William schrieb am Sonntag, 12. Dezember 2021 um 16:05:50 UTC+1:
> On Sunday, December 12, 2021 at 7:12:55 AM UTC-4, WM wrote:

> > 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".

Yes, of course. But what is the infinite content remaining in all infinite endsegments?

Regards, WM

Jim Burns schrieb am Sonntag, 12. Dezember 2021 um 14:05:45 UTC+1:
> On 12/12/2021 6:22 AM, WM wrote:
> > Jim Burns schrieb

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

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

Therefore it is in C.

> > But there is none.
> > Therefore the definitions define one and the same C.
> >
All infinite endsegments belong to the collection C of all these endsegments which
together with their predecessors have an infinite intersection.

Regards, WM

On 12/12/2021 1:38 PM, WM wrote:
> William schrieb am Sonntag, 12. Dezember 2021 um 16:05:50 UTC+1:
>> On Sunday, December 12, 2021 at 7:12:55 AM UTC-4, WM wrote:
>
>>> 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".
>
> Yes, of course. But what is the infinite content remaining in all infinite endsegments?
>
> Regards, WM

Each Endsegment is a fixed infinite set, E(k) = {k,k+1,k+2,...}

no "content" was added, nor removed from any Endsegment.

The Intersection of sets, is the set of elements that are in both (or more) sets.

my, you are having a LOT of TROUBLE with this...

On 12/12/2021 1:47 PM, WM wrote:
> Jim Burns schrieb am Sonntag, 12. Dezember 2021 um 14:05:45 UTC+1:
>> On 12/12/2021 6:22 AM, WM wrote:
>>> Jim Burns schrieb
>
>>>>>>> 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.
>> Let E" be the first end segment after E'
>> E" is not among
>> E' together with the predecessors of E'
>
> Irrelevant.
>>
>> But E" is among "all these endsegments which together
>> with their predecessors have an infinite intcersection".
>
> Therefore it is in C.
>
>>> But there is none.
>>> Therefore the definitions define one and the same C.
>>>
> All infinite endsegments belong to the collection C of all these endsegments which
> together with their predecessors have an infinite intersection.

no. the intersection is empty.

>
> Regards, WM

On 12/12/2021 2:47 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 12. Dezember 2021 um 14:05:45 UTC+1:
>> On 12/12/2021 6:22 AM, WM wrote:
>>> Jim Burns schrieb

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

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

>> But E" is among "all these endsegments which together
>> with their predecessors have an infinite intcersection".
>
> Therefore it is in C.

Thus, your two definitions of C differ by E".
Equivocation bad.

>>> But there is none.
>>> Therefore the definitions define one and the same C.
>
> All infinite endsegments belong to the collection C of
> all these endsegments which together with their
> predecessors have an infinite intersection.

Each collection of an end segment and its predecessors
has a last end segment. For each of those collections,
its intersection is infinite.

The collection C of all those end-segments-at-the-end-
-of-a-collection-with-infinite-intersection does not have
a last end segment. C is not any of those collections.
The intersection of C is empty.

When we do not equivocate (Equivocating bad.) we can see
that there is no conflict that needs to be resolved.

On Sunday, December 12, 2021 at 3:38:16 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 12. Dezember 2021 um 16:05:50 UTC+1:
>> Every element of the Peano set of natural numbers is "lost".
> Yes, of course.

So all natural numbers are "lost"

>But what is the infinite content remaining in all infinite endsegments?

There is no "infinite content remaining in all infinite endsegments" . Such content would have to consist of "dark" elements and the set of natural numbers is a Peano set which does not contain "dark" elements.

--
William Hughes

William schrieb am Sonntag, 12. Dezember 2021 um 22:26:41 UTC+1:
> On Sunday, December 12, 2021 at 3:38:16 PM UTC-4, WM wrote:
> > William schrieb am Sonntag, 12. Dezember 2021 um 16:05:50 UTC+1:
> >> Every element of the Peano set of natural numbers is "lost".
> > Yes, of course.
> So all natural numbers are "lost"
> >But what is the infinite content remaining in all infinite endsegments?
> There is no "infinite content remaining in all infinite endsegments" .

They all are infinite but without infinite content? How that? What makes them infinite?

> Such content would have to consist of "dark" elements

Hit!

> and the set of natural numbers is a Peano set which does not contain "dark" elements.

If you respect quantifiers, you cannot deny that ℕ_P is the collection:

∀n ∈ ℕ_P: |F(n)|/|E(n)| = 0.
∀n ∈ ℕ_P: |E(n)| - |F(n)| = ℵo.

Regards, WM

Jim Burns schrieb am Sonntag, 12. Dezember 2021 um 21:43:21 UTC+1:
> On 12/12/2021 2:47 PM, WM wrote:

> When we do not equivocate (Equivocating bad.) we can see
> that there is no conflict that needs to be resolved.

∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.

Regards, WM

On Monday, December 13, 2021 at 5:46:43 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 12. Dezember 2021 um 22:26:41 UTC+1:
> > On Sunday, December 12, 2021 at 3:38:16 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 12. Dezember 2021 um 16:05:50 UTC+1:
> > >> Every element of the Peano set of natural numbers is "lost".
> > > Yes, of course.
> > So all natural numbers are "lost"
> > >But what is the infinite content remaining in all infinite endsegments?
> > There is no "infinite content remaining in all infinite endsegments" .
> They all are infinite but without infinite content?

Piffle. There is a different infinite set for each endsegment. Each endsegment has infinite content There is no one set that works for
*all*l endsegments
- William Hughes

On 12/13/2021 4:52 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 12. Dezember 2021 um 21:43:21 UTC+1:

>> When we do not equivocate (Equivocating bad.) we can see
>> that there is no conflict that needs to be resolved.
>
> ∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.

Even better...

∀j ∈ ∪{FISON'}, ∀k ∈ ∪{FISON'},
|F(j)| < |E(k)|

We can see that there is no conflict.

You may ask:
Even when j approaches the end of ∪{FISON'} ?
There is no conflict because
there is no end of ∪{FISON'}

fredag 10 december 2021 kl. 17:13:25 UTC+1 skrev WM:
> 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

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

Non-sequitor. Sets can have same cardinality but be different.

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

Correct, but the intersection of all endsegments is still empty.

>k is not a fixed number. But if you are confused by the notation then use the expression:

It is a fixed number, k is not equal to k+1 or k-1 etc

>C is the collection of all endsegments with infinite intersection.

Do you mean the collections of sets of endsegments with infinite intersection? Because otherwise it makes no sense.

William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
> On Monday, December 13, 2021 at 5:46:43 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 12. Dezember 2021 um 22:26:41 UTC+1:
"
> > > >But what is the infinite content remaining in all infinite endsegments?
> > > There is no "infinite content remaining in all infinite endsegments" ..
> > They all are infinite but without infinite content?
> Piffle. There is a different infinite set for each endsegment.

That is a silly argument. Different infinite sets do not change the fact that all contents is infinite and is there remaining from the big bang E(1) = ℕ and will never be reduced to less than ℵo elements in infinite endsegments.

> Each endsegment has infinite content There is no one set that works for
> *all*l endsegments

Of course the contents is dark. Therefore it does not "work". But the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1). Any doubts?

Regards, WM

Jim Burns schrieb am Montag, 13. Dezember 2021 um 16:39:27 UTC+1:
> On 12/13/2021 4:52 AM, WM wrote:

> > ∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.
> Even better...
>
> ∀j ∈ ∪{FISON'}, ∀k ∈ ∪{FISON'},
> |F(j)| < |E(k)|
>
> We can see that there is no conflict.
>
> You may ask:
> Even when j approaches the end of ∪{FISON'} ?
> There is no conflict because
> there is no end of ∪{FISON'}

Again the argument of con-artists.
There are all n ∈ ℕ such that none remains.
There is nothing beyond all n ∈ ℕ. If all n were definable, then the E(n) would be exhausted. Then ~∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.

If there really was no end, then there would be no endsegments (and no complete Cantor-list and no complete enumeration of infinite sets). Alas, if we we accept this completeness, then the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1). Infinite sets have not lost all numbers. Any doubts?

By the way: Consider a ruler with all unit fractions between 0 and 1 marked (assuming that all unit fractions are existing and could be marked). Du you believe that if both of us had such a ruler, our rulers could deviate by one or more marks? This shows that all "Cantor-bijections" concern only a very small potentially infinite initial segment of the bijected sets.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 14. Dezember 2021 um 06:15:32 UTC+1:
> fredag 10 december 2021 kl. 17:13:25 UTC+1 skrev WM:

> > C is the collection of all endsegments with infinite intersection.
> >
> >Then they are the same set and then they must have the same intersection.. Contradiction.
>
> Non-sequitor. Sets can have same cardinality but be different.

Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
>
> >The collection of all endsegments with infinite intersection is not finite. You accepted this above. Further it is obvious
>
> Correct, but the intersection of all endsegments is still empty.

The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
>
> >C is the collection of all endsegments with infinite intersection.
>
> Do you mean the collections of sets of endsegments with infinite intersection?

C is the collection of all endsegments which have infinite intersection. All have infinite contents which is there remaining from the big bang E(1) = ℕ and will never be reduced to less than ℵo elements in infinite endsegments.

Regards, WM

On 12/14/2021 3:28 AM, WM wrote:
> William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
>> On Monday, December 13, 2021 at 5:46:43 AM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 12. Dezember 2021 um 22:26:41 UTC+1:
> "
>>>>> But what is the infinite content remaining in all infinite endsegments?
>>>> There is no "infinite content remaining in all infinite endsegments" .
>>> They all are infinite but without infinite content?
>> Piffle. There is a different infinite set for each endsegment.
>
> That is a silly argument. Different infinite sets do not change the fact that all contents is infinite and is there remaining from the big bang E(1) = ℕ and will never be reduced to less than ℵo elements in infinite endsegments.

you present bad math.
E(1) is different from E(2), both are infinite sets and they are different, as 1 is not an element of E(2), but is an element of E(1).

You still do not understand that Sets are fixed, and they do not change. There is no "remaining", no "reduced" that is lazy uninformed thinking.

>
>> Each endsegment has infinite content There is no one set that works for
>> *all*l endsegments
>
> Of course the contents is dark. Therefore it does not "work". But the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1). Any doubts?

You have no Proof, you have not presented any Proof that there are dark numbers, it is only your imagination and bad math.

>
> Regards, WM

On 12/14/2021 3:53 AM, WM wrote:
> Jim Burns schrieb am Montag, 13. Dezember 2021 um 16:39:27 UTC+1:
>> On 12/13/2021 4:52 AM, WM wrote:
>
>>> ∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.
>> Even better...
>>
>> ∀j ∈ ∪{FISON'}, ∀k ∈ ∪{FISON'},
>> |F(j)| < |E(k)|
>>
>> We can see that there is no conflict.
>>
>> You may ask:
>> Even when j approaches the end of ∪{FISON'} ?
>> There is no conflict because
>> there is no end of ∪{FISON'}
>
> Again the argument of con-artists.
> There are all n ∈ ℕ such that none remains.

nope. all that says is that "n is an element of the set of natural numbers", nothing else, "such that none remains" is BS.

> There is nothing beyond all n ∈ ℕ.

of course there is real numbers, rational numbers, ...

> If all n were definable, then the E(n) would be exhausted. Then ~∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.

your "definable" with your beeps, flashes, raps, taps, hourse hoofs, is silly.

On 12/14/2021 3:59 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 14. Dezember 2021 um 06:15:32 UTC+1:
>> fredag 10 december 2021 kl. 17:13:25 UTC+1 skrev WM:
>
>>> C is the collection of all endsegments with infinite intersection.
>>>
>>> Then they are the same set and then they must have the same intersection. Contradiction.
>>
>> Non-sequitor. Sets can have same cardinality but be different.
>
> Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
>>
>>> The collection of all endsegments with infinite intersection is not finite. You accepted this above. Further it is obvious
>>
>> Correct, but the intersection of all endsegments is still empty.
>
> The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).

n ∉ E(n+1) - only says that "n is not an element of Endsegment(n+1)" that is all

Your L above is simply a FISON(n) but sets do not lose or gain elements.

and sets do not loose or gain elements, *you treat sets as variables* which is not in Mathematics.

>>> C is the collection of all endsegments with infinite intersection.
>>
>> Do you mean the collections of sets of endsegments with infinite intersection?
>
> C is the collection of all endsegments which have infinite intersection.

so C does not include the intersection of all endsegments.

> All have infinite contents which is there remaining from the big bang E(1) = ℕ and will never be reduced to less than ℵo elements in infinite endsegments.
>
> Regards, WM

On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
> William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:

> > Each endsegment has infinite content There is no one set that works for
> > *all*l endsegments
> Of course the [elements of any single set are] dark.

And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.

--
William Hughes

On 12/14/2021 4:53 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 13. Dezember 2021 um 16:39:27 UTC+1:
>> On 12/13/2021 4:52 AM, WM wrote:

>>> ∀n ∈ ℕ_def: |F(n)|/|E(n)| = 0.
>>
>> Even better...
>>
>> ∀j ∈ ∪{FISON'}, ∀k ∈ ∪{FISON'},
>> |F(j)| < |E(k)|
>>
>> We can see that there is no conflict.
>>
>> You may ask:
>> Even when j approaches the end of ∪{FISON'} ?
>> There is no conflict because
>> there is no end of ∪{FISON'}

> If there really was no end,

Define
an order of the elements in any FISONs by FISON-inclusion.
j =< k <-> j ∈ {0,...,k}

| Assume m is the second end of ∪{FISON'}
| | m ∈ ∪{FISON'}
| ∀k ∈ ∪{FISON'}, k ∈ {0,...,m}
| | However, m+1 ∈ ∪{FISON'} and m+1 ∉ {0,...,m}
| ~∀k ∈ ∪{FISON'}, k ∈ {0,...,m}
| Contradiction.
| | Therefore,
| m is not the second end of ∪{FISON'}

Generalizing,
there really is no end of ∪{FISON'}

> If there really was no end,
> then there would be no endsegments

Define the end segment E(k) of ∪{FISON'} so that,
for each j, j ∈ E(k) <->
j ∈ ∪{FISON'} ∧ k =< j

The question of whether there would be end segment E(k)
can be avoided by talking about the numbers that would be
in E(k) if E(k) "really" existed.
We know how to do that. See above: instead of saying
| j ∈ E(k)
say
| j ∈ ∪{FISON'} ∧ k =< j

For example, define
j is in the intersection of all end segments <->
∀k ∈ ∪{FISON'} ( j ∈ ∪{FISON'} ∧ k =< j )

For j to be in the intersection of all end segments,
k ∈ {0,...,j} would need to be true for
all k ∈ ∪{FISON'}

However, j+1 ∉ {0,...,j}
So, k ∈ {0,...,j} is NOT true for all k ∈ ∪{FISON'}

j is not in the intersection of all end segments
The intersection of all end segments is empty.

tisdag 14 december 2021 kl. 11:00:04 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 14. Dezember 2021 um 06:15:32 UTC+1:
> > fredag 10 december 2021 kl. 17:13:25 UTC+1 skrev WM:
>
> > > C is the collection of all endsegments with infinite intersection.
> > >
> > >Then they are the same set and then they must have the same intersection. Contradiction.
> >
> > Non-sequitor. Sets can have same cardinality but be different.
> Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
> >
> > >The collection of all endsegments with infinite intersection is not finite. You accepted this above. Further it is obvious
> >
> > Correct, but the intersection of all endsegments is still empty.
> The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
> >
> > >C is the collection of all endsegments with infinite intersection.
> >
> > Do you mean the collections of sets of endsegments with infinite intersection?
> C is the collection of all endsegments which have infinite intersection. All have infinite contents which is there remaining from the big bang E(1) = ℕ and will never be reduced to less than ℵo elements in infinite endsegments.
>
> Regards, WM

>Infinite sets like FISON and endsegments cannot differ by more than finite subsets.

FISONs are finite sets

infinite sets can differ by infinite subsets and have nothing in comon. So you are all wrong.

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

L=N

ℕ\L=ℕ\ℕ={}

so the intersection is empty as stated.

>C is the collection of all endsegments which have infinite intersection

Intersection WITH WHAT!?

William schrieb am Dienstag, 14. Dezember 2021 um 15:49:09 UTC+1:
> On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
> > William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
>
> > > Each endsegment has infinite content There is no one set that works for
> > > *all*l endsegments
> > Of course the [elements of any single set are] dark.
>
> And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.

Infinite sets can be exhausted according to Cantor. An example is the set of all unit fractions which has a fixed quantity of elements: |ℕ| =/= |ℕ| + 1. If you doubt this, then consider a ruler where all unit fractions carry marks. Two such rulers will not differ by any mark.
All of them can be subtracted such that none remains. This can be done one by one for defined numbers. All definable unit fractions have a last one. But all unit fractions have no last one although being a fixed quantity. This proves dark unit fractions.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 15. Dezember 2021 um 06:11:58 UTC+1:
> tisdag 14 december 2021 kl. 11:00:04 UTC+1 skrev WM:

>
> >Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
> FISONs are finite sets
>
> infinite sets can differ by infinite subsets and have nothing in comon.

That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.

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

AS long as the endsegments are infinite, they have not lost all natnumbers.

Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.

> >C is the collection of all endsegments which have infinite intersection
> Intersection WITH WHAT!?

The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.

Regards, WM

Jim Burns schrieb am Mittwoch, 15. Dezember 2021 um 00:41:43 UTC+1:
> On 12/14/2021 4:53 AM, WM wrote:

> Generalizing,
> there really is no end of ∪{FISON'}

But there is an end of the set of all natnumbers.

If you doubt this, then consider a ruler where all unit fractions carry marks. Two such rulers will not differ by any mark.
All of them can be subtracted such that none remains. This can be done one by one for defined numbers. All definable unit fractions have a last one. But all unit fractions have no last one although being a fixed quantity. This proves dark unit fractions.

> For example, define
> j is in the intersection of all end segments <->
> ∀k ∈ ∪{FISON'} ( j ∈ ∪{FISON'} ∧ k =< j )

You cannot define it because it is a dark number. You cannot define the last natnumber although completeness and linearity would prove its existence.

> j is not in the intersection of all end segments
> The intersection of all end segments is empty.

Yes, because not all are infinite. All infinite endsegments contain natnumbers, don't they?

Regards, WM

On 12/15/2021 4:55 AM, WM wrote:
> William schrieb am Dienstag, 14. Dezember 2021 um 15:49:09 UTC+1:
>> On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
>>> William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
>>
>>>> Each endsegment has infinite content There is no one set that works for
>>>> *all*l endsegments
>>> Of course the [elements of any single set are] dark.
>>
>> And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.
>
> Infinite sets can be exhausted according to Cantor. An example is the set of all unit fractions which has a fixed quantity of elements: |ℕ| =/= |ℕ| + 1. If you doubt this, then consider a ruler where all unit fractions carry marks. Two such rulers will not differ by any mark.
> All of them can be subtracted such that none remains. This can be done one by one for defined numbers. All definable unit fractions have a last one. But all unit fractions have no last one although being a fixed quantity. This proves dark unit fractions.
>
> Regards, WM

from the one who uses Fixed Sets as Variables...

Mark Ants
Carry a Mark Ants
Ruler Ants
All of Ants Subtracted such that none Remain
Fractioned Ants

On 12/15/2021 5:06 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 15. Dezember 2021 um 00:41:43 UTC+1:
>> On 12/14/2021 4:53 AM, WM wrote:
>
>> Generalizing,
>> there really is no end of ∪{FISON'}

<snip crap>

>
>> For example, define
>> j is in the intersection of all end segments <->
>> ∀k ∈ ∪{FISON'} ( j ∈ ∪{FISON'} ∧ k =< j )
>
> You cannot define it because it is a dark number.

Wrong

>
>> j is not in the intersection of all end segments
>> The intersection of all end segments is empty.
>
> Yes, because not all are infinite.

wrong.

> All infinite endsegments contain natnumbers, don't they?

what do you think ? or did you "EXHAUST" them all outta the EXHAUST PIPE ?

>
> Regards, WM