On Monday, April 26, 2021 at 3:43:01 PM UTC-3, WM wrote:

> If dark numbers were significantly different from natnumbers, they would not be natnumbers. The only but important
> difference is that the are not connected by FISONs to the origin 0 or 1.

And since all natural numbers are connected by FISONS to the origin (trivial induction),
dark numbers are not natnumbers.

--
William Hughes

William has brought this to us :
> On Monday, April 26, 2021 at 3:43:01 PM UTC-3, WM wrote:
>
>> If dark numbers were significantly different from natnumbers, they would not
>> be natnumbers. The only but important difference is that the
>> are not connected by FISONs to the origin 0 or 1.
>
> And since all natural numbers are connected by FISONS to the origin (trivial
> induction), dark numbers are not natnumbers.

Yes, and two is the only even prime in the naturals. It is not "dark"
no matter how WM has tried to define "dark". All natural numbers are
defined (ants) have continued decimal expansions (ants) are
instantiated (ants) realized (ants) etcetera.

William schrieb am Montag, 26. April 2021 um 23:32:40 UTC+2:
> On Monday, April 26, 2021 at 3:43:01 PM UTC-3, WM wrote:
>
> > If dark numbers were significantly different from natnumbers, they would not be natnumbers. The only but important
> > difference is that the are not connected by FISONs to the origin 0 or 1..
> And since all natural numbers are connected by FISONS to the origin (trivial induction),
> dark numbers are not natnumbers.

Then there are not more natnumbers than are definable by unit fractions in the interval [1/n, 1] - their cardinal number is not larger than all natnumbers.

But not all natnumbers have the same properties. For instance, there are some less than 100 and some larger than 100, some having only one prime factor, others having two or more.

Well, trivial induction may be assumed as a fundamental property of natnumbers. But it shows that every natnumber that is subject to induction is insufficient to complete the set |N. If there is complete set then almost all natnumbers cannot be connected by FISONs to the origin. Of course you may *define* that every natnumber can be connected by a FISON to the origin, but then you have to concede that almost all of these FISONs cannot be determined. That is an irrefutable truth.

Regards, WM

On Tuesday, April 27, 2021 at 7:57:28 AM UTC-3, WM wrote:

> Of course you may *define* that every natnumber can be connected by a FISON to the origin,

This is in fact done. The fact that every natnumber can be connected by a FISON to the origin
is part of the definition of natnumber (it is easy to show this is equivalent to induction).

> but then you have to concede that almost all of these FISONs cannot be determined.

Whatever this means, it does not change the fact that every natnumber can be connected by a FISON to the origin

--
William Hughes

William schrieb am Dienstag, 27. April 2021 um 16:23:43 UTC+2:
> On Tuesday, April 27, 2021 at 7:57:28 AM UTC-3, WM wrote:
>
> > Of course you may *define* that every natnumber can be connected by a FISON to the origin,
> This is in fact done. The fact that every natnumber can be connected by a FISON to the origin
> is part of the definition of natnumber (it is easy to show this is equivalent to induction).
> > but then you have to concede that almost all of these FISONs cannot be determined.
> Whatever this means, it does not change the fact that every natnumber can be connected by a FISON to the origin

No, the claim that it "can be connected" is wrong. It may be assumed to be connected. But this cannot be done. Every natnumber that you can determine, identify, or represent by digits belongs to a finite set. If all natnumbers can be exhausted (as set theorists claim, see below), then almost all cannot be exhausted in a linear procedure like counting, because exhausting all would imply counting a last one.

===========================
Hausdorff: Before that (axiom of choice) it was usual to argue as follows: From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element aω+1, and so on. This procedure must come to an end, because beyond the set W of ordinal numbers which are mapped on elements of A, there are greater numbers; these obviously cannot be mapped on elements of A.

This is nonsense. It would supply a procedure for well-ordering every set.

Nevertheless there are some contemporary logicians who persist to endorse this method. In MathOverflow, answering the question [Wilhelm: "Endorsers of the method of well-ordering reported by Hausdorff?", MathOverflow (11 Mar 2018)] Emil Jeřábek counterfactually claimed: "This does not violate any Peano axioms. It is a perfectly valid and commonly used construction. [...] Peano axioms are axioms of natural numbers. The sequence here is not indexed by natural numbers, but by ordinals, so Peano axioms are irrelevant." And Joel David Hamkins boasted: "I endorse this method."

Cast in the same mould, Cohen suggests "we repeat this process countably many times" . And Hrbacek and Jech ponder, in order to increase the cardinality of a model, "if this procedure is iterated aleph_2 times".

Regards, WM

On Tuesday, April 27, 2021 at 4:02:34 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 27. April 2021 um 16:23:43 UTC+2:
> > On Tuesday, April 27, 2021 at 7:57:28 AM UTC-3, WM wrote:
> >
> > > Of course you may *define* that every natnumber can be connected by a FISON to the origin,
> > This is in fact done. The fact that every natnumber can be connected by a FISON to the origin
> > is part of the definition of natnumber (it is easy to show this is equivalent to induction).
> > > but then you have to concede that almost all of these FISONs cannot be determined.
> > Whatever this means, it does not change the fact that every natnumber can be connected by a FISON to the origin
> No, the claim that it "can be connected" is wrong. It may be assumed to be connected. But this cannot be done. Every natnumber that you can determine, identify, or represent by digits belongs to a finite set.

Let C(t) be the set of all natnumbers that have been explicitly connected at time t. Then the following two statements are true.

i. At any time t, there is a natural number n(t) that is not an element of C(t) (note that n(t) may change if t changes)

ii. Any natural number, n, can be an element of C(k) for some time k.

--
William Hughes

On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:
> ii. Any natural number, n, can be an element of C(k) for some time k.

To make things clearer

ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)

--
William Hughes

On 4/27/2021 5:57 AM, WM wrote:
> William schrieb am Montag, 26. April 2021 um 23:32:40 UTC+2:
>> On Monday, April 26, 2021 at 3:43:01 PM UTC-3, WM wrote:
>>
>>> If dark numbers were significantly different from natnumbers, they would not be natnumbers. The only but important
>>> difference is that the are not connected by FISONs to the origin 0 or 1.
>> And since all natural numbers are connected by FISONS to the origin (trivial induction),
>> dark numbers are not natnumbers.
>
> Then there are not more natnumbers than are definable by unit fractions in the interval [1/n, 1] - their cardinal number is not larger than all natnumbers.
>
> But not all natnumbers have the same properties. For instance, there are some less than 100 and some larger than 100, some having only one prime factor, others having two or more.

1 All dark numbers have identically the same properties.
2 There is no such thing as a single dark number.
3 Rejected Dark Numbers do not know that they are rejected.
4. (see THE ANT LIST 4.0 for more properties)

>
> Well, trivial induction may be assumed as a fundamental property of natnumbers. But it shows that every natnumber that is subject to induction is insufficient to complete the set |N. If there is complete set then almost all natnumbers cannot be connected by FISONs to the origin. Of course you may *define* that every natnumber can be connected by a FISON to the origin, but then you have to concede that almost all of these FISONs cannot be determined. That is an irrefutable truth.
>
> Regards, WM
>

On 4/27/2021 2:02 PM, WM wrote:
> William schrieb am Dienstag, 27. April 2021 um 16:23:43 UTC+2:
>> On Tuesday, April 27, 2021 at 7:57:28 AM UTC-3, WM wrote:
>>
>>> Of course you may *define* that every natnumber can be connected by a FISON to the origin,
>> This is in fact done. The fact that every natnumber can be connected by a FISON to the origin
>> is part of the definition of natnumber (it is easy to show this is equivalent to induction).
>>> but then you have to concede that almost all of these FISONs cannot be determined.
>> Whatever this means, it does not change the fact that every natnumber can be connected by a FISON to the origin
>
> No, the claim that it "can be connected" is wrong. It may be assumed to be connected. But this cannot be done. Every natnumber that you can determine, identify, or represent by digits belongs to a finite set. If all natnumbers can be exhausted (as set theorists claim, see below), then almost all cannot be exhausted in a linear procedure like counting,

counting works.

> because exhausting all would imply counting a last one.

your conclusion is wrong, there is no last one,
and those people are STILL COUNTING.

(Its a job, and its doesn't pay well either.)

>
> ============================
>
> Hausdorff: Before that (axiom of choice) it was usual to argue as follows: From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element aω+1, and so on. This procedure must come to an end,

no is does not have to come to an end, there is no last element, If you
state in this case it does, it ends at k, so it ends, because you
stopped at k.

<snip>

William schrieb am Mittwoch, 28. April 2021 um 02:07:10 UTC+2:
> On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:

> ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)

No, that is not true. For a simpler picture use the unit fractions and the interval (0, 1] containing all unit fractions with no doubt. But the interval [1/n, 1] will not contain all unit fractions, how far the time k_n may have advanced. In the interval (0, 1/n) there will remain ℵo unit fractions forever - that is not a matter of time or ressources. It is an unavoidable and unchangeable principle. It could be used as an axiom replacing Zermelo's axiom of infinity.

Regards, WM

Sergio schrieb am Mittwoch, 28. April 2021 um 03:08:55 UTC+2:
> On 4/27/2021 2:02 PM, WM wrote:

> > Hausdorff: Before that (axiom of choice) it was usual to argue as follows: From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element aω+1, and so on. This procedure must come to an end,

> no is does not have to come to an end, there is no last element, If you
> state in this case it does, it ends at k, so it ends, because you
> stopped at k.

Right! Therefore there remain alwaysw ℵo numbers uncounted. They are dark.

Regards, WM

On 4/28/2021 7:59 AM, WM wrote:
> William schrieb am Mittwoch, 28. April 2021 um 02:07:10 UTC+2:
>> On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:
>
>> ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)
>
> No, that is not true. For a simpler picture use the unit fractions and the interval (0, 1] containing all unit fractions

that is not what he is working on.

Solve it directly as is, and without your "smoke and mirrors" unit
fractions.

On 4/28/2021 8:02 AM, WM wrote:
> Sergio schrieb am Mittwoch, 28. April 2021 um 03:08:55 UTC+2:
>> On 4/27/2021 2:02 PM, WM wrote:
>
>>> Hausdorff: Before that (axiom of choice) it was usual to argue as follows: From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element aω+1, and so on. This procedure must come to an end,
>
>> no is does not have to come to an end, there is no last element, If you
>> state in this case it does, it ends at k, so it ends, because you
>> stopped at k.
>

>>> because exhausting all would imply counting a last one.

>>your conclusion is wrong, there is no last one,
>>and those people are STILL COUNTING.

>>(Its a job, and its doesn't pay well either.)

> Right! Therefore there remain alwaysw ℵo numbers uncounted. They are dark.
>
> Regards, WM
>

there is no "last one"

"dark" is meaningless.

On Wednesday, April 28, 2021 at 9:59:26 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 28. April 2021 um 02:07:10 UTC+2:
> > On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:
>
> > ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)
> No, that is not true. For a simpler picture use the unit fractions and the interval (0, 1] containing all unit fractions with no doubt. But the interval [1/n, 1] will not contain all unit fractions, how far the time k_n may have advanced. In the interval (0, 1/n) there will remain ℵo unit fractions forever

Let the set of unit fractions that are not elements of [1/n,1] at time k_n be R(k_n). Then the following two statements are true.

i. R(k_n) is infinite
ii For every natural number n, 1/n is not an element of R(k_n)

i.e. R(k_n) always has an infinite number of elements, but there is no single element which is in R(k_n) for every n.

--
William Hughes

On Wednesday, April 28, 2021 at 7:50:59 AM UTC-7, William wrote:
> On Wednesday, April 28, 2021 at 9:59:26 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 28. April 2021 um 02:07:10 UTC+2:
> > > On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:
> >
> > > ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)
> > No, that is not true. For a simpler picture use the unit fractions and the interval (0, 1] containing all unit fractions with no doubt. But the interval [1/n, 1] will not contain all unit fractions, how far the time k_n may have advanced. In the interval (0, 1/n) there will remain ℵo unit fractions forever
> Let the set of unit fractions that are not elements of [1/n,1] at time k_n be R(k_n). Then the following two statements are true.
>
> i. R(k_n) is infinite
> ii For every natural number n, 1/n is not an element of R(k_n)
>
> i.e. R(k_n) always has an infinite number of elements, but there is no single element which is in R(k_n) for every n.

And there you have a key to the answer to your original question:

Since the whole premise of 'dark numbers' is based on reversing-the-quantifiers ('quantifier dyslexia'), all you need is a characterization of even prime using
for every ... there exists ...

Then, reversing those quantifiers
there exists ... for every ...

will give you a necessary affirmative for the existence of an even dark prime.

William schrieb am Mittwoch, 28. April 2021 um 16:50:59 UTC+2:
> On Wednesday, April 28, 2021 at 9:59:26 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 28. April 2021 um 02:07:10 UTC+2:
> > > On Tuesday, April 27, 2021 at 5:53:30 PM UTC-3, William wrote:
> >
> > > ii Any natural number, n, can be an element of C(k_n) for some time k_n (if n changes then the time may change)
> > No, that is not true. For a simpler picture use the unit fractions and the interval (0, 1] containing all unit fractions with no doubt. But the interval [1/n, 1] will not contain all unit fractions, how far the time k_n may have advanced. In the interval (0, 1/n) there will remain ℵo unit fractions forever
> Let the set of unit fractions that are not elements of [1/n,1] at time k_n be R(k_n). Then the following two statements are true.
>
> i. R(k_n) is infinite
> ii For every natural number n, 1/n is not an element of R(k_n)
>
> i.e. R(k_n) always has an infinite number of elements,

R(k_n) always has an infinite number of elements.
Many numbers will leave R(k_n).
No numbers will enter R(k_n).
Infinitely many numbers will stay in R(k_n) from the start till the end.
That means infinitely many of its elements are in R(k_n) forever.

> but there is no single element which is in R(k_n) for every n.

There are infinitely many elements in R(k_n) for every n since R(k_n) always has an infinite number of elements. This set is shrinking but never less than infinite. None of its permanent elements can be reresented by digits or instantiated in any other way. They are not definable as individuals. They are dark numbers.

Regards, WM

On Wednesday, April 28, 2021 at 4:18:34 PM UTC-3, WM wrote:
> Infinitely many numbers will stay in R(k_n) from the start till the end.

There is no end. There are always more steps at which more elements of R(k_n) will be removed.
There are always more numbers to remove and always more steps at which numbers are removed.

--
William Hughes

William schrieb am Mittwoch, 28. April 2021 um 21:42:48 UTC+2:
> On Wednesday, April 28, 2021 at 4:18:34 PM UTC-3, WM wrote:
>
> > Infinitely many numbers will stay in R(k_n) from the start till the end.
> There is no end.

If the set is complete, then there is an end. Compare the set of unit fractions collectively completely removed when the interval (0, 1] is removed.

But when you deny the end, then we have to use "forever". An infinite set will stay in R(k_n) forever. Therefore your statement "but there is no single element which is in R(k_n) for every n" is wrong.

> There are always more steps at which more elements of R(k_n) will be removed.

Of course. The set is potentially infinite, i.e., always finite. Nevertheless almost all elements (aleph_0) will remain in R(k_n).
> There are always more numbers to remove and always more steps at which numbers are removed.

Of course. The set of definable natnumbers is potentially infinite.

Regards, WM