On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
> > On Friday, September 10, 2021 at 5:35:51 PM UTC-3, WM wrote:
> > > William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
>
> > > [The endless sequence] consist of only finite initial subsequences.
> >
> > An endless sequence consists of an infinite number of finite sequences. An infinite sequence has properties that no finite sequence has.
> >
> > You are fond of the argument:
> Every infinite endsegment has an infinite intersection with all its predecessors. If the infinite sequence has only infinite endsegments, then this cannot change and does not change.
And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n). Nothing new, nothing to see here.
> If you claim that a sequence of only infinite endsegments has an empty intersection, then this is nonsense and not useful to continue this discussion.
The intersection over *INFINITE* (do you see the difference?) intersections is *EMPTY* because the minimal point k of *ANY* end segment E(k) is *NOT CONTAINED* in any other end segment m with m > k. Again, nothing new, nothing nonsensical here. The only one here who seems to have any difficulty with this is the great Professor Muckenheim himself. In order to gratify yourself, you engage in lies, misdirections, switching of quantifiers and circular logic. You are a disgrace to your profession, and the sooner you get out, the better it will be for all concerned, including yourself.

On 9/11/2021 9:58 AM, Serg io wrote:
> On 9/10/2021 5:51 PM, Jim Burns wrote:
>> On 9/10/2021 12:52 PM, WM wrote:
>>> Greg Cunt schrieb
>>> am Freitag, 10. September 2021 um 18:49:34 UTC+2:

>>>> Actually, the intersection of (the set of) all endsegments
>>>> is empty.
>>>
>>> Actually it is the minimum endsegment.
>>> If all endsegments are infinite,
>>
>> If all end segments are infinite,
>> then each end segment is followed by at least one end segment.
>
> yes and yes
>
>> Therefore,
>> if all end segments are infinite,
>> then there is no last end segment.
>
> agree
>
>>
>> If the intersection of all end segments is an end segment,
>> then there is a last end segment.
>
> why would the intersection be an end segment ?

It is an if-then statement.

A statement of the form
| if Assumption, then Conclusion

is true
when and only when a statement of the form
| Assumption and not Conclusion

is false.

There are other natural-language uses of "if...then..."
but this is how it is very commonly understood in
mathematics.

Notice that
| Assumption and not Conclusion

is false whenever
-- Conclusion is true, or
-- Assumption is false.

It's this second clause that causes problems.
Some important different uses of "if...then..." are not
intended to consider Assumption being false.
This leads to some funny, odd sentences that _sound_ false
but that the material conditional declares true.
("if A then B" defined as "not( A and not B )")

https://en.wikipedia.org/wiki/Paradoxes_of_material_implication

If I avoid "if" and "then", then I expect your objection to go away.

Either the intersection of all end segments is not an end segment,
or there is a last end segment.

> there is no last endsegment.

There is no last end segment.

One of the consequences of there being no last end segment
is that the intersection of all end segments is empty.
( You don't need that proved to you.
( But you aren't really who I'm addressing.

How I am showing that that's an unavoidable consequence is
by assuming that it's NOT an unavoidable consequence and
showing how that leads to nonsensical, impossible situations.
And so, I consider what it would be like for there to be
a last end segment.

I haven't claimed that there is a last end segment.
I have claimed that _if_ there is a last end segment,
_then_ something something.

>> If there is no last end segment.
>> then the intersection of all end segments is not an end segment,
>
> provide a showing or proof.
> (ez to prove it is empty)

Even easier to show it follows from the preceding claim

| If the intersection of all end segments is an end segment,
| then there is a last end segment.

( Proof of preceding claim:
( It's the intersection.

Write "the intersection is an end segment" as "Assumption"
and "there is a last end segment" as "Conclusion".

| If Assumption, then Conclusion

is equivalent to

| not( Assumption and not Conclusion )

| not( not Conclusion and Assumption )

| not( (not Conclusion) and not (not Assumption) )

| If not Conclusion, then not Assumption

| If there is no last end segment,
| then the intersection of all end segments is not an end segment.

>> Therefore,
>> if all end segments are infinite,
>> then the intersection of all end segments is not an end segment.
>
> you skip a step, show the intersection is empty,
> then show it cannot be an endsegment.

I'm not skipping a step.
I'm showing that _from_ not-an-end-segment, we get empty.
I need not-an-end-segment first.

( Because it's _closed upwards_
( the intersection is an end segment or empty.
( Because _not an end segment_ it's empty.

( "set B closed upwards" ==
( "for x < y, if x in B, then y in B"
( ( equivalently...
( "for x < y, not( x in B and y not-in B )"

>>
>> The intersection of closed upwards sets is closed upwards.
>>
>> The intersection of all end segments (which are closed upwards) is
>> either an end segment (if it's not empty),
>> or the empty set (if it's empty).
>>
>> If all end segments are infinite,
>> then the intersection of all end segments is
>> either an end segment or the empty set, and is not an end segment.
>>
>> Therefore,
>> if all end segments are infinite,
>> then the intersection of all end segments is the empty set.
>>
>>> If all endsegments are infinite,
>>> the minimum endsegment is infinite too.
>>
>> If all end segments are infinite,
>> then the intersection of all end segments is the empty set.
>>
>

On 9/11/2021 10:52 AM, Serg io wrote:
> On 9/11/2021 8:55 AM, Loy Gue wrote:

>> biden, "we need to protect the vaccinated from the unvaccinated".
>>
>> *boooong!!*
>>
>
> Joe Biden the Fuck-Up.

The Asshole Party 2022 theme:
| Vote for the Assholes.
| Joe Biden can't stop us from killing ourselves.

The Asshole Party doesn't have anything better to run on.

Who knows? It might work. Sleep well tonight.

On 9/11/2021 8:17 AM, FromTheRafters wrote:
> WM pretended :
>> Greg Cunt schrieb am Freitag, 10. September 2021 um 22:57:57 UTC+2:
>>> On Friday, September 10, 2021 at 10:29:35 PM UTC+2, WM wrote:
>>
>>>>>> stepwise loss ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.
>>>> it is the definition of the endsegments.
>>> "it"? What?
>>
>> See first line above.
>>
>>>> If the intersection is not infinite, then there must be a first infinite endsegment with finite intersection because  of stepwise loss ∀k ∈ ℕ:
>>>> E(k+1) = E(k) \ {k}
>>> ?
>>> Well, little wonder. Since the claim is false.
>>
>> The claim is the definition of endsegments. If you cannot accept it, then you have in mind other objects which you erroneously denote as endsegments.
>
> Where in the definition is the 'stepwise' part necessary? It is just your mental block again.
>
> *ALL* k in N have FISONs ending with k and infinite endsegments starting with k + 1 -- there is no need to add a stepwise character to the mental
> process of considering *ALL* of something.

agree.

WM seems to only knows one line of really simplistic math, E(k+1) = E(k) \ {k}

On 9/11/2021 11:02 AM, Jim Burns wrote:
> On 9/11/2021 9:58 AM, Serg io wrote:
>> On 9/10/2021 5:51 PM, Jim Burns wrote:
>>> On 9/10/2021 12:52 PM, WM wrote:
>>>> Greg Cunt schrieb
>>>> am Freitag, 10. September 2021 um 18:49:34 UTC+2:
>
>>>>> Actually, the intersection of (the set of) all endsegments
>>>>> is empty.
>>>>
>>>> Actually it is the minimum endsegment.
>>>> If all endsegments are infinite,
>>>
>>> If all end segments are infinite,
>>> then each end segment is followed by at least one end segment.
>>
>> yes and yes
>>
>>> Therefore,
>>> if all end segments are infinite,
>>> then there is no last end segment.
>>
>> agree
>>
>>>
>>> If the intersection of all end segments is an end segment,
>>> then there is a last end segment.
>>
>> why would the intersection be an end segment ?
>
> It is an if-then statement.

of course,
but you have not shown "the intersection of all end segments is an end segment" is true,
and you have not shown that if true that it causes "there is a last end segment"

<snip>

>
> Either the intersection of all end segments is not an end segment,
> or there is a last end segment.

I dont see how that follows,
the intersection can be an endsegment(which is an infinite sequence), or not an endsegment(a finite sequence), or a null set (a degenerate endsegment
(?)) (three outcomes)

but none of that determines a "last" endsegment

unless you can show the linkage, and then also define "last endsegment"

I find it easier to prove the intersection of all endsegments is empty,

PROOF
assume m is in the intersection
however E(m+1) does not contain m
therefore m is not in the intersection
therefore the assumption is wrong
therefore the intersection of all endsegments is empty.

>
>> there is no last endsegment.
>
> There is no last end segment.
>
> One of the consequences of there being no last end segment
> is that the intersection of all end segments is empty.

> ( You don't need that proved to you.
> ( But you aren't really who I'm addressing.
>
> How I am showing that that's an unavoidable consequence is
> by assuming that it's NOT an unavoidable consequence and
> showing how that leads to nonsensical, impossible situations.
> And so, I consider what it would be like for there to be
> a last end segment.
>
> I haven't claimed that there is a last end segment.
> I have claimed that _if_ there is a last end segment,
> _then_ something something.
>
>>> If there is no last end segment.
>>> then the intersection of all end segments is not an end segment,
>>
>> provide a showing or proof.
>> (ez to prove it is empty)
>
> Even easier to show it follows from the preceding claim
>
> | If the intersection of all end segments is an end segment,
> | then there is a last end segment.
>
> ( Proof of preceding claim:
> ( It's the intersection.
>
> Write "the intersection is an end segment" as "Assumption"
> and "there is a last end segment" as "Conclusion".
>
> | If Assumption, then Conclusion
>
> is equivalent to
>
> | not( Assumption and not Conclusion )
>
> | not( not Conclusion and Assumption )
>
> | not( (not Conclusion) and not (not Assumption) )
>
> | If not Conclusion, then not Assumption
>
> | If there is no last end segment,
> | then the intersection of all end segments is not an end segment.
>
>>> Therefore,
>>> if all end segments are infinite,
>>> then the intersection of all end segments is not an end segment.
>>
>> you skip a step, show the intersection is empty,
>> then show it cannot be an endsegment.
>
> I'm not skipping a step.
> I'm showing that _from_ not-an-end-segment, we get empty.
> I need not-an-end-segment first.
>
> ( Because it's _closed upwards_
> ( the intersection is an end segment or empty.
> ( Because _not an end segment_ it's empty.
>
> ( "set B closed upwards" ==
> ( "for x < y, if x in B, then y in B"
> (
> ( equivalently...
> ( "for x < y, not( x in B and y not-in B )"
>
>>>
>>> The intersection of closed upwards sets is closed upwards.
>>>
>>> The intersection of all end segments (which are closed upwards) is
>>> either an end segment (if it's not empty),
>>> or the empty set (if it's empty).
>>>
>>> If all end segments are infinite,
>>> then the intersection of all end segments is
>>> either an end segment or the empty set, and is not an end segment.
>>>
>>> Therefore,
>>> if all end segments are infinite,
>>> then the intersection of all end segments is the empty set.
>>>
>>>> If all endsegments are infinite,
>>>> the minimum endsegment is infinite too.
>>>
>>> If all end segments are infinite,
>>> then the intersection of all end segments is the empty set.
>>>
>>
>

On 9/11/2021 12:55 PM, Serg io wrote:
> On 9/11/2021 11:02 AM, Jim Burns wrote:

>> Either the intersection of all end segments is not an end segment,
>> or there is a last end segment.
>
> I dont see how that follows,
> the intersection can be
> an endsegment(which is an infinite sequence),
> or not an endsegment(a finite sequence),
> or a null set (a degenerate endsegment (?))
> (three outcomes)
>
> but none of that determines a "last" endsegment
> unless you can show the linkage,
> and then also define "last endsegment"

If we order by inclusion (as we have been),
( A =< B iff A sub B ),
the intersection INT(C) of the collection C comes after each of
the sets which are members of C.

This is a consequence of what an intersection is.
Each element in the intersection is in each set
being intersected.
Each element in a subset is in the set it is a subset of.

Consider the claim
| Either the intersection of all end segments is not an end segment,
| or there is a last end segment.

There are two cases.
(i)
The intersection of all end segments is not an end segment.

The claim is true.

(ii)
The intersection of all end segments is an end segment.

Then the intersection is an end segment after all end segments.
It is the last end segment.
And so, there is a last end segment,
and the claim is true.

In both cases, the claim is true.
Therefore,
either the intersection of all end segments is not an end segment,
or there is a last end segment.

On 9/11/2021 9:05 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 11. September 2021 um 00:51:17 UTC+2:

>> if all end segments are infinite,
>> then there is no last end segment.
>
> Then there is an infinite set of natural nunbers
> remaining in all endsegments forever.

No, there are no forever-numbers.
A natural number which cannot be counted to
is not a natural number.

More quantifier dyslexia from you.

> This is an infinite intersection.

An end segment is a non-empty, closed upwards set.

The intersection of closed upwards sets is closed upwards.

The intersection of end segments is either
the empty set (which is vacuously closed upwards), or
some end segment (being closed upwards and non-empty).

If the intersection is an end segment, it is the last
end segment.

Therefore,
the intersection of end segments is
either the empty set or the last end segment.

If there is no last end segment, then the intersection is empty.

On Saturday, September 11, 2021 at 8:28:41 PM UTC+2, Jim Burns wrote:
> On 9/11/2021 9:05 AM, WM wrote:
> >
> > [...] there is an infinite set of natural numbers remaining in all endsegments forever.
> >
> No, there are no forever-numbers.

I'm quite sure that he will get it THIS TIME!!!

I mean, since the following quite complicated argument did not convince him

For any natural number n: n is not in the endsegment E(n+1) and hence not in the intersection of (the set of) all endsegments; Hence no natural number is in the intersection of (the set of) all endsegments.

your rather direct arguments will certainly achieve this feat.

On Saturday, September 11, 2021 at 11:41:39 AM UTC-7, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 8:28:41 PM UTC+2, Jim Burns wrote:
> > On 9/11/2021 9:05 AM, WM wrote:
> > >
> > > [...] there is an infinite set of natural numbers remaining in all endsegments forever.
> > >
> > No, there are no forever-numbers.
> I'm quite sure that he will get it THIS TIME!!!
>
> I mean, since the following quite complicated argument did not convince him
>
> For any natural number n: n is not in the endsegment E(n+1) and hence not in the intersection of (the set of) all endsegments; Hence no natural number is in the intersection of (the set of) all endsegments.
>
> your rather direct arguments will certainly achieve this feat.

What's an endsegment?

The usual notion of "no infinitely descending chains", is only for framing
the infinity in the middle, that of course only from one or the other of the
ends, that the vertical asymptote at the origin only holds up one hand:
is for drawing past scale, that effectively for there being points (0, oo),
and (0, 0), and (0, oo), there is (oo, oo), holding up R+ ^2.

Here the "frontsegment" is [0,1], and "endsegment" is [oo-1, oo].

On Saturday, September 11, 2021 at 11:57:26 AM UTC-7, Ross A. Finlayson wrote:
> On Saturday, September 11, 2021 at 11:41:39 AM UTC-7, Greg Cunt wrote:
> > On Saturday, September 11, 2021 at 8:28:41 PM UTC+2, Jim Burns wrote:
> > > On 9/11/2021 9:05 AM, WM wrote:
> > > >
> > > > [...] there is an infinite set of natural numbers remaining in all endsegments forever.
> > > >
> > > No, there are no forever-numbers.
> > I'm quite sure that he will get it THIS TIME!!!
> >
> > I mean, since the following quite complicated argument did not convince him
> >
> > For any natural number n: n is not in the endsegment E(n+1) and hence not in the intersection of (the set of) all endsegments; Hence no natural number is in the intersection of (the set of) all endsegments.
> >
> > your rather direct arguments will certainly achieve this feat.
> What's an endsegment?
>
> The usual notion of "no infinitely descending chains", is only for framing
> the infinity in the middle, that of course only from one or the other of the
> ends, that the vertical asymptote at the origin only holds up one hand:
> is for drawing past scale, that effectively for there being points (0, oo),
> and (0, 0), and (0, oo), there is (oo, oo), holding up R+ ^2.
>
> Here the "frontsegment" is [0,1], and "endsegment" is [oo-1, oo].

Some people "define" 0^0 as 1, instead of zero, because x^0 = 1,
not because 0^x =/= 0. It's the "usual" convention, for example
much like right-hand rule. (It's where the terms are in the exponents
for the arithmetic in the exponents under powers usually in terms of
the readily available tractability of the algebraic cancellation in terms.)

Here there is a notion that the direct sum of sets, accumulates their
intersection or as it were, and, some people, and, the usual definition,
have that it's that the direct sum of sets N is empty set, while, because
structurally instead it works out that it's zero the difference that the
(infinite) direct sum, of N, is itself, is among reasons why number theory
has a point at infinity.

So, if you might look into the _definition_ of the direct products and
sums of sets, over an inductive set like the naturals model or as model
the naturals, then you can understand that underneath the definition,
there is a branch in conventions, why mathematics can talk about for
example "point at infinity" in the numbers, both quite usually, and,
under terms, also the alternate branch.

(As what the existence of the terms imply.)

On 9/11/2021 1:57 PM, Ross A. Finlayson wrote:
> On Saturday, September 11, 2021 at 11:41:39 AM UTC-7, Greg Cunt wrote:
>> On Saturday, September 11, 2021 at 8:28:41 PM UTC+2, Jim Burns wrote:
>>> On 9/11/2021 9:05 AM, WM wrote:
>>>>
>>>> [...] there is an infinite set of natural numbers remaining in all endsegments forever.
>>>>
>>> No, there are no forever-numbers.
>> I'm quite sure that he will get it THIS TIME!!!
>>
>> I mean, since the following quite complicated argument did not convince him
>>
>> For any natural number n: n is not in the endsegment E(n+1) and hence not in the intersection of (the set of) all endsegments; Hence no natural number is in the intersection of (the set of) all endsegments.
>>
>> your rather direct arguments will certainly achieve this feat.
>
> What's an endsegment?
>
> The usual notion of "no infinitely descending chains", is only for framing
> the infinity in the middle, that of course only from one or the other of the
> ends, that the vertical asymptote at the origin only holds up one hand:
> is for drawing past scale, that effectively for there being points (0, oo),
> and (0, 0), and (0, oo), there is (oo, oo), holding up R+ ^2.
>
> Here the "frontsegment" is [0,1], and "endsegment" is [oo-1, oo].
>

what about midsegment ?
and backsegment ?
or startsegment ?
or unsegment?

On Saturday, September 11, 2021 at 9:34:11 PM UTC+2, Serg io wrote:

> what about midsegment ?
> and backsegment ?
> or startsegment ?
> or unsegment?

Please don't forget about the

antsegment

!

On Saturday, September 11, 2021 at 12:50:20 PM UTC-7, Greg Cunt wrote:
> On Saturday, September 11, 2021 at 9:34:11 PM UTC+2, Serg io wrote:
>
> > what about midsegment ?
> > and backsegment ?
> > or startsegment ?
> > or unsegment?
> Please don't forget about the
>
> antsegment
>
> !

I read you as "anti-segment".

This is writing the same numbers as 0+ 0, 1, 2, 3, ..., oo- 0, 1, 2, 3, ....

I.e. it would be incomplete to deny the numbers their symmetry.

I.e., 0 + 0, 1, 2, 3, ..., 3, 2, 1, 0 -oo.

With that being the compass closing, there are all sorts of
usual simplest arguments what make guarantees.

Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
> On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> > William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
> > > On Friday, September 10, 2021 at 5:35:51 PM UTC-3, WM wrote:
> > > > William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
> >
> > > > [The endless sequence] consist of only finite initial subsequences.
> > >
> > > An endless sequence consists of an infinite number of finite sequences. An infinite sequence has properties that no finite sequence has.
> > >
> > > You are fond of the argument:
> > Every infinite endsegment has an infinite intersection with all its predecessors. If the infinite sequence has only infinite endsegments, then this cannot change and does not change.
> And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).

And there is no finite n leading to an infinite intersection. But there are onkly finite n. Hece tere is no empty intersection.

> > If you claim that a sequence of only infinite endsegments has an empty intersection, then this is nonsense and not useful to continue this discussion.
> The intersection over *INFINITE* (do you see the difference?)

I see that some fools believe that finite n could cause infinite sets. One should not admit these fools to mathematics.

Regards, WM

Jim Burns schrieb am Samstag, 11. September 2021 um 18:02:11 UTC+2:

> Write "the intersection is an end segment" as "Assumption"
> and "there is a last end segment" as "Conclusion".

The intersection cannot be less than the smallest endsegment. If all endsegments are infinite, then the smallest endsegment is infinite too.

Regards, WM

Jim Burns schrieb am Samstag, 11. September 2021 um 20:28:41 UTC+2:
> On 9/11/2021 9:05 AM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 11. September 2021 um 00:51:17 UTC+2:
>
> >> if all end segments are infinite,
> >> then there is no last end segment.
> >
> > Then there is an infinite set of natural numbers
> > remaining in all endsegments forever.
> No, there are no forever-numbers.

All endsegments are infinite. What do they contain?

Regards, WM

Greg Cunt schrieb am Samstag, 11. September 2021 um 20:41:39 UTC+2:
> On Saturday, September 11, 2021 at 8:28:41 PM UTC+2, Jim Burns wrote:

> > No, there are no forever-numbers.
> I'm quite sure that he will get it THIS TIME!!!

All endsegments are infinite. What do they consist of? What of their contents is not in all predecessors?

Regards, WM

On Saturday, September 11, 2021 at 5:22:50 PM UTC-3, WM wrote:
> And there is no finite n leading to a [finite] intersection. But there are onkly finite n.

So what? We are talking about a property of the sequence. There are only finite n. But the sequence is made from
and infinite number of finite n. No finite number of finite n can lead to a finite intersection. But an infinite number of finite n can.

--
William Hughes

On 9/11/2021 3:27 PM, WM wrote:
> Jim Burns schrieb am Samstag, 11. September 2021 um 20:28:41 UTC+2:
>> On 9/11/2021 9:05 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 11. September 2021 um 00:51:17 UTC+2:
>>
>>>> if all end segments are infinite,
>>>> then there is no last end segment.
>>>
>>> Then there is an infinite set of natural numbers
>>> remaining in all endsegments forever.
>> No, there are no forever-numbers.
>
> All endsegments are infinite. What do they contain?
>
> Regards, WM
>

What if an endsegment has a leak in it ?
does the contents run out ?
Does it drip out, step by step ?

On 9/11/2021 3:25 PM, WM wrote:
> Jim Burns schrieb am Samstag, 11. September 2021 um 18:02:11 UTC+2:
>
>> Write "the intersection is an end segment" as "Assumption"
>> and "there is a last end segment" as "Conclusion".
>
> The intersection cannot be less than the smallest endsegment.

conjecture, prove it.

>If all endsegments are infinite,

by definition the are.

> then the smallest endsegment is infinite too.

there is no smallest.

All endsegments have the same number of elements => Aleph_null.

>
> Regards, WM
>

On Saturday, September 11, 2021 at 10:28:05 PM UTC+2, WM wrote:

> All endsegments are infinite. What do they contain?

Natural numbers?

In fact, if E is a endsegment, then there's a natural number k such that E = {k, k+1, k+2, ...}.

On Saturday, September 11, 2021 at 5:25:50 PM UTC-3, WM wrote:

> The intersection cannot be less than the smallest endsegment .

If a "smallest endsegment" exists. If there is no last endseqment there is no "smallest endsegment". (Note That E_(n+1) is a proper subset of E_n even though both sets have the same cardinality.)

--
William Hughes

On Saturday, September 11, 2021 at 10:22:50 PM UTC+2, WM wrote:

> finite n could [form] infinite sets.

Indeed! If there are infinitely many such n.

For example Ax(x e IN -> x is finite) & IN is infinite. (Assuming that IN is defined due to von Neumann.)

On Saturday, September 11, 2021 at 10:26:15 AM UTC+2, Ross A. Finlayson wrote:

> Please pick a new name if you're not using your real one,

You may call me Great Cunt, if you like.

On Saturday, September 11, 2021 at 10:25:50 PM UTC+2, WM wrote:

> The intersection cannot be less than the smallest endsegment. If all endsegments are infinite, then the smallest endsegment is infinite too.

Yeah, and if wishes were horses, beggars would ride.

Hint: There IS NO /smallest endsegment/.