WM wrote :
> William schrieb am Freitag, 10. Dezember 2021 um 18:27:29 UTC+1:
>> On Friday, December 10, 2021 at 12:32:46 PM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 9. Dezember 2021 um 18:58:59 UTC+1:
>>>> On Thursday, December 9, 2021 at 12:41:47 PM UTC-4, WM wrote:
>>>>
>>>>> The potentially infinite set {E(1), E(2), ..., E(k)} with not fixed k
>>>> "Potentially infinite set" is nonsense.
>>> Then use the collection C of all endsegments which together have an
>>> infinite intersection.
>> No such thing as "the" collection. This is something that changes, so is not
>> a set.
>
> Therefore it is called a collection.

Is "Collection Theory" going to be the foundation of Muckymath? Are
there axioms for "CT" so we can test if your "collection" adheres to
the axioms?

On Saturday, December 11, 2021 at 12:17:43 PM UTC+1, WM wrote:
> Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
> > On 12/10/2021 11:24 AM, WM wrote:
> > > Jim Burns schrieb
> > > am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
> > >> On 12/9/2021 11:41 AM, WM wrote:
> >
> > >>> One and the same set cannot yield
> > >>> and infinite intersection and an empty intersection
> > We've proved that any number of times.
> > As a consequence, C = {END}
> Your "proofs" are contradicted by the fact that a collection C with infinite intersection cannot be the same as a collection or set {END} with empty intersection. Therefore they are worthless and deleted.
> > There is no n which is in all end segments.
> > See E(n+1).
> But there are infinitely many n which are in all endsegments.
> > Therefore, the answer to
> > "Is the intersection of all end segments empty?"
> > is _yes_
> > while the answer to the different question
> > "Is any end segment empty?"
> > is _no_
> Sober minds are not able to tolerate that contradiction.
>
Try half a fifth of Glenfiddich. It cures what ails you.

on 12/11/2021, WM supposed :
> Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>> On 12/10/2021 11:24 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
>>>> On 12/9/2021 11:41 AM, WM wrote:
>>
>>>>> One and the same set cannot yield
>>>>> and infinite intersection and an empty intersection
>
>> We've proved that any number of times.
>> As a consequence, C = {END}
>
> Your "proofs" are contradicted by the fact that a collection C with infinite
> intersection cannot be the same as a collection or set {END} with empty
> intersection. Therefore they are worthless and deleted.
>
>> There is no n which is in all end segments.
>> See E(n+1).
>
> But there are infinitely many n which are in all endsegments.
>
True. However, there is no specific element which is in all
endsegments. Damn, if we only had a last element we could claim it was
in all endsegments. We don't though, so we have to rely upon Muckymath
where the "Axiom of Because I Said So" reigns supreme.

On 12/11/2021 6:27 AM, FromTheRafters wrote:
> WM wrote :
>> William schrieb am Freitag, 10. Dezember 2021 um 18:27:29 UTC+1:
>>> On Friday, December 10, 2021 at 12:32:46 PM UTC-4, WM wrote:
>>>> William schrieb am Donnerstag, 9. Dezember 2021 um 18:58:59 UTC+1:
>>>>> On Thursday, December 9, 2021 at 12:41:47 PM UTC-4, WM wrote:
>>>>>> The potentially infinite set {E(1), E(2), ..., E(k)} with not fixed k
>>>>> "Potentially infinite set" is nonsense.
>>>> Then use the collection C of all endsegments which together have an infinite intersection.
>>> No such thing as "the" collection. This is something that changes, so is not a set.
>>
>> Therefore it is called a collection.
>
> Is "Collection Theory" going to be the foundation of Muckymath? Are there axioms for "CT" so we can test if your "collection" adheres to the axioms?

Ants collect things all the time!

Collection Theory Ants that collect adherent Axioms

note: to have a collection, one must write them all down on the SAME piece of paper...

On Saturday, December 11, 2021 at 7:05:51 AM UTC-4, WM wrote:
> William schrieb am Freitag, 10. Dezember 2021 um 18:27:29 UTC+1:
> > On Friday, December 10, 2021 at 12:32:46 PM UTC-4, WM wrote:
> > > William schrieb am Donnerstag, 9. Dezember 2021 um 18:58:59 UTC+1:
> > > > On Thursday, December 9, 2021 at 12:41:47 PM UTC-4, WM wrote:
> > > >
> > > > > The potentially infinite set {E(1), E(2), ..., E(k)} with not fixed k
> > > > "Potentially infinite set" is nonsense.
> > > Then use the collection C of all endsegments which together have an infinite intersection.
> > No such thing as "the" collection. This is something that changes, so is not a set.
> Therefore it is called a collection.

Indeed, a collection, something that changes. No matter how many times it changes it is finite.
(Finite but unbounded makes sense for something that changes),

> The collection of all endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is not the set of all endsegments

Yes it is. (Note this is not " the collection C of all endsegments which together have an infinite intersection", it is the set of all endsegments such that the endsegment and its predecessors have infinite intersection. Note further that this is a set as well as a collection)

---
William Hughes

JVR schrieb am Samstag, 11. Dezember 2021 um 13:35:45 UTC+1:
> On Saturday, December 11, 2021 at 12:17:43 PM UTC+1, WM wrote:
> > Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:

> > > Therefore, the answer to
> > > "Is the intersection of all end segments empty?"
> > > is _yes_

That requires that no natural number remains in all endsegments.

> > > while the answer to the different question
> > > "Is any end segment empty?"
> > > is _no_

That requires that infinitely many natural numbers remain in all endsegments. And they are not exchanged.

> > Sober minds are not able to tolerate that contradiction.
> >
> Try half a fifth of Glenfiddich. It cures what ails you.

That may be the only cure, to accept that infinite endsegments contain infinitely many numbers none of which is in all infinite endsegments. Alas I am a teetotaller.

Regards, WM

FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:37:24 UTC+1:
> on 12/11/2021, WM supposed :
> > Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
> >> On 12/10/2021 11:24 AM, WM wrote:
> >>> Jim Burns schrieb
> >>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
> >>>> On 12/9/2021 11:41 AM, WM wrote:
> >>
> >>>>> One and the same set cannot yield
> >>>>> and infinite intersection and an empty intersection
> >
> >> We've proved that any number of times.
> >> As a consequence, C = {END}
> >
> > Your "proofs" are contradicted by the fact that a collection C with infinite
> > intersection cannot be the same as a collection or set {END} with empty
> > intersection. Therefore they are worthless and deleted.
> >
> >> There is no n which is in all end segments.
> >> See E(n+1).
> >
> > But there are infinitely many n which are in all endsegments.
> >
> True. However, there is no specific element which is in all
> endsegments.

Infinitely many remaining in all infinite endsegments from the first E(1) on imply that finitely many and therefore also specific natnumbers remain.

> Damn, if we only had a last element we could claim it was
> in all endsegments.

It is sufficient to know that all infinite definable endsegments have an infinite intersection. It is impossible to define endsegments which have not. Try it! You can only state that the infinite set of all endsegments has an empty intersection. But that proves that you cannot define the endsegments required to get the empty intersection.

Regards, WM

FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:27:46 UTC+1:
> WM wrote :

> >> No such thing as "the" collection. This is something that changes, so is not
> >> a set.
> >
> > Therefore it is called a collection.
> Is "Collection Theory" going to be the foundation of Muckymath? Are
> there axioms for "CT" so we can test if your "collection" adheres to
> the axioms?

We need no axioms. We need only the answer to the question: Why are there no empty endsegments when "n ∉ E(n+1)" removes every content? Only what is removed from an endsegment is removed from the intersection.

The answer shows whether you are a mathematician or a fool.

Regards, WM

William schrieb am Samstag, 11. Dezember 2021 um 16:37:05 UTC+1:
> On Saturday, December 11, 2021 at 7:05:51 AM UTC-4, WM wrote:

> Indeed, a collection, something that changes. No matter how many times it changes it is finite.
> (Finite but unbounded makes sense for something that changes),

That is the set of endsegments with infinite intersection, for instance.
>
> > The collection of all endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is not the set of all endsegments
>
> Yes it is. (Note this is not " the collection C of all endsegments which together have an infinite intersection", it is the set of all endsegments such that the endsegment and its predecessors have infinite intersection. Note further that this is a set as well as a collection)

If this is the set of all endsegments, then there are no successors remaining. There are no successors to all endsegments. Then the intersection over all depends only on predecessors. This intersection is infinite. But the intersection over all is also empty.
Contradition in case of one and the same set.

Regards, WM

On 12/11/2021 12:18 PM, WM wrote:
> FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:27:46 UTC+1:
>> WM wrote :
>
>>>> No such thing as "the" collection. This is something that changes, so is not
>>>> a set.
>>>
>>> Therefore it is called a collection.
>> Is "Collection Theory" going to be the foundation of Muckymath? Are
>> there axioms for "CT" so we can test if your "collection" adheres to
>> the axioms?
>
> We need no axioms. We need only the answer to the question: Why are there no empty endsegments when "n ∉ E(n+1)" removes every content? Only what is removed from an endsegment is removed from the intersection.
>
> The answer shows whether you are a mathematician or a fool.
>
> Regards, WM

What "element" is in the Intersection of all Endsegments ?

We know that "element" ∉ E("element" + 1), therefore "element" is cannot be in the Intersection.

There are no elements in the Intersection of all Endsegments.

Is the Intersection of all Endsegments, an Endsegment itself ? No, not at all.

On 12/11/2021 12:03 PM, WM wrote:
> JVR schrieb am Samstag, 11. Dezember 2021 um 13:35:45 UTC+1:
>> On Saturday, December 11, 2021 at 12:17:43 PM UTC+1, WM wrote:
>>> Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>
>>>> Therefore, the answer to
>>>> "Is the intersection of all end segments empty?"
>>>> is _yes_
>
> That requires that no natural number remains in all endsegments.

Wrong. It requires that no natural number is in all endsegments.

>
>>>> while the answer to the different question
>>>> "Is any end segment empty?"
>>>> is _no_
>
> That requires that infinitely many natural numbers remain in all endsegments. And they are not exchanged.

there are no "exchanged", "removed", "remain".
Endsegments are fixed sets.

>
>>> Sober minds are not able to tolerate that contradiction.
>>>
>> Try half a fifth of Glenfiddich. It cures what ails you.
>
> That may be the only cure, to accept that infinite endsegments contain infinitely many numbers none of which is in all infinite endsegments.

if no number is common to all endsegments, the intersection of all endsegments is empty

Alas I am a teetotaller.
>
> Regards, WM

On 12/11/2021 12:11 PM, WM wrote:
> FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:37:24 UTC+1:
>> on 12/11/2021, WM supposed :
>>> Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>>>> On 12/10/2021 11:24 AM, WM wrote:
>>>>> Jim Burns schrieb
>>>>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
>>>>>> On 12/9/2021 11:41 AM, WM wrote:
>>>>
>>>>>>> One and the same set cannot yield
>>>>>>> and infinite intersection and an empty intersection
>>>
>>>> We've proved that any number of times.
>>>> As a consequence, C = {END}
>>>
>>> Your "proofs" are contradicted by the fact that a collection C with infinite
>>> intersection cannot be the same as a collection or set {END} with empty
>>> intersection. Therefore they are worthless and deleted.
>>>
>>>> There is no n which is in all end segments.
>>>> See E(n+1).
>>>
>>> But there are infinitely many n which are in all endsegments.
>>>
>> True. However, there is no specific element which is in all
>> endsegments.
>
> Infinitely many remaining in all infinite endsegments from the first E(1) on imply that finitely many and therefore also specific natnumbers remain.

However, there is no specific element which is in all endsegments. the intersetion of all endsegments is empty.

>
>> Damn, if we only had a last element we could claim it was
>> in all endsegments.
>
> It is sufficient to know that all infinite definable endsegments have an infinite intersection. It is impossible to define endsegments which have not. Try it! You can only state that the infinite set of all endsegments has an empty intersection. But that proves that you cannot define the endsegments required to get the empty intersection.

wrong approach.

You tell me, which number is common to all endsegments ?

if you can name one, the intersection is not empty.
if you cannot name any, the intersection is empty.

>
> Regards, WM

WM wrote on 12/11/2021 :
> FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:27:46 UTC+1:
>> WM wrote :
>
>>>> No such thing as "the" collection. This is something that changes, so is
>>>> not a set.
>>>
>>> Therefore it is called a collection.
>> Is "Collection Theory" going to be the foundation of Muckymath? Are
>> there axioms for "CT" so we can test if your "collection" adheres to
>> the axioms?
>
> We need no axioms. We need only the answer to the question: Why are there no
> empty endsegments

Because endsegments are non-empty.

Wow, it really was that easy. :D

WM used his keyboard to write :
> FromTheRafters schrieb am Samstag, 11. Dezember 2021 um 13:37:24 UTC+1:
>> on 12/11/2021, WM supposed :
>>> Jim Burns schrieb am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>>>> On 12/10/2021 11:24 AM, WM wrote:
>>>>> Jim Burns schrieb
>>>>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
>>>>>> On 12/9/2021 11:41 AM, WM wrote:
>>>>>>> One and the same set cannot yield
>>>>>>> and infinite intersection and an empty intersection
>>>> We've proved that any number of times.
>>>> As a consequence, C = {END}
>>>
>>> Your "proofs" are contradicted by the fact that a collection C with
>>> infinite intersection cannot be the same as a collection or set {END} with
>>> empty intersection. Therefore they are worthless and deleted.
>>>
>>>> There is no n which is in all end segments.
>>>> See E(n+1).
>>>
>>> But there are infinitely many n which are in all endsegments.
>>>
>> True. However, there is no specific element which is in all
>> endsegments.
>
> Infinitely many remaining in all infinite endsegments from the first E(1) on
> imply that finitely many and therefore also specific natnumbers remain.

You're babbling again.

>> Damn, if we only had a last element we could claim it was
>> in all endsegments.
>
> It is sufficient to know that all infinite definable endsegments

Snipped more babbling.

Serg io schrieb am Samstag, 11. Dezember 2021 um 19:31:03 UTC+1:

> What "element" is in the Intersection of all Endsegments ?
>
> We know that "element" ∉ E("element" + 1), therefore "element" is cannot be in the Intersection.

What "element" is not removed from all endsegments?
If there are infinitely many elements in all Endsegments, then they are remaining from the first endsegment. Then at least one is remaining from the first endsegment.

Regards, WM

On 12/11/2021 1:14 PM, WM wrote:
> Serg io schrieb am Samstag, 11. Dezember 2021 um 19:31:03 UTC+1:
>
>> What "element" is in the Intersection of all Endsegments ?
>>
>> We know that "element" ∉ E("element" + 1), therefore "element" is cannot be in the Intersection.
>
> What "element" is not removed from all endsegments?

there are no elements common to all endsegments, do you agree ?

> If there are infinitely many elements in all Endsegments, then they are remaining from the first endsegment.

nope. each endsegment is a fixed non-changing set, do you agree ?

> Then at least one is remaining from the first endsegment.

under what conditions?

>
> Regards, WM

On Saturday, December 11, 2021 at 2:29:49 PM UTC-4, WM wrote:
> William schrieb am Samstag, 11. Dezember 2021 um 16:37:05 UTC+1:
> > On Saturday, December 11, 2021 at 7:05:51 AM UTC-4, WM wrote:
>
> > > The collection of all endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is not the set of all endsegments
> >
> > Yes it is. (Note this is not " the collection C of all endsegments which together have an infinite intersection", it is the set of all endsegments such that the [single] endsegment and its predecessors have infinite intersection. Note further that this is a set as well as a collection)
> If this is the set of all endsegments [,E,], then there are no successors remaining.

This is trivially true.

>There are no successors to all endsegments. Then the intersection over all depends only on predecessors.

But not to predecessors of a single endsegment. E contains an infinite number of endsegments
For any natural number k the set {E(1),E(2),...,E(k)} contains a finite number of endegments. The intersetction over
a finite set of endsegments is infinite, the intersetction over an infinite number of endsegments is null.

--
William Hughes

On 12/11/2021 6:17 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>> On 12/10/2021 11:24 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
>>>> On 12/9/2021 11:41 AM, WM wrote:

>>>>> One and the same set cannot yield
>>>>> and infinite intersection and an empty intersection

>>> We use the collection C of all these endsegments which
>>> together with their predecessors have an infinite
>>> intersection.

>> cC = { E' ∈ {END} | ℵ₀ =< ∩{before E'} }

>>>> An end segment and all of its _predecessors_ have
>>>> an infinite intersection.
>>
>> ∀E' ∈ {END}, ℵ₀ =< ∩{before E'}

>> We've proved that any number of times.
>> As a consequence, C = {END}
>
> Your "proofs" are contradicted by the fact that
> a collection C with infinite intersection

That's not C.
You:
>>> We use the collection C of all these endsegments which
>>> together with their predecessors have an infinite
>>> intersection.

You have equivocated on what C is. Which is a fallacy.

> Your "proofs" are contradicted by the fact that
> a collection C with infinite intersection

No, see above.
C is the collection of E' with ℵ₀ =< |∩{before E'}|

ℵ₀ =< |∩C| is a different claim.

> a collection C with infinite intersection
> cannot be the same as a collection or set {END}
> with empty intersection.

None of the end-segment collections with infinite
intersection are {END} so that's all right, then.

( Let {RANDO} ⊆ {END} be a collection of end segments
( and ℵ₀ =< |∩{RANDO}|
( ( Assume k ∈ ∩{RANDO}
( E(k+1) ∉ {RANDO} but E(k+1) ∈ {END}
( {RANDO} ≠ {END}

On 12/11/2021 5:01 PM, Jim Burns wrote:
> On 12/11/2021 6:17 AM, WM wrote:
>> Jim Burns schrieb
>> am Freitag, 10. Dezember 2021 um 19:44:02 UTC+1:
>>> On 12/10/2021 11:24 AM, WM wrote:
>>>> Jim Burns schrieb
>>>> am Freitag, 10. Dezember 2021 um 14:37:25 UTC+1:
>>>>> On 12/9/2021 11:41 AM, WM wrote:
>
>>>>>> One and the same set cannot yield
>>>>>> and infinite intersection and an empty intersection
>
>>>> We use the collection C of all these endsegments which
>>>> together with their predecessors have an infinite
>>>> intersection.
>
>>> cC  =  { E' ∈ {END} | ℵ₀ =< ∩{before E'} }
>
>>>>> An end segment and all of its _predecessors_ have
>>>>> an infinite intersection.
>>>
>>> ∀E' ∈ {END}, ℵ₀ =< ∩{before E'}
>
>>> We've proved that any number of times.
>>> As a consequence, C = {END}
>>
>> Your "proofs" are contradicted by the fact that
>> a collection C with infinite intersection
>
> That's not C.
> You:
>>>> We use the collection C of all these endsegments which
>>>> together with their predecessors have an infinite
>>>> intersection.
>
> You have equivocated on what C is. Which is a fallacy.

Fallacy bad.

>> Your "proofs" are contradicted by the fact that
>> a collection C with infinite intersection
>
> No, see above.
> C is the collection of E' with  ℵ₀ =< |∩{before E'}|
>
> ℵ₀ =< |∩C| is a different claim.
>
>> a collection C with infinite intersection
>> cannot be the same as a collection or set {END}
>> with empty intersection.
>
> None of the end-segment collections with infinite
> intersection are {END}  so that's all right, then.
>
> ( Let {RANDO} ⊆ {END} be a collection of end segments
> ( and  ℵ₀ =< |∩{RANDO}|
> (
> ( Assume k ∈ ∩{RANDO}
> ( E(k+1) ∉ {RANDO}  but  E(k+1) ∈ {END}
> ( {RANDO} ≠ {END}
>
>

William schrieb am Samstag, 11. Dezember 2021 um 21:06:27 UTC+1:
> On Saturday, December 11, 2021 at 2:29:49 PM UTC-4, WM wrote:

> > If this is the set E of all endsegments E(n) then there are no successors remaining.
> This is trivially true.
> >There are no successors to all endsegments. Then the intersection over all depends only on predecessors.
> But not to predecessors of a single endsegment. E contains an infinite number of endsegments

All of them are predecessors.

> For any natural number k the set {E(1),E(2),...,E(k)} contains a finite number of endegments. The intersetction over
> a finite set of endsegments is infinite, the intersetction over an infinite number of endsegments is null.

All endsegments belong to finite sets {E(1),E(2),...,E(k)}. Therefore there is nothing remaining to reduce the infinite intersection.

The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
The intersection of E(1) = ℕ.
The intersection of all endsegments is empty.
The intersection of all infinite endsegments is infinite because infinitely many numbers have not been lost by definition.

Regards, WM

Jim Burns schrieb am Samstag, 11. Dezember 2021 um 23:01:18 UTC+1:
> On 12/11/2021 6:17 AM, WM wrote:

> > Your "proofs" are contradicted by the fact that
> > a collection C with infinite intersection
> That's not C.
> You:
> >>> We use the collection C of all these endsegments which
> >>> together with their predecessors have an infinite
> >>> intersection.
> You have equivocated on what C is.

No. Your claim requires to find an infinite endsegment by which the two definitions differ. But there is none. Therefore the definitions define one and the same C.

> None of the end-segment collections with infinite
> intersection are {END} so that's all right, then.

But there are no further infinite endsegments beyond C.

The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
The intersection of E(1) = ℕ.
The intersection of all endsegments is empty.
The intersection of all infinite endsegments is infinite because infinitely many numbers have not been lost by definition.

But all must be lost by
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
in single steps!

You refuse to accept this equation, you refuse to accept inclusion monotony, you refuse to accept the pigeon hole principle, ...
You refuse to accept mathematics.

Regards, WM

On 12/12/2021 6:22 AM, WM wrote:
> Jim Burns schrieb
> am Samstag, 11. Dezember 2021 um 23:01:18 UTC+1:
>> On 12/11/2021 6:17 AM, WM wrote:

>>> Your "proofs" are contradicted by the fact that
>>> a collection C with infinite intersection
>>
>> That's not C.
>> You:
>>>>> We use the collection C of all these endsegments which
>>>>> together with their predecessors have an infinite
>>>>> intersection.
>>
>> You have equivocated on what C is.
>
> No.

Show a statement-path using only reliable, valid steps
which starts at
| For each end segment E,
| E' is in C if and only if
| E' together with the predecessors of E'
| have an infinite intersection

and ends at
| C has an infinite intersection

Quantifier shift is not a reliable, valid step.

> Your claim requires to find an infinite endsegment
> by which the two definitions differ.

Let E" be the first end segment after E'
E" is not among
E' together with the predecessors of E'

But E" is among "all these endsegments which together
with their predecessors have an infinite intcersection".

> But there is none.
> Therefore the definitions define one and the same C.
>
>> None of the end-segment collections with infinite
>> intersection are {END} so that's all right, then.
>
> But there are no further infinite endsegments beyond C.

Right.
And there are end segments beyond {before E'}
So, you equivocated. Which is a fallacy.
(Fallacy bad.)

On 12/12/2021 5:12 AM, WM wrote:
> William schrieb am Samstag, 11. Dezember 2021 um 21:06:27 UTC+1:
>> On Saturday, December 11, 2021 at 2:29:49 PM UTC-4, WM wrote:
>
>>> If this is the set E of all endsegments E(n) then there are no successors remaining.
>> This is trivially true.
>>> There are no successors to all endsegments. Then the intersection over all depends only on predecessors.
>> But not to predecessors of a single endsegment. E contains an infinite number of endsegments
>
> All of them are predecessors.
>
>> For any natural number k the set {E(1),E(2),...,E(k)} contains a finite number of endegments. The intersetction over
>> a finite set of endsegments is infinite, the intersetction over an infinite number of endsegments is null.
>
> All endsegments belong to finite sets {E(1),E(2),...,E(k)}.

wrong, you stopped at k !

> Therefore there is nothing remaining to reduce the infinite intersection.

only if you stop at k

>
> The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).

no, stop mutilating the math. Try again

> The intersection of E(1) = ℕ.

no, E(1) = ℕ (intersection requires at least 2 sets.)

> The intersection of all endsegments is empty.

true. all endsegments are infinite fixed sets they do not change

> The intersection of all infinite endsegments is infinite because infinitely many numbers have not been lost by definition.

Wrong. conflicts with above

>
> Regards, WM
>

On Sunday, December 12, 2021 at 7:12:55 AM UTC-4, WM wrote:
> William schrieb am Samstag, 11. Dezember 2021 um 21:06:27 UTC+1:
a finite set of endsegments is infinite, the intersetction over an infinite number of endsegments is null.
> All endsegments belong to finite sets {E(1),E(2),...,E(k)}.

And all ensegments belong to an infinite set. The fact that a set , A, contains endsegments does not tell you if A is finite (infinite intersection) , or if A is infinite (null intersection)

> <snip> The intersection of all infinite endsegments is

null, because no natural number is in every endsegment. (Recall that the Peano set of natural numbers does not contain "dark"
elements;)

> infinite because infinitely many numbers have not been lost

False, Every element of the Peano set of natural numbers is "lost".

--
William Hughes

On 12/12/2021 5:22 AM, WM wrote:
> Jim Burns schrieb am Samstag, 11. Dezember 2021 um 23:01:18 UTC+1:
>> On 12/11/2021 6:17 AM, WM wrote:
>
>>> Your "proofs" are contradicted by the fact that
>>> a collection C with infinite intersection
>> That's not C.
>> You:
>>>>> We use the collection C of all these endsegments which
>>>>> together with their predecessors have an infinite
>>>>> intersection.
>> You have equivocated on what C is.
>
> No. Your claim requires to find an infinite endsegment by which the two definitions differ. But there is none. Therefore the definitions define one and the same C.
>

no, use math and Prove it.

>> None of the end-segment collections with infinite
>> intersection are {END} so that's all right, then.
>
> But there are no further infinite endsegments beyond C.
>
> The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).

wrong.

> The intersection of E(1) = ℕ.

wrong.

> The intersection of all endsegments is empty.

true.

> The intersection of all infinite endsegments is infinite because infinitely many numbers have not been lost by definition.
>

Wrong.

> But all must be lost by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> in single steps!

you show no such steps, and that is not how one uses that equation!

>
> You refuse to accept this equation, you refuse to accept inclusion monotony, you refuse to accept the pigeon hole principle, ...
> You refuse to accept mathematics.

I refuse your math, anyone who took Algebra 1 will tell you that your math is wrong.

Are all endsegments fixed sets that do not change ?

>
> Regards, WM