On Wednesday, 11 August 2021 at 16:43:15 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 11. August 2021 um 16:32:32 UTC+2:
> > On Wednesday, 11 August 2021 at 11:04:20 UTC-3, WM wrote:
> > > The antinomy is that all endsegments which can finish a set shall be sufficient to form an infinite set (actually infinite!) and there have an empty intersection. Note that only such endsegments are available if ℕ_def = ℕ.
> > Boring... It has been proven to you many times that ℕ_def = ℕ. That you do not understand or accept the proof is immaterial.
> I understand that the theory this "proof" is based upon is inconsistent, because it is this same theory which claims that intersections are indepedent of the order of endsegments.

More Mucke-shit. You don't even pretend to think anymore. (But I think we know why that is.) The order has nothing to do with the intersection, but the *NUMBER* of end segments that are being intersected. Your posts are becoming increasingly annoying and incoherent. Why don't you just fade into the woodwork?

On Wednesday, August 11, 2021 at 9:43:15 PM UTC+2, WM wrote:

> intersections are indepedent of the order of <bla>.

Indeed!

Hint: INTERSECTION M is what it is, ***completely*** independent of any order defined on M, you silly crank!

Hint: It only depends on the CONTENT of M.

Holy Shit!

On 8/11/2021 10:24 PM, Greg Cunt wrote:
> On Wednesday, August 11, 2021 at 9:43:15 PM UTC+2,
> WM wrote:

>> intersections are indepedent of the order of <bla>.
>
> Indeed!
>
> Hint: INTERSECTION M is what it is, ***completely*** independent
> of any order defined on M, you silly crank!
> Hint: It only depends on the CONTENT of M.

In this case, the order of the end segments is by inclusion.
So, in this case, it's not possible to speak of their order
without at the same time speaking of their content.

----
Because of the order/content of the end segments,
the intersection of a collection of end segments
equals the last end segment in that collection
-- but not if no last end segment exists.

Last end segment, intersection is the last end segment.

Because of the order/content of the end segments and
because all {0,...,m} are finite,
no number m is in the intersection of a collection of
end segments unless there is a last end segment and
m is in it.

No last end segment, intersection is empty.

----
I'm thinking about WM's appendix and...
Last end segment, intersection is the last end segment.
No last end segment, intersection is empty.

WM rejects No Last End Segment.
Instead, he defines into existence an appendix with the
property that it makes the intersection empty.

That's nonsense, I know.
But this shines a light on how Wolfgang Mückenheim thinks
mathematics should work -- one defines into existence things
with the required properties -- _any_ required properties.

This explains why he refers to unique prime factorization as
being a _defined_ property of the natural numbers.
As he sees it, defining into existence things with the
unique-prime-factorization property is the only way to get
the unique-prime-factorization property true without
exception over some infinite domain.

I think it also explains why he keeps adding dark numbers
despite our repeated insistence that they be left out.
He thinks it's standard operating procedure to just *poof*
more things into the domain if you need them or want them.

tisdag 10 augusti 2021 kl. 13:36:58 UTC+2 skrev WM:
> Gus Gassmann schrieb am Dienstag, 10. August 2021 um 12:59:05 UTC+2:
> > On Tuesday, 10 August 2021 at 06:49:39 UTC-3, WM wrote:
>
> > > There are two opinions: Either all endsegments which are definable as a last one have infinite intersections with all their predecessors
> > >
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵo (*)
> > >
> > > can also be used in an infinite set such that
> > >
> > > E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.
> > Gibberish.
> You cannot even understand this simple thought?
> > > Or the infinite set must contain more endsegments than are in (*).
> > *WHAT* infinite set?
> The set intersected above: E(1), E(2), E(3), ...
> > > The first opinion is wrong because (1) an infinite set of endsegments contains more endsegments than every finite set. And (2) the intersection is independent of the order.
> > >
> > > There it is refuted that only all endsegments which in finite sets leave infinite intersections can constitute an infinite set yielding an empty intersection. Are you really unable to understand these two simple arguments?
> > The third, majority opinion is that WM has lost his marbles.
> Among fools of matheology this may be a majority opinion.
>
> Regards, WM
mathematics, stop with this "matheology" thing, it only stamps you as a crank.

On Thursday, August 12, 2021 at 6:04:06 AM UTC+2, Jim Burns wrote:

> In this case <bla>

Once more, again:

Hint: INTERSECTION M is what it is, ***completely*** independent of any order defined on M. It only depends on the CONTENT of M. [ A set doesen't have an order "per se" - even if there is a subset relation between the sets in M. ]

__________________________________________________________________

Concerning the CONTENT of the sets M: in "our case" there's a nice theorem which I've concocted recently:

| Let I a nonempty set and (A_i)_(i e I) a family of subsets of a set A.
| | Then Intersection {A_i : i e I} = { }, if Ax e A: Ei e I: x !e A_i.

In our case we have I = IN, A = IN and (A_i)_(i e I) = (E(n))_(n e IN) with E(n) = {m e IN : m >= n} for all n e IN.

The "crucial property" Ax e A: Ei e I: x !e A_i holds. (In our case we have An e IN: Ek e IN: n !e A_k.)

Note that this property would hold too if the sets E(n) were notch sets.

William schrieb am Mittwoch, 11. August 2021 um 22:19:51 UTC+2:

> He said staring at one true statement and one false statement.

If all indices are used to enumerate the endsegments, then not all endsegments can contain aleph_0 natnumbers. That is very simple to understand. Never a student of mine has failed since I invented this argument.

Regards, WM

Greg Cunt schrieb am Donnerstag, 12. August 2021 um 04:24:34 UTC+2:
> On Wednesday, August 11, 2021 at 9:43:15 PM UTC+2, WM wrote:
>
> > intersections are indepedent of the order
>
> Indeed!
>
> Hint: INTERSECTION M is what it is, ***completely*** independent of any order defined on M,
>
> Hint: It only depends on the CONTENT of M.

Therefore the intersection of all definable endsegments, that are those which appear in
∀k ∈ ℕ_def ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo ,
will never yield another result than this:
|∩{E(k) | k ∈ ℕ_def}| = ℵo .
Regards, WM

Jim Burns schrieb am Donnerstag, 12. August 2021 um 06:04:06 UTC+2:

> No last end segment, intersection is empty.

Every definable endsegment is a last endsegment.

> I'm thinking about WM's appendix and...
> Last end segment, intersection is the last end segment.
> No last end segment, intersection is empty.
>
> WM rejects No Last End Segment.

No, but I reject that last endsegments can yield a set with no last endsegment - other than potentially infinite. But that has always a last one.

> Instead, he defines into existence an appendix with the
> property that it makes the intersection empty.

Every actually infinite set is larger than any finite set. But definable endsegments belong to finite sets by definition.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 12. August 2021 um 06:51:48 UTC+2:
> tisdag 10 augusti 2021 kl. 13:36:58 UTC+2 skrev WM:

> > Among fools of matheology this may be a majority opinion.
> >
> mathematics, stop with this "matheology" thing,

No! Never!! Unless you withdraw the belief that more than countably many real numbers could be well-ordered, i.e., distinguished as individuals.

Regards, WM

On Thursday, August 12, 2021 at 9:29:08 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 11. August 2021 um 22:19:51 UTC+2:
>
> > He said staring at one true statement and one false statement.
> If all indices are used to enumerate the endsegments, then not all endsegments can contain aleph_0 natnumbers

He said staring at a simple counterexample.

Every element of |N_F is used in E= (E(1), E(2), E(3),...)
For every element i of |N_F E(i) has cardinality aleph_0

--
William Hughes

William schrieb am Donnerstag, 12. August 2021 um 15:45:34 UTC+2:

> Every element of |N_F is used in E= (E(1), E(2), E(3),...)
> For every element i of |N_F E(i) has cardinality aleph_0

Important however is that all elements are used. Think of Bijektion, Surjection and such stuff.

Regards, WM

On Thursday, August 12, 2021 at 9:48:33 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 12. August 2021 um 15:45:34 UTC+2:
>
> > Every element of |N_F is used in E= (E(1), E(2), E(3),...)
> > For every element i of |N_F E(i) has cardinality aleph_0
> Important however is that all elements are used.

He said staring at an example where every element of |N_F is used,

--
William Hughes

On 8/12/2021 8:36 AM, WM wrote:
> Jim Burns schrieb am Donnerstag, 12. August 2021 um 06:04:06 UTC+2:
>
>
>> No last end segment, intersection is empty.
>
> Every definable endsegment is a last endsegment.

there is no last endsegment.

>
>> I'm thinking about WM's appendix and...
>> Last end segment, intersection is the last end segment.
>> No last end segment, intersection is empty.
>>
>> WM rejects No Last End Segment.
>
> No, but I reject that last endsegments can yield a set with no last endsegment - other than potentially infinite. But that has always a last one.

only if *you stop* counting.

>
>> Instead, he defines into existence an appendix with the
>> property that it makes the intersection empty.
>
> Every actually infinite set is larger than any finite set.

duh...

>But definable endsegments belong to finite sets by definition.

your "definable", is not math, but imagination.

it creates "External Observer Dependent Math" which is a crock

>
> Regards, WM
>

On 8/12/2021 8:39 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 12. August 2021 um 06:51:48 UTC+2:
>> tisdag 10 augusti 2021 kl. 13:36:58 UTC+2 skrev WM:
>
>>> Among fools of matheology this may be a majority opinion.
>>>
>> mathematics, stop with this "matheology" thing,
>
> No! Never!! Unless you withdraw the belief that more than countably many real numbers could be well-ordered, i.e., distinguished as individuals.
>
> Regards, WM
>

"distinguished as individuals" is not a math term, and requires an
External Observer Dependent Math, which is a crock.

On 8/12/2021 8:29 AM, WM wrote:
> William schrieb am Mittwoch, 11. August 2021 um 22:19:51 UTC+2:
>
>> He said staring at one true statement and one false statement.
>
> If all indices are used to enumerate the endsegments, then not all endsegments can contain aleph_0 natnumbers.

obviously wrong.

indices have nothing to do with aleph_0 in endsegments.

> That is very simple to understand. Never a student of mine has failed since I invented this argument.

your students completely forgot your class after the final.

>
> Regards, WM
>

"invented" is not a term used in Math.

We use Proofs.

google Mathatical Proofs

On Thursday, August 12, 2021 at 3:36:29 PM UTC+2, WM wrote:

> endsegments belong to finite sets

Sure. Each and every set belongs to a finite set.

Hint: If A is a set then it belongs to the set {A}.

On Thursday, August 12, 2021 at 3:31:42 PM UTC+2, WM wrote:

> the intersection of all [...] endsegments [is]

{ } .

Errr... Sure. So what?

On Thursday, August 12, 2021 at 3:36:29 PM UTC+2, WM wrote:

> Every [...] infinite set is larger than any finite set.

Exactamundo, Mückenheim!

Another incredible insight!

On 8/12/2021 6:44 AM, Greg Cunt wrote:
> On Thursday, August 12, 2021 at 6:04:06 AM UTC+2,
> Jim Burns wrote:

>> In this case <bla>
>
> Once more, again:
>
> Hint: INTERSECTION M is what it is, ***completely***
> independent of any order defined on M.

The definition of the intersection of M does not mention
any order on M.

If M is ordered by inclusion, as the end segments are,
then another way to say
| L is in M, and L is a subset of each element of M
is
| L is last in M.

If L is in M, and L is a subset of each element of M,
then L is the intersection of M.

If L is last in M,
then L is the intersection of M,
because M is ordered by inclusion.

> It only depends on the CONTENT of M.
> [ A set doesen't have an order "per se" -
> even if there is a subset relation between the sets in M. ]

_In this case_ we have given the end segments the
order by inclusion. There is no logical requirement
that we give them that order, or any order, but we have
given them that order.

We can express the same thought about intersections and
elements which are subsets of each element without mentioning
any order. As long as we don't mind it taking longer and
being less clear.

On 8/12/2021 9:36 AM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 12. August 2021 um 06:04:06 UTC+2:

>> No last end segment, intersection is empty.
>
> Every definable endsegment is a last endsegment.

You have changed what "last"means so that there are two kinds
of "last", those that change and those that do not change.

For you, WM, I will say it this way:

No [unchanging] last end segment, intersection is empty.

>> I'm thinking about WM's appendix and...
>> Last end segment, intersection is the last end segment.
>> No last end segment, intersection is empty.
>>
>> WM rejects No Last End Segment.
>
> No, but I reject that last endsegments

....[changing] last end segments..

> can yield a set with no last endsegment

....no [unchanging] last end segment...

> - other than potentially infinite.
> But that has always a last one.

....a [changing] last one...

Once we make clear that you have changing and unchanging
end segments, we see that you are saying:

| Every definable end segment is a [changing] last end segment.

| I (WM) reject that [changing] last end segments can yield
| a set with no [unchanging] last end segments.
| - other than potentially infinite.
| But that has always a [changing] last one.

How you manage to hold both these propositions in your
head at the same time without your head imploding is by
obscuring the difference between changing and unchanging.
This is a textbook case of equivocation, a fallacy.

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

>> Instead, he defines into existence an appendix with the
>> property that it makes the intersection empty.
>
> Every actually infinite set is larger than any finite set.
> But definable endsegments belong to finite sets by definition.

How I read
| E(k) belongs to a finite set
is
| The set {E(1),...,E(k)} of definable end segments between
| E(1) and definable E(k) is finite.

That's true. This is also true:
| No set {E(1),...,E(k)} of definable end segments between
| E(1) and definable E(k) contains each definable end segment.

Because
Every definable end segment is a [changing] last end segment.
No definable end segment is an [unchanging] last end segment.
Each {E(1),...,E(k)} can be expanded to include more
definable end segments.

William schrieb am Donnerstag, 12. August 2021 um 15:54:10 UTC+2:
> On Thursday, August 12, 2021 at 9:48:33 AM UTC-4, WM wrote:
> > William schrieb am Donnerstag, 12. August 2021 um 15:45:34 UTC+2:
> >
> > > Every element of |N_F is used in E= (E(1), E(2), E(3),...)
> > > For every element i of |N_F E(i) has cardinality aleph_0
> > Important however is that all elements are used.
> He said staring at an example where every element of |N_F is used,

Every is not enough. Every has almost all as successors.

Regards, WM

Jim Burns schrieb am Donnerstag, 12. August 2021 um 20:16:47 UTC+2:
> On 8/12/2021 6:44 AM, Greg Cunt wrote:
> > On Thursday, August 12, 2021 at 6:04:06 AM UTC+2,
> > Jim Burns wrote:

> > Hint: INTERSECTION M is what it is, ***completely***
> > independent of any order defined on M.
> The definition of the intersection of M does not mention
> any order on M.

Therefore all endsegments satisfying
∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
can be combined to
|∩{E(k) : k ∈ ℕ_def}| = ℵo .

Regards, WM

On 8/14/2021 9:16 AM, WM wrote:
> William schrieb am Donnerstag, 12. August 2021 um 15:54:10 UTC+2:
>> On Thursday, August 12, 2021 at 9:48:33 AM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 12. August 2021 um 15:45:34 UTC+2:
>>>
>>>> Every element of |N_F is used in E= (E(1), E(2), E(3),...)
>>>> For every element i of |N_F E(i) has cardinality aleph_0
>>> Important however is that all elements are used.
>> He said staring at an example where every element of |N_F is used,
>
> Every is not enough. Every has almost all as successors.
>
> Regards, WM
>

Every Ants

All Ants

Almost Ants

Successor Ants

Not Enough Ants

Every Ant has a successor

Almost Every Ant has a successor

All Successor Ants

..
..
..
..

On 8/13/2021 1:23 PM, Jim Burns wrote:
> On 8/12/2021 9:36 AM, WM wrote:
>> Jim Burns schrieb
>> am Donnerstag, 12. August 2021 um 06:04:06 UTC+2:
>
>>> No last end segment, intersection is empty.
>>
>> Every definable endsegment is a last endsegment.
>
> You have changed what "last"means so that there are two kinds
> of "last", those that change and those that do not change.
>
> For you, WM, I will say it this way:
>
> No [unchanging] last end segment, intersection is empty.
>
>>> I'm thinking about WM's appendix and...
>>> Last end segment, intersection is the last end segment.
>>> No last end segment, intersection is empty.
>>>
>>> WM rejects No Last End Segment.
>>
>> No, but I reject that last endsegments
>
> ...[changing] last end segments..
>
>> can yield a set with no last endsegment
>
> ...no [unchanging] last end segment...
>
>> - other than potentially infinite.
>> But that has always a last one.
>
> ...a [changing] last one...
>
> Once we make clear that you have changing and unchanging
> end segments, we see that you are saying:
>
> | Every definable end segment is a [changing] last end segment.
>
> | I (WM) reject that [changing] last end segments can yield
> | a set with no [unchanging] last end segments.
> | - other than potentially infinite.
> | But that has always a [changing] last one.
>
> How you manage to hold both these propositions in your
> head at the same time without your head imploding is by
> obscuring the difference between changing and unchanging.
> This is a textbook case of equivocation, a fallacy.
>
> https://en.wikipedia.org/wiki/Equivocation
>
>>> Instead, he defines into existence an appendix with the
>>> property that it makes the intersection empty.
>>
>> Every actually infinite set is larger than any finite set.
>> But definable endsegments belong to finite sets by definition.
>
> How I read
> | E(k) belongs to a finite set
> is
> | The set {E(1),...,E(k)} of definable end segments between
> | E(1) and definable E(k) is finite.
>
> That's true. This is also true:
> | No set {E(1),...,E(k)} of definable end segments between
> | E(1) and definable E(k) contains each definable end segment.
>
> Because
> Every definable end segment is a [changing] last end segment.
> No definable end segment is an [unchanging] last end segment.
> Each {E(1),...,E(k)} can be expanded to include more
> definable end segments.
>

how about using cLast and uLast ?

Jim Burns schrieb am Freitag, 13. August 2021 um 20:23:30 UTC+2:
> On 8/12/2021 9:36 AM, WM wrote:
> > Jim Burns schrieb
> > am Donnerstag, 12. August 2021 um 06:04:06 UTC+2:
>
> >> No last end segment, intersection is empty.
> >
> > Every definable endsegment is a last endsegment.
> You have changed what "last"means so that there are two kinds
> of "last", those that change and those that do not change.

The endsegments satifying
∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
dco not change. But their being last does change if we add another endsegment.
>
> For you, WM, I will say it this way:
>
> No [unchanging] last end segment, intersection is empty.

This can only be realized by endsegments which are never last.
> >>
> >> WM rejects No Last End Segment.
> >
> > No, but I reject that last endsegments
> ...[changing] last end segments..
> > can yield a set with no last endsegment

> ...no [unchanging] last end segment...
> > - other than potentially infinite.
> > But that has always a last one.
> ...a [changing] last one...
>
> Once we make clear that you have changing and unchanging
> end segments, we see that you are saying:
>
> | Every definable end segment is a [changing] last end segment.
>
> | I (WM) reject that [changing] last end segments can yield
> | a set with no [unchanging] last end segments.
> | - other than potentially infinite.
> | But that has always a [changing] last one.
>
> How you manage to hold both these propositions in your
> head at the same time without your head imploding is by
> obscuring the difference between changing and unchanging.
> This is a textbook case of equivocation, a fallacy.

Every definable endsegment is the last one of its Finite initial segment. It can never make a set larger than finite.
>
> > Every actually infinite set is larger than any finite set.
> > But definable endsegments belong to finite sets by definition.
> How I read
> | E(k) belongs to a finite set
> is
> | The set {E(1),...,E(k)} of definable end segments between
> | E(1) and definable E(k) is finite.
>
> That's true. This is also true:
> | No set {E(1),...,E(k)} of definable end segments between
> | E(1) and definable E(k) contains each definable end segment.

That is true because the collection is potentially infinite.
>
> Because
> Every definable end segment is a [changing] last end segment.
> No definable end segment is an [unchanging] last end segment.
> Each {E(1),...,E(k)} can be expanded to include more
> definable end segments.

Of course. What is difficult with this notion? There is only one important point: You must not confuse pot. and act. as set theorists deplorably deliberately do.

Regards, WM