3, WM wrote:
> > [...] Even in an infinite set of red apples all apples are red.
>
> Yes, but the *SET* is not red.

Here we talk about extrapolating height and width of a figure, not about the resulting set of FISONs.

Regards, WM

William schrieb am Mittwoch, 13. April 2022 um 16:48:43 UTC+2:
> On Wednesday, April 13, 2022 at 11:13:02 AM UTC-3, WM wrote:
> > Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
> > > Crank Wolfgang Mückenheim, aka WM wrote:
> > > > There is no element a in A which is not in B when an element b exist in B which is not in A.
> > > This is not even a statement about "inclusion monotony", it is only
> > > stating strict inclusion.
> > That is what endsegments obey and what we abbreviate by inclusion monotony here.
> > > Hardly a counter argument to the fact that
> > > there is no element that is in every endsegment (which is trivial).
> > The statement is: There is one and the same element in every infinite endsegment.
> >
> > Example:
> > A = {2, 3, 4,...}
> > B = {1, 2, 3, 4, ...}
> > Every element of A is in B since there exists b = 1 in B which is not in A.
> Lets create the intersection I. 2 is in A and B maybe 2 is in I.
> B = {1, 2, 3, 4, ...}
> A = { 2, 3, 4, ...}
> C = {3, 4, 5 ...}
>
Rather confused. But not required. For every set of definable endsegments there is an infinite intersection.

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

No problem to find a common element.

Regards, WM

On 4/13/2022 12:48 PM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:47:29 UTC+2:
>> On Wednesday, 13 April 2022 at 10:41:36 UTC-3, WM wrote:
>>> [...] Even in an infinite set of red apples all apples are red.
>>
>> Yes, but the *SET* is not red.
>
> Here we talk about extrapolating height and width of a figure, not about the resulting set of FISONs.
>
> Regards, WM

you dodge again, Deceiver.

On Wednesday, 13 April 2022 at 14:45:03 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:45:40 UTC+2:
> > On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
> > > Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > > > On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
> > > > >
> > > > > There are at least 100 elements per endsegment. They are in the intersection.
> > > >
> > > > Quantifier shift.
> > > Correct in this case of inclusion monotony.
> > Bullshit. Quantifier shift always needs a proof.
> Here is no shift but a correct statement. If you wish to attack it, find a counter example.

Your statement, like 99.9% of all your stuff here in the last 20 years, is utter bullshit. I do not have to try to give you a nonexistent counterexample (which itself depends on a quantifier shift). But let me ask you, for the record: Have you stopped beating *your* wife?

On 4/13/2022 12:44 PM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:45:40 UTC+2:
>> On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
>>> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
>>>> On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
>>>>>
>>>>> There are at least 100 elements per endsegment. They are in the intersection.
>>>>
>>>> Quantifier shift.
>>> Correct in this case of inclusion monotony.
>> Bullshit. Quantifier shift always needs a proof.
>
> Here is no shift but a correct statement. If you wish to attack it, find a counter example.
>
> Regards, WM

False. It is a False Statement.

The intersection of all endsegments is empty, because k is not in E(k+1). Got it ?

On 4/13/2022 12:52 PM, WM wrote:
> William schrieb am Mittwoch, 13. April 2022 um 16:48:43 UTC+2:
>> On Wednesday, April 13, 2022 at 11:13:02 AM UTC-3, WM wrote:
>>> Python schrieb am Mittwoch, 13. April 2022 um 16:01:28 UTC+2:
>>>> Crank Wolfgang Mückenheim, aka WM wrote:
>>>>> There is no element a in A which is not in B when an element b exist in B which is not in A.
>>>> This is not even a statement about "inclusion monotony", it is only
>>>> stating strict inclusion.
>>> That is what endsegments obey and what we abbreviate by inclusion monotony here.
>>>> Hardly a counter argument to the fact that
>>>> there is no element that is in every endsegment (which is trivial).
>>> The statement is: There is one and the same element in every infinite endsegment.
>>>
>>> Example:
>>> A = {2, 3, 4,...}
>>> B = {1, 2, 3, 4, ...}
>>> Every element of A is in B since there exists b = 1 in B which is not in A.
>> Lets create the intersection I. 2 is in A and B maybe 2 is in I.
>> B = {1, 2, 3, 4, ...}
>> A = { 2, 3, 4, ...}
>> C = {3, 4, 5 ...}
>>
> Rather confused. But not required. For every set of definable endsegments there is an infinite intersection.
>
> ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
>
> No problem to find a common element.

then name it.

Call it k. however k is not in E(k+1), and this holds for all k.

therefore the intersetion of all endsegments is empty

>
> Regards, WM

On Wednesday, April 13, 2022 at 2:42:48 PM UTC-3, WM wrote

>We know inclusion-monotony. Therefore the set must exist.

Piffle

It cannot be named. Therefore dark numbers must exist.

Piffle. The natural numbers are a Peano set. This set cannot contain "dark numbers"

> >
> > You never get a set that works.
> But I know that either there is no actual infinity and hence no endsegments, or there are endsegments proving by your argument that dark numbers must exist - or this simple mathematics fails in your world: There is no element a in A which is not in B when an element b exists in B which is not in A..
> > Indeed, But what you need is "There is an a(A) in A and in B when an element b exist in B which is not in A." When you change A you may have to change a(A).
> No problem: Then the first A

There is no "first" A. There is just a endless stream of "Whoops, that set is not the first A"

> .. A definable endsegment is, by definition, an endsegment which has infinitely many natnumbers in common with all definable endsegments: |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

Piffle. Your set , ℕ_def, is a finite set. |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ . So what? It is trivial to note that a finite set of endsegments has infinite intersection. You need a set like ℕ_p, the Peano set of natural numbers, which has cardinality ℵ₀ and no "dark elements" so
|∩{E(k) : k ∈ ℕ_p}| = 0.

--
William Hughes

It happens that WM formulated :
> FromTheRafters schrieb am Mittwoch, 13. April 2022 um 16:32:25 UTC+2:
>> WM laid this down on his screen :
>>> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
>>>> On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
>>>>>
>>>>> There are at least 100 elements per endsegment. They are in the
>>>>> intersection.
>>>>
>>>> Quantifier shift.
>>>
>>> Correct in this case of inclusion monotony.
>>>>
>>>> For each and every endsegment: there are at least 100 elements in it.
>>>> (true)
>>>>
>>>> WM concludes (by a quantifier shift):
>>>>
>>>> There are at least 100 elements, which are in each and every endsegement.
>>>> (false)
>>>>
>>>> "Hence": They (!) are in the intersection (of all endsegments). for a
>>>
>>> Try to find a counter example for an inclusion-monotonic sequence of sets
>>> (without assuming the contrary).
>> How is your inclusion monotonic sequence structured? It looks to me
>> like you are starting at infinity and generating them in reverse order
>> to claim inclusion monotonicity when actually they spring forth as
>> complementary sets from the FISON generation. The FISONs are inclusion
>> monotonic not the endsegments.
>
> Both can be named inclusion-monotonic, one ascending, the other descending.

Descending from infinity?

On Wednesday, April 13, 2022 at 2:52:50 PM UTC-3, WM wrote:
For every set of definable

Piffle. To you a definable set of endsegments is a finite set. So what? We know that every finite set of endsegments has an infinite intersection.

--
Willliam Hughes

horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 20:05:03 UTC+2:
> On Wednesday, 13 April 2022 at 14:45:03 UTC-3, WM wrote:

> > > > > > There are at least 100 elements per endsegment. They are in the intersection.
> > If you wish to attack it, find a counter example.
> I do not have to try to give you a nonexistent counterexample

You cannot give a counter example, becaue there is none.
Every infinite endsegment receives its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

Regards, WM

William schrieb am Mittwoch, 13. April 2022 um 20:59:53 UTC+2:
> On Wednesday, April 13, 2022 at 2:42:48 PM UTC-3, WM wrote
> >We know inclusion-monotony. Therefore the set must exist.
> Piffle
> It cannot be named. Therefore dark numbers must exist.
> Piffle. The natural numbers are a Peano set. This set cannot contain "dark numbers"

But it does. Every infinite endsegment receives its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀ (*)
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

> > There is no element a in A which is not in B when an element b exists in B which is not in A.
> > > Indeed, But what you need is "There is an a(A) in A and in B when an element b exist in B which is not in A." When you change A you may have to change a(A).
> > No problem: Then the first A
> There is no "first" A. There is just a endless stream of "Whoops, that set is not the first A"

This endless stream (it is potentially infinite) does what I just described. Never a set will show up with no common elements with all predecessors. Provable by (*).
>
> > .. A definable endsegment is, by definition, an endsegment which has infinitely many natnumbers in common with all definable endsegments: |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
>
> Piffle. Your set , ℕ_def, is a finite set. |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

Of course. Define an endsegment individually which does not belong to a finite set. Fail.

> It is trivial to note that a finite set of endsegments has infinite intersection.

Then we agree. Individually definable means finite n.
Collectively definable means not individually definable or dark.

> You need a set like ℕ_p, the Peano set of natural numbers, which has cardinality ℵ₀

Wrong as you can see by (*).

> and no "dark elements" so
> |∩{E(k) : k ∈ ℕ_p}| = 0.

Impossible without dark elements.

Regards, WM

On 4/13/2022 3:07 PM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 20:05:03 UTC+2:
>> On Wednesday, 13 April 2022 at 14:45:03 UTC-3, WM wrote:
>
>>>>>>> There are at least 100 elements per endsegment. They are in the intersection.
>>> If you wish to attack it, find a counter example.
>> I do not have to try to give you a nonexistent counterexample
>
> You cannot give a counter example, becaue there is none.
> Every infinite endsegment receives its contents from the first one
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

Wrong.

k is not in E(k+1) therefore it is not in the intersection of all endsegments.
Fail.

>
> Regards, WM Always Wrong WM Never Learns.

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, ...

Regards, WM

William schrieb am Mittwoch, 13. April 2022 um 21:07:30 UTC+2:
> On Wednesday, April 13, 2022 at 2:52:50 PM UTC-3, WM wrote:
> For every set of definable
> Piffle. To you a definable set of endsegments is a finite set. So what?

Have you every defined an n that was not finite?

> We know that every finite set of endsegments has an infinite intersection..

And now you know why that is only true for finite sets of endsegments. It is true in all cases which you can check. You cannot check the individual endsegments in infinite sets. You can only notice the absence of natnumbers in the intersection by definition.

But you should be able to note that infinite inclusion-monotonic sets would not have an empty intersection.

Regards, WM

On Wednesday, April 13, 2022 at 5:15:01 PM UTC-3, WM wrote:

> Wrong as you can see by (*).

Nope. (*) proves something about every *element* of the set ℕ_p. It shows nothing about the set ℕ_p. In particular ℕ_p has cardinality ℵ₀
and

|∩{E(k) : k ∈ ℕ_p}| = 0.

--
William Hughes

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 you should be able to note that infinite inclusion-monotonic sets would not have an empty intersection.

Nope, this is trivially false for a set with no miminum. An infinite set may not have a mininum. e.g. the Peano set of endsegments, does not have a minimum

--
William Hughes

onsdag 13 april 2022 kl. 15:38:06 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
> > >
> > > There are at least 100 elements per endsegment. They are in the intersection.
> >
> > Quantifier shift.
> Correct in this case of inclusion monotony.

Nope

> >
> > For each and every endsegment: there are at least 100 elements in it. (true)
> >
> > WM concludes (by a quantifier shift):
> >
> > There are at least 100 elements, which are in each and every endsegement. (false)
> >
> > "Hence": They (!) are in the intersection (of all endsegments). for a
>
> Try to find a counter example for an inclusion-monotonic sequence of sets (without assuming the contrary).

No one has to because it is an irrelevant property here.

>
> Regards, WM

onsdag 13 april 2022 kl. 15:41:36 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 12. April 2022 um 14:58:50 UTC+2:
> > tisdag 12 april 2022 kl. 14:17:27 UTC+2 skrev WM:
> > > William schrieb am Montag, 11. April 2022 um 21:40:49 UTC+2:
> > > > On Monday, April 11, 2022 at 3:28:51 PM UTC-3, WM wrote:
> > > > >.. Then a finite number of n
> > > >
> > > > Nope, n is not finite, n is the infinite cardinality ℵo. A set of symbols with cardinality ℵo can distinguish the set of FISONs which has cardinality ℵo (like any Peano set). The set is infinite, the *elements* of the set are finite.
> > > Impossible for FISONs. Note that in
> > > o
> > > oo
> > > ooo
> > > ...
> > > height ℵ₀ implies width ℵ₀.
> > Here you go trying to extrapolate from the finite, the infinite.
> Of course.

YOU CANNOT DO THAT!

>There is no other was to do *mathematics* of the infinite. Even in an infinite set of red apples all apples are red.

Faulty comparison.

> >
> > IT DOES NOT WORK!
>
> Then all your claims are invalid too.

Nope, you CAN argue about infinite sets, you just cannot say "becuase it is true for finite, it is true for infinite"

>
> Regards, WM

On Wednesday, 13 April 2022 at 17:07:30 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 20:05:03 UTC+2:
> > On Wednesday, 13 April 2022 at 14:45:03 UTC-3, WM wrote:
>
> > > > > > > There are at least 100 elements per endsegment. They are in the intersection.
> > > If you wish to attack it, find a counter example.
> > I do not have to try to give you a nonexistent counterexample
> You cannot give a counter example, becaue there is none.

Bullshit! The intersection of infinitely many (different) end segments is provably empty. That you wish it to be nonempty is well known, but wishing does not make it so. You have also never responded to the example I gave you: The intersection of the sequence of intervals (0, 1/n) for n in IN. This sequence is inclusion monotone, and yet the intersection is empty. If you think otherwise, give an element you think is in the intersection.

On Wednesday, 13 April 2022 at 17:22:05 UTC-3, WM wrote:
> William schrieb am Mittwoch, 13. April 2022 um 21:07:30 UTC+2:
> > On Wednesday, April 13, 2022 at 2:52:50 PM UTC-3, WM wrote:
> > For every set of definable
> > Piffle. To you a definable set of endsegments is a finite set. So what?
> Have you every defined an n that was not finite?

Yes: Let n = aleph_2 if the Riemann hypothesis is true, aleph_3 otherwise. There!

Of course every *natural number* n is finite. And there still are infinitely many of them; aleph_0 to be precise. All of this is old hat, and that your mental deficiencies do not allow you to process this is your problem alone.

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|.

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|.
So then, tell us: which is the first infinite FISON? And then explain why it is not an IISON. Likewise: What would a finite end segment look like? Name one!

On Wednesday, April 13, 2022 at 7:45:03 PM UTC+2, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:45:40 UTC+2:
> > On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
> > > Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > > > On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
> > > > >
> > > > > There are at least 100 elements per endsegment. They are in the intersection.
> > > >
> > > > Quantifier shift.
> > > >
> > > Correct in this case of inclusion monotony.

Nonsense, you psychotic asshole..

> > Bullshit. Quantifier shift always needs a proof.
> >
> 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, 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 endsegments, namely 0.

You are dumb like shit, man.

On Wednesday, April 13, 2022 at 7:45:03 PM UTC+2, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 13. April 2022 um 16:45:40 UTC+2:
> > On Wednesday, 13 April 2022 at 10:38:06 UTC-3, WM wrote:
> > > Fritz Feldhase schrieb am Dienstag, 12. April 2022 um 14:42:31 UTC+2:
> > > > On Tuesday, April 12, 2022 at 2:20:46 PM UTC+2, WM wrote:
> > > > >
> > > > > There are at least 100 elements per endsegment. They are in the intersection.
> > > >
> > > > Quantifier shift.
> > > >
> > > Correct in this case of inclusion monotony.

Nonsense, you psychotic asshole..

> > Bullshit. Quantifier shift always needs a proof.
> >
> 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, 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.

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!

In particular ℕ_p has cardinality ℵ₀
> and
> |∩{E(k) : k ∈ ℕ_p}| = 0.

*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.

Regards, WM

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.
> >
> > But you should be able to note that infinite inclusion-monotonic sets would not have an empty intersection.
> Nope, this is trivially false for a set with no miminum.

The minimum is infinite for infinite endsegments. It is larger than any finite number of natural numbers. Therefore there are infinitely many natural numbers inside.

Regards, WM