FromTheRafters schrieb am Donnerstag, 7. Oktober 2021 um 14:29:38 UTC+2:
> WM formulated on Thursday :
> > zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
> >>> Everyone who says that all natural numbers are used up as indices of
> >>> endsegments but infinitely many are remaining as contents of all
> >>> endsegments (which used up all already) say that there is an infinite set
> >>> |N (of indices) which is followed by an infinite set |N (of contents)..
> >> This is pure nonsense, no one says anything like this
> >
> > They claim that there is a bijection between the set |N and the set of
> > endsegments. That means that all natural numbers n are used, they are mapped
> > on endsegments E(n). None is remaining.
> >
> >>> Every endsegment has infinite contents. But all endsegments exhaust |N.
> >> Doesn't change the fact that the intersection is empty because no element is
> >> in all sets.
> >
> > Try a bit of logic!
> > (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. (2) Every infinite endsegment has an
> > infinite intersection with all other infinite endsegments. This is proved by
> > inclusion monotony.
> Show us what inclusion monotony has to do with your fantasy.

Every natnumber of an endsegment is in all predecessors. If only nonempty successors exist, then they all have numbers 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.

Regards, WM

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 you silly crank:

1, ..., n, ...

EOD

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.

Of course, how would you know?!

> <bullshit deleted>

EOD

On Thursday, October 7, 2021 at 3:48:06 PM UTC+2, Greg Cunt wrote:
> 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, you silly crank:
>
> 1, ..., n, ...

Using the usual language of set theory:

Ex e ORD: x = n

is a simple theorem in the present context.

> EOD

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 or myckemyth?

WM pretended :
> FromTheRafters schrieb am Donnerstag, 7. Oktober 2021 um 14:29:38 UTC+2:
>> WM formulated on Thursday :
>>> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
>>>>> Everyone who says that all natural numbers are used up as indices of
>>>>> endsegments but infinitely many are remaining as contents of all
>>>>> endsegments (which used up all already) say that there is an infinite set
>>>>>> N (of indices) which is followed by an infinite set |N (of contents).
>>>> This is pure nonsense, no one says anything like this
>>>
>>> They claim that there is a bijection between the set |N and the set of
>>> endsegments. That means that all natural numbers n are used, they are
>>> mapped on endsegments E(n). None is remaining.
>>>
>>>>> Every endsegment has infinite contents. But all endsegments exhaust |N.
>>>> Doesn't change the fact that the intersection is empty because no element
>>>> is in all sets.
>>>
>>> Try a bit of logic!
>>> (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. (2) Every infinite endsegment has an
>>> infinite intersection with all other infinite endsegments. This is proved
>>> by inclusion monotony.
>> Show us what inclusion monotony has to do with your fantasy.
>
> Every natnumber of an endsegment is in all predecessors.

Are there any natural numbers in an endsegment's set?

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

They have a size 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. That step-by-step process you are
obsessed with is a real PITA.

On 10/7/2021 4:19 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
>>> Everyone who says that all natural numbers are used up as indices of endsegments but infinitely many are remaining as contents of all endsegments (which used up all already) say that there is an infinite set |N (of indices) which is followed by an infinite set |N (of contents).
>> This is pure nonsense, no one says anything like this
>
> They claim that there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.

no worries, just buy another set of |N on amazon, they are very cheap

>
>>> Every endsegment has infinite contents. But all endsegments exhaust |N.
>> Doesn't change the fact that the intersection is empty because no element is in all sets.
>
> Try a bit of logic!
> (1) If every endsegment has infinite contents,

that is pointless, every endsegment is infinite.

> then infinitely many natural numbers cannot be mapped on endsegments.

does not follow at all.

>There is no bijection between the set |N and the set of endsegments.

wrong.

On 10/7/2021 8:06 AM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:53:29 UTC+2:
>> On Thursday, October 7, 2021 at 11:19:55 AM UTC+2, WM wrote:
>>
>> Let's assume that
>>> there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.

just recycle them, using them once is wasteful

>> Right. Hence now the set IN would be
>
> completely used. No natnumber remains after all used ones. Otherwise it would no be a mapping from |N to the set of endsegments. Links total rechts eindeutig.
>
> Regards, WM
>

Used Ants

Completely used Ants

On 10/7/2021 7:29 AM, FromTheRafters wrote:
> WM formulated on Thursday :
>> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
>>>> Everyone who says that all natural numbers are used up as indices of endsegments but infinitely many are remaining as contents of all endsegments
>>>> (which used up all already) say that there is an infinite set |N (of indices) which is followed by an infinite set |N (of contents).
>>> This is pure nonsense, no one says anything like this
>>
>> They claim that there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on
>> endsegments E(n). None is remaining.
>>
>>>> Every endsegment has infinite contents. But all endsegments exhaust |N.
>>> Doesn't change the fact that the intersection is empty because no element is in all sets.
>>
>> Try a bit of logic!
>> (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. (2) Every infinite endsegment has an infinite intersection with all other infinite endsegments. This is proved by
>> inclusion monotony.
>
> Show us what inclusion monotony has to do with your fantasy.

Exclusion Polyphony Ants

torsdag 7 oktober 2021 kl. 11:19:55 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
> > >Everyone who says that all natural numbers are used up as indices of endsegments but infinitely many are remaining as contents of all endsegments (which used up all already) say that there is an infinite set |N (of indices) which is followed by an infinite set |N (of contents).
> > This is pure nonsense, no one says anything like this
> They claim that there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.
> > >Every endsegment has infinite contents. But all endsegments exhaust |N..
> > Doesn't change the fact that the intersection is empty because no element is in all sets.
> Try a bit of logic!
> (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.
> (2) Every infinite endsegment has an infinite intersection with all other infinite endsegments. This is proved by inclusion monotony.
>
> Regards, WM

>They claim that there is a bijection between the set |N and the set of endsegments.

Wchich there is

>That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.

Meaningless noise

>Try a bit of logic!

I use it! How about you fucking try some for once in your life?

>(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

>(2) Every infinite endsegment has an infinite intersection with all other infinite endsegments. This is proved by inclusion monotony.

Empty assertion

You failed doing anything logical here.

torsdag 7 oktober 2021 kl. 12:25:51 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:05:20 UTC+2:
>
> > False, the intersection of ALL endsegments is EMPTY
> Yes. But all infinite endsegments have an infinite intersection together. 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.
>
> Regards, WM
only finite ones do you moron

On 10/7/2021 5:23 AM, WM wrote:
> Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
>> On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
>>
>>> I know that
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
>>> in connection with
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> excludes
>>> Lim E(n) = { }
>> Great. Can you PROVE what you "know" (i.e. believe)?
>>
> The proof stands above. Further we have inclusion monotony. Try to learn and understand these things.

Above is not a proof. Can you show inclusion monotony must apply ? Try to learn and understand these things.

>
> All infinite endsegments

All endsegments ARE infinite by definition.

>have an infinite intersection together.

wrong, the intersection is empty.

> The counter-example would be an infinite endsegment

All endsegments ARE infinite by definition.

> which has not an infinite intersection with all other infinite endsegments.

easily done. What element is a member of ALL endsegments ? none.

> That can be reduced to two infinite endsegments without an infinite common intersection.

which two ?

>
> Regards, WM
>

On 10/7/2021 5:48 AM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:38:07 UTC+2:
>> On Thursday, October 7, 2021 at 12:23:19 PM UTC+2, WM wrote:
>>> Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
>>>> On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
>>>>>
>>>>> I know that
>>>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
>>>>> in connection with
>>>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>>>> excludes
>>>>> Lim E(n) = { }
>>>>>
>>>> Great. Can you PROVE what you "know" (i.e. believe)?
>>>>
>>> The proof stands above.
>> There is no proof,
>
> Try to learn to read.
>>
>> I asked for a PROOF for this claim.
>
> Since you cannot understand my proof, try this:
> (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.

nope. The intersection of a finite number of endsegments is infinite.

> (2) During the steps of the sequence, every natural number is lost from the endsegments. Therefore the endsegments become empty.

nope. The intersection of a finite number of endsegments is infinite.

>
> Matheologians accept (1) but don't accept (2) without any substantial reason.
>
> Regards, WM
>

you fail to understand that each endsegment is a fixed set of numbers and does not change.

there is no step by step process with intersection, or other set functions.

On 10/7/2021 5:18 AM, WM wrote:
> Serg io schrieb am Dienstag, 5. Oktober 2021 um 21:24:58 UTC+2:
>
>> ∩{E(1), E(2), ...} = { } is empty, not infinite.
>
> Yes. But presently only infinite endsegments are concerned. For them we have
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
>
> Regards, WM
>
>

so you agree that ∩{E(1), E(2), ...} = { } is empty, and not infinite.

stupider than a bag of rocks, zelos...@gmail.com wrote:

>> extreme crank zelos...@gmail.com wrote:
>>
>> > tisdag 5 oktober 2021 kl. 16:22:08 UTC+2 skrev Hilton Blome:
>> >> in america for instance, the CDC, a deeply evil private org, is
>> >> using Fluoridated water mostly in black communities, to keep their
>> >> IQ low. The same in Europe, ie in czech and polish communities. They
>> >> have concentration layers right now.
>> >
>> > are you insane or just mentally retarded?
>> you are a known fucking imbecile around here. Go fuck yourself other
>> places. Cretin.
>
> I am known to smack down cranks, so I take you're one of them

you just proved you are that imbecile, idiot. CDC is a private org with
patents in both viruses and vaccines, and are running pharmakia, you
fucking stupid. Educate yourself.

On 10/7/2021 5:10 AM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 18:38:41 UTC+2:
>> "Every end segment is not empty" is decidedly not the same thing as "there is a nonempty intersection"
>
> If you deny mathematics, you may believe that.

"Every end segment is not empty" => applies to endsegments.

"there is a nonempty intersection" => applies to intersections

they are not the same thing at all.

>
>> , your moronic application of inclusion monotony notwithstanding.
>
> A sequence of non-empty sets which never gain new elements but only lose elements, cannot have an empty intersection. That is basic mathematics.

applies only to finite sets. It does not apply to infinite sets

you do not know infinity.

>
>> Since there is no set of infinite cardinality common to all end segments
>
> All infinite endsegments have an infinite intersection.

wrong.

>Try to find a counter example,

sure, YOU name an element that is in the intersection of ALL each and every endsegment. FAIL.

> that is an infinite endsegment without an infinite intersection with all other infinite endsegments.

scrambled eggs

>
> Regards, WM
>

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.

> But all infinite endsegments have an infinite intersection together.

Wrong. the intersection of ALL endsegments is EMPTY.

> 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.

silly. 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.

try to read that slowly and let it sink in.

>
> Regards, WM
>

On Thursday, 7 October 2021 at 17:48:14 UTC-3, Serg io wrote:
> On 10/7/2021 5:10 AM, WM wrote:
[...]
> > All infinite endsegments have an infinite intersection.
> wrong.

Not even wrong. Intentionally ambiguous and misleading. Pairwise intersections of *any* two end segments have an intersection of cardinality aleph_0. Even finite intersections do, but infinite intersections do not. Of course, WM does not accept that, because he is too stupid to *learn* Cantor's work, and he doesn't want to simply memorize *this one* --- unlike *all* the other proofs he has ever written down.

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_

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.

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.

>> 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, and there is
no largest FISON. But that's _your_ problem, not ours.

Each member of only the FISONs is a FISON.
Each FISON is a member of all the FISONs.
If that were not so, that would be _our_ problem.
(I don't think this worries anyone.)

[1]
I am calling a totally-ordered collection _finite_ iff
the collection has a first and a last, and, for each split,
a crossing-pair exists. Clearly, each {0,...,k} described
above satisfies these conditions.

----
0 has a successor. Each successor has a successor.
Successors do not change.

We could take "Successors do not change" as part of the
initial description of them from which we'll reason, or
we could look at some widely-accepted example of 'successor',
such as for the decimal numerals.

| (numeral)# --> (numeral)+1
| | where
| 0# --> 1
| 1# --> 2
| 2# --> 3
| 3# --> 4
| 4# --> 5
| 5# --> 6
| 6# --> 7
| 7# --> 8
| 8# --> 9
| 9# --> #0

That successor does not change.

----
No FISON-ender can ever be a not-FISON-ender.
There would need to be a first j in {0,...,k}
that _will change_ but all i in {0,...,j-1} will not
change, and successors do not change, so j can't.
Contradiction.

No not-FISON-ender can ever be a FISON-ender.
There would need to be a first j in {0,...,k}
that _has changed_ but all i in {0,...,j-1} have not
changed, and successors do not change, so j can't.
Contradiction.

----
There is no largest FISON.

Assume k is a FISON-ender.
{0,...,k} exists.
k+1 exists.

Define
{0,...,k+1} = {0,...,k}U{k+1}

with the order '<' between i,j in {0,...,k}
inherited from [0,...,k}, and,
for each j in {0,...,k}, (j < k+1) and ~(k+1 < j)

{0,...,k+1} is totally-ordered, has a first 0 and a last k+1,
and, for each split, there is a crossing-pair j,j+1.

{0,...,k+1} is a FISON. k+1 is a FISON-ender.

For each FISON {0,...,k}, {0,...,k+1} is larger.

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

v>
>
> Regards, WM
>

Gus Gassmann schrieb am Freitag, 8. Oktober 2021 um 01:18:13 UTC+2:
> On Thursday, 7 October 2021 at 17:48:14 UTC-3, Serg io wrote:
> > On 10/7/2021 5:10 AM, WM wrote:
> [...]
> > > All infinite endsegments have an infinite intersection.
> > wrong.
> Intentionally ambiguous and misleading.

No, clear and obviously true.

> Pairwise intersections of *any* two end segments have an intersection of cardinality aleph_0.

We do not need pairwise intersections, but only all infinite endsegments. A sequence of decreasing endsegments with not smaller than infinite terms has an intersection of ℵo elements. That is basic mathematics.

> 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.

Regards, 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

Serg io schrieb am Donnerstag, 7. Oktober 2021 um 19:54:08 UTC+2:
> On 10/7/2021 5:48 AM, WM wrote:
> > (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.
> nope. The intersection of a finite number of endsegments is infinite.
> > (2) During the steps of the sequence, every natural number is lost from the endsegments. Therefore the endsegments become empty.
> nope. The intersection of a finite number of endsegments is infinite.
> >
> > Matheologians accept (1) but don't accept (2) without any substantial reason.
> >
> each endsegment is a fixed set of numbers and does not change.

But the terms of the sequence do: E(1), E(2), E(3), ...
>
> there is no step by step process with intersection, or other set functions.

The terms of the sequence are: E(1), E(1) ∩ E(2), E(1) ∩ E(2) ∩ E(3), ...

Regards, WM

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.

> > 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.

Regards, WM

torsdag 7 oktober 2021 kl. 21:33:11 UTC+2 skrev Darreld Focht:
> stupider than a bag of rocks, zelos...@gmail.com wrote:
>
> >> extreme crank zelos...@gmail.com wrote:
> >>
> >> > tisdag 5 oktober 2021 kl. 16:22:08 UTC+2 skrev Hilton Blome:
> >> >> in america for instance, the CDC, a deeply evil private org, is
> >> >> using Fluoridated water mostly in black communities, to keep their
> >> >> IQ low. The same in Europe, ie in czech and polish communities. They
> >> >> have concentration layers right now.
> >> >
> >> > are you insane or just mentally retarded?
> >> you are a known fucking imbecile around here. Go fuck yourself other
> >> places. Cretin.
> >
> > I am known to smack down cranks, so I take you're one of them
> you just proved you are that imbecile, idiot. CDC is a private org with
> patents in both viruses and vaccines, and are running pharmakia, you
> fucking stupid. Educate yourself.

I am clearly more educated than you when you are a conspiratard

fredag 8 oktober 2021 kl. 12:18:43 UTC+2 skrev WM:
> Gus Gassmann schrieb am Freitag, 8. Oktober 2021 um 01:18:13 UTC+2:
> > On Thursday, 7 October 2021 at 17:48:14 UTC-3, Serg io wrote:
> > > On 10/7/2021 5:10 AM, WM wrote:
> > [...]
> > > > All infinite endsegments have an infinite intersection.
> > > wrong.
> > Intentionally ambiguous and misleading.
>
> No, clear and obviously true.
> > Pairwise intersections of *any* two end segments have an intersection of cardinality aleph_0.
> We do not need pairwise intersections, but only all infinite endsegments. A sequence of decreasing endsegments with not smaller than infinite terms has an intersection of ℵo elements. That is basic mathematics.
> > 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.
>
> Regards, WM
there is no suddenness, and you DO despise mathematics because it doesn't conform to your insanity.