On Wednesday, August 18, 2021 at 4:28:42 PM UTC+2, FromTheRafters wrote:
> WM formulated the question :
> >
> > IN is the limit of the sequence of FISONs.

Indeed!

> The sequence of FISONs has a limit?

Sure. It's IN.

lim_(n->oo) {1, ..., n} = IN .

See: https://en.wikipedia.org/wiki/Set-theoretic_limit

On 8/18/2021 6:51 AM, WM wrote:
> Greg Cunt schrieb am Mittwoch, 18. August 2021 um 02:08:49 UTC+2:
>> if the natural numbers are defined due to von Neuman (which is the standard approach in ZFC these days) the set of FISONs is IDENTICAL with IN.
>
> ZFC is a big blunder. It is hard to understand that a hyper-intelligent person like on Neumann fell prey to this nonsense. Obviously some kind of herd instinct.
>>
>> Now IN -as you certainly know- is the "prototype" of an "actually infinite" set in ZFC. So, no, it's not "easy to contradict for actual infinity",
>
> But here it has been achieved: The definable natural numbers have a discernible well-ordering. Would they complete the indexing of the rationals, then we could find out which interval was completed first.

remove the word "definable" and I would agree.

using an ill defined "definable" eliminates it from serious math, and puts it into Joke Math.

>
> Regards, WM
>
>

On 8/18/2021 6:47 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 18. August 2021 um 07:36:05 UTC+2:
>> tisdag 17 augusti 2021 kl. 14:18:12 UTC+2 skrev 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
>
>> You assume that N_def is finite and this is a fault assumption, all "definable", whatever that means, endsegments is still N
>
> The definable natural numbers have a discernible well-ordering. Would they complete the i ndexing of the rationals, then we could find out which interval is completed first.

there is no "completed first"

>
> Regards, WM
>

On 8/18/2021 9:12 AM, WM wrote:
> Greg Cunt schrieb am Mittwoch, 18. August 2021 um 15:03:01 UTC+2:
>> On Wednesday, August 18, 2021 at 1:47:28 PM UTC+2, WM wrote:
>>
>>> The [...] natural numbers [are] well-order[ed by <=]. Would they complete the indexing of the rationals, then we could find out which interval is completed first.
>>
>> No, we "would't", since we can't.
>>
>> Actually, ***no*** intervall "is completed first",
>
> If all intervals are completed, then one of them must be completed first

nope. each interval is an infinite set of rationals.

> because there is a linear sequence of natnumbers which are used one after the other,

it is an infinite, there is no last one

> and one natnumber cannot complete more than one interval - if completion is possible at all.

Wrong again.

Back to basics, which natural number "completes" an interval ?

>
> Regards, WM
>

On Wednesday, 18 August 2021 at 12:20:46 UTC-3, Sergio wrote:
> On 8/18/2021 6:51 AM, WM wrote:
> > Greg Cunt schrieb am Mittwoch, 18. August 2021 um 02:08:49 UTC+2:
> >> if the natural numbers are defined due to von Neuman (which is the standard approach in ZFC these days) the set of FISONs is IDENTICAL with IN.
> >
> > ZFC is a big blunder. It is hard to understand that a hyper-intelligent person like on Neumann fell prey to this nonsense. Obviously some kind of herd instinct.
> >>
> >> Now IN -as you certainly know- is the "prototype" of an "actually infinite" set in ZFC. So, no, it's not "easy to contradict for actual infinity",
> >
> > But here it has been achieved: The definable natural numbers have a discernible well-ordering. Would they complete the indexing of the rationals, then we could find out which interval was completed first.
> remove the word "definable" and I would agree.
>
> using an ill defined "definable" eliminates it from serious math, and puts it into Joke Math.

Or "Mythomatics"... WM specializes in this category.

On Wednesday, August 18, 2021 at 7:34:26 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 17. August 2021 um 22:44:16 UTC+2:
> > On Tuesday, August 17, 2021 at 3:50:50 PM UTC-4, WM wrote:
> >
> > > The set of FISONs is not larger than every FISON because it does not contain anything larger than every FISON.
> > Given any elementof the set of FISONs x, I can find an element of the set of FISONS k(x) such that k(x) > x.
> Yes.
> > Thus for every element x of the set of FISONs the cardinality of the *set of FISONs* is greater than the cardinality of x.
> There is no cardinality of potentially infinite sets.

True, because the term "Potentiallyl infinite set" is nonsense. A set is something that does not change. The set of FISONs does not change.

==
William Hughes

On Wednesday, August 18, 2021 at 7:43:08 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 17. August 2021 um 22:54:58 UTC+2:
> > > When indexing all positive rationals, then for each interval (k, k+1] we can determine the first index issued to this interval. This requires at least as many indexes as there are intervals: aleph_0. According to set theory every interval is finished, i.e., completely indexed. But we cannot determine the order how this happens
> > because the term "order" has no meaning here.
> Why not? For every definable index it has meaning. If it has no meaning later, then there are no definable indices.

Nope, The fact that it has meaning for every index you can write down, does not mean it has meaning for every index you cannot write down.

> > Either no intervals are finished or all intervals are finished.
> This cannot happen by one index. At least aleph_0 indices are required. But between finishing one interval and all intervals no order and no indices can be distinguished.
> > There is no order in which the intervals are completed.
> If they were completed by definable indices, then there would be an order.. But there is none.
> > Note that for an element of the set |N_F the term order does have meaning.
> Of course!
> > This is true even for elements of |N_F that you cannot write down.
> That's why "dark" means more than not writable.

The set |N_F does not contain "dark" elements.
--
William Hughes

FromTheRafters schrieb am Mittwoch, 18. August 2021 um 16:28:42 UTC+2:
> WM formulated the question :
> > William schrieb am Dienstag, 17. August 2021 um 22:44:16 UTC+2:
> >> On Tuesday, August 17, 2021 at 3:50:50 PM UTC-4, WM wrote:
> >>
> >>> The set of FISONs is not larger than every FISON because it does not
> >>> contain anything larger than every FISON.
> >> Given any elementof the set of FISONs x, I can find an element of the set of
> >> FISONS k(x) such that k(x) > x.
> >
> > Yes.
> >
> >> Thus for every element x of the set of FISONs the cardinality of the *set of
> >> FISONs* is greater than the cardinality of x.
> >
> > There is no cardinality of potentially infinite sets. Cardinality is a
> > property of actually infinite sets, i.e. sets which are complete. You cannot
> > find a natural number outside of |N. Otherwise nobody could claim a bijection
> > between natural numbers and fractions. But you can always find a larger
> > natural number than n.
> >
> >> So what if I cannot find a single element of the *set of FISONs* that works
> >> for every *element of the set of FISONs*?
> >
> > It does not exist. |N is the limit of the sequence of FISONs.
> The sequence of FISONs has a limit?

Yes, like the sequence (1/n) has the limit 0.

Regards, WM

Sergio schrieb am Mittwoch, 18. August 2021 um 17:20:46 UTC+2:
> On 8/18/2021 6:51 AM, WM wrote:

> > But here it has been achieved: The natural numbers have a discernible well-ordering. Would they complete the indexing of the rationals, then we could find out which interval was completed first.
> remove the word "definable" and I would agree.

Done.

Regards, WM

On Wednesday, August 18, 2021 at 10:12:35 AM UTC-4, WM wrote:
> there is a linear sequence of natnumbers which are used one after the other,

Nope you cannot "complete" all intervals by a stepwise process. The fact that the set |N_F is totally ordered (linear) does not change this. To work with infinite sets you need to use tools that are appropriate. Stepwise processes are not appropriate for infinite sets.

--
William Hughes

William schrieb am Mittwoch, 18. August 2021 um 18:13:59 UTC+2:
> On Wednesday, August 18, 2021 at 7:43:08 AM UTC-4, WM wrote:

> > Why not? For every definable index it has meaning. If it has no meaning later, then there are no definable indices.
> Nope, The fact that it has meaning for every index you can write down, does not mean it has meaning for every index you cannot write down.

Correct. It has no meaning for dark indices.

> > > Either no intervals are finished or all intervals are finished.
> > This cannot happen by one index. At least aleph_0 indices are required. But between finishing one interval and all intervals no order and no indices can be distinguished.
> > > There is no order in which the intervals are completed.
> > If they were completed by definable indices, then there would be an order. But there is none.
> > > Note that for an element of the set |N_F the term order does have meaning.
> > Of course!
> > > This is true even for elements of |N_F that you cannot write down.
> > That's why "dark" means more than not writable.
> The set |N_F does not contain "dark" elements.

But if it indexes all fraction, then it contains elements that cannot be ordered.

Regards, WM

On Wednesday, August 18, 2021 at 6:23:09 PM UTC+2, WM wrote:

> <bla bla bla> then it contains elements that cannot be ordered.

Actually, every set can (even) be well-ordered.

See: https://en.wikipedia.org/wiki/Well-ordering_theorem

On Wednesday, August 18, 2021 at 12:23:09 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 18. August 2021 um 18:13:59 UTC+2:
> > On Wednesday, August 18, 2021 at 7:43:08 AM UTC-4, WM wrote:
>
> > > Why not? For every definable index it has meaning. If it has no meaning later, then there are no definable indices.
> > Nope, The fact that it has meaning for every index you can write down, does not mean it has meaning for every index you cannot write down.
> Correct. It has no meaning for dark indices.

The set |N_F does not contain "dark" elements.

> > > > Either no intervals are finished or all intervals are finished.
> > > This cannot happen by one index. At least aleph_0 indices are required. But between finishing one interval and all intervals no order and no indices can be distinguished.
> > > > There is no order in which the intervals are completed.
> > > If they were completed by definable indices, then there would be an order. But there is none.
> > > > Note that for an element of the set |N_F the term order does have meaning.
> > > Of course!
> > > > This is true even for elements of |N_F that you cannot write down.
> > > That's why "dark" means more than not writable.
> > The set |N_F does not contain "dark" elements.
> But if it indexes all fraction, then it contains elements that cannot be ordered.

Nope there is no element of U_F that cannot be indexed by |N_F and no element of |N_F that cannot be ordered.

--
William Hughes

On 8/17/2021 4:01 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 17. August 2021 um 20:11:24 UTC+2:

>>> 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.
>
> Correct. Most are and remain without index.

Let p/q be a positive rational in lowest terms, gcd(p,q) = 1.

Define
index(p/q) = q*q*p*p/rad(p)

rad(p) == the radical of p

https://en.wikipedia.org/wiki/Radical_of_an_integer

| In number theory, the radical of a positive integer n is defined
| as the product of the distinct prime numbers dividing n.
| Each prime factor of n occurs exactly once as a factor of
| this product:

....
index(97) = 97
index(2/7) = 98
index(11/3) = 99
index(1/10) = 100
index(101) = 101
....

No integer-ended interval has any rationals in it without index.
No integer-ended interval has a last-indexed rational in it.

>> All of the rationals in all of the intervals are indexed.
>
> That's just what I mean.
>
>> These are different claims.
>
> No, I mean exactly that. In every interval there is a first
> rational indexed. This requires at least aleph_0 rationals
> for aleph_0 intervals. Further all intervals are completed:
> "All of the rationals in all of the intervals are indexed."
> This requires another aleph_0 indexes.
> But we know that no interval is indexed before any other.
> We cannot count the indexes any longer. We cannot find out
> their order as we could with the first index per interval.
> They are dark.

We know the order of the definable finite indexes.
For each definable index, there are finitely-many definable
indexes before it and infinitely many definable finite indexes
after it. There is no last definable index.

We know many ways to map each definable positive rational
so that, for each rational, a unique index exists, and,
for each index, a unique rational exists. Pick one way.

The definable positive rationals have a certain, often-used
order _in the line_
'p < q' == "p is before q in the line"

The way we picked to map each definable positive rational to
its own definable finite index is a different order, with its
own properties. I'll call it the order _in the list_
'p << q' == "p is before q in the list"

For any two definable positive rationals p,q, there are
infinitely-many definable positive rationals between them
_in the line_
{ r in Q+ | p < r < q } is infinite.

For any two definable positive rationals p,q, there are
finitely-many definable positive rationals between them
_in the list_
{ r in Q+ | p << r << q } is finite.

----
All of these are definable indexes and rationals.
I guess that you would say that none of them are dark.
However, these non-dark indexes and rationals already have
the properties that you explain by introducing dark numbers.

| No interval has all of its rationals indexed before
| any other interval has all of its rationals indexed.
-- me

Apparently you like what I said there. You've been quoting it.

You think the quote implies the existence of your dark numbers.
No, it doesn't imply that.

Each rational is finitely-indexed.
Each interval contains infinitely-many rationals.

If all of the rationals in [p,q] were indexed before
all of the rationals in [r,s], there would need to be
rationals later in [r,s] after the infinitely-many in [p,q].
But then, these later rationals in [r,s] would have to be
infinitely-indexed. They're all finitely-indexed.

Therefore,
no interval has all of its rationals indexed before
any other interval has all of its rationals indexed.

To be clear, these rationals are definable positive rationals
and these indexes are definable finite indexes.
Your dark numbers are without necessity.

On 8/17/2021 8:18 AM, WM wrote:
> zelos...@gmail.com schrieb
> am Dienstag, 17. August 2021 um 09:20:48 UTC+2:

>> 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

Each rational is finitely-indexed.
Each interval contains infinitely-many rationals.

If all of the rationals in [p,q] were indexed before
all of the rationals in [r,s], there would need to be
rationals later in [r,s] after the infinitely-many in [p,q].
But then, these later rationals in [r,s] would have to be
infinitely-indexed. They're all finitely-indexed.

Therefore,
no interval has all of its rationals indexed before
any other interval has all of its rationals indexed.

To be clear, these rationals are definable positive rationals
and these indexes are definable finite indexes.
Your dark numbers are without necessity.

On 8/18/2021 11:20 AM, Sergio wrote:
> On 8/18/2021 6:51 AM, WM wrote:

>> But here it has been achieved: The definable natural numbers have
>> a discernible well-ordering. Would they complete the indexing of the
>> rationals, then we could find out which interval was completed first.
>
> remove the word "definable" and I would agree.
>
> using an ill defined "definable" eliminates it from serious math,
> and puts it into Joke Math.

Removing "definable" doesn't fix it.

Each rational is finitely-indexed.
Each interval contains infinitely-many rationals.

If all of the rationals in [p,q] were indexed before
all of the rationals in [r,s], there would need to be
rationals later in [r,s] after the infinitely-many in [p,q].
But then, these later rationals in [r,s] would have to be
infinitely-indexed. They're all finitely-indexed.

Therefore,
no interval has all of its rationals indexed before
any other interval has all of its rationals indexed.

To be clear, these rationals are definable positive rationals
and these indexes are definable finite indexes.
WM's dark numbers are without necessity.

On 8/17/2021 3:01 PM, WM wrote:
> Jim Burns schrieb am Dienstag, 17. August 2021 um 20:11:24 UTC+2:
>> On 8/17/2021 8:36 AM, WM wrote:
>
>>>>> "No interval has all of its rationals indexed before
>>>>> any other interval has all of its rationals indexed."
>>>>> Jim Burns
>
>> 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.
>
> If in every interval at least one fractions is indexed, then aleph_0 indexes must have been supplied.
>>
>> _Up to_ any finite index, no more than finitely-many positive
>> rationals in no more than finitely-many integer-ended intervals
>> are indexed.
>
> That is true. Therefore the rest cannot be distinguished by counting.
>>
>> 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).
>
> But not by a finite number of definable indexes?!
>
>>> 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.
>
> Correct. Most are and remain without index.

wrong, there is no "most"

>
>> That is apparently what you mean by "completed".
>
> Not at all. I explicitly excluded a last indexed rational. I only used the term completed by what you understand by completed.
>
>> All of the rationals in all of the intervals are indexed.
>
> That's just what I mean.
>
>> These are different claims.
>
> No, I mean exactly that. In every interval there is a first rational indexed. This requires at least aleph_0 rationals for aleph_0 intervals.

wrong. requires aleph_0 rationals for an interval

> Further all intervals are completed: "All of the rationals in all of the intervals are indexed." This requires another aleph_0 indexes.

wrong. all rationals in the interval are indexed each by a real number.

> But we know that no interval is indexed before any other. We cannot count the indexes any longer.

wrong. all rationals are indexed by the real numbers.

> We cannot find out their order

wrong. all rationals are indexed by the real numbers.

> as we could with the first index per interval.

wrong. all rationals are indexed by the real numbers.

> They are dark.

wrong again.

>
> Regards, WM
>

7 wrongs in one post, you are degenerating

On 8/18/2021 10:12 AM, WM wrote:
> Greg Cunt schrieb
> am Mittwoch, 18. August 2021 um 15:03:01 UTC+2:
>> On Wednesday, August 18, 2021 at 1:47:28 PM UTC+2, WM wrote:

>>> The [...] natural numbers [are] well-order[ed by <=].
>>> Would they complete the indexing of the rationals,
>>> then we could find out which interval is completed first.
>>
>> No, we "would't", since we can't.
>> Actually, ***no*** intervall "is completed first",
>
> If all intervals are completed, then one of them must be
> completed first because there is a linear sequence of
> natnumbers which are used one after the other, and one
> natnumber cannot complete more than one interval
> - if completion is possible at all.

True in some cases:
A claim that something is after each item in an unending
sequence. (See omega.)

Contradictory:
A claim that something is after the last item in an unending
sequence. (See grossone -- or, better, don't.)

On 8/17/2021 8:18 AM, WM wrote:

> 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

Each rational is finitely-indexed.
Each interval contains infinitely-many rationals.

If all of the rationals in [p,q] were indexed before
all of the rationals in [r,s], there would need to be
rationals later in [r,s] after the infinitely-many in [p,q].
But then, these later rationals in [r,s] would have to be
infinitely-indexed. They're all finitely-indexed.

Therefore,
no interval has all of its rationals indexed before
any other interval has all of its rationals indexed.

To be clear, these rationals are definable positive rationals
and these indexes are definable finite indexes.
Your dark numbers are without necessity.

Jim Burns

Greg Cunt schrieb am Mittwoch, 18. August 2021 um 18:31:58 UTC+2:

> Actually, every set can (even) be well-ordered.
>
> See: https://en.wikipedia.org/wiki/Well-ordering_theorem

Counterfactual foolish nonsense.

Regards, WM

On 8/17/2021 8:16 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 16. August 2021 um 23:46:29 UTC+2:

>> 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

Each rational is finitely-indexed.
Each interval contains infinitely-many rationals.

If all of the rationals in [p,q] were indexed before
all of the rationals in [r,s], there would need to be
rationals later in [r,s] after the infinitely-many in [p,q].
But then, these later rationals in [r,s] would have to be
infinitely-indexed. They're all finitely-indexed.

Therefore,
no interval has all of its rationals indexed before
any other interval has all of its rationals indexed.

To be clear, these rationals are definable positive rationals
and these indexes are definable finite indexes.
Your dark numbers are without necessity.

Jim Burns

William schrieb am Mittwoch, 18. August 2021 um 18:32:37 UTC+2:
> On Wednesday, August 18, 2021 at 12:23:09 PM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 18. August 2021 um 18:13:59 UTC+2:
> > > On Wednesday, August 18, 2021 at 7:43:08 AM UTC-4, WM wrote:
> >
> > > > Why not? For every definable index it has meaning. If it has no meaning later, then there are no definable indices.
> > > Nope, The fact that it has meaning for every index you can write down, does not mean it has meaning for every index you cannot write down.
> > Correct. It has no meaning for dark indices.
> The set |N_F does not contain "dark" elements.

How do you explain that many indexes can be issued in order at the beginning while no index can be distinguished after all intervals have been defiled?
>
> > > The set |N_F does not contain "dark" elements.
> > But if it indexes all fraction, then it contains elements that cannot be ordered.
> Nope there is no element of U_F that cannot be indexed by |N_F and no element of |N_F that cannot be ordered.
How do you explain that no index can be ordered after all intervals have been defiled?

Regards, WM

Jim Burns schrieb am Mittwoch, 18. August 2021 um 20:26:59 UTC+2:

> | No interval has all of its rationals indexed before
> | any other interval has all of its rationals indexed.
> -- me
>
> Apparently you like what I said there. You've been quoting it.
>
> You think the quote implies the existence of your dark numbers.
> No, it doesn't imply that.

How do you explain that after having indexed every interval at least one time no order of the further indexing can be found?
>
> Therefore,
> no interval has all of its rationals indexed before
> any other interval has all of its rationals indexed.

But you think that all rationals of all intervals are indexed?

> To be clear, these rationals are definable positive rationals
> and these indexes are definable finite indexes.

Then define the indexes completing the intervals - whatever you understand by completing. But don't try to claim that all incomplete intervals are completed by one index.

Regards, WM

Jim Burns schrieb am Mittwoch, 18. August 2021 um 20:58:14 UTC+2:
> On 8/18/2021 10:12 AM, WM wrote:

> > If all intervals are completed, then one of them must be
> > completed first because there is a linear sequence of
> > natnumbers which are used one after the other, and one
> > natnumber cannot complete more than one interval
> > - if completion is possible at all.
> True in some cases:
> A claim that something is after each item in an unending
> sequence. (See omega.)

You say all can be completed. You know you cannot find out what is completed first. But it is clear that not all can be completed by one index. So there are at least aleph_0 indexes which cannot be distimngusihed or defined.

Regards, WM

Jim Burns schrieb am Mittwoch, 18. August 2021 um 21:07:09 UTC+2:
> On 8/17/2021 8:18 AM, WM wrote:
>
> > 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
> Each rational is finitely-indexed.

Impossible. All aleph_0 finite indexes are already issued to defile aleph_0 intervals.
> If all of the rationals in [p,q] were indexed before
> all of the rationals in [r,s], there would need to be
> rationals later in [r,s] after the infinitely-many in [p,q].
> But then, these later rationals in [r,s] would have to be
> infinitely-indexed. They're all finitely-indexed.

Impossible. All aleph_0 finite indexes are already issued to defile aleph_0 intervals.
>
> Therefore,
> no interval has all of its rationals indexed before
> any other interval has all of its rationals indexed.

It is sufficient to index one rational in every interval to issue aleph_0 indexes.

> To be clear, these rationals are definable positive rationals
> and these indexes are definable finite indexes.

For aleph_0 intervals less than aleph_0 indices are required?

> Your dark numbers are without necessity.

For aleph_0 intervals less than aleph_0 indices are required to defile them?
For aleph_0 intervals less than aleph_0 indices are required to complete them?

Regards, WM