On Monday, August 16, 2021 at 11:00:17 AM UTC-4, WM wrote:
> William schrieb am Montag, 16. August 2021 um 16:54:23 UTC+2:
> > On Monday, August 16, 2021 at 10:29:42 AM UTC-4, WM wrote:
> >
> > > Every definable natnumber is the last one of a FISON.
> > There is no element of |N_F that is not the "last one of a FISION".
> That means every definable element is followed by infinitely many undefinable elements

Nope, infinitely many elements that are the "last one of a FISON". The fact that every element of |N_F that can be written down is the last one of a FISON does not mean that an element of |N_F that is the last one of a FISON can be written downe.

--
William Hughes

On 8/16/2021 8:03 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 16. August 2021 um 05:32:18 UTC+2:
>> On 8/15/2021 4:53 PM, WM wrote:

>> For each infinite end segment E,
>> there is another infinite end segment EE such that
>> EE is a proper subset of E.
>
> Yes.

Thank you.
If you change your mind later, I'll prove it for you then.

> But not every infinite endsegment must be definable.

Every infinite end segment is infinite and an end segment.
Every infinite end segment is in the set of all infinite
end segments.

> Further EE can be the last of a set of infinite endsegments.

Not if that set is the set of ALL infinite end segments.
This is my point.

Note that it is the intersection of the set of ALL infinite
end segments which we are discussing. EE cannot be the last
in THAT set.

> Then it is the intersection of this set.
>
>> Infinite end segment E is NOT a subset of
>> infinite end segment EE.
>
> Unless

There is no "unless" possible here.
EE is a proper subset of E.
That means there is something in E which is not in EE.
And that means that E is not a subset of EE.

This is purely a matter of what "subset" and "proper subset" mean.

> Unless EE is the intersection of a set of endsegments.
> Then it is the last one.
>
>>> Every set of infinite endsegments has an infinite intersection.
>>
>> Assume that IAIES, the Intersection of All Infinite End
>> Segments, is an infinite end segment.
>
> It is with no doubt the last of the intersected endsegments.

It would be, if it existed.
This creates a logical problem for the intersection of the set
of all infinite end segments, since no last infinite end segment
exists in that set.

>> On the one hand, there is another infinite end segment EE
>> such that EE is a proper subset of IAIES, and
>> infinite end segment IAIES *IS NOT* a subset of
>> infinite end segment EE.
>
> If all infinite endsegments had been intersected,
> then EE is the last infinite endsegment of the set.

IAIES, the intersection of the set SAIES of all infinite
end segments, is _after_ each element of SAIES (by inclusion),
whatever IAIES may be.

If IAIES was an element of SAIES, then IAIES would be
the _last_ element of SAIES.

No element of SAIES is the _last_ element of SAIES.
( For each E in SAIES, there is EE after it in SAIES. )

Therefore,
IAIES is NOT an element of SAIES.

The intersection of all infinite end segments is NOT
an infinite end segment.

>> On the other hand, because IAIES is the Intersection of
>> All Infinite End Segments, IAIES is a subset of each
>> infinite end segment and IAIES *IS* a subset of
>> infinite end segment EE,
>
> Not necessarily a proper subset.
>
>> Contradiction.
>
> Why? EE is simply the last endsegment of the set.

IAIES is the last element of the set, if IAIES is in SAIES.
But EE is the hypothetical infinite-end-segment proper-subset
of IAIE which we know exists if IAIES is in SAIES.

EE isn't IAIES. It would come _after_ IAIES, the _last_ in SAIES.
EE would be _post-ultimate_ if IAIES were in SAIES.

"Post-ultimate" is a contradiction.
IAIES is not in SAIES.

>> Therefore,
>> IAIES, the Intersection of All Infinite End Segments,
>> is not an infinite end segment.
>
> Non sequitur.
>
> Regards, WM
>

William schrieb am Montag, 16. August 2021 um 17:18:09 UTC+2:
> On Monday, August 16, 2021 at 11:00:17 AM UTC-4, WM wrote:
> > William schrieb am Montag, 16. August 2021 um 16:54:23 UTC+2:
> > > On Monday, August 16, 2021 at 10:29:42 AM UTC-4, WM wrote:
> > >
> > > > Every definable natnumber is the last one of a FISON.
> > > There is no element of |N_F that is not the "last one of a FISION".
> > That means every definable element is followed by infinitely many undefinable elements
> Nope, infinitely many elements that are the "last one of a FISON".

Possible for potential infinity, but impossible for actual infinity. All FISONs are finite. They cannot contain or distinguish more than finitely many elements. Pigeon hole principle.

> The fact that every element of |N_F that can be written down is the last one of a FISON does not mean that an element of |N_F that is the last one of a FISON can be written down.

A sequence of elements of a FISON:

o
oo
ooo
oooo
ooooo
....

There is no actually infinite set of FISONs possible by the pigeon hole principle. Further every FISON is followed by aleph_0 elements which cannot all belong to FISONs because FISONs are finite.

Regards, WM

Jim Burns schrieb am Montag, 16. August 2021 um 19:55:41 UTC+2:
> On 8/16/2021 8:03 AM, WM wrote:

> > Further EE can be the last of a set of infinite endsegments.
> Not if that set is the set of ALL infinite end segments.
> This is my point.

It is wrong. EE an be infinite and can be the last infinite.
>
> Note that it is the intersection of the set of ALL infinite
> end segments which we are discussing. EE cannot be the last
> in THAT set.

Why not? EE is the last with no doubt.

> EE is a proper subset of E.

and is infinite.

> That means there is something in E which is not in EE.
> And that means that E is not a subset of EE.

It does not mean that EE is finite.
>
> It would be, if it existed.
> This creates a logical problem for the intersection of the set
> of all infinite end segments, since no last infinite end segment
> exists in that set.

The set is mainly dark. So there is no problem.

> The intersection of all infinite end segments is NOT
> an infinite end segment.

It is with no doubt.

> > Why? EE is simply the last endsegment of the set.
> IAIES is the last element of the set, if IAIES is in SAIES.

EE is the last infinite endsegment. Note that the intersection is always the last endsegment.

Regards, WM

On Monday, August 16, 2021 at 4:46:49 PM UTC-4, WM wrote:
> William schrieb am Montag, 16. August 2021 um 17:18:09 UTC+2:
> > On Monday, August 16, 2021 at 11:00:17 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 16. August 2021 um 16:54:23 UTC+2:
> > > > On Monday, August 16, 2021 at 10:29:42 AM UTC-4, WM wrote:
> > > >
> > > > > Every definable natnumber is the last one of a FISON.
> > > > There is no element of |N_F that is not the "last one of a FISION".
> > > That means every definable element is followed by infinitely many undefinable elements
> > Nope, infinitely many elements that are the "last one of a FISON".
> Possible for potential infinity

"Potential infinity" is a nonsensical term. There is no such thing.

> but impossible for actual infinity. All FISONs are finite.

Correct, there is no *element of the set of FISONs* that is not finite, i.e. an *element of the set of FISONs* contains only enough elements to distinguish a finite number of FISONs. So what?
The *set of FISONs* contains an infinite number of elements.

This is the original Pink Elephant.

Every element of the set S has property P
Look! Over there! A Pink Elephant
Set S has property P.

--
William Hughes

On Monday, August 16, 2021 at 4:56:27 PM UTC-4, WM wrote:
> EE is the last infinite endsegment.

This is nonsense. There is no such thing as "the last infinite endsegment".

--
William Hughes

On 8/16/2021 4:56 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 16. August 2021 um 19:55:41 UTC+2:
>> On 8/16/2021 8:03 AM, WM wrote:

>>> Further EE can be the last of a set of infinite endsegments.
>>
>> Not if that set is the set of ALL infinite end segments.
>> This is my point.
>
> It is wrong. EE an be infinite and can be the last infinite.

No, it can't be last in the set SAIES. This is my point.

<JB<WM<JB>>>
>>
>> For each infinite end segment E,
>> there is another infinite end segment EE such that
>> EE is a proper subset of E.
>
> Yes.

Thank you.
If you change your mind later, I'll prove it for you then.

</JB<WM<JB>>>

Proof.

Consider E(k) and E(k+1) = E(k)\{k}

If a last element m exists in E(k+1),
then that same m exists in E(k) and is last in E(k).

Thus,
if E(k+1) is finite, then E(k) is finite.

This is equivalent to
If E(k) is not finite, then E(k+1) is not finite.

----
If you don't like that proof, I've got more.
You have been trying for years to go from the finite
to the infinite in a single step.

You are confused about "finite" and "infinite".
But, even though you are confused, we are not.
It doesn't fly. It will never fly.

>> Note that it is the intersection of the set of ALL infinite
>> end segments which we are discussing. EE cannot be the last
>> in THAT set.
>
> Why not? EE is the last with no doubt.

For each infinite end segment E,
there is another infinite end segment EE such that
EE is a proper subset of E.

EE is an infinite end segment.
There is another infinite end segment
which is a proper subset of EE.
EE is not last in SAIES.

>> EE is a proper subset of E.
>
> and is infinite.
>
>> That means there is something in E which is not in EE.
>> And that means that E is not a subset of EE.
>
> It does not mean that EE is finite.

For any two _sets_ B and C,
if B is a proper subset of C,
then C is NOT a subset of B.
This is about _sets_

Take that fact about sets and apply it to a set C which
supposedly is a subset of each set in SAIES
but also for which a proper subset B of C is in SAIES.
That is a contradictory set of conditions on C.
Therefore, no such set C exists.

What is it that doesn't exist?
A set equal to the intersection of SAIES which is also
an element of SAIES doesn't exist.

Therefore, the intersection of all infinite end segments
is not an infinite end segment.

>> It would be, if it existed.
>> This creates a logical problem for the intersection of the set
>> of all infinite end segments, since no last infinite end segment
>> exists in that set.
>
> The set is mainly dark. So there is no problem.
>
>> The intersection of all infinite end segments is NOT
>> an infinite end segment.
>
> It is with no doubt.
>
>>> Why? EE is simply the last endsegment of the set.
>>
>> IAIES is the last element of the set, if IAIES is in SAIES.
>
> EE is the last infinite endsegment.
> Note that the intersection is always the last endsegment.

IAIES is the intersection. IAIES is last.
EE is after the last. "After the last" is contradictory.

Thus, the intersection IAIES is not an infinite end segment,
which it would be if it were in SAIES.

On Monday, August 16, 2021 at 2:46:29 PM UTC-7, Jim Burns wrote:
> On 8/16/2021 4:56 PM, WM wrote:
> > Jim Burns schrieb
> > am Montag, 16. August 2021 um 19:55:41 UTC+2:
> >> On 8/16/2021 8:03 AM, WM wrote:
>
> >>> Further EE can be the last of a set of infinite endsegments.
> >>
> >> Not if that set is the set of ALL infinite end segments.
> >> This is my point.
> >
> > It is wrong. EE an be infinite and can be the last infinite.
> No, it can't be last in the set SAIES. This is my point.
>
> <JB<WM<JB>>>
> >>
> >> For each infinite end segment E,
> >> there is another infinite end segment EE such that
> >> EE is a proper subset of E.
> >
> > Yes.
>
> Thank you.
> If you change your mind later, I'll prove it for you then.
> </JB<WM<JB>>>
>
> Proof.
>
> Consider E(k) and E(k+1) = E(k)\{k}
>
> If a last element m exists in E(k+1),
> then that same m exists in E(k) and is last in E(k).
>
> Thus,
> if E(k+1) is finite, then E(k) is finite.
>
> This is equivalent to
> If E(k) is not finite, then E(k+1) is not finite.
>
> ----
> If you don't like that proof, I've got more.
> You have been trying for years to go from the finite
> to the infinite in a single step.
>
> You are confused about "finite" and "infinite".
> But, even though you are confused, we are not.
> It doesn't fly. It will never fly.
> >> Note that it is the intersection of the set of ALL infinite
> >> end segments which we are discussing. EE cannot be the last
> >> in THAT set.
> >
> > Why not? EE is the last with no doubt.
> For each infinite end segment E,
> there is another infinite end segment EE such that
> EE is a proper subset of E.
> EE is an infinite end segment.
> There is another infinite end segment
> which is a proper subset of EE.
> EE is not last in SAIES.
> >> EE is a proper subset of E.
> >
> > and is infinite.
> >
> >> That means there is something in E which is not in EE.
> >> And that means that E is not a subset of EE.
> >
> > It does not mean that EE is finite.
> For any two _sets_ B and C,
> if B is a proper subset of C,
> then C is NOT a subset of B.
> This is about _sets_
>
> Take that fact about sets and apply it to a set C which
> supposedly is a subset of each set in SAIES
> but also for which a proper subset B of C is in SAIES.
> That is a contradictory set of conditions on C.
> Therefore, no such set C exists.
>
> What is it that doesn't exist?
> A set equal to the intersection of SAIES which is also
> an element of SAIES doesn't exist.
>
> Therefore, the intersection of all infinite end segments
> is not an infinite end segment.
> >> It would be, if it existed.
> >> This creates a logical problem for the intersection of the set
> >> of all infinite end segments, since no last infinite end segment
> >> exists in that set.
> >
> > The set is mainly dark. So there is no problem.
> >
> >> The intersection of all infinite end segments is NOT
> >> an infinite end segment.
> >
> > It is with no doubt.
> >
> >>> Why? EE is simply the last endsegment of the set.
> >>
> >> IAIES is the last element of the set, if IAIES is in SAIES.
> >
> > EE is the last infinite endsegment.
> > Note that the intersection is always the last endsegment.
> IAIES is the intersection. IAIES is last.
> EE is after the last. "After the last" is contradictory.
>
> Thus, the intersection IAIES is not an infinite end segment,
> which it would be if it were in SAIES.

Yeah, must have already been last.

måndag 16 augusti 2021 kl. 16:16:53 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 16. August 2021 um 14:40:59 UTC+2:
> > måndag 16 augusti 2021 kl. 14:07:16 UTC+2 skrev WM:
> > > Gus Gassmann schrieb am Montag, 16. August 2021 um 13:52:05 UTC+2:
> > > > On Monday, 16 August 2021 at 08:02:37 UTC-3, WM wrote:
> > > > [...]
> > > > > Of course. But the definable endsegments are not all endsegments, because all definable endsegments have aleph_0 successors.
> > > > Boohoo! In ZFC every natural number, and hence every end segment, is definable.
> > > That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> > > Proof: Every definable number has aleph_0 successors.
> > > Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
> >
> > >That is the reason why ZFC is inconsistent. It is an obviously wrong claim.
> > WRONG!
> >
> > You have not once shown ANY contradiction in it!
> > >Not every number has aleph_0 successors because |N contains only numbers and if they are collectively subtracted from |N by |N \ |N, then nothing remains.
> > WRONG AGAIN!
> Fact.
>
> You are a believer in nonsense. But they have modified your brain such that you cannot understand. You can only parrot: "not once shown ANY contradiction". Your judgement is therefore irrelevant.
>
> Regards, WM

It is not a fact, it is your personal opinion based on absolutely nothing.

I understand mathematics far better than you because you're the one so stupid as to think there is a last natural number. Which is pure insanity.

I repeat that phrase because you never HAVE shown ANYTHING contradictory. You assert it based on pre-concieved ideas of what you think it SHOULD be rather than what it IS

>Of course. Every definable element is the last one of a finite initial segment. How could it be else?

Because it is so.

If you define |∩{E(k) : k ∈ ℕ_def}| = ℵo by this property, then your N_def = N because N has the exact same property.

On Tuesday, August 17, 2021 at 9:20:48 AM UTC+2, zelos...@gmail.com wrote:

> If you define |∩{E(k) : k ∈ ℕ_def}| = ℵo by this property,

???

You can't define N_def by this property, since this property holds for, say, N_def = {1} as well as for, say, N_def = {2}, etc.

> then your N_def = N because N has the exact same property.

Nope.

|∩{E(k) : k ∈ ℕ}| = 0 .

William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
> On Monday, August 16, 2021 at 4:46:49 PM UTC-4, WM wrote:

> > > Nope, infinitely many elements that are the "last one of a FISON".
> > Possible for potential infinity
> "Potential infinity" is a nonsensical term. There is no such thing.

There are many examples. The set of known prime numbers is potentially infinite.

> > but impossible for actual infinity. All FISONs are finite.
> Correct, there is no *element of the set of FISONs* that is not finite, i.e. an *element of the set of FISONs* contains only enough elements to distinguish a finite number of FISONs. So what?
> The *set of FISONs* contains an infinite number of elements.

No. The set is a single object. Of course it has other properties than its elements. But it cannot increase their properties. All FISONs are finite by the pigeon hole principle.
>
Regards, WM

Jim Burns schrieb am Montag, 16. August 2021 um 23:46:29 UTC+2:
> On 8/16/2021 4:56 PM, WM wrote:
> > Jim Burns schrieb
> > am Montag, 16. August 2021 um 19:55:41 UTC+2:
> >> On 8/16/2021 8:03 AM, WM wrote:
>
> >>> Further EE can be the last of a set of infinite endsegments.
> >>
> >> Not if that set is the set of ALL infinite end segments.
> >> This is my point.
> >
> > It is wrong. EE an be infinite and can be the last infinite.
> No, it can't be last in the set SAIES. This is my point.

This point is wrong. Also the set of dark endsegments may contain infinite endsegments. It must contain such endsegments because otherwise finite endsegments could become visible.
>
> >> For each infinite end segment E,
> >> there is another infinite end segment EE such that
> >> EE is a proper subset of E.
> >
> > Yes.
>
> Thank you.

> If you change your mind later, I'll prove it for you then.

> Proof.
>
> Consider E(k) and E(k+1) = E(k)\{k}
>
> If a last element m exists in E(k+1),
> then that same m exists in E(k) and is last in E(k).

The last aleph_0 elements in every endsegment are dark.
>
> Thus,
> if E(k+1) is finite, then E(k) is finite.

Here you try to distinguish dark endsegments. This is impossible. Almost all endsegments and almost all natural numbers containd in them are dark. This has been proven once and for all by a very simple idea:

"No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 17. August 2021 um 09:20:48 UTC+2:
> >Of course. Every definable element is the last one of a finite initial segment. How could it be else?
> Because it is so.
>
> If you define |∩{E(k) : k ∈ ℕ_def}| = ℵo by this property, then your N_def = N because N has the exact same property.

No, only the definable endsegments have this property. Most however are undefinable. Like the natural numbers: "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns

Regards, WM

On Tuesday, 17 August 2021 at 09:16:28 UTC-3, WM wrote:
> Jim Burns schrieb am Montag, 16. August 2021 um 23:46:29 UTC+2:
> >
[...]
> >
>
> "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns

Another classic example of Muckenheim insanity: You are throwing a quote of Jim Burns back at the original author, expecting that he now agrees with your nonsense!

William schrieb am Montag, 16. August 2021 um 23:33:39 UTC+2:
> On Monday, August 16, 2021 at 4:56:27 PM UTC-4, WM wrote:
>
> > EE is the last infinite endsegment.
> This is nonsense. There is no such thing as "the last infinite endsegment".

There is no last one by order because most are dark, even most of the infinite endsegments. But there is a last one by cardinality because we assume that
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
holds for all endsegments. There is even a last not empty one, E(G), having cardinality 1, and a very last one which is empty:
E(G+1) = E(ω) = E(G) \ {G} = { }
where G is the Grossone, the last natural number.

Regards, WM

Gus Gassmann schrieb am Dienstag, 17. August 2021 um 14:25:12 UTC+2:
> On Tuesday, 17 August 2021 at 09:16:28 UTC-3, WM wrote:
> > Jim Burns schrieb am Montag, 16. August 2021 um 23:46:29 UTC+2:
> > >
> [...]
> > >
> >
> > "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns
> You are throwing a quote of Jim Burns back at the original author,

Yes, I did so in order to make JB realize what he has said and to understand what it means. Would be nice whether you agreed too; or do you believe that the aleph_0 unit intervals are completed by indexes which can be distinguished?

Regards, WM

On Tuesday, 17 August 2021 at 09:37:04 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 17. August 2021 um 14:25:12 UTC+2:
> > On Tuesday, 17 August 2021 at 09:16:28 UTC-3, WM wrote:
> > > Jim Burns schrieb am Montag, 16. August 2021 um 23:46:29 UTC+2:
> > > >
> > [...]
> > > >
> > >
> > > "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns
> > You are throwing a quote of Jim Burns back at the original author,
> Yes, I did so in order to make JB realize what he has said and to understand what it means. Would be nice whether you agreed too; or do you believe that the aleph_0 unit intervals are completed by indexes which can be distinguished?

You are full of shit, but I leave it to Jim Burns to continue his endeavors at educating the uneducatable.

On Tuesday, August 17, 2021 at 8:06:37 AM UTC-4, WM wrote:
> William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
> > Correct, there is no *element of the set of FISONs* that is not finite, i.e. an *element of the set of FISONs* contains only enough elements to distinguish a finite number of FISONs. So what?
> > The *set of FISONs* contains an infinite number of elements.
> No. The set is a single object. Of course it has other properties than its elements. But it cannot increase their properties. All [elements of the set of FISONs] are finite by the pigeon hole principle.

So what? No *element of the set of FISONs* is infinite (duh). Knowing that every *element of S* had property P does not mean *the set S* has property P. *The set of FISONs" is infinite.

--
William Hughes

On Tuesday, August 17, 2021 at 8:31:42 AM UTC-4, WM wrote:
> William schrieb am Montag, 16. August 2021 um 23:33:39 UTC+2:
> > On Monday, August 16, 2021 at 4:56:27 PM UTC-4, WM wrote:
> >
> > > EE is the last infinite endsegment.
> > This is nonsense. There is no such thing as "the last infinite endsegment".
> There is no last one by order

Correct, and as no endsegment is not ordered, there is no last endsegment.

because most are dark,

Nope, no element of |N_F is dark (dark to you means more than just "cannot be written down")..

--
William Hughes

On 8/17/2021 8:36 AM, WM wrote:
> Gus Gassmann schrieb
> am Dienstag, 17. August 2021 um 14:25:12 UTC+2:
>> On Tuesday, 17 August 2021 at 09:16:28 UTC-3,
>> WM wrote:

>>> "No interval has all of its rationals indexed before
>>> any other interval has all of its rationals indexed."
>>> Jim Burns
>>
>> You are throwing a quote of Jim Burns back at the
>> original author,
>
> Yes, I did so in order to make JB realize what he has said
> and to understand what it means.

To review, I am answering your claim
<WM>
> Since in the natural order of |N, used by Cantor, never two
> consecutive sets of cardinality ℵo can exist, this means that
> up to this step only finitely many natural numbers can have
> been applied. - For ℵo intervals! Therefore it is impossible
> that Cantor's enumeration of the positive rationals is complete.
</WM>

There are no _consecutive_ sets of aleph_0 cardinality
in the enumeration.

_Up to_ any finite index, no more than finitely-many positive
rationals in no more than finitely-many integer-ended intervals
are indexed.

Including _all_ the finite indexes and _all_ the positive
rationals, infinitely-many positive rationals in infinitely-many
integer-ended intervals are indexed (that is, all of them).

> Would be nice whether you agreed too; or do you believe that
> the aleph_0 unit intervals are completed by indexes which
> can be distinguished?

No integer-ended interval has a last-indexed rational
in it. That is apparently what you mean by "completed".
So, none of them are ever (WM) "complete".

All of the rationals in all of the intervals are indexed.
These are different claims.

On 8/17/2021 7:06 AM, WM wrote:
> William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
>> On Monday, August 16, 2021 at 4:46:49 PM UTC-4, WM wrote:
>
>>>> Nope, infinitely many elements that are the "last one of a FISON".
>>> Possible for potential infinity
>> "Potential infinity" is a nonsensical term. There is no such thing.
>
> There are many examples. The set of known prime numbers is potentially infinite.

wrong, it is infinite.
Can you prove The set of known prime numbers is finite? NO.

>
>>> but impossible for actual infinity. All FISONs are finite.
>> Correct, there is no *element of the set of FISONs* that is not finite, i.e. an *element of the set of FISONs* contains only enough elements to distinguish a finite number of FISONs. So what?
>> The *set of FISONs* contains an infinite number of elements.
>
> No. The set is a single object. Of course it has other properties than its elements. But it cannot increase their properties. All FISONs are finite by the pigeon hole principle.

use Math instead. You dont use math much at all.
>>
> Regards, WM
>
>

On 8/17/2021 7:16 AM, WM wrote:
> Jim Burns schrieb am Montag, 16. August 2021 um 23:46:29 UTC+2:
>> On 8/16/2021 4:56 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Montag, 16. August 2021 um 19:55:41 UTC+2:
>>>> On 8/16/2021 8:03 AM, WM wrote:
>>
>>>>> Further EE can be the last of a set of infinite endsegments.
>>>>
>>>> Not if that set is the set of ALL infinite end segments.
>>>> This is my point.
>>>
>>> It is wrong. EE an be infinite and can be the last infinite.
>> No, it can't be last in the set SAIES. This is my point.
>
> This point is wrong. Also the set of dark endsegments may contain infinite endsegments.

wrong. there is no set of dark endsegments, because you cannot name the elements.

>It must contain such endsegments because otherwise finite endsegments could become visible.

wrong. endsegments are infinite.

>>
>>>> For each infinite end segment E,
>>>> there is another infinite end segment EE such that
>>>> EE is a proper subset of E.
>>>
>>> Yes.
>>
>> Thank you.
>
>> If you change your mind later, I'll prove it for you then.
>
>> Proof.
>>
>> Consider E(k) and E(k+1) = E(k)\{k}
>>
>> If a last element m exists in E(k+1),
>> then that same m exists in E(k) and is last in E(k).
>
> The last aleph_0 elements in every endsegment are dark.

Wrong. there are no "last" elements.

>>
>> Thus,
>> if E(k+1) is finite, then E(k) is finite.
>
> Here you try to distinguish dark endsegments. This is impossible. Almost all endsegments and almost all natural numbers containd in them are dark. This has been proven once and for all by a very simple idea:

Wrong.

>
> "No interval has all of its rationals indexed before any other interval has all of its rationals indexed." Jim Burns

each interval has an infinity of rationals in it, there is no "before", so his statement is true.

>
> Regards, WM
>

now you rely on other posters statements, that you do not understand, to lend legs to your flaky math, fail

On Tuesday, 17 August 2021 at 15:14:52 UTC-3, Sergio wrote:
> On 8/17/2021 7:06 AM, WM wrote:
> > William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
> >> On Monday, August 16, 2021 at 4:46:49 PM UTC-4, WM wrote:
> >
> >>>> Nope, infinitely many elements that are the "last one of a FISON".
> >>> Possible for potential infinity
> >> "Potential infinity" is a nonsensical term. There is no such thing.
> >
> > There are many examples. The set of known prime numbers is potentially infinite.
> wrong, it is infinite.
> Can you prove The set of known prime numbers is finite? NO.

Ah, Wait now! The largest known prime is a Mersenne prime, M(82,589,933) = 2^(82,589,933) - 1. So, yes, the set of known prime numbers is a subset of FISON(M(82,589,933)), so it *is* finite. (Now I know and you know, and *WM* knows, that *he* can't prove that, so I'll give you half a point...)

On 8/17/2021 1:45 PM, Gus Gassmann wrote:
> On Tuesday, 17 August 2021 at 15:14:52 UTC-3, Sergio wrote:
>> On 8/17/2021 7:06 AM, WM wrote:
>>> William schrieb am Montag, 16. August 2021 um 23:26:53 UTC+2:
>>>> On Monday, August 16, 2021 at 4:46:49 PM UTC-4, WM wrote:
>>>
>>>>>> Nope, infinitely many elements that are the "last one of a FISON".
>>>>> Possible for potential infinity
>>>> "Potential infinity" is a nonsensical term. There is no such thing.
>>>
>>> There are many examples. The set of known prime numbers is potentially infinite.
>> wrong, it is infinite.
>> Can you prove The set of known prime numbers is finite? NO.
>
> Ah, Wait now! The largest known prime is a Mersenne prime, M(82,589,933) = 2^(82,589,933) - 1. So, yes, the set of known prime numbers is a subset of FISON(M(82,589,933)), so it *is* finite. (Now I know and you know, and *WM* knows, that *he* can't prove that, so I'll give you half a point...)
>

you can download that prime all written out, so ala WM, it would be a "defined" number by the one who downloads it, and is looking at it.

https://www.mersenne.org/primes/digits/M82589933.zip

here is the part I like;

"...2603343236007708672514133111649086783353633849942687758706757692769523079277096599021779607049140938
4229567653642016971337377070302028747457966475634475908072057942969309063678532125186840903373004947
3571854394960973849347019380128171183833656448472295090537235964395950264065333642350034869102640051..."

However, is the front end still "defined" if you reading the back end, and do not remember the front end anymore ?

so this number would only be "partially definable", or "as viewing definable" or "in process potentially definable" ?