William schrieb am Mittwoch, 18. August 2021 um 18:21:17 UTC+2:
> 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.

Yes! Why?

As long as the natural numbers are definable we can follow the steps from 1 to that natural number. Where this is no longer possible they have ceased to be definable. But Cantor's claim and that of matheology is ist that every rational gets an index that can be stepwise reached from the origin 1.

This claim has turned out wrong.

Regards, WM

Greg Cunt formulated the question :
> 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

Oh, *that* limit.

WM explained :
> 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

So, the sequence of FISONs, as a series, has a sequence of partial sums
which converges to a number in the set of natural numbers?

On 8/18/2021 3:31 PM, WM wrote:
> 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.

The Axiom of Choice is obviously true,
the Well Ordering Principle is obviously false,
and who can tell about Zorn's Lemma?

This is a standard math geek joke.
What's funny is that they are each equally true or
equally false, even if many intuitions say otherwise.

Perhaps you (WM) would prefer

-- For each cut B,A of the interval [0,1], a point x exists
such that each non-x point in B is before x and each non-x
point in A is after x.
-- The set of all points in [0,1] exists.
-- If set B exists, then the powerset Pwr(B) of B exists.
-- If a collection C exists of pairwise disjoint sets,
then the union Union(C) of C exists.

Perhaps someone will need to correct me, but I think those
leads to the axiom of choice and, from AC, to well-ordering.
(I want to provide support for an inaccessible cardinal.)

Jim Burns schrieb am Mittwoch, 18. August 2021 um 22:12:53 UTC+2:
> On 8/18/2021 3:31 PM, WM wrote:
> > 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.
> The Axiom of Choice is obviously true,

It is nonsense. Only if we refrain from believing in dark numbers, then the axiom is obviously true because there are no uncountable sets.

> the Well Ordering Principle is obviously false,
> and who can tell about Zorn's Lemma?
>
> This is a standard math geek joke.
> What's funny is that they are each equally true or
> equally false, even if many intuitions say otherwise.

You cannot choose any index that completes the enumeration of any interval. But you claim that every index n can be analysed and addressed by its FISON.

Regards, WM

FromTheRafters schrieb am Mittwoch, 18. August 2021 um 22:07:31 UTC+2:
> WM explained :
> > 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.

> So, the sequence of FISONs, as a series, has a sequence of partial sums
> which converges to a number in the set of natural numbers?

No. 0 is not a natural number, neither is oemga.

Regards, WM

On 8/18/2021 3:45 PM, WM wrote:
> 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?

Every rational in every interval has a finite index.

( There are many ways to do this. Here's one:
( Let p/q be in lowest terms.
( 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
( ...

q*q*p*p/rad(p) is finite.

> How do you explain that after having indexed every interval
> at least one time no order of the further indexing can be found?

Every rational in every interval has a finite index.
For example, q*q*p*p/rad(p) is finite.

After infinitely-many rationals have been indexed,
All finitely-indexed rationals have been indexed.
That would be all the rationals, since all rationals
are finitely-indexed. (See q*q*p*p/rad(p).)

There is no further indexing left to be done.

>> 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?

For p/q in lowest terms, define
index(p/q) = q*q*p*p/rad(p)

rad(p) == the radical of p

>> 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.

What I understand is that we cannot, even in principle,
step through an infinite set one-by-one.
What we can do is reason about each of infinitely-many,
starting with a description true of each, and proceeding by
truth-preserving inferences from the description.

"The indexes completing the intervals" looks like a
description of things that don't exist, like natural numbers
in the ratio of sqrt(2).

On 8/18/2021 3:53 PM, WM wrote:
> 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.

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
....

q*q*p*p/rad(p) is finite.

WM expressed precisely :
> FromTheRafters schrieb am Mittwoch, 18. August 2021 um 22:07:31 UTC+2:
>> WM explained :
>>> 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.
>
>> So, the sequence of FISONs, as a series, has a sequence of partial sums
>> which converges to a number in the set of natural numbers?
>
> No. 0 is not a natural number, neither is oemga.

Is |N a natural number?

Jim Burns schrieb am Mittwoch, 18. August 2021 um 22:50:40 UTC+2:
> On 8/18/2021 3:45 PM, WM wrote:

> > How do you explain that after having indexed every interval
> > at least one time no order of the further indexing can be found?
> Every rational in every interval has a finite index.

But you cannot discern most, let alone count to it.

> After infinitely-many rationals have been indexed,
> All finitely-indexed rationals have been indexed.

But you cannot know these indices because you cannot distinguish the completed intervals when not yet all have been completed.

> That would be all the rationals, since all rationals
> are finitely-indexed. (See q*q*p*p/rad(p).)
>
> There is no further indexing left to be done.

So you have indexed at least one rational in aleph_0 intervals and have used aleph_0 natural numbers. What do you use for completing the intervals?

> >> 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.

> What I understand is that we cannot, even in principle,
> step through an infinite set one-by-one.

Above you said that all indexes are finite. What finite index cannot be stepped to? And why?

> What we can do is reason about each of infinitely-many,

Each finite index can be stepped to in principle: 1, 2, 3, ..., n.
>
> "The indexes completing the intervals" looks like a
> description of things that don't exist,

Why then do you claim that all rationals of all intervals can be indexed by finite numbers?

> like natural numbers
> in the ratio of sqrt(2).

Who has claimed that? Nobody. But you have most recently uttered tha foolish claim that infinitely many rationals in infinitely many intervals can be indexed by finite numbers although you would need all already for the first round.

Regards, WM

Jim Burns schrieb am Mittwoch, 18. August 2021 um 23:01:34 UTC+2:
> On 8/18/2021 3:53 PM, WM wrote:
> > 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.
> Let p/q be a positive rational in lowest terms, gcd(p,q) = 1.

What a nonsense! You can cover only definable numbers in that way!

Explain why every index is finite, each rational is finitely-indexed, and every interval is completed but you cannot find out which is the first index completing an interval.

Regards, WM

On 8/18/2021 3:49 PM, WM wrote:
> 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.)

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.)

> You say all can be completed.
> You know you cannot find out what is completed first.

I know that none of the intervals is complete before any of
the other intervals. If [r,s] is still incomplete at an
index that [p,q] is complete, then not all of the rationals
in [r,s] are finitely-indexed.

index(p/q) = q*q*p*p/rad(p) is finite.

> 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.

You know, it was just a few days ago that you (WM) were using
your claim that some intervals completed before others to
"prove" that dark numbers exist.[1]

Now, you're using the negation of that to prove that dark
numbers exist. It's funny how some things change
but some things don't change.

[1]
Newsgroups: sci.logic
Subject: Cantor's incomplete enumeration
Date: Sat, 14 Aug 2021 05:59:17 -0700 (PDT)
From: Christa M <transfinity01@gmail.com>
Message-ID: <b6d8bcbb-1d43-4e2f-9370-3b4ee000449an@googlegroups.com>

On 8/18/2021 3:02 PM, WM wrote:
> William schrieb am Mittwoch, 18. August 2021 um 18:21:17 UTC+2:
>> 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.
>
> Yes! Why?
>
> As long as the natural numbers are definable we can follow the steps from 1 to that natural number. Where this is no longer possible they have ceased to be definable. But Cantor's claim and that of matheology is ist that every rational gets an index that can be stepwise reached from the origin 1.
>
> This claim has turned out wrong.
>
> Regards, WM
>

what is wrong about it ?

On 8/18/2021 2:49 PM, WM wrote:
> 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.

true.

>You know you cannot find out what is completed first.

true

> But it is clear that not all can be completed by one index.

wrong. Show your work.

> So there are at least aleph_0 indexes which cannot be distimngusihed or defined.

wrong again.

>
> Regards, WM
>

On 8/18/2021 3:36 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 18. August 2021 um 22:12:53 UTC+2:
>> On 8/18/2021 3:31 PM, WM wrote:
>>> 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.
>> The Axiom of Choice is obviously true,
>
> It is nonsense. Only if we refrain from believing in dark numbers, then the axiom is obviously true because there are no uncountable sets.
>
>> the Well Ordering Principle is obviously false,
>> and who can tell about Zorn's Lemma?
>>
>> This is a standard math geek joke.
>> What's funny is that they are each equally true or
>> equally false, even if many intuitions say otherwise.
>
> You cannot choose any index that completes the enumeration of any interval. But you claim that every index n can be analysed and addressed by its FISON.
>
> Regards, WM
>

Your admission that Dark Numbers are a Belief, and not fact does help see your point of view.

On 8/18/2021 2:38 PM, WM wrote:
> 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?

in what context ?

>>
>>>> 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?

in what context ?

>
> Regards, WM
>

On 8/18/2021 4:06 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 18. August 2021 um 22:50:40 UTC+2:
>> On 8/18/2021 3:45 PM, WM wrote:
>
>>> How do you explain that after having indexed every interval
>>> at least one time no order of the further indexing can be found?
>> Every rational in every interval has a finite index.
>
> But you cannot discern most, let alone count to it.

that is conjecture, show that you cannot.

>
>> After infinitely-many rationals have been indexed,
>> All finitely-indexed rationals have been indexed.
>
> But you cannot know these indices because you cannot distinguish the completed intervals when not yet all have been completed.

no makie sense, you are missing context.

>
>> That would be all the rationals, since all rationals
>> are finitely-indexed. (See q*q*p*p/rad(p).)
>>
>> There is no further indexing left to be done.
>
> So you have indexed at least one rational in aleph_0 intervals and have used aleph_0 natural numbers. What do you use for completing the intervals?

who used aleph_0 natural numbers up ?

On Wednesday, August 18, 2021 at 10:07:31 PM UTC+2, FromTheRafters wrote:
> WM explained :
> > FromTheRafters schrieb am Mittwoch, 18. August 2021 um 16:28:42 UTC+2:
> >> WM formulated:
> > > >
> > > > 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.

Ah?

> So, the sequence of FISONs [...] converges to a number in the set of natural numbers?

C'mon, that's not fair. We all know that WM is an idiot. But h e r e there's nothing (!) to complain, I's say.

1. "N is the limit of the sequence of FISONs." (true)

2. "The sequence (1/n) has the limit 0." (true)

3. "... like ..." (true, in a certain sense. After all, a set theoretic limit is a "limit" - sort of. There are formal similarities between the notion of /set-theoretic limit/ and the "limit" defined for sequences of real numbers).

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

On 8/18/2021 5:10 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 18. August 2021 um 23:01:34 UTC+2:
>> On 8/18/2021 3:53 PM, WM wrote:
>>> 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.

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

>>> All aleph_0 finite indexes are already issued
>>> to defile aleph_0 intervals.
>>
>> Let p/q be a positive rational in lowest terms, gcd(p,q) = 1.
>
> What a nonsense! You can cover only definable numbers in that way!

For each definable positive rational, a unique definable
finite index exists.
For each definable finite index, a unique definable
positive rational exists.

> Explain why every index is finite, each rational is finitely-indexed,
> and every interval is completed but you cannot find out which is
> the first index completing an interval.

No finite index can complete an interval with infinitely-many
indexed rationals.

On Wednesday, August 18, 2021 at 10:37:55 PM UTC+2, WM wrote:

> 0 is not a natural number

Really?

Hint: "In der DIN-Norm 5473 wird die Null zu den natürlichen Zahlen gezählt." ["In DIN standard 5473, zero is counted as a natural number."]

On Wednesday, August 18, 2021 at 11:07:07 PM UTC+2, WM wrote:
> Jim Burns schrieb am Mittwoch, 18. August 2021 um 22:50:40 UTC+2:
> >
> > Every rational in every interval has a finite index.
> >
> But you cannot [...] count [them all].

Indeed, only Chuck Norris could do that!

Remember: Chuck Norris counted to infinity - twice.

On Wednesday, August 18, 2021 at 3:38:35 PM UTC-4, WM wrote:
> 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?

Nothing to explain as there is no problem in distinguishing elements of |N_F that cannot be written down.

--
William Hughes

On 8/18/2021 1:47 PM, Jim Burns wrote:
> 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.

agree

>
> If all of the rationals in [p,q] were indexed before
> all of the rationals in [r,s],

the initial assumption

> there would need to be
> rationals later in [r,s] after the infinitely-many in [p,q].

yup, that would have to occur, but I don't see how it could.

> But then, these later rationals in [r,s] would have to be
> infinitely-indexed. They're all finitely-indexed.

so someone stopped at k in [r,s]

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

agree.

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

almost the same way to look at it =>, [p,q] has an infinite number of rationals so does [r,s], if you are filling these up and you stop, you still have
infinitely more to go in each. One does not "complete" (which is stopping) before the other, as they never get "completed" both still have infinities
more to go.

WM likes to wiggle around by "stopping" in an infinite count, and make conjecture with non math words, complete, full, last, etc.
I would love it if he could prove stuff using math, but no.

> WM's dark numbers are without necessity.
>

true

>The definable natural numbers have a discernible well-ordering

Whatever "discernible" means.

but N is well-ordered so again, N_def=N

>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.

Or, the issue is you and you're an idiot.

Jim Burns schrieb am Mittwoch, 18. August 2021 um 23:23:07 UTC+2:
> On 8/18/2021 3:49 PM, WM wrote:

> > You say all can be completed.
> > You know you cannot find out what is completed first.
> I know that none of the intervals is complete before any of
> the other intervals.

(1) You claim that all intervals can be completed.
(2) You know that not all intervals can be completed by one index.
(3) So there are at least two indices required to complete all intervals.
(4) You cannot find out their order by their FISONs.
(5) But you claim that only indices having FISONs are existing.

This is a contradiction outside of matheology.

> If [r,s] is still incomplete at an
> index that [p,q] is complete, then not all of the rationals
> in [r,s] are finitely-indexed.

You need not prove (4). It is obvious without your proof. So this is only an attempt to blur the straight argument.
>
> You know, it was just a few days ago that you (WM) were using
> your claim that some intervals completed before others to
> "prove" that dark numbers exist.[1]

Of course. If (1) is assumed, then (3) is unavoidable. Then (4) proves dark indices.
>
> Now, you're using the negation of that to prove that dark
> numbers exist.

No. Go through (1) to (5) above. Explain your position, but don't try to prove (1) to (4). It is not doubted.

Regards, WM