On Monday, 11 October 2021 at 11:30:49 UTC-3, Greg Cunt wrote:
> On Monday, October 11, 2021 at 4:10:15 PM UTC+2, Gus Gassmann wrote:
> > On Sunday, 10 October 2021 at 17:37:09 UTC-3, WM wrote:
> > >
> > > [...] Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo.
> > >
> > Bullshit. {1, 3, 5, 7, ..., 2, 4, 6, 8,... } is well-ordering of the natural numbers of order type ℵo + ℵo.
> I guess you meant omega + omega here.

Touche'!

> Anyway, WM's claim "ℵo + ℵo is larger than ℵo" is clearly bullshit, since ℵo + ℵo = ℵo-
>
> Without any doubt:
> > You clearly have no idea about infinity, well-orders, the natural numbers, set, or really *ANYTHING* that has to do with mathematics. You are a demented fool.

FromTheRafters schrieb am Sonntag, 10. Oktober 2021 um 23:40:55 UTC+2:
> WM brought next idea :

> > I am not interested in the intersection. You said that all endsegments are
> > infinite. So they all contain natnumbers, don't they?
> No, they are all proper subsets of the natural numbers.

Don't such subsets contain natnumbers?

Regards, WM

zelos...@gmail.com schrieb am Montag, 11. Oktober 2021 um 06:53:46 UTC+2:

> It proves that ALL natural numbers are missing in some of the endsegments,

All natnumbers cannot be missing unless all are missing together in an empty endsegment.

> that is all it needs to do to show the intersection is emtpy!

All infinite endegments contain infinitely many natnumbers infinitely many of which are in all infinite endsegments. Otherwise there were one or more infinite endsegments not sharing infinitely many natnumbers with all other infinite endsegments.

Regards, WM

Greg Cunt schrieb am Montag, 11. Oktober 2021 um 14:21:35 UTC+2:
> On Monday, October 11, 2021 at 4:07:13 AM UTC+2, Serg io wrote:
> > On 10/10/2021 3:33 PM, WM wrote bullshit.
> >
> > The topic is "the intersection of all endsegments is empty" you said it was infinite,
> > then you agreed it was empty, then you said it was infinite, make up your mind or [...]
>
> Well. it depends on the day of week and/or time of day.

Np, I said that all infinite endsegments have an infinite intersections while all endsegments have an empty intersection.

All infinite endegments contain infinitely many natnumbers infinitely many of which are in all infinite endsegments. Otherwise there were one or more infinite endsegments not sharing infinitely many natnumbers with all other infinite endsegments.

Regards, WM

WM expressed precisely :
> FromTheRafters schrieb am Sonntag, 10. Oktober 2021 um 23:40:55 UTC+2:
>> WM brought next idea :
>
>>> I am not interested in the intersection. You said that all endsegments are
>>> infinite. So they all contain natnumbers, don't they?
>> No, they are all proper subsets of the natural numbers.
>
> Don't such subsets contain natnumbers?

I don't know.

{1, 2, 3, 4, 5, 6, 7,...16}

Contains some natural numbers.

{17, 18, 19,...}

Does '19' connect back to the initial element (one or zero) through
finite iterations of the predecessor function?

Gus Gassmann schrieb am Montag, 11. Oktober 2021 um 16:10:15 UTC+2:
> On Sunday, 10 October 2021 at 17:37:09 UTC-3, WM wrote:
> > [...] Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo.
>
> Bullshit. {1, 3, 5, 7, ..., 2, 4, 6, 8,... } is well-ordering of the natural numbers of order type ℵo + ℵo.

Order type is not ℵo but ω.

Two consecutive sets in ω cannot have cardinality ℵo both.
If all endsegments have card ℵo then the set of their indices is finite.

Regarda, WM

Greg Cunt schrieb am Montag, 11. Oktober 2021 um 16:30:49 UTC+2:

> Anyway, WM's claim "ℵo + ℵo is larger than ℵo" is clearly bullshit, since ℵo + ℵo = ℵo-

Two consecutive sets in ω cannot have cardinality ℵo both.
If all endsegments have card ℵo then the set of their indices is finite.

Regards, WM

zelos...@gmail.com schrieb am Montag, 11. Oktober 2021 um 06:56:10 UTC+2:
> söndag 10 oktober 2021 kl. 19:46:46 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
> > > fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
> >
> > > >The contents of endsegments is what is not used to index its predecessors.
> > > Irrelevant.
> > Chuckle. Your only way to pretend ignorance?
> > > >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.
> > > Meaningless empty assertion.
> > Meaningful and very important for mathematicians who can think.
>
> >Chuckle. Your only way to pretend ignorance?
> Going like Gerbil now?

Don't know. Who is Gerbil?

> What I am saying is that it is IRRELEVANT

Two consecutive sets in ω cannot have cardinality ℵo both.
If all endsegments have card ℵo then the set of their indices is finite.

Regards, WM

Gus Gassmann schrieb am Sonntag, 10. Oktober 2021 um 22:34:03 UTC+2:
> On Sunday, 10 October 2021 at 14:25:09 UTC-3, WM wrote:
> [...]
> > The contents of an endsegment cannot be used to index its predecessors:
> >
> > E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
> > E(1), E(2) = {2, 3, 4, 5, 6, ...}
> > E(1), E(2), E(3) = {3, 4, 5, 6, ...}
> > E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
> > The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).
> >
> > If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.
> Of course, *NO* end segment has more than finitely many predecessors.

But all infinite endsegment have not more than finitely many infinite successors, because aleph_0 natnumbers are remaining as contents and cannot become indices.

Regards, WM

On 10/11/2021 2:10 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 10. Oktober 2021 um 22:34:03 UTC+2:
>> On Sunday, 10 October 2021 at 14:25:09 UTC-3, WM wrote:
>> [...]
>>> The contents of an endsegment cannot be used to index its predecessors:
>>>
>>> E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
>>> E(1), E(2) = {2, 3, 4, 5, 6, ...}
>>> E(1), E(2), E(3) = {3, 4, 5, 6, ...}
>>> E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
>>> The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).
>>>
>>> If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.
>> Of course, *NO* end segment has more than finitely many predecessors.
>
> But all infinite endsegment

all endsegments are by definition infinite, MR LAMO.

> have not more than finitely many infinite successors,

what does "not more than finitely many infinite" mean ? IT smells like BS

> because aleph_0 natnumbers are remaining as contents and cannot become indices.

that is silly, you restrict yourself to one set of numbers for everything, just stupid.

>
> Regards, WM
>

On 10/11/2021 1:58 PM, WM wrote:
> Greg Cunt schrieb am Montag, 11. Oktober 2021 um 16:30:49 UTC+2:
>
>> Anyway, WM's claim "ℵo + ℵo is larger than ℵo" is clearly bullshit, since ℵo + ℵo = ℵo-
>
> Two consecutive sets in ω cannot have cardinality ℵo both.

wrong, sure they can.

> If all endsegments have card ℵo then the set of their indices is finite.

wrong again!

>
> Regards, WM
>

Why you always wrong ?

On Monday, 11 October 2021 at 15:57:57 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Oktober 2021 um 16:10:15 UTC+2:
> > On Sunday, 10 October 2021 at 17:37:09 UTC-3, WM wrote:
> > > [...] Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo.
> >
> > Bullshit. {1, 3, 5, 7, ..., 2, 4, 6, 8,... } is well-ordering of the natural numbers of order type ℵo + ℵo.
> Order type is not ℵo but ω.

Yes. Greg C pointed that out already.

> Two consecutive sets in ω cannot have cardinality ℵo both.
> If all endsegments have card ℵo then the set of their indices is finite.

That is still gibberish. The set of *ALL* indices never precedes any one end segment. You are therefore again switching quantifiers around to suit your nefarious purposes. You now *SHIT* about mathematics!

On Monday, 11 October 2021 at 16:10:20 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 10. Oktober 2021 um 22:34:03 UTC+2:
> > On Sunday, 10 October 2021 at 14:25:09 UTC-3, WM wrote:
> > [...]
> > > The contents of an endsegment cannot be used to index its predecessors:
> > >
> > > E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
> > > E(1), E(2) = {2, 3, 4, 5, 6, ...}
> > > E(1), E(2), E(3) = {3, 4, 5, 6, ...}
> > > E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
> > > The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).
> > >
> > > If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.
> > Of course, *NO* end segment has more than finitely many predecessors.
> But all infinite endsegment have not more than finitely many infinite successors, because aleph_0 natnumbers are remaining as contents and cannot become indices.

Garbled nonsense, handwaving and quantifier dyslexia. I'd call that the Muckenheim trifecta.

On Monday, October 11, 2021 at 8:58:49 PM UTC+2, WM wrote:

> If all endsegments have card ℵo then the set of their indices is finite.

No, you silly crank. The index set of the sequence (E_n)_(n e IN) is infinite (since IN is infinite).

Still card(E_n) = aleph_0 BY DEFINITION, since E_n := {m e IN : m >= n} for all n e IN.

Why are you always wrong, Mückenheim?

On Monday, October 11, 2021 at 8:58:49 PM UTC+2, WM wrote:

> If all endsegments have card ℵo then the set of their indices is finite.

No, you silly crank. The index set of the sequence (E_n)_(n e IN) is infinite (since IN is infinite).

Still for all n e IN: card(E_n) = aleph_0 BY DEFINITION, since E_n := {m e IN : m >= n} for all n e IN.

Why are you always wrong, Mückenheim?

On Monday, 11 October 2021 at 20:02:44 UTC-3, Greg Cunt wrote:
> On Monday, October 11, 2021 at 8:58:49 PM UTC+2, WM wrote:
>
> > If all endsegments have card ℵo then the set of their indices is finite.
> No, you silly crank. The index set of the sequence (E_n)_(n e IN) is infinite (since IN is infinite).
>
> Still for all n e IN: card(E_n) = aleph_0 BY DEFINITION, since E_n := {m e IN : m >= n} for all n e IN.
>
> Why are you always wrong, Mückenheim?

That's axiomatic. In that regard he is very much like Smullyan's red knight in "Alice in Puzzle Land": He believes *exactly* those things that are wrong. (And with such zeal and fervor, too.)

On 10/11/2021 2:06 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 11. Oktober 2021 um 06:56:10 UTC+2:
>> söndag 10 oktober 2021 kl. 19:46:46 UTC+2 skrev WM:
>>> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
>>>> fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
>>>
>>>>> The contents of endsegments is what is not used to index its predecessors.
>>>> Irrelevant.
>>> Chuckle. Your only way to pretend ignorance?
>>>>> 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.
>>>> Meaningless empty assertion.
>>> Meaningful and very important for mathematicians who can think.
>>
>>> Chuckle. Your only way to pretend ignorance?
>> Going like Gerbil now?
>
> Don't know. Who is Gerbil?
>
>> What I am saying is that it is IRRELEVANT
>
> Two consecutive sets in ω cannot have cardinality ℵo both.

totally wrong, that marble is gone!

HINT: E(1) and E(2) are two sets and both have Card ℵo

> If all endsegments have card ℵo then the set of their indices is finite.

Totally wrong, another marble gone!

>
> Regards, WM
>
>

as we age, sometimes math gets just too hard, the equations are not fluid anymore, the chalk on the board has difficulty with the intergral signs, too
many mistakes too...

On 10/11/2021 1:49 PM, FromTheRafters wrote:
> WM expressed precisely :
>> FromTheRafters schrieb am Sonntag, 10. Oktober 2021 um 23:40:55 UTC+2:
>>> WM brought next idea :
>>
>>>> I am not interested in the intersection. You said that all endsegments are infinite. So they all contain natnumbers, don't they?
>>> No, they are all proper subsets of the natural numbers.
>>
>> Don't such subsets contain natnumbers?
>
> I don't know.
>
> {1, 2, 3, 4, 5, 6, 7,...16}
>
> Contains some natural numbers.
>
> {17, 18, 19,...}
>
> Does '19' connect back to the initial element (one or zero) through finite iterations of the predecessor function?

One can back up their set to the dock and get containers of natnumbers.

Also you can buy them on eBay.

this is for a set of natural numbers used for painting;

https://www.ebay.com/b/Vintage-Paint-By-Number/10900/bn_7023240329?mkevt=1&mkcid=1&mkrid=711-53200-19255-0&campid=5337590776&customid=&toolid=10001

I like the HEX painting sets myself.

On 10/11/2021 1:46 PM, WM wrote:
> Greg Cunt schrieb am Montag, 11. Oktober 2021 um 14:21:35 UTC+2:
>> On Monday, October 11, 2021 at 4:07:13 AM UTC+2, Serg io wrote:
>>> On 10/10/2021 3:33 PM, WM wrote bullshit.
>>>
>>> The topic is "the intersection of all endsegments is empty" you said it was infinite,
>>> then you agreed it was empty, then you said it was infinite, make up your mind or [...]
>>
>> Well. it depends on the day of week and/or time of day.
>
> Np, I said that all infinite endsegments have an infinite intersections while all endsegments have an empty intersection.
>
> All infinite endegments contain infinitely many natnumbers infinitely many of which are in all infinite endsegments. Otherwise there were one or more infinite endsegments not sharing infinitely many natnumbers with all other infinite endsegments.
>
> Regards, WM
>

lets clarify,

let
A = infinite
B = endsegment(s)
C = Intersection
D = natnumbers

here is translation;

"Np, I said that all A B have an A C while all B have an empty C."

"All A B contain A many D A many of which are in all A B. Otherwise there were one or more A B not sharing A many D with all other A B."

Much Better!!

måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 11. Oktober 2021 um 06:53:46 UTC+2:
>
> > It proves that ALL natural numbers are missing in some of the endsegments,
> All natnumbers cannot be missing unless all are missing together in an empty endsegment.
> > that is all it needs to do to show the intersection is emtpy!
> All infinite endegments contain infinitely many natnumbers infinitely many of which are in all infinite endsegments. Otherwise there were one or more infinite endsegments not sharing infinitely many natnumbers with all other infinite endsegments.
>
> Regards, WM

>All natnumbers cannot be missing unless all are missing together in an empty endsegment.

Each and everyone is missing in some endsegment, not in the same endsegment you fucking imbecile.

for any natural number k, it is not in E(k+1)

ergo it is missing!

This applies to EACH AND EVERY naturalnumber!

>All infinite endegments contain infinitely many natnumbers infinitely many of which are in all infinite endsegments.

False, NO element is in ALL endsegment, that is why the intersection is emtpy. This is an empty assertion of yours.

>Otherwise there were one or more infinite endsegments not sharing infinitely many natnumbers with all other infinite endsegments.

No endsegment share all with the rest

måndag 11 oktober 2021 kl. 21:06:55 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 11. Oktober 2021 um 06:56:10 UTC+2:
> > söndag 10 oktober 2021 kl. 19:46:46 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
> > > > fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
> > >
> > > > >The contents of endsegments is what is not used to index its predecessors.
> > > > Irrelevant.
> > > Chuckle. Your only way to pretend ignorance?
> > > > >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.
> > > > Meaningless empty assertion.
> > > Meaningful and very important for mathematicians who can think.
> >
> > >Chuckle. Your only way to pretend ignorance?
> > Going like Gerbil now?
> Don't know. Who is Gerbil?
> > What I am saying is that it is IRRELEVANT
> Two consecutive sets in ω cannot have cardinality ℵo both.
> If all endsegments have card ℵo then the set of their indices is finite.
>
> Regards, WM
John Gabriel.

No one says that you dishonest shit.

zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 06:59:24 UTC+2:
> måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:

> No endsegment share all with the rest

All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 07:00:25 UTC+2:
> måndag 11 oktober 2021 kl. 21:06:55 UTC+2 skrev WM:

> > > What I am saying is that it is IRRELEVANT
> > Two consecutive sets in ω cannot have cardinality ℵo both.
> > If all endsegments have card ℵo then the set of their indices is finite.

> No one says that you dishonest shit.

But it is fact. omega cannot be divided in two consecutive infinite sets. If it is divided, then the stop is at a finite position.

Regards, WM

onsdag 13 oktober 2021 kl. 13:15:14 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 07:00:25 UTC+2:
> > måndag 11 oktober 2021 kl. 21:06:55 UTC+2 skrev WM:
>
> > > > What I am saying is that it is IRRELEVANT
> > > Two consecutive sets in ω cannot have cardinality ℵo both.
> > > If all endsegments have card ℵo then the set of their indices is finite.
> > No one says that you dishonest shit.
> But it is fact. omega cannot be divided in two consecutive infinite sets. If it is divided, then the stop is at a finite position.
>
> Regards, WM
no one says that again, you are making it up on your own you dishonest crank.

onsdag 13 oktober 2021 kl. 13:11:34 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 06:59:24 UTC+2:
> > måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:
>
> > No endsegment share all with the rest
> All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
>
> Regards, WM
incorrect, they share NOTHING with all of them. No matter what element you say is in all of them, I can find an endsegment missing it.

That is why the intersection is empty