On 12/8/2021 2:40 PM, Chris M. Thomasson wrote:
> On 12/8/2021 9:15 AM, WM wrote:
>> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:34:06
>> UTC+1:
>>> onsdag 8 december 2021 kl. 11:42:17 UTC+1 skrev WM:
>>>> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:11
>>>> UTC+1:
>>>>
>>>>> There is no such thing as "potential infinite"
>>>> You are wrong. By endsegments you see the difference between pot and
>>>> act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
>>>> infinite but less than the sets which have an empty intersection
>>>> |∩{E(1), E(2), ...}| = 0, like the set of all endsegments. They are
>>>> actually infinite. That is much more than all finite sets.
>>
>>> Nope, I am right because you cannot define what your "potential
>>> infinite" even fucking mathematically mean for sets.
>>
>> It means finite but without a last element.
> [...]
>
> Huh? That's infinite. Look...
>
> { 0 }
> { { 0 }, 1 }
> { { { 0 }, 1 }, 2 }
> ...
>
> This is: infinity!

Oh, look... I can even index them! lol:

[0] = { 0 }
[1] = { { 0 }, 1 }
[2] = { { { 0 }, 1 }, 2 }
....

;^)

Jim Burns wrote on 12/8/2021 :
> On 12/8/2021 1:02 PM, WM wrote:
>
>> For all FISONs the set of successors is infinite.
>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
> ℕ = ∪{FISON'}
>
> ∀n ∈ ℕ: |ℕ\{0,...,n}| >= ℵ₀
>
>> Note that finite is less than infinite.
>> So we can state:
>> Every FISON cover less than half of all natural numbers.
>
> "Less than half than half of all natural numbers"
> needs to be translated.
>
> How about this?
> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|
>
>> This means that there is an infinite intersection of
>> all successors.
>
> This means that, for any natural number k,
> there are ways that what follows k can be totally-ordered
> (NOT steppable-and-two-ended ways)
> and there are ways theycannot be totally-ordered
> (steppable-and-two-ended ways).
>
>> Here quantifier exchange is correct.
>
> Here quantifier shift is incorrect.
>
> A quantifier shift is ALWAYS an unreliable step.
>
> Because we are looking for statement-paths for which
> each step is reliable, a quantifier shift is always
> incorrect for such a path.
>
> That doesn't rule out some other reliable-step path that
> doesn't have a quantifier shift. However, proclaiming
> that here is correct doesn't make a path reliable.
> Compare to:
> Proclaiming that your checking account is not overdrawn
> doesn't put money in your account.

Uh oh, Don't go there -- remember McDuck?

On 12/8/2021 6:16 PM, FromTheRafters wrote:
> Jim Burns wrote on 12/8/2021 :
>> On 12/8/2021 1:02 PM, WM wrote:

>>> Here quantifier exchange is correct.
>>
>> Here quantifier shift is incorrect.
>>
>> A quantifier shift is ALWAYS an unreliable step.
>>
>> Because we are looking for statement-paths for which
>> each step is reliable, a quantifier shift is always
>> incorrect for such a path.
>>
>> That doesn't rule out some other reliable-step path that
>> doesn't have a quantifier shift. However, proclaiming
>> that here is correct doesn't make a path reliable.
>> Compare to:
>> Proclaiming that your checking account is not overdrawn
>> doesn't put money in your account.
>
> Uh oh, Don't go there -- remember McDuck?

Well, you know what they say,
that that is is that that isn't isn't but
that that isn't isn't that that is and
that that is isn't that that isn't.

If McDuck is overdrawn, he's overdrawn.

On 12/8/2021 5:16 PM, FromTheRafters wrote:
> Jim Burns wrote on 12/8/2021 :
>> On 12/8/2021 1:02 PM, WM wrote:
>>
>>> For all FISONs the set of successors is infinite.
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>>
>> ℕ = ∪{FISON'}
>>
>> ∀n ∈ ℕ: |ℕ\{0,...,n}| >= ℵ₀
>>
>>> Note that finite is less than infinite.
>>> So we can state:
>>> Every FISON cover less than half of all natural numbers.
>>
>> "Less than half than half of all natural numbers"
>> needs to be translated.
>>
>> How about this?
>> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|
>>
>>> This means that there is an infinite intersection of
>>> all successors.
>>
>> This means that, for any natural number k,
>> there are ways that what follows k can be totally-ordered
>> (NOT steppable-and-two-ended ways)
>> and there are ways theycannot be totally-ordered
>> (steppable-and-two-ended ways).
>>
>>> Here quantifier exchange is correct.
>>
>> Here quantifier shift is incorrect.
>>
>> A quantifier shift is ALWAYS an unreliable step.
>>
>> Because we are looking for statement-paths for which
>> each step is reliable, a quantifier shift is always
>> incorrect for such a path.
>>
>> That doesn't rule out some other reliable-step path that
>> doesn't have a quantifier shift. However, proclaiming
>> that here is correct doesn't make a path reliable.
>> Compare to:
>> Proclaiming that your checking account is not overdrawn
>> doesn't put money in your account.
>
> Uh oh, Don't go there -- remember McDuck?

Ants that think they are McDuck
McAnts

onsdag 8 december 2021 kl. 18:15:48 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:34:06 UTC+1:
> > onsdag 8 december 2021 kl. 11:42:17 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:11 UTC+1:
> > >
> > > > There is no such thing as "potential infinite"
> > > You are wrong. By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but less than the sets which have an empty intersection |∩{E(1), E(2), ...}| = 0, like the set of all endsegments. They are actually infinite. That is much more than all finite sets.
> > Nope, I am right because you cannot define what your "potential infinite" even fucking mathematically mean for sets.
> It means finite but without a last element.
> > >The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite
> > with cardinality aleph-0
> No, the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0. It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.
>
> Regards, WM

>It means finite but without a last element.

That is nonsense for sets. If it has no last element, it means it is infinite, ergo not finite.

>No

Yes

>the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0

is also aleph-0 and equal to the set of endsegments such that |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

>It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

That set is FINITE, it is not "potential", it has a last element, namely E(k)

On 12/8/2021 6:01 PM, Serg io wrote:
> On 12/8/2021 5:16 PM, FromTheRafters wrote:
>> Jim Burns wrote on 12/8/2021 :
>>> On 12/8/2021 1:02 PM, WM wrote:
>>>
>>>> For all FISONs the set of successors is infinite.
>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>>>
>>> ℕ = ∪{FISON'}
>>>
>>> ∀n ∈ ℕ: |ℕ\{0,...,n}| >= ℵ₀
>>>
>>>> Note that finite is less than infinite.
>>>> So we can state:
>>>> Every FISON cover less than half of all natural numbers.
>>>
>>> "Less than half than half of all natural numbers"
>>> needs to be translated.
>>>
>>> How about this?
>>> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|
>>>
>>>> This means that there is an infinite intersection of
>>>> all successors.
>>>
>>> This means that, for any natural number k,
>>> there are ways that what follows k can be totally-ordered
>>> (NOT steppable-and-two-ended ways)
>>> and there are ways theycannot be totally-ordered
>>> (steppable-and-two-ended ways).
>>>
>>>> Here quantifier exchange is correct.
>>>
>>> Here quantifier shift is incorrect.
>>>
>>> A quantifier shift is ALWAYS an unreliable step.
>>>
>>> Because we are looking for statement-paths for which
>>> each step is reliable, a quantifier shift is always
>>> incorrect for such a path.
>>>
>>> That doesn't rule out some other reliable-step path that
>>> doesn't have a quantifier shift. However, proclaiming
>>> that here is correct doesn't make a path reliable.
>>> Compare to:
>>> Proclaiming that your checking account is not overdrawn
>>> doesn't put money in your account.
>>
>> Uh oh, Don't go there -- remember McDuck?
>
> Ants that think they are McDuck
> McAnts
>

The spiders are starting to get just a bit jealous of the ants fame. I
mean, ants only have 6 legs, vs 8 for us spiders! Ant Bastards! ;^) lol.

zelos...@gmail.com schrieb am Donnerstag, 9. Dezember 2021 um 06:32:05 UTC+1:
> onsdag 8 december 2021 kl. 18:15:48 UTC+1 skrev WM:

> > > >The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite
> > > with cardinality aleph-0
> > No, the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0. It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

> >It means finite but without a last element.
> That is nonsense for sets. If it has no last element, it means it is infinite, ergo not finite.

The collection of all finite sets {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but it is less than the set of all endsegments {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0.
That means we have two consecutive infinities.

> >the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0
> is also aleph-0 and equal to the set of endsegments such that |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

The result of the intersection is very different. For two identical sets it must be the same however. This shows that these two sets are not the same.

> >It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.
> That set is FINITE,

If
|∩{E(1), E(2), ..., E(k)}| = ℵ₀
then
|∩{E(1), E(2), ..., E(k+1)}| = ℵ₀

> it is not "potential", it has a last element, namely E(k)

Which k is that? Whatever you tell, E(k+1) belongs to the set too.

Regards, WM

William schrieb am Mittwoch, 8. Dezember 2021 um 20:20:55 UTC+1:
> On Wednesday, December 8, 2021 at 2:37:54 PM UTC-4, WM wrote:

> > > > > E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.
> > > > True, a Peano set, but not an ℵ₀ set
> > > A Peano set has cardinality aleph_0.
> > No.
> Yes. This is a fundamental property of a Peano set.

That has been said before people understood the difference between definable and undefinable.
> > {E(1), E(2), ..., E(k)} with not fixed k
> is something that changes and is not a set.
> Potentially infinite set is nonsense.

If
|∩{E(1), E(2), ..., E(k)}| = ℵ₀
then
|∩{E(1), E(2), ..., E(k+1)}| = ℵ₀

So there is no last endsegment. The collection satisfying |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite. But it is not the set satisfying |∩{E(1), E(2), ...}| = 0. The collection is only a subset or subcollection of the set of all endsegments. But there are not two consecutive ℵ₀-sets possible in ℕ. Can you understand this?

Regards, WM

Jim Burns schrieb am Mittwoch, 8. Dezember 2021 um 22:08:07 UTC+1:
> On 12/8/2021 1:02 PM, WM wrote:
>
> > Every FISON cover less than half of all natural numbers.
> "Less than half than half of all natural numbers"
> needs to be translated.
>
> How about this?
> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|

That is not what I said. |{0,...,n+n}| is not necessary. This is the point:

∀n ∈ ℕ⁺ : |{0,...,n}| < |{n+1, n+2, ...}|

> > This means that there is an infinite intersection of
> > all successors.
> > Here quantifier exchange is correct.
> Here quantifier shift is incorrect.
>
> A quantifier shift is ALWAYS an unreliable step.

I have explained that in sci.logic.

> Because we are looking for statement-paths for which
> each step is reliable, a quantifier shift is always
> incorrect for such a path.
>
I have shown a path that is not only relible but obvious:

If
|∩{E(1), E(2), ..., E(k)}| = ℵ₀
then
|∩{E(1), E(2), ..., E(k+1)}| = ℵ₀

So there is no last endsegment. The collection satisfying |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite. But it is not the set satisfying |∩{E(1), E(2), ...}| = 0. The collection is only a subset or subcollection of the set of all endsegments. But there are not two consecutive ℵ₀-sets possible in ℕ. Or do you deny this?

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 12:52:01 UTC+1:
> On Wednesday, 8 December 2021 at 07:39:43 UTC-4, WM wrote:

> > Potential infinity has nothing to do with reality.
> Did you really say this??? Are you now so deranged that you can't even keep your attributions straight any more?

Reality is existing completely, finite or, very unlikely, infinite, but not potential. Potential infinity is a process of realizing further steps. Potential infinity is not an attribute of reality.

Regards, WM

On Thursday, December 9, 2021 at 6:14:30 AM UTC-4, WM wrote:

> <snip> The [set] satisfying |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} . Note you can not write down every element of E, but this does not change the fact that E has cardinality aleph_0. There is no element of that Peano set of natural numbers |N_P which is contained in every element of E.

--
William Hughes

On 12/9/2021 4:48 AM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 12:52:01 UTC+1:
>> On Wednesday, 8 December 2021 at 07:39:43 UTC-4, WM wrote:
>
>>> Potential infinity has nothing to do with reality.
>> Did you really say this??? Are you now so deranged that you can't even keep your attributions straight any more?
>
> Reality is existing completely, finite or, very unlikely, infinite, but not potential. Potential infinity is a process of realizing further steps. Potential infinity is not an attribute of reality.
>
> Regards, WM
>

Wrong, corrected => Potential infinity is a process of *not* realizing further steps.

On 12/9/2021 4:24 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 8. Dezember 2021 um 22:08:07 UTC+1:
>> On 12/8/2021 1:02 PM, WM wrote:
>>
>>> Every FISON cover less than half of all natural numbers.
>> "Less than half than half of all natural numbers"
>> needs to be translated.
>>
>> How about this?
>> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|
>
> That is not what I said. |{0,...,n+n}| is not necessary. This is the point:
>
> ∀n ∈ ℕ⁺ : |{0,...,n}| < |{n+1, n+2, ...}|
>
>>> This means that there is an infinite intersection of
>>> all successors.
>>> Here quantifier exchange is correct.
>> Here quantifier shift is incorrect.
>>
>> A quantifier shift is ALWAYS an unreliable step.
>
> I have explained that in sci.logic.

Quantifier shifting is what Trolls use.

>
>> Because we are looking for statement-paths for which
>> each step is reliable, a quantifier shift is always
>> incorrect for such a path.
>>
> I have shown a path that is not only relible but obvious:

<snip rest of wrong stuff>

William schrieb am Donnerstag, 9. Dezember 2021 um 14:06:16 UTC+1:
> On Thursday, December 9, 2021 at 6:14:30 AM UTC-4, WM wrote:
>
> > <snip> The [set] satisfying |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
> E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} .

No. There is a big difference between ℵ₀ and 0. One and the same set cannot yield and infinite intersection and an empty intersection

> Note you can not write down every element of E,

That is irrelevant. The potentially infinite set is defined by its infinite intersection. The complete set has an empty intersection.

> but this does not change the fact that E has cardinality aleph_0. There is no element of that Peano set of natural numbers |N_P which is contained in every element of E.

Correct. But there are infinitely many natural numbers contained in every endsegment of the potentially infinite set {E(1), E(2), ..., E(k)} with not fixed k, i.e., in every set {E(1), E(2), ..., E(k)} where k is a writable natural number.

Regards, WM

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.
--
William Hughes

torsdag 9 december 2021 kl. 11:01:52 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 9. Dezember 2021 um 06:32:05 UTC+1:
> > onsdag 8 december 2021 kl. 18:15:48 UTC+1 skrev WM:
>
> > > > >The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite
> > > > with cardinality aleph-0
> > > No, the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0. It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.
> > >It means finite but without a last element.
> > That is nonsense for sets. If it has no last element, it means it is infinite, ergo not finite.
> The collection of all finite sets {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but it is less than the set of all endsegments {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0.
> That means we have two consecutive infinities.
> > >the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0
> > is also aleph-0 and equal to the set of endsegments such that |∩{E(1), E(2), ..., E(k)}| = ℵ₀.
> The result of the intersection is very different. For two identical sets it must be the same however. This shows that these two sets are not the same.
> > >It is much larger than the potentially infinite oo-set {E(1), E(2), ...., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.
> > That set is FINITE,
> If
> |∩{E(1), E(2), ..., E(k)}| = ℵ₀
> then
> |∩{E(1), E(2), ..., E(k+1)}| = ℵ₀
> > it is not "potential", it has a last element, namely E(k)
> Which k is that? Whatever you tell, E(k+1) belongs to the set too.
>
> Regards, WM

>The collection of all finite sets {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite

Correct, which means it is of cardinality aleph-0 AT LEAST

>but it is less than the set of all endsegments {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0.

Nope, those have the same cardinality.

>That means we have two consecutive infinities.

As your claim before was wrong this one is too :)

>The result of the intersection is very different.

Because one collection is finite and another is infinite.

>For two identical sets it must be the same however. This shows that these two sets are not the same.

No one said they are the same set, only that their CARDINALITY is the same!

>Which k is that? Whatever you tell, E(k+1) belongs to the set too.

False, {E(1), E(2), ..., E(k)} does not contain E(k+1) :)

It is finite one with k elements.

On 12/9/2021 11:41 AM, WM wrote:
> William schrieb
> am Donnerstag, 9. Dezember 2021 um 14:06:16 UTC+1:
>> On Thursday, December 9, 2021 at 6:14:30 AM UTC-4,
>> WM wrote:

>>> <snip> The [set] satisfying
>>> |∩{E(1), E(2), ..., E(k)}| = ℵ₀
>>> is
>>
>> E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} .
>
> No.
> There is a big difference between ℵ₀ and 0.

The empty set has 0 ends.
An infinite sequence has 1 end.
A finite sequence has 2 ends.

----
We describe and then reason from the description.
This allows us to reason reliably about one of
however-many we have described, even if they're
infinitely-many.

Describe a finite sequence.

It has a first and a last. 2 ends.
It is _steppable_

Describe _steppable_ S
Totally ordered.
Each non-empty closed-downward subset contains a last.
Each non-empty closed-upward subset contains a first.

Describe an infinite sequence.

It contains _all_ the finite sequences -- wait, not quite.
It contains all of _certain_ finite sequences.

Each of these finite sequences has a fixed _first_
( For natural numbers, the first is 0
( For end segments, the first is ℕ

In each of these finite sequence, there is a fixed
_successor_ relationship between the last of any
closed-downward subset and the first of its complement.
( For natural numbers, it's our familiar '+1'
( For end segments, E(k+1) = E(k)\{k}

( Successor and first. This is enough information about
( these finite sequences to derive
( -- the uniqueness of FISON {0,...,k} for FISON-ender k
( -- the total-order of FISON-membership j ∈ {0,...,k}
( -- the well-order of ℕ = ∪{FISON'}
( -- many other properties of ℕ

Each of these certain finite sequences, these FISONs,
has the same fixed first 0
Their union has that first.

For _all_ of these FISONs unioned together, there can't
be a second end, a last. This can be derived from our
familiar '+1' and or selection of these special finite
sequences. k+1 ∉ {0,...,k}

ℕ has a first but no last. 1 end.
If an end segment had a last, it would also be
last of ℕ. So an end segment has only 1 end.

The intersection of end segments is a photographic
negative of the union of FISONs. The union has no last.
By a similar argument, the intersection has no first.

But we are working in ∪{FISON'}. The only subset
of ∪{FISON'} with no first is the empty set {}.

{} has no first and no last. 0 ends.

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

An end segment and all of its _predecessors_ have
an infinite intersection.

An end segment and all of its _successors_ have
an empty intersection.

Different collections with different intersections.
So?

zelos...@gmail.com schrieb am Freitag, 10. Dezember 2021 um 06:22:53 UTC+1:
> torsdag 9 december 2021 kl. 11:01:52 UTC+1 skrev WM:

> >The collection of all finite sets {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite
> Correct, which means it is of cardinality aleph-0 AT LEAST
> >but it is less than the set of all endsegments {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0.
> Nope, those have the same cardinality.

Then they are the same set and then they must have the same intersection. Contradiction.

> >The result of the intersection is very different.
> Because one collection is finite and another is infinite.

The collection of all endsegments with infinite intersection is not finite. You accepted this above. Further it is obvious.

> >For two identical sets it must be the same however. This shows that these two sets are not the same.
> No one said they are the same set, only that their CARDINALITY is the same!

The cardinality of an initial segment of a well-ordered set of cardinality ℵ₀ cannot be ℵ₀ unless the initial segment is the whole set.

> >Which k is that? Whatever you tell, E(k+1) belongs to the set too.

> False, {E(1), E(2), ..., E(k)} does not contain E(k+1) :)

k is not a fixed number. But if you are confused by the notation then use the expression:
C is the collection of all endsegments with infinite intersection.

Regards, WM

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
> An end segment and all of its _predecessors_ have
> an infinite intersection.

We use the collection C of all these endsegments which together with their predecessors have an infinite intersection.
>
> An end segment and all of its _successors_ have
> an empty intersection.

The set E of all endsegments has an empty intersection.
>
> Different collections with different intersections.

What however is the difference between C and E? The claim that there is no difference cannot be true because identical sets have identical intersections. It is obviously false. This proves by the photographic negative that your related claim "There is no set beyond all FISONs" is also wrong.

And why are there no empty endsegments when "n ∉ E(n+1)" removes every content?

Regards, WM

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. They are not finitiely many but they are not the complete set of all ℵo Endsegments having empty intersection.

Regards, WM

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.
Does not change the fact that the set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set, something that does not change

--
William Hughes

On 12/10/2021 10:24 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
>> An end segment and all of its _predecessors_ have
>> an infinite intersection.
>
> We use the collection C of all these endsegments which together with their predecessors have an infinite intersection.
>>
>> An end segment and all of its _successors_ have
>> an empty intersection.
>
> The set E of all endsegments has an empty intersection.
>>
>> Different collections with different intersections.
>
> What however is the difference between C and E? The claim that there is no difference cannot be true because identical sets have identical intersections. It is obviously false. This proves by the photographic negative that your related claim "There is no set beyond all FISONs" is also wrong.
>
> And why are there no empty endsegments when "n ∉ E(n+1)" removes every content?
>
> Regards, WM
>

collection C Ants

set E Ants

photographic negative Ants

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
>>
>> An end segment and all of its _predecessors_ have
>> an infinite intersection.
>
> We use the collection C of all these endsegments which
> together with their predecessors have an infinite
> intersection.

Okay. Define...

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

{before E'} = { E" ∈ {END} | E" ⊇ E' }

{END} = { ℕ/F | F ∈ {FISON'} }

ℕ = ∪{FISON'}

>> 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}

>> An end segment and all of its _successors_ have
>> an empty intersection.
>
> The set E of all endsegments has an empty intersection.

True. ∩{END} = {}
That's a consequence of my last claim there,
∀E' ∈ {END}, ∩{after E'} = {}

since {END} = {after ℕ}

{after E} = { E' ∈ {END} | E ⊇ E' }

>> Different collections with different intersections.
>
> What however is the difference between C and E?

No difference. (C = E = {END})

My claim is for {before E'} and {after E'}
(E' ∈ {END})
Those are different collections.
And they have different intersections.
∩{before E'} = E'
∩{after E'} = {}

> The claim that there is no difference

is someone else's claim, possibly. I don't know whose.

> The claim that there is no difference cannot be true
> because identical sets have identical intersections.
> It is obviously false. This proves by the photographic
> negative that your related claim
> "There is no set beyond all FISONs"
> is also wrong.

My claim is that there is no FISON beyond all other
FISONs. Because k+1 ∉ {0,...,k}

I'm nearly convinced you're not doing that on purpose.

> And why are there no empty endsegments
> when "n ∉ E(n+1)" removes every content?

For each end segment,
there is a first end, it is not empty.
but there is not a second end, it is not finite.
Each end segment is infinite.

There is no n which is in all end segments.
See E(n+1).
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_

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.

> Does not change the fact that the set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set, something that does not change

The collection of all endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is not the set of alle endsegments which has an empty intersection ∩{E(1), E(2), E(3) ...} = { }.

The intersection is emptied by further endsegments, not by delusions of infinity.

Regards, WM

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.

Regards, WM