On Friday, 8 October 2021 at 07:18:43 UTC-3, WM wrote:
> Gus Gassmann schrieb am Freitag, 8. Oktober 2021 um 01:18:13 UTC+2:
> > On Thursday, 7 October 2021 at 17:48:14 UTC-3, Serg io wrote:
> > > On 10/7/2021 5:10 AM, WM wrote:
> > [...]
> > > > All infinite endsegments have an infinite intersection.
> > > wrong.
> > Intentionally ambiguous and misleading.
>
> No, clear and obviously true.
> > Pairwise intersections of *any* two end segments have an intersection of cardinality aleph_0.
> We do not need pairwise intersections, but only all infinite endsegments. A sequence of decreasing endsegments with not smaller than infinite terms has an intersection of ℵo elements. That is basic mathematics.
> > Even finite intersections do, but infinite intersections do not.
> All infinitely many infinite endsegments do. Fact is however that they are only a potentially infinite collection, [...]

Boohoo. "I am right, because... But even if that weren't true (it isn't!), I would still be right because..." Well, you're not.

You are an incorrigible arsehole.

On 10/8/2021 9:35 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 8. Oktober 2021 um 12:18:06 UTC+2:
>> On 10/7/2021 5:49 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 5. Oktober 2021 um 20:57:40 UTC+2:
>>>> On 10/5/2021 4:35 AM, WM wrote:

>>>> I think that you (WM) claim that potentially infinite
>>>> collections _change_
>>>
>>> It is the definition:
>>
>> Then the collection of all and only the FISON-enders
>> is NOT _potentially infinite_
>
> It is.
> All natural numbers have the smallest transfinite cardinality ℵo.
> All FISONs (containing all FISON-enders) have smaller cardinality.
>
>> My reason for accepting the collection of all and only
>> the FISON-enders as "potentially infinite" was that we would
>> understand that to mean "totally ordered, does NOT have
>> both a first and a last, and each split has a crossing-pair".
>>
>> However, a FISON-ender is never not a FISON-ender, and
>> a not-FISON-ender is never a FISON-ender. If change is part of
>> the definition of a potentially infinite collection, then
>> the collection of FISON-enders is not potentially infinite.
>
> The definition is this:
> No largest element, but cardinality smaller than ℵo.

Then change is NOT part of the definition of "potentially infinite".

Whether you call the collection of all and only FISON-enders
potentially infinite, or you don't, we are justified in claiming
k ends a FISON, knowing only that k is a member of that collection,
and in claiming that k is a member, knowing only that k is a FISON.

Also, the collection of all and only the FISON-enders is not
potentially infinite under this definition, either. Nothing is.

As you say above, aleph_0 is the smallest infinite cardinality.
A cardinality smaller than aleph_0 is not infinite == finite.
A totally-ordered finite collection has a largest element.
Nothing is (this version of) potentially infinite.

>> We can reason about each member of the collection of FISON-enders,
>> and calling their collection potentially infinite or denying it's
>> potentially infinite won't change our reasoning, since we're
>> reasoning from the existence, for k, of a totally-ordered {0,...,k}
>> from 0 to k in which each split has a crossing-pair j,j+1.
>
> It is like Cantor's example:
> "potentially infinite ... a finite in the process of change
> having in each of its current states a finite size; like,
> for instance, the temporal duration since the beginning of
> the world, which, when measured in some time-unit, for instance
> a year, is finite in every moment, but always growing beyond
> all finite limits, without ever becoming really infinitely large."
> [G. Cantor, letter to I. Jeiler (13 Oct 1895)]" [Cantor]

A FISON {0,...,k} can be used to describe the process of
matching an unmatched sheep to the next unmatched natural
number, which halts at k by running out of unmatched sheep.

It's a process. I suppose some things change as it proceeds,
which sheep is currently being matched, or whatever.
Whether {0,...,k} is a FISON does not change.

These unchanging things are useful for describing changing things.

_All_ of the counting processes that halt by running out of
things-being-counted have some properties in common.
( As noted, a totally ordered {0,...,k} exists in which,
( for each split, a crossing-pair j,j+1 exists.

We can reason truth-preserving-ly from those common properties
to further common properties.

That's not the same as applying the counting process to
FISON-enders the way that we apply it to sheep.
We don't claim that we can do that -- but we still can
reason truth-preserving-ly from and to common properties.

In fact, in order for this counting process to operate securely,
it must be impossible to run out of FISON-enders the way we would
run out of sheep. Declaring this impossible does not contradict
anything _we_ have claimed: "These are properties each
FISON-ender has".

>>>> However, the collection of all and only FISON-enders
>>>> _does not change_ A FISON exists which k ends, or
>>>> a FISON does NOT exist which k ends, and which it is
>>>> does not change.
>>>
>>> That is impossible. All FISONs are finite. None is infinite.
>>> Would they not change, then there was a largest finite FISON.
>>
>> And yet, each is finite[1], they do not change,
>
> No FISON changes

This is equivalent to:
The set of all and only the FISON-enders does not change.

>> and there is
>> no largest FISON.
>
> Therefore the position "last one" does change.

No, I didn't mean "the largest FISON changes".
There is no largest FISON. There is no last one.

Compare to
There is NO right triangle such that the square of the length
of its longest side is NOT equal to the sum of the squares of
of the lengths of the two remaining sides.

That's NOT a claim that _we have not found_ such a right triangle.
That's a claim that _it does not exist_ found or not found.

We justify that claim about found and not-found right triangles
starting from simple, obvious statements true of a right
triangle, just because it's a right triangle, whether it's found
or whether it's not found.

The same for FISONs: there is no last FISON, found or not found.

>> But that's _your_ problem, not ours.
>
> That is not a problem but a fact.
> The limit of FISONs is {1, 2, 3, ...}.
> But every FISON is much, much smaller.
>
>> Each FISON is finite.
>> Each FISON does not change.
>> Each FISON is followed by more FISONs.
>> Denying these claims leads to contradictions.
>
> Correct.
> But finally:
> Between all FISONs and |N there is a huge gap of dark numbers.

There is a certain genre of question-answer-exchange that
I have come to expect. It might typically go...
Me:
"Blahblahblah is true of FISON-ender k"
You:
"What about dark numbers?"
Me:
"Is it a FISON-ender?"
You:
"It's a dark number. We can't see."
Me:
"If the dark number is a FISON-ender, then we can reason
from {0,...,k} existing. If the dark number not a FISON-ender,
it cannot be a counter-example, because we do not make
a claim here about non-FISON-enders."

> Between all FISONs and |N there is a huge gap of dark numbers.

So what?

On 10/8/2021 11:58 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
>> fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
>
>>> Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
>
>> ANd htat is yet anotehr empty assertion of yours with no proof.
>
> The contents of endsegments is what is not used to index its predecessors.
> If all endsegments have infinite contents,

by definition endsegments are sets that have an infinite number of elements. did you miss that ?

> then infinitely many numbers are not used to index all predecessors, i.e., all endsegments.

you have an if then statement, If above is always true, therefor you are stating that

"infinitely many numbers are not used to index all predecessors"

predecessors to what ?

> The complement, used to index them, is finite.

complement of what ?

index what ?

>
> Regards, WM
>

On 10/8/2021 5:55 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 12:48:12 UTC+2:
>> fredag 8 oktober 2021 kl. 12:29:50 UTC+2 skrev WM:
>
> If there is an empty intersection, then there are two endsegments having an empty intersection. The brainless trick using the intersection of sets like {a, b}, {b, c}, {c, a} is not compatible with inclusion monotony.
>> It demonstrates that things can have an intersection with every member but the combined intersection is emtpy which is what happens here
>
> No that cannot happen here.
>
>> and your inclusion monotony does not change it.
>
> Of course it does. It excludes that elements can be acquired. Elements can only be lost. As lomg as not all elements are lost, they are in all predecessors too.
>
> Regards, WM
>

there is no "acquiring" elements and no "losing" elements, as each endsegments is fixed infinite set of natural numbers, and they do not change.

you are skipping over required steps, statements, in Math, because you don't know them.

On Friday, October 8, 2021 at 12:58:43 PM UTC-4, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
> > fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
>
> > > Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
> > ANd htat is yet anotehr empty assertion of yours with no proof.
> The contents of endsegments is what is not used to index its predecessors.
> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
>
> Regards, WM
I wonder if your problem relates to: https://en.wikipedia.org/wiki/Burali-Forti_paradox
which I bumped into reading Russell's theory of types in his and Whitehead's Principia Mathematica.

zelos...@gmail.com wrote:

>> >> > are you insane or just mentally retarded?
>> >> you are a known fucking imbecile around here. Go fuck yourself other
>> >> places. Cretin.
>> >
>> > I am known to smack down cranks, so I take you're one of them
>> you just proved you are that imbecile, idiot. CDC is a private org with
>> patents in both viruses and vaccines, and are running pharmakia, you
>> fucking stupid. Educate yourself.
>
> I am clearly more educated than you when you are a conspiratard

You certainly are not qualified stating the above, not sensing already
what is going on in center for death control, CDC, based on the given
above facts, not theories. You are an imbecile. I only need to mention
that mass murderer Brevick, what's his name, killing children. Completely
conspiratards. Killing children, are you *fucking_insane*??

zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:31:36 UTC+2:
> fredag 8 oktober 2021 kl. 12:53:00 UTC+2 skrev WM:

> >∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)

> >You cannot find any endsegment not covered by (*).
> Nor do I need to.

In order to refute suddenness you would need at least infinitely many endsegments not covered by (*). But there is not even a single one.
> >Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
> There is nothing sudden there.

All endsegments are infinite but the limit is empty. That is suddenness.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:41:22 UTC+2:
> fredag 8 oktober 2021 kl. 12:55:48 UTC+2 skrev WM:
.. As long as not all elements are lost, they are in all predecessors too.

> It does not change the fact that the intersection is empty.

That would contradict mathematics:
Inclusion monotony and ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> And yes, it can happen here because the intersection of all endsegments IS empty

Yes, but not the intersection of all infinite endsegments.

> and it is extremely easy to prove it

It is extremely easy to prove that the intersection of infinite endsegments is infinite.

Regards, WM

On 10/9/2021 8:31 AM, Timothy Golden wrote:
> On Friday, October 8, 2021 at 12:58:43 PM UTC-4, WM wrote:
>> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
>>> fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
>>
>>>> Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
>>> ANd htat is yet anotehr empty assertion of yours with no proof.
>> The contents of endsegments is what is not used to index its predecessors.
>> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
>>
>> Regards, WM
>
> I wonder if your problem relates to: https://en.wikipedia.org/wiki/Burali-Forti_paradox
> which I bumped into reading Russell's theory of types in his and Whitehead's Principia Mathematica.
>
https://en.wikipedia.org/wiki/Burali-Forti_paradox ;

"Modern axioms for formal set theory such as ZF and ZFC circumvent this antinomy by not allowing the construction of sets using terms like "all sets
with the property P {\displaystyle P} P", as is possible in naive set theory and as is possible with Gottlob Frege's axioms – specifically Basic Law V –
in the "Grundgesetze der Arithmetik." Quine's system New Foundations (NF) uses a different solution. Rosser (1942) showed that in the original version
of Quine's system "Mathematical Logic" (ML), an extension of New Foundations, it is possible to derive the Burali-Forti paradox, showing that this
system was contradictory. Quine's revision of ML following Rosser's discovery does not suffer from this defect, and indeed was subsequently proved
equiconsistent with NF by Hao Wang. "

On Saturday, 9 October 2021 at 10:48:59 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:31:36 UTC+2:
> > fredag 8 oktober 2021 kl. 12:53:00 UTC+2 skrev WM:
>
> > >∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > >You cannot find any endsegment not covered by (*).
> > Nor do I need to.
> In order to refute suddenness you would need at least infinitely many endsegments not covered by (*). But there is not even a single one.
> > >Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
> > There is nothing sudden there.
> All endsegments are infinite but the limit is empty. That is suddenness.

Yeah well, deal with it. sgn(1/n) = 1 for every n in n and yet sgn(0) = 0. That's no less "sudden". (Do you actually think that there is a mathematical definition or quantification of "suddenness"?)

On 10/9/2021 8:48 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:31:36 UTC+2:
>> fredag 8 oktober 2021 kl. 12:53:00 UTC+2 skrev WM:
>
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
>
>>> You cannot find any endsegment not covered by (*).
>> Nor do I need to.
>
> In order to refute suddenness you would need at least infinitely many endsegments not covered by (*). But there is not even a single one.
>>> Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
>> There is nothing sudden there.
>
> All endsegments are infinite but the limit is empty.

corrected;

All endsegments are infinite but the intersection of all endsegments is empty.

>That is suddenness.

time is not a factor in math operations.

>
> Regards, WM
>

On 10/9/2021 8:51 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:41:22 UTC+2:
>> fredag 8 oktober 2021 kl. 12:55:48 UTC+2 skrev WM:
> . As long as not all elements are lost, they are in all predecessors too.
>
>> It does not change the fact that the intersection is empty.
>
> That would contradict mathematics:
> Inclusion monotony and ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>>
>> And yes, it can happen here because the intersection of all endsegments IS empty
>
> Yes, but not the intersection of all infinite endsegments.
>
>> and it is extremely easy to prove it
>
> It is extremely easy to prove that the intersection of infinite endsegments is infinite.
>
> Regards, WM
>

how to prove there are no elements in intersection of all endsegments ?

PROOF;

for any k, where k is a natural number

k is not in the endsegments E(k+1), E(k+2), ...

therefore k is not in the intersection of all endsegments.

since we chose ANY k, and proved there are no such k's in the intersection of all endsegments, then there are no natural numbers in the intersection.

EZ CHEESY

On Saturday, 9 October 2021 at 10:48:59 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:31:36 UTC+2:
> > fredag 8 oktober 2021 kl. 12:53:00 UTC+2 skrev WM:
>
> > >∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo (*)
> > >You cannot find any endsegment not covered by (*).
> > Nor do I need to.
> In order to refute suddenness you would need at least infinitely many endsegments not covered by (*). But there is not even a single one.
> > >Therefore the function has only infinite terms but the limit is zero. That is as sudden as possible a decrease from infinity to zero.
> > There is nothing sudden there.
> All endsegments are infinite but the limit is empty. That is suddenness.

Yeah well, deal with it. sgn(1/n) = 1 for every n in |N and yet sgn(0) = 0. That's no less "sudden". (Do you actually think that there is a mathematical definition or quantification of "suddenness"?)

On 10/5/2021 7:08 PM, Greg Cunt wrote:
> On Tuesday, October 5, 2021 at 2:46:05 PM UTC+2, Serg io wrote:
>
>> a subtle difference there;
>>
>> | Let n e {1, 2, 3}. Then n ...
>>
>> n is constant or parameter (member of the set)
>
> Nope. You have to differentiate between the term (name/constant/paramter, whatever) - "n" in this case - and the objects the term refers to, namely a number, i. e. a member of the set {1, 2, 3}, n in this case.
>
> Of course, in the context of math we usually call certain/specific numbers "constants" too. So please don't mix that up. I'm talking about the language here (not the mathematical objects).
>
> Again:
>
> | Let n e {1, 2, 3}. Then n ...
>
> Here, "n" refers to a CERTAIN number in {1, 2, 3}, though we don't and can't KNOW which one. [Morover, *I* wouldn't call n a "constant" here - imho its not specific enough to be called that way.]
>
> Please note the difference: IN THIS CONTEXT "n" is a constant (or "parameter", if you like).
>
> On the other hand, in the statement
>
> | For all n in IN: n >= 0.
>
> "n" is (used as) a variable.
>
> Or even in a statement like
>
> | n e IN --> n >= 0,
>
> or some algebraic "equations", etc.
>

sometimes I feel like a #2 pencil that has been sharpened (!)

Now I'm ready for n-dimentional Euclidean space (had a course in it, but could not visualize it, so it was a very hard course)

On Saturday, October 9, 2021 at 10:18:09 AM UTC-4, Serg io wrote:
> On 10/9/2021 8:31 AM, Timothy Golden wrote:
> > On Friday, October 8, 2021 at 12:58:43 PM UTC-4, WM wrote:
> >> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
> >>> fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
> >>
> >>>> Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
> >>> ANd htat is yet anotehr empty assertion of yours with no proof.
> >> The contents of endsegments is what is not used to index its predecessors.
> >> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
> >>
> >> Regards, WM
> >
> > I wonder if your problem relates to: https://en.wikipedia.org/wiki/Burali-Forti_paradox
> > which I bumped into reading Russell's theory of types in his and Whitehead's Principia Mathematica.
> >
> https://en.wikipedia.org/wiki/Burali-Forti_paradox ;
>
>
> "Modern axioms for formal set theory such as ZF and ZFC circumvent this antinomy by not allowing the construction of sets using terms like "all sets
> with the property P {\displaystyle P} P", as is possible in naive set theory and as is possible with Gottlob Frege's axioms – specifically Basic Law V –
> in the "Grundgesetze der Arithmetik." Quine's system New Foundations (NF) uses a different solution. Rosser (1942) showed that in the original version
> of Quine's system "Mathematical Logic" (ML), an extension of New Foundations, it is possible to derive the Burali-Forti paradox, showing that this
> system was contradictory. Quine's revision of ML following Rosser's discovery does not suffer from this defect, and indeed was subsequently proved
> equiconsistent with NF by Hao Wang. "

It's on ordinals, but the naturals aren't far off. Some of WM's concerns seem nearby.
Set theory I suppose is to blame. Sets as elements; elements as sets.

Gus Gassmann schrieb am Samstag, 9. Oktober 2021 um 20:30:28 UTC+2:
> On Saturday, 9 October 2021 at 10:48:59 UTC-3, WM wrote:

> > All endsegments are infinite but the limit is empty. That is suddenness..
> Yeah well, deal with it. sgn(1/n) = 1 for every n in |N and yet sgn(0) = 0. That's no less "sudden".

So it is. But sgn(x) is not expected to become zero by losing elements. For endsegments however we have to observe ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Regards, WM

Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:

> PROOF;
> for any k, where k is a natural number
> k is not in the endsegments E(k+1), E(k+2), ...
> therefore k is not in the intersection of all endsegments.
> since we chose ANY k,

No! Every endsegment is infinite. For infinitely many k your proof fails.

Regards, WM

On Sunday, 10 October 2021 at 14:10:25 UTC-3, WM wrote:
> Gus Gassmann schrieb am Samstag, 9. Oktober 2021 um 20:30:28 UTC+2:
> > On Saturday, 9 October 2021 at 10:48:59 UTC-3, WM wrote:
>
> > > All endsegments are infinite but the limit is empty. That is suddenness.
> > Yeah well, deal with it. sgn(1/n) = 1 for every n in |N and yet sgn(0) = 0. That's no less "sudden".
> So it is. But sgn(x) is not expected to become zero by losing elements. For endsegments however we have to observe ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

And for every n we have to observe that sgn(1/n) is positive. Your rationalizations are meaningless drivel.

söndag 10 oktober 2021 kl. 19:15:01 UTC+2 skrev WM:
> Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:
>
> > PROOF;
> > for any k, where k is a natural number
> > k is not in the endsegments E(k+1), E(k+2), ...
> > therefore k is not in the intersection of all endsegments.
> > since we chose ANY k,
> No! Every endsegment is infinite. For infinitely many k your proof fails.
>
> Regards, WM
are you seriously this stupid?

It is ANY k that means it applies to ALL of them, NONE does it fail for.

timba...@gmail.com schrieb am Samstag, 9. Oktober 2021 um 15:31:18 UTC+2:

> I wonder if your problem relates to: https://en.wikipedia.org/wiki/Burali-Forti_paradox
> which I bumped into reading Russell's theory of types in his and Whitehead's Principia Mathematica.

No, it is not Burali-Forti. But it is as simple.

The contents of an endsegment cannot be used to index its predecessors:

E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
E(1), E(2) = {2, 3, 4, 5, 6, ...}
E(1), E(2), E(3) = {3, 4, 5, 6, ...}
E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).

If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.

Regards, WM

Serg io schrieb am Samstag, 9. Oktober 2021 um 06:02:49 UTC+2:
> On 10/8/2021 11:58 AM, WM wrote:

> > The contents of endsegments is what is not used to index its predecessors.
> > If all endsegments have infinite contents,
> by definition endsegments are sets that have an infinite number of elements.

The contents of an endsegment cannot be used to index its predecessors:

E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
E(1), E(2) = {2, 3, 4, 5, 6, ...}
E(1), E(2), E(3) = {3, 4, 5, 6, ...}
E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).

If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.

Regards, WM

fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 8. Oktober 2021 um 13:42:19 UTC+2:
> > fredag 8 oktober 2021 kl. 12:58:42 UTC+2 skrev WM:
>
> > > Fact by the properties of endsegments. If all have an infinite contents, then an infinite set of natural numbers is not available as indices.
> > ANd htat is yet anotehr empty assertion of yours with no proof.
> The contents of endsegments is what is not used to index its predecessors.
> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
>
> Regards, WM

>The contents of endsegments is what is not used to index its predecessors.

Irrelevant.

>If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.

Meaningless empty assertion.

You are conflating shit as always

Jim Burns schrieb am Freitag, 8. Oktober 2021 um 21:57:31 UTC+2:
> On 10/8/2021 9:35 AM, WM wrote:

> > The definition is this:
> > No largest element, but cardinality smaller than ℵo.
> Then change is NOT part of the definition of "potentially infinite".

Without change there would be a largest FISON because every number smallerthan ℵo is finite.
>
> Whether you call the collection of all and only FISON-enders
> potentially infinite, or you don't, we are justified in claiming
> k ends a FISON, knowing only that k is a member of that collection,
> and in claiming that k is a member, knowing only that k is a FISON.
>
> Also, the collection of all and only the FISON-enders is not
> potentially infinite under this definition, either. Nothing is.
>
> As you say above, aleph_0 is the smallest infinite cardinality.
> A cardinality smaller than aleph_0 is not infinite == finite.
> A totally-ordered finite collection has a largest element.
> Nothing is (this version of) potentially infinite.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

This shows all FISONs are finite. But without changing, there would be a largest one. There is none. Therefore the set is potentially infinite.

Regards, WM

zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:21:59 UTC+2:
> söndag 10 oktober 2021 kl. 19:15:01 UTC+2 skrev WM:
> > Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:
> >
> > > PROOF;
> > > for any k, where k is a natural number
> > > k is not in the endsegments E(k+1), E(k+2), ...
> > > therefore k is not in the intersection of all endsegments.
> > > since we chose ANY k,
> > No! Every endsegment is infinite. For infinitely many k your proof fails.
> >
> It is ANY k that means it applies to ALL of them, NONE does it fail for.

What are the infinitely many elements which remain in every infinite endsegment? Does the proof prove that they disappear such that none remains. How can all the endsegments remain infinite?

Regards, WM

zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
> fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:

> >The contents of endsegments is what is not used to index its predecessors.
> Irrelevant.

Chuckle. Your only way to pretend ignorance?

> >If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
> Meaningless empty assertion.

Meaningful and very important for mathematicians who can think.

Regards, WM