Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:42:53 UTC+2:
> On Tuesday, 5 October 2021 at 05:12:00 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:49:30 UTC+2:
> > > On Monday, 4 October 2021 at 18:23:38 UTC-3, WM wrote:
> >
> > > > Do you know any endsegment that is not belonging to one of the following sets?
> > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > > > No. Therefore all infinite endsegments have an infinite intersection. Inclusion monotony is always true.
> > > You don't know what inclusion monotony is
> > I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo.
> Nonsense. You think inclusion monotony for a finite number of sets and inclusion monotony for an in finite number of sets behave in the same way.

There is nothing that behaves. If an ednsegment contains a natural number, then this number is in all predecessors. Since there are no empty successors, every endsegment has at least one element in common with wll endsegments..

> > Do you know any endsegment that is not covered by (*)? This proves that all endsegments (which can be known) have infinite intersection.
> And here you go, slipping this innocent "can be known" into your sentence..

If all can be known, then this phrase is not evil, is it?

> You do this so casually -- and knowingly! -- to misdirect.

No, I do it in order to apply only those which are accepted in ZFC.

> > Look: Every infinite endsegment has an infinite intersection with itself and with all its predecessors. But you claim that there are no other endsegments.
> *YOU* claim that, nobody else.

You claim that all are infinite. So all have an infinite intersection with all its predecessors and with all its successors. Otherwise there would be a first one with less than infinitely many natnumbers.

> Of course there are infinitely many end segments that are not in these finite intersections that you are talking about.

Name only one that is not in any set covered by

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

> intersect {E(k): k < n} =/= intersect {E(k): k <= n}.

That is irrelevant since all are infinite.
>
> But every one of these intersections is over a finite set. To get intersect {E(k): k in |N} you have to use *LIMITS* (about which of course you have no clue).

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) = { }
for all infinite endsegments.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 15:23:56 UTC+2:
> tisdag 5 oktober 2021 kl. 10:06:49 UTC+2 skrev WM:

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

> >If no endsegment exists outside of (*), then all are covered by (*).
> It is true for all endsegments but that does not mean what you want.

It is true for all endsegments. That is what I claim.

> >From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on. If all endsegments have infinite contents, then infinite contents is in all of them.
> Correct but that doesn't mean that the infinite intesrection of all is not empty.

Just that is fact because the infinite intersection concerns only all endsegments for which the intersection is infinite.

> >I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo
> No one is surpressing anything. It just doesn't fucking mean what you want it to!

It means that all endsegments are infinite and have an infinite intersection. More is not claimed. The intersection is restricted to infinity by

∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Regards, WM

Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 15:34:51 UTC+2:
> On Tuesday, 5 October 2021 at 05:06:49 UTC-3, WM wrote:
> > From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on.
> From the fact that E(3) is not empty you try to deduce that E(3) is a subset of E(4), E(5), etc.???

Not this, but the other way round.

> > If all endsegments have infinite contents, then infinite contents is in all of them.
> This, of course, is equally untenable.

An infinite subset of all infinite endsegments is a subset of all infinite endsegments.

Show an infinite endsegment that has no infinite common contents with any other infinite endsegment.

Regards, WM

On 10/3/2021 7:54 AM, WM wrote:
> Serg io schrieb am Samstag, 2. Oktober 2021 um 18:58:21 UTC+2:
>
>> An Endsegment, E(k), is only shorthand for this E(k) = {k,k+1,k+2,...} a fixed (by k) infinite set of natural numbers that does not change.
>>
>> There are no empty endsegment.
>
> All existing infinite endsegments have an infinite intersection with each other.
>
> Regards, WM
>

Wrong again. Use Mathematics.

1 the intersection of a finite # of endsegments is infinite.

2 the intersection of all endsegments is null { } zero, nada, zip...

On 10/3/2021 7:51 AM, WM wrote:
> Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 18:13:59 UTC+2:
>> On Friday, October 1, 2021 at 8:04:37 PM UTC+2, WM wrote:
>>
>>> Why are the endsegments *not* empty?
>> Because they (i.e. all of them) are infinite sets BY DEFINITION (and hence not empty).
>
> Then they (i.e. all of them) have an infinite intersection with all infinite endsegments:
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo.

wrong. k is finite. you stopped at k, k is not all, Fail.
you get an F. for Fibbing.

On 10/5/2021 10:44 AM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:42:53 UTC+2:
>> On Tuesday, 5 October 2021 at 05:12:00 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 00:49:30 UTC+2:
>>>> On Monday, 4 October 2021 at 18:23:38 UTC-3, WM wrote:
>>>
>>>>> Do you know any endsegment that is not belonging to one of the following sets?
>>>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>>>>> No. Therefore all infinite endsegments have an infinite intersection. Inclusion monotony is always true.
>>>> You don't know what inclusion monotony is
>>> I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo.
>> Nonsense. You think inclusion monotony for a finite number of sets and inclusion monotony for an in finite number of sets behave in the same way.
>
> There is nothing that behaves. If an ednsegment contains a natural number,

ALL ENDSEGMENTS CONTAIN NATURAL NUMBERS BY DEFINITION.

> then this number is in all predecessors.

no. choose 1, is that in all endsegments ? no. fail.

> Since there are no empty successors, every endsegment has at least one element in common with wll endsegments.

which one is that ?

>
>>> Do you know any endsegment that is not covered by (*)? This proves that all endsegments (which can be known) have infinite intersection.
>> And here you go, slipping this innocent "can be known" into your sentence.
>
> If all can be known, then this phrase is not evil, is it?
>
>> You do this so casually -- and knowingly! -- to misdirect.
>
> No, I do it in order to apply only those which are accepted in ZFC.
>
>>> Look: Every infinite endsegment has an infinite intersection with itself and with all its predecessors. But you claim that there are no other endsegments.
>> *YOU* claim that, nobody else.
>
> You claim that all are infinite. So all have an infinite intersection with all its predecessors and with all its successors. Otherwise there would be a first one with less than infinitely many natnumbers.

you have confused yourself.

Serg io schrieb am Dienstag, 5. Oktober 2021 um 17:58:19 UTC+2:
> On 10/3/2021 7:51 AM, WM wrote:

> > Then they (i.e. all of them) have an infinite intersection with all infinite endsegments:
> >
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo. (*)
> k is finite. you stopped at k, k is not all.

What endsegment of the infinite intersection ∩{E(1), E(2), ...} is not covered by a set in (*)? Please name one.

Regards, WM

On Tuesday, 5 October 2021 at 12:52:56 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 15:34:51 UTC+2:
> > On Tuesday, 5 October 2021 at 05:06:49 UTC-3, WM wrote:
> > > From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on.
> > From the fact that E(3) is not empty you try to deduce that E(3) is a subset of E(4), E(5), etc.???
> Not this, but the other way round.

Please *FUCKING* learn to express yourself clearly. "[A]ll endsegments from the first one on" means "E(1) and E(2) and E(3) and ... " or at best if one is generous "E(n), E(n+1), E(n+2), ...".

> > > If all endsegments have infinite contents, then infinite contents is in all of them.
> > This, of course, is equally untenable.
> An infinite subset of all infinite endsegments is a subset of all infinite endsegments.

Erm, yes, but "Every end segment is not empty" is decidedly not the same thing as "there is a nonempty intersection", your moronic application of inclusion monotony notwithstanding. Since there is no set of infinite cardinality common to all end segments, this is once more a "ex falso quodlibet". You are a master of those.

On 10/5/2021 9:34 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 15:25:50 UTC+2:
>> tisdag 5 oktober 2021 kl. 10:03:10 UTC+2 skrev WM:
>
>>>> There is nothign that says they cannot have infinite cardinality.
>>>> And they all demonstrably do have infinite cardinality because each one is in bijection with N
>>> And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible. Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite. In the limit n --> ω we get:
>>>
>>> Lim n = ω
>>> Lim E(n) = E(ω) = { }
>>> Lim |E(n)| = 0
>>
>>> And all endsegments are also in bijection with |N. But two consecutive infinite sets in |n are impossible.
>> Who says otherwise? Not me!
>
> 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).

scrambled eggs.

>>> Therefore the set of infinite endsegments cannot exhaust all natnumbers as indices. It is not actually infinite.
>> The intersection is still emtpy.
>
> Every endsegment has infinite contents. But all endsegments exhaust |N.

wrong, Every endsegment has infinite number of elements.

"contents" and "exhaust" are not math terms.

>> In the limit n --> ω we get:
>>
>>> Lim n = ω
>> Can be said
>>> Lim E(n) = E(ω) = { }
>> E is not defined for omega.
>
> The limit is defined if Lim n = ω.

what do you mean by that ? ( you forgot parts of the equation with Limit)

>>
>>> Lim |E(n)| = 0
>>
>> incorrect, it is aleph_0
>
> This is impossible if the limit of E(n) is empty and

all E(n) are defined as infinite. None are empty.

>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

you stop at k again.

>
> Regards, WM
>

On 10/5/2021 6:31 AM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 12:44:16 UTC+2:
>> On Monday, 4 October 2021 at 23:43:34 UTC-3, Serg io wrote:
>>
>>> Let n be a natural number. Any problems with that statement ?
>> Yes, you made it, not he.
>
> Of course I often use this phrase. But I know what it means: Any natural number can be inserted at position of n.
>
> Regards, WM
>

I have a 17, can that be inserted at the position of n ?

On 10/5/2021 2:54 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 06:53:57 UTC+2:
>> måndag 4 oktober 2021 kl. 23:16:48 UTC+2 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 4. Oktober 2021 um 07:36:05 UTC+2:
>>>> fredag 1 oktober 2021 kl. 19:51:03 UTC+2 skrev WM:
>>>>> If I say it is from the set of natural number it is by definition a natural number.
>>>>> It is not your competence to define what natural numbers are.
>>>>> Please mark the place of n on the following scale:
>>>>>
>>>>> ...1...10...100...1000...10000...
>>>>>
>>>> why?
>>>
>>> To prove your claim.
>>>> Again, I don't need to do that, it is somewhere
>>> A number is not somewhere but has a fixed place.
>
>>> To prove your claim.
>>
>> Of course it is at a specific fixed place, but my ability to place it does not demonstrate it. It is somewhere, even if I cannot point out where :)
>
> My ability to place it everywhere, at as many places as I like, however, shows that n is not a number.
>
> Regards, WM
>

you dont "place it", you exposed the "dark number" hiding in n's position!

Outting a Dark Number which was concealed by timid n, who was paid to do so.

Is there no Shame ?

On 10/5/2021 4:35 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 5. Oktober 2021 um 00:42:23 UTC+2:

>> A natural number which cannot be counted to,
>> even in principle, is not a natural number.
>
> Then there is no set of natural numbers but only
> a potentially infinite collection, namely the ends of FISONs.
> These are the only alternatives.

I think that what you mean by the terms you use is,
for a _totally-ordered_ collection B,

| B is _finite_ iff
| B contains both a first and a last, and
| for each split of B, there is a crossing-pair,

(That is my own definition of "finite", but you seem to agree)

| B is _potentially infinite_ iff
| B does NOT contain both a first and a last.

I'm not sure whether you include the requirement
| for each split of B, there is a crossing-pair.

| B is _actually infinite_ iff
| B contains both a first and a last
| BUT not all splits have a crossing-pair.

----
I think that you (WM) claim that potentially infinite
collections _change_

( That argues for including the requirement
( | for each split of B, there is a crossing-pair.

I have no complaint with you calling the collection of
all and only FISON-enders "potentially infinite".
There is no last FISON-ender, and
for each split of B, there is a crossing-pair j,j+1.

This permits us to reason about k, even if the only thing
we know about k is that it is in the collection of
all and only FISON-enders. "k is a FISON-ender ... "

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.

The support for the collection not changing leads back to
the successor. There are various definitions of it, but
the most commonly used is probably for decimal numerals.
| (numeral)# --> (numeral)+1
| | where
| 0# --> 1
| 1# --> 2
| 2# --> 3
| 3# --> 4
| 4# --> 5
| 5# --> 6
| 6# --> 7
| 8# --> 8
| 9# --> #0

That successor does not change.

0 has a successor.
Each successor has a successor.
There may be things without successors,
but they aren't in FISONs.
There may be things with changing successors,
but they aren't in FISONs.

A collection {0,...,k}, beginning with 0,
ending with k, and, for each split of B,
there is a crossing-pair j,j+1
contains only unchanging things.

Whether such a collection {0,...,k} exists
or does not exist does not change.

The collection of all and only FISON-enders
does not change.

On 10/5/2021 11:04 AM, WM wrote:
> Serg io schrieb am Dienstag, 5. Oktober 2021 um 17:58:19 UTC+2:
>> On 10/3/2021 7:51 AM, WM wrote:
>
>>> Then they (i.e. all of them) have an infinite intersection with all infinite endsegments:
>>>
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo. (*)
>> k is finite. you stopped at k, k is not all.
>
> What endsegment of the infinite intersection ∩{E(1), E(2), ...} is not covered by a set in (*)? Please name one.
>
> Regards, WM
>

∩{E(1), E(2), ...} = { } is empty, not infinite.

If you think not, Please name one element.

On Monday, October 4, 2021 at 11:23:38 PM UTC+2, WM wrote:

> I can disprove [what] you believe.

No, you can't, idiot.

On Tuesday, October 5, 2021 at 10:16:43 AM UTC+2, WM wrote:
> Greg Cunt schrieb am Dienstag, 5. Oktober 2021 um 05:06:45 UTC+2:
> > On Sunday, October 3, 2021 at 1:37:39 PM UTC+2, WM wrote:
> > > Greg Cunt schrieb am Samstag, 2. Oktober 2021 um 02:32:38 UTC+2:
> > > >
> > > > Please mark the place of Schnirelmann's constant on the following "scale",
> > > > 1, 2, 3, 4, 5, 6, 7, ...
> > > >
> > > If it is a constant [...], then it has a place there [...]. n has no place there.
> > >
> > Nope. Since n is a natural number it certainly has "a [certain/fixed] place there"
> >
> I can place n at 1, at 2, at 3, at any place I like.

No, you can't.

If we introduce "n" with the phrase "Let n e IN", then WE DON'T KNOW the referent of "n", but WE DO KNOW that "n" refers to a CERTAIN natural number - i.e. a FIXED denotation. So you can't just stipulate (afterwards) that n = 1, or stipulate that n = 2, etc.

If you wan't "n" to denote a SPECIFIC number _of your choice_, you have to introduce "n" the following way: "Let n = 1", or "Let n = 2", etc.

> This is not possible with a natural number like 7 for instance, and it is not possible with a hitherto unknown number.

Exactly. Hence it is not possible with n either.

On Tuesday, October 5, 2021 at 2:46:05 PM UTC+2, Serg io wrote:

> a subtle difference there;
>
> | Let n e {1, 2, 3}. Then n ...
>
> n is constant or parameter (member of the set)

Nope. You have to differentiate between the term (name/constant/paramter, whatever) - "n" in this case - and the objects the term refers to, namely a number, i. e. a member of the set {1, 2, 3}, n in this case.

Of course, in the context of math we usually call certain/specific numbers "constants" too. So please don't mix that up. I'm talking about the language here (not the mathematical objects).

Again:

| Let n e {1, 2, 3}. Then n ...

Here, "n" refers to a CERTAIN number in {1, 2, 3}, though we don't and can't KNOW which one. [Morover, *I* wouldn't call n a "constant" here - imho its not specific enough to be called that way.]

Please note the difference: IN THIS CONTEXT "n" is a constant (or "parameter", if you like).

On the other hand, in the statement

| For all n in IN: n >= 0.

"n" is (used as) a variable.

Or even in a statement like

| n e IN --> n >= 0,

or some algebraic "equations", etc.

On Tuesday, October 5, 2021 at 3:26:27 PM UTC+2, zelos...@gmail.com wrote:
> tisdag 5 oktober 2021 kl. 09:54:23 UTC+2 skrev WM:
> >
> > My ability to place it everywhere, at as many places as I like, however, shows that n is not a number.

As usual his "ability" to do things which cannot be done leads to disaster.

Let n be a natural number. (*)

Then Mückenheim can place it at n+1 too. With other words, *he* can enforce n = n+1. Hmmm... not specific enough?

Ok, let's be more specific: Mückenheim can place it at, say, 1. Hence n = 1 [though this seem to contradict his claim that "n is not a number"].. Since n was an arbitrary natural number (by (*)) this means that for all natural numbers k: k = 1.

> I [...] cannot place it anywhere.

Same, same.

Of course ,Chuck Norris could do it too.

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. belief)?

Hint: It's called /delusion/.

"A delusion is a fixed belief that is not amenable to change in light of conflicting evidence. As a pathology, it is distinct from a belief based on false or incomplete information, confabulation, dogma, illusion, or some other misleading effects of perception, as individuals with those beliefs /are/ able to change or readjust their beliefs upon reviewing the evidence."

See: https://en.wikipedia.org/wiki/Delusion

On Tuesday, October 5, 2021 at 6:04:45 PM UTC+2, WM wrote:

> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo. (*)
>
> What <bla bla bla> by a set in (*)?

Do formulas contain sets as elements in mückenmath?

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)?

Hint: It's called /delusion/.

"A delusion is a fixed belief that is not amenable to change in light of conflicting evidence. As a pathology, it is distinct from a belief based on false or incomplete information, confabulation, dogma, illusion, or some other misleading effects of perception, as individuals with those beliefs /are/ able to change or readjust their beliefs upon reviewing the evidence."

See: https://en.wikipedia.org/wiki/Delusion

tisdag 5 oktober 2021 kl. 16:22:08 UTC+2 skrev Hilton Blome:
> On 05.Oct.2021, Serg io wrote:
>
> >> Only an empty endsegment can cause an emty intersection.
> >> Regards, WM
> >
> > there are no empty endsegments by definition.
> except is used thereafter in tensors and everything. Give me a break.
>
> 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?

tisdag 5 oktober 2021 kl. 17:49:02 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 5. Oktober 2021 um 15:23:56 UTC+2:
> > tisdag 5 oktober 2021 kl. 10:06:49 UTC+2 skrev WM:
>
> > > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > >If no endsegment exists outside of (*), then all are covered by (*).
> > It is true for all endsegments but that does not mean what you want.
> It is true for all endsegments. That is what I claim.
> > >From the fact that an endsegment has contents, I can deduce that this contents is remaining in all endsegments from the first one on. If all endsegments have infinite contents, then infinite contents is in all of them.
> > Correct but that doesn't mean that the infinite intesrection of all is not empty.
> Just that is fact because the infinite intersection concerns only all endsegments for which the intersection is infinite.
> > >I know it very precisely. Matheologians try to suppress it. If not being forced, they never talk about this taboo
> > No one is surpressing anything. It just doesn't fucking mean what you want it to!
> It means that all endsegments are infinite and have an infinite intersection. More is not claimed. The intersection is restricted to infinity by
>
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
>
> Regards, WM

False, the intersection of ALL endsegments is EMPTY

>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 jumbled mess of shite except you.

>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.

>This is impossible if the limit of E(n) is empty and

Nope, 2 different limits you idiot

extreme crank zelos...@gmail.com wrote:

> tisdag 5 oktober 2021 kl. 16:22:08 UTC+2 skrev Hilton Blome:
>> On 05.Oct.2021, Serg io wrote:
>>
>> >> Only an empty endsegment can cause an emty intersection.
>> >> Regards, WM
>> >
>> > there are no empty endsegments by definition.
>> except is used thereafter in tensors and everything. Give me a break.
>>
>> 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.

onsdag 6 oktober 2021 kl. 21:31:51 UTC+2 skrev Hilton Blome:
> extreme crank zelos...@gmail.com wrote:
>
> > tisdag 5 oktober 2021 kl. 16:22:08 UTC+2 skrev Hilton Blome:
> >> On 05.Oct.2021, Serg io wrote:
> >>
> >> >> Only an empty endsegment can cause an emty intersection.
> >> >> Regards, WM
> >> >
> >> > there are no empty endsegments by definition.
> >> except is used thereafter in tensors and everything. Give me a break.
> >>
> >> 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 :)