fredag 8 oktober 2021 kl. 12:29:50 UTC+2 skrev WM:
> Serg io schrieb am Donnerstag, 7. Oktober 2021 um 22:52:06 UTC+2:
> > On 10/7/2021 5:25 AM, WM wrote:
> > > zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:05:20 UTC+2:
> > >
> > >> False, the intersection of ALL endsegments is EMPTY
> > >
> > > Yes.
> > so you agree, the intersection of ALL endsegments is EMPTY.
> Yes, I said so above.
> > > But all infinite endsegments have an infinite intersection together.
> > Wrong. the intersection of ALL endsegments is EMPTY.
> Yes, but the intersection of all infinite endsegments ist infinite. It is not possible that a sequence of shrinking but always infinite contents has an empty intersection. Inclusion monotony is basic to mathematics.
> > > The counter-example would be an infinite endsegment which has not an infinite intersection with all other infinite endsegments. That can be reduced to two infinite endsegments without an infinite common intersection.
> > Just assume there is an element k, in the intersection of all endsegments, However this element k cannot be in the next endsegment E(k+1)
> > conflict, so k is not in the intersection of all.
> Then there are two endsegments not both containing k.
> >
> > try to read that slowly and let it sink in.
> I have seen this at first glance. If there is an empty intersection, then there are two endsegments having an empty intersection. The brainless trick using the intersection of sets like {a, b}, {b, c}, {c, a} is not compatible with inclusion monotony.
>
> Regards, WM

>It is not possible that a sequence of shrinking but always infinite contents has an empty intersection

False, it is very possible if no element is in all of the members.

>I have seen this at first glance. If there is an empty intersection, then there are two endsegments having an empty intersection. The brainless trick using the intersection of sets like {a, b}, {b, c}, {c, a} is not compatible with inclusion monotony.

It demonstrates that things can have an intersection with every member but the combined intersection is emtpy which is what happens here and your inclusion monotony does not change it.

zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:46:34 UTC+2:
> fredag 8 oktober 2021 kl. 12:18:43 UTC+2 skrev WM:

> > > Even finite intersections do, but infinite intersections do not.
> > All infinitely many infinite endsegments do. Fact is however that they are only a potentially infinite collection, because the complementary FISONs are also a potentially infinite collection: E(n) = ℕ \ {1, 2, 3, ..., n-1}.
> > > Of course, WM does not accept that, because he
> > does not despise mathematics. Mathematics is based on inclusion monotony. Should the intersection that is infinite up to every infinite endsegment
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > disappear suddenly? That is a foolish claim, in particular because
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> >
> > But only matheologians need this suddenness. Fact is that the natural numbers that above every definable n fill the set ℕ \ {1, 2, 3, ..., n} are dark. This removes all contradictions with mathematics.
> >
> there is no suddenness,

Wrong. ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
shows that the intersection for all endsegments is infinite. You cannot find any endsegment not covered by (*). Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:48:12 UTC+2:
> fredag 8 oktober 2021 kl. 12:29:50 UTC+2 skrev WM:

If there is an empty intersection, then there are two endsegments having an empty intersection. The brainless trick using the intersection of sets like {a, b}, {b, c}, {c, a} is not compatible with inclusion monotony.
> It demonstrates that things can have an intersection with every member but the combined intersection is emtpy which is what happens here

No that cannot happen here.

> and your inclusion monotony does not change it.

Of course it does. It excludes that elements can be acquired. Elements can only be lost. As lomg as not all elements are lost, they are in all predecessors too.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 7. Oktober 2021 um 19:06:17 UTC+2:
> torsdag 7 oktober 2021 kl. 11:19:55 UTC+2 skrev WM:
> >
> >(1) If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
> Empty assertion

Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.

Regards, WM

FromTheRafters schrieb am Donnerstag, 7. Oktober 2021 um 16:58:36 UTC+2:
> WM pretended :

> > If only nonempty
> > successors exist, then they all have numbers in common.
> They have a size in common.

This size is caused numbers, infinitely many of which they have in common.

> > (1) During the steps of the sequence (E(n)), every natural number n is lost
> > from the intersection ∩{E(1), E(2), ..., E(n)}. Therefore the intersections
> > become empty. (2) During the steps of the sequence (E(n)), every natural
> > number n is lost from the endsegment E(n). Therefore the endsegments become
> > empty.
> >
> > Matheologians accept (1) but don't accept (2) without any substantial reason.
> That is because you and the other matheologians haven't finished
> building the number line yet.

The numberline is ready:
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
Can you see ∀k ∈ ℕ?

Regards, WM

Gus Gassmann schrieb am Donnerstag, 7. Oktober 2021 um 16:11:06 UTC+2:
> On Thursday, 7 October 2021 at 10:10:02 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Donnerstag, 7. Oktober 2021 um 14:05:12 UTC+2:
> > > On Thursday, 7 October 2021 at 06:19:55 UTC-3, WM wrote:
> > > > If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
> > > By your "symmetry" argument you claimed/proved a few days ago that there is.
> > Of course there is a bijection. But no natural numbers are remaining such that all endsegments in this mapping could have infinite contents.
> OK. You made the two statements:
>
> 1. There is no bijection between the set |N and the set of endsegments.
> 2. Of course there is a bijection.
>
> Can you clarify for each statement: Is that in mathematics

1) is must be wrong, because 2) is correct in mathematics and matheology.
But that implies that not all endsegments are infinite.

Regards, WM

fredag 8 oktober 2021 kl. 12:53:00 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:46:34 UTC+2:
> > fredag 8 oktober 2021 kl. 12:18:43 UTC+2 skrev WM:
>
> > > > Even finite intersections do, but infinite intersections do not.
> > > All infinitely many infinite endsegments do. Fact is however that they are only a potentially infinite collection, because the complementary FISONs are also a potentially infinite collection: E(n) = ℕ \ {1, 2, 3, ..., n-1}.
> > > > Of course, WM does not accept that, because he
> > > does not despise mathematics. Mathematics is based on inclusion monotony. Should the intersection that is infinite up to every infinite endsegment
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > > disappear suddenly? That is a foolish claim, in particular because
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
> > >
> > > But only matheologians need this suddenness. Fact is that the natural numbers that above every definable n fill the set ℕ \ {1, 2, 3, ...., n} are dark. This removes all contradictions with mathematics.
> > >
> > there is no suddenness,
>
> Wrong. ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> shows that the intersection for all endsegments is infinite. You cannot find any endsegment not covered by (*). Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
>
> Regards, WM

>Wrong.

I am correct, unlike you.

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

does not mean what you want it to mean.

>shows that the intersection for all endsegments is infinite.

False, it shows that any finite intersection gives a set with infinite cardinality.

it shows nothing about the intersection of ALL endsegments.

>You cannot find any endsegment not covered by (*).

Nor do I need to.

>Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.

There is nothing sudden there.

fredag 8 oktober 2021 kl. 12:55:48 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:48:12 UTC+2:
> > fredag 8 oktober 2021 kl. 12:29:50 UTC+2 skrev WM:
>
> If there is an empty intersection, then there are two endsegments having an empty intersection. The brainless trick using the intersection of sets like {a, b}, {b, c}, {c, a} is not compatible with inclusion monotony.
> > It demonstrates that things can have an intersection with every member but the combined intersection is emtpy which is what happens here
> No that cannot happen here.
> > and your inclusion monotony does not change it.
> Of course it does. It excludes that elements can be acquired. Elements can only be lost. As lomg as not all elements are lost, they are in all predecessors too.
>
> Regards, WM

It does not change the fact that the intersection is empty.

And yes, it can happen here because the intersection of all endsegments IS empty and it is extremely easy to prove it and you are an idiot for not getting it

fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 7. Oktober 2021 um 19:06:17 UTC+2:
> > torsdag 7 oktober 2021 kl. 11:19:55 UTC+2 skrev WM:
> > >
> > >(1) If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
> > Empty assertion
> Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
>
> Regards, WM

False, your stated facts do not support your claims, that is why it is empty.

ANd htat is yet anotehr empty assertion of yours with no proof.

fredag 8 oktober 2021 kl. 13:05:24 UTC+2 skrev WM:
> Gus Gassmann schrieb am Donnerstag, 7. Oktober 2021 um 16:11:06 UTC+2:
> > On Thursday, 7 October 2021 at 10:10:02 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Donnerstag, 7. Oktober 2021 um 14:05:12 UTC+2:
> > > > On Thursday, 7 October 2021 at 06:19:55 UTC-3, WM wrote:
> > > > > If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
> > > > By your "symmetry" argument you claimed/proved a few days ago that there is.
> > > Of course there is a bijection. But no natural numbers are remaining such that all endsegments in this mapping could have infinite contents.
> > OK. You made the two statements:
> >
> > 1. There is no bijection between the set |N and the set of endsegments.
> > 2. Of course there is a bijection.
> >
> > Can you clarify for each statement: Is that in mathematics
> 1) is must be wrong, because 2) is correct in mathematics and matheology.
> But that implies that not all endsegments are infinite.
>
> Regards, WM
Here we see your flawed logic and inability to use logic.

2 does not in anyway imply that

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:48:06 UTC+2:
> On Thursday, October 7, 2021 at 3:03:22 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:44:51 UTC+2:
> Let n e IN.
> > > Hint: n has exactly the position n "on the ordinal line".
> > >
> > Alas there is no position n [...]
>
> Sure there is:
>
> 1, ..., n, ...

n > 1? n e IN ? Has 1 lost its state as a natural number after it has been degraded as a prime number?
>
Regards, WM

On Friday, October 8, 2021 at 1:48:41 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:48:06 UTC+2:
Let n e IN.

> > 1, ..., n, ...
> >
> n > 1 ?

No, dumbo.

When we write, say,

{1, ..., n}

in math, this does not exclude n to be 1.

FUCK OFF, MÜCKENHEIM!

You are too dumb for any rational discussion.

EOD

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:53:05 UTC+2:
> On Thursday, October 7, 2021 at 3:02:25 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:42:06 UTC+2:
> > > On Thursday, October 7, 2021 at 11:27:46 AM UTC+2, WM wrote:
> Let n e IN.
> > > > I can place n at 1, at 2, at 3, at any place I like.
> > > >
> > > No, you can't.
> Hint: It's a mathematical no-go to define a constant two times in one and the same context.

f(n) = n^2

must be restricted to one value? Which one?

Regards, WM

Jim Burns schrieb am Freitag, 8. Oktober 2021 um 12:18:06 UTC+2:
> On 10/7/2021 5:49 AM, WM wrote:
> > Jim Burns schrieb
> > am Dienstag, 5. Oktober 2021 um 20:57:40 UTC+2:
> >> On 10/5/2021 4:35 AM, WM wrote:
>
> >> I think that you (WM) claim that potentially infinite
> >> collections _change_
> >
> > It is the definition:
> Then the collection of all and only the FISON-enders
> is NOT _potentially infinite_

It is. All natural numbers have the smallest transfinite cardinality ℵo.
All FISONs (containing all FISON-enders) have smaller cardinality.
>
> My reason for accepting the collection of all and only
> the FISON-enders as "potentially infinite" was that we would
> understand that to mean "totally ordered, does NOT have
> both a first and a last, and each split has a crossing-pair".
>
> However, a FISON-ender is never not a FISON-ender, and
> a not-FISON-ender is never a FISON-ender. If change is part of
> the definition of a potentially infinite collection, then
> the collection of FISON-enders is not potentially infinite.

The definition is this: No largest element, but cardinality smaller than ℵo.
>
> We can reason about each member of the collection of FISON-enders,
> and calling their collection potentially infinite or denying it's
> potentially infinite won't change our reasoning, since we're
> reasoning from the existence, for k, of a totally-ordered {0,...,k}
> from 0 to k in which each split has a crossing-pair j,j+1.

It is like Cantor's example: "potentially infinite ... a finite in the process of change having in each of its current states a finite size; like, for instance, the temporal duration since the beginning of the world, which, when measured in some time-unit, for instance a year, is finite in every moment, but always growing beyond all finite limits, without ever becoming really infinitely large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]" [Cantor]

> >> However, the collection of all and only FISON-enders
> >> _does not change_ A FISON exists which k ends, or
> >> a FISON does NOT exist which k ends, and which it is
> >> does not change.
> >
> > That is impossible. All FISONs are finite. None is infinite.
> > Would they not change, then there was a largest finite FISON.
> And yet, each is finite[1], they do not change,

No FISON changes

> and there is
> no largest FISON.

Therefore the position "last one" does change.

> But that's _your_ problem, not ours.

That is not a problem but a fact. The limit of FISONs is {1, 2, 3, ...}. But every FISON is much, much smaller.

> Each FISON is finite.
> Each FISON does not change.
> Each FISON is followed by more FISONs.
> Denying these claims leads to contradictions.

Correct. But finally: Between all FISONs and |N there is a huge gap of dark numbers.

Regards, WM

On 10/8/2021 5:58 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 7. Oktober 2021 um 19:06:17 UTC+2:
>> torsdag 7 oktober 2021 kl. 11:19:55 UTC+2 skrev WM:
>>>
>>> (1) If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
>> Empty assertion
>
> Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
>
> Regards, WM
>

that is silly. Math is just not your thing.

On 10/8/2021 6:54 AM, Greg Cunt wrote:
> On Friday, October 8, 2021 at 1:48:41 PM UTC+2, WM wrote:
>> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:48:06 UTC+2:
> Let n e IN.
>
>>> 1, ..., n, ...
>>>
>> n > 1 ?
>
> No, dumbo.
>
> When we write, say,
>
> {1, ..., n}
>
> in math, this does not exclude n to be 1.
>
> FUCK OFF, MÜCKENHEIM!
>
> You are too dumb for any rational discussion.
>
> EOD
>

WM explores the areas for not (dark) rational discussion.

On 10/8/2021 8:23 AM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:53:05 UTC+2:
>> On Thursday, October 7, 2021 at 3:02:25 PM UTC+2, WM wrote:
>>> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:42:06 UTC+2:
>>>> On Thursday, October 7, 2021 at 11:27:46 AM UTC+2, WM wrote:
>> Let n e IN.
>>>>> I can place n at 1, at 2, at 3, at any place I like.
>>>>>
>>>> No, you can't.
>> Hint: It's a mathematical no-go to define a constant two times in one and the same context.
>
> f(n) = n^2
>
> must be restricted to one value? Which one?
>
> Regards, WM
>

f(n) is a function, dependent upon n .

This is very fundamental and clear to all mathematicians, and high school students.

why are you stating it must be restricted to one value ? Please clarify.

On 10/8/2021 8:35 AM, WM wrote:
> Jim Burns schrieb am Freitag, 8. Oktober 2021 um 12:18:06 UTC+2:
>> On 10/7/2021 5:49 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 5. Oktober 2021 um 20:57:40 UTC+2:
>>>> On 10/5/2021 4:35 AM, WM wrote:
>>
>>>> I think that you (WM) claim that potentially infinite
>>>> collections _change_
>>>
>>> It is the definition:
>> Then the collection of all and only the FISON-enders
>> is NOT _potentially infinite_
>
> It is. All natural numbers have the smallest transfinite cardinality ℵo.

no. Each natural number has a cardinality of 1. Fail.

On 10/8/2021 5:40 AM, WM wrote:
> Serg io schrieb am Donnerstag, 7. Oktober 2021 um 19:49:54 UTC+2:
>> On 10/7/2021 5:23 AM, WM wrote:
>
>> Can you show inclusion monotony must apply ?
>
> By the definition of endsegment they never acquire elements but only lose elements. That shows that all elements of an endsegment are in all preceding endsegments. That is inclusion monotony.

nope. endsegments are a fixed set of numbers, they do not lose or acquire elements ever.

This is a trip up point for you, a miss thinking.

>
>>> The counter-example would be an infinite endsegment
>>> which has not an infinite intersection with all other infinite endsegments.
>>> That can be reduced to two infinite endsegments without an infinite common intersection.
>> which two ?
>
> They must be found by those who claim an empty intersection of infinite endsegments.

you are the one stating "That can be reduced to two infinite endsegments without an infinite common intersection."

Which two ?

>
> Regards, WM
>

On 10/8/2021 5:52 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:46:34 UTC+2:
>> fredag 8 oktober 2021 kl. 12:18:43 UTC+2 skrev WM:
>
>>>> Even finite intersections do, but infinite intersections do not.
>>> All infinitely many infinite endsegments do. Fact is however that they are only a potentially infinite collection, because the complementary FISONs are also a potentially infinite collection: E(n) = ℕ \ {1, 2, 3, ..., n-1}.
>>>> Of course, WM does not accept that, because he
>>> does not despise mathematics. Mathematics is based on inclusion monotony. Should the intersection that is infinite up to every infinite endsegment
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
>>> disappear suddenly? That is a foolish claim, in particular because
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>>>
>>> But only matheologians need this suddenness. Fact is that the natural numbers that above every definable n fill the set ℕ \ {1, 2, 3, ..., n} are dark. This removes all contradictions with mathematics.
>>>
>> there is no suddenness,
>
> Wrong. ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> shows that the intersection for all endsegments is infinite. You cannot find any endsegment not covered by (*). Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
>
> Regards, WM
>

wrong. Try again.

On Friday, October 8, 2021 at 3:53:54 PM UTC+2, Serg io wrote:
> On 10/8/2021 8:23 AM, WM wrote:
> >
> > f(n) = n^2
> >
> > must be restricted to one value? Which one?
> >
> f(n) is a function, dependent upon n .

Moreover in THIS context "n" would be a variable.

NO one (except WM) would first state

Let n e IN. And then define f with

f(n) = n^2 .

Actually, we would rather state, say,

f(m) = m^2 for all m e IN .

You see, "m" is a bound variable here.

WM ist just too dumb for ANY math. He does not even have a command of the basics.

____________________________

Of course, the following might be part of a proof:

Let n e IN. Now we consider the function f: IN ---> IN defined with f(m) = m^2 for all m e IN. Then we get especially f(n) = n^2 and f(1) = 1^2 = 1. Hence <etc.>

WM is too dumb to get that we may substitute "m" in "f(m)" and "m^2" by "n" JUST BECAUSE n is a natural number (i.e. n e IN). (Same with "1", which denotes the natural number 1 btw.)

On Friday, October 8, 2021 at 3:55:37 PM UTC+2, Serg io wrote:
> On 10/8/2021 8:35 AM, WM wrote:

> All natural numbers have the smallest transfinite cardinality ℵo.

Nonsense.

THE SET of all natural numbers have the smallest transfinite cardinality ℵo.

> no. Each natural number has a cardinality of 1.

I guess you meant the singleton sets, containing natural numbers: {0}, {1}, {2}, ...

If IN is defined tue to von Neumann, we would have

An e IN: card(n) = n ,

since then 0 = {}

and An e IN \ {0}: n = {0, ..., n-1} .

> Fail.

zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
> fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:

> > Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.

> ANd htat is yet anotehr empty assertion of yours with no proof.

The contents of endsegments is what is not used to index its predecessors.
If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.

Regards, WM

Greg Cunt wrote on 10/8/2021 :
> On Friday, October 8, 2021 at 1:48:41 PM UTC+2, WM wrote:
>> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 15:48:06 UTC+2:
> Let n e IN.
>
>>> 1, ..., n, ...
>>>
>> n > 1 ?
>
> No, dumbo.
>
> When we write, say,
>
> {1, ..., n}
>
> in math, this does not exclude n to be 1.
>
> FUCK OFF, MÜCKENHEIM!
>
> You are too dumb for any rational discussion.
>
> EOD

A hslf-wit should be perfect for such a discussion.

WM formulated the question :

> The definition is this: No largest element, but cardinality smaller than ℵo.

That's a good definition for muckymath stuff.