William schrieb am Freitag, 17. Dezember 2021 um 16:06:28 UTC+1:
> On Friday, December 17, 2021 at 6:02:54 AM UTC-4, WM wrote:
> > William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
>
> > > There is just one "step", an operator applied to the set
> > > (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)

> > That is possible however for all endsegments E(n) with definable indexes n.

> No it is not possible, The state "at infinity" may not have the same properties as the state at finite steps.

Maybe. But with definable indexes this state will never be reached. They will never change their properties. Only dark indexes can do so.

And if other properties may appear, then Cantor's enumeration of the fractions and his diagonalization may not have the same properties too "in the infinite". Not even the state at infinity of the ordered set of natural numbers may continue to have the properties known at finite steps. May be the set turns out empty "at infinity". May be it turns out subcountable or uncountable. Maybe it tells the observer: All matheologians are fools. You never know.

Regards, WM

zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 13:27:08 UTC+1:
> fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:

>
> But with that, congrats, you have L=N :)
>
> because n ∉ E(n+1) is true for ALL n

What contain the non-empty endsegments?

> >But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
> Who says any endsegment is empty?

What do non-empty endsegments contain when all n are gone?

Regards, WM

On Friday, December 17, 2021 at 6:06:34 PM UTC-4, WM wrote:
> William schrieb am Freitag, 17. Dezember 2021 um 16:06:28 UTC+1:
> > On Friday, December 17, 2021 at 6:02:54 AM UTC-4, WM wrote:
> > > William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
> >
> > > > There is just one "step", an operator applied to the set
> > > > (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)
> > > That is possible however for all endsegments E(n) with definable indexes n.
> > No it is not possible, The state "at infinity" may not have the same properties as the state at finite steps.
> Maybe.

Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.

Note that there is no such thing as a "dark" natural number. (No the set of natural numbers that you cannot write down are not "dark",
eg: every element of this set is the largest element of a FISON).

--
William Hughes

On 12/17/2021 4:06 PM, WM wrote:
> William schrieb am Freitag, 17. Dezember 2021 um 16:06:28 UTC+1:
>> On Friday, December 17, 2021 at 6:02:54 AM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
>>
>>>> There is just one "step", an operator applied to the set
>>>> (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)
>
>>> That is possible however for all endsegments E(n) with definable indexes n.
>
>> No it is not possible, The state "at infinity" may not have the same properties as the state at finite steps.
>
> Maybe. But with definable indexes this state will never be reached. They will never change their properties. Only dark indexes can do so.

wrong thinking. With your "definable" number you always stop at some number k, you always remain finite.

>
> And if other properties may appear, then Cantor's enumeration of the fractions and his diagonalization may not have the same properties too "in the infinite".

may ? then prove it, one way or the other.

Hint: this is trivial. His diagonalization have same properties "in the infinite".

> Not even the state at infinity of the ordered set of natural numbers may continue to have the properties known at finite steps.

such as ? anything ??

William schrieb am Samstag, 18. Dezember 2021 um 01:31:46 UTC+1:

> Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.

Why not? All definable steps can be looked at.
>
> Note that there is no such thing as a "dark" natural number.

That is your self-contradiction. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps because they are followed by dark steps.

> (No the set of natural numbers that you cannot write down are not "dark",
> eg: every element of this set is the largest element of a FISON).

Every such element indexes a step that can be looked at. Alas for all of them we have
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Regards, WM

Serg io schrieb am Samstag, 18. Dezember 2021 um 02:47:02 UTC+1:
> With your "definable" number you always stop at some number k, you always remain finite.

Of course. Try to find an infinite definable element in a FISON.

Regards, WM

On Saturday, December 18, 2021 at 8:08:48 AM UTC-4, WM wrote:
> William schrieb am Samstag, 18. Dezember 2021 um 01:31:46 UTC+1:
>
> > Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.
> Why not?

Let the set which is produced by applying the operator to {E(1). E(2), E(3),...} be S.

You are considering a process, call it T, with steps at all natural numbers . Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))

There is no reason why S should be have any connection to the T(k) Knowing that the T(k) share a common property does not mean that S must share this property.

>All definable steps can be looked at.

Looking at some of the T(k) (the T(k) for which k can be written down) tells you nothing about S.

--
William Hughes
..

William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
> On Saturday, December 18, 2021 at 8:08:48 AM UTC-4, WM wrote:
> > William schrieb am Samstag, 18. Dezember 2021 um 01:31:46 UTC+1:
> >
> > > Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.
> > Why not?
> Let the set which is produced by applying the operator to {E(1). E(2), E(3),...} be S.
>
> You are considering a process, call it T, with steps at all natural numbers .

At all *definable* natural numbers. The endsegments are infinite, their intersection is infinite.

> Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
>
> There is no reason why S should be have any connection to the T(k) Knowing that the T(k) share a common property does not mean that S must share this property.

If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers. S would be an infinite set.

> >All definable steps can be looked at.
> Looking at some of the T(k) (the T(k) for which k can be written down) tells you nothing about S.

That's because undefinable numbers empty the infinite set by ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. It is not empty as long as any natnumber is contained. Why should it be empty, if all endsegments contained natnumbers inherited from the first one?

Regards, WM

On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
> William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
l > > Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
> >
> > There is no reason why S should be have any connection to the T(k). Knowing that the T(k) share a common property does not mean that S must share this property.
> If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.

What happens to "they", the T(k), does not allow you to determine S. Knowing whether or not the T(k) have "lost all their numbers"
does not allow you to say anything about S.

--
William Hughes

On 12/18/2021 9:29 AM, WM wrote:
> William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
>> On Saturday, December 18, 2021 at 8:08:48 AM UTC-4, WM wrote:
>>> William schrieb am Samstag, 18. Dezember 2021 um 01:31:46 UTC+1:
>>>
>>>> Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.
>>> Why not?
>> Let the set which is produced by applying the operator to {E(1). E(2), E(3),...} be S.
>>
>> You are considering a process, call it T, with steps at all natural numbers .
>
> At all *definable* natural numbers. The endsegments are infinite, their intersection is infinite.

no. the intersection of all endsegments is empty.

>
>> Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))

what operator ?

>>
>> There is no reason why S should be have any connection to the T(k) Knowing that the T(k) share a common property does not mean that S must share this property.
>
> If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.

your sets are leaky. You use sets as variables, fail.

> S would be an infinite set.
>
>>> All definable steps can be looked at.
>> Looking at some of the T(k) (the T(k) for which k can be written down) tells you nothing about S.
>
> That's because undefinable numbers empty the infinite set by ∀k ∈ ℕ: E(k+1) = E(k) \ {k}.

Sets do not empty. that equation only states the relation of E(k) to E(k+1).
you are lazy at math.

>It is not empty as long as any natnumber is contained. Why should it be empty, if all endsegments contained natnumbers inherited from the first one?

you have a fundamental misunderstanding about sets.

>
> Regards, WM

On 12/18/2021 7:29 AM, WM wrote:
> William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
>> On Saturday, December 18, 2021 at 8:08:48 AM UTC-4, WM wrote:
>>> William schrieb am Samstag, 18. Dezember 2021 um 01:31:46 UTC+1:
>>>
>>>> Certainly. Determining the value of the operator applied to the set. cannot be done by looking at the states at finite steps.
>>> Why not?
>> Let the set which is produced by applying the operator to {E(1). E(2), E(3),...} be S.
>>
>> You are considering a process, call it T, with steps at all natural numbers .
>
> At all *definable* natural numbers. The endsegments are infinite, their intersection is infinite.
>
>> Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
>>
>> There is no reason why S should be have any connection to the T(k) Knowing that the T(k) share a common property does not mean that S must share this property.
>
> If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers. S would be an infinite set.

[...]

were infinite? They are infinite, and no end segments in infinity...

William schrieb am Samstag, 18. Dezember 2021 um 16:49:01 UTC+1:
> On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
> > William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
> l
> > > Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
> > >
> > > There is no reason why S should be have any connection to the T(k). Knowing that the T(k) share a common property does not mean that S must share this property.
> > If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.
> What happens to "they", the T(k), does not allow you to determine S.

I agree. But all endegments E(k) resulting from a step T(k) have a non-empty intersection.

> Knowing whether or not the T(k) have "lost all their numbers"
> does not allow you to say anything about S.

I talk about all endsegments which have not lost all their numbers but, on the contrary, contain infinitely many. These endsegments have an infinite intersection with all their predecessors. But since there is no last such endsegment, they all are predecessors.

Regards, WM

Chris M. Thomasson schrieb am Sonntag, 19. Dezember 2021 um 02:15:09 UTC+1:
> On 12/18/2021 7:29 AM, WM wrote:

> > If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers. S would be an infinite set.
> [...]
>
> were infinite? They are infinite,

Then they are not empty but contain natural numbers in common with all other endsegments. Their intersetion is not empty.

Regards, WM

WM wrote:

>> What happens to "they", the T(k), does not allow you to determine S.
>
> I agree. But all endegments E(k) resulting from a step T(k) have a
> non-empty intersection.

beautiful, fair enough. Now read this and tell me what you think.

When Mans Laws Fail Natural Law Prevails by Brother Alexis Bugnolo
https://www.bitchute.com/video/2paVx6QfXmZw/

On Sunday, December 19, 2021 at 5:51:32 AM UTC-4, WM wrote:
> William schrieb am Samstag, 18. Dezember 2021 um 16:49:01 UTC+1:
> > On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
> > > William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
> > l
> > > > Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
> > > >
> > > > There is no reason why S should be have any connection to the T(k). Knowing that the T(k) share a common property does not mean that S must share this property.
> > > If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.
> > What happens to "they", the T(k), does not allow you to determine S.
> I agree.
>But all endegments

Knowing what happens at finite steps does not allow you to say anything about S.

--
William Hughes

fredag 17 december 2021 kl. 23:10:59 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 13:27:08 UTC+1:
> > fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:
>
> >
> > But with that, congrats, you have L=N :)
> >
> > because n ∉ E(n+1) is true for ALL n
> What contain the non-empty endsegments?
> > >But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
> > Who says any endsegment is empty?
> What do non-empty endsegments contain when all n are gone?
>
> Regards, WM

what do you mean "when all n are gone"? endsegments are only defined for natural numbers so your question is non-sense

On 12/19/2021 1:52 AM, WM wrote:
> Chris M. Thomasson schrieb am Sonntag, 19. Dezember 2021 um 02:15:09 UTC+1:
>> On 12/18/2021 7:29 AM, WM wrote:
>
>>> If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers. S would be an infinite set.
>> [...]
>>
>> were infinite? They are infinite,
>
> Then they are not empty but contain natural numbers in common with all other endsegments. Their intersetion is not empty.

[0] 0
/ \
/ \
/ \
/ \
/ \
[1] 1 2
/ \ / \
/ \ / \
[2] 3 4 5 6
/ \ / \ / \ / \
[3] 7 8 9 10 11 12 13 14
.................................

It goes on forever. There is no end. There are an infinite number of
levels, however, no leafs in sight... ;^)

Want to get to the number 12, we go: (0, 2, 5, 12), or, starting at the
root, go (R, L, R) RLR... a finite path to 12. Want to get to a big
number, go all rights... (0, 2, 6, 14, ...). Notice a pattern? You can
never reach infinity, even though the tree is infinite in and of itself...

starting at root, 9 = LRL, and 10 = LRR

;^)

William schrieb am Sonntag, 19. Dezember 2021 um 16:48:45 UTC+1:
> On Sunday, December 19, 2021 at 5:51:32 AM UTC-4, WM wrote:
> > William schrieb am Samstag, 18. Dezember 2021 um 16:49:01 UTC+1:
> > > On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
> > > > William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
> > > l
> > > > > Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
> > > > >
> > > > > There is no reason why S should be have any connection to the T(k). Knowing that the T(k) share a common property does not mean that S must share this property.
> > > > If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.
> > > What happens to "they", the T(k), does not allow you to determine S.
> > I agree.
> >But all endegments
> Knowing what happens at finite steps does not allow you to say anything about S.

In mathematics that is the only way to say anything about S. Matheology is different. But even in matheology the following should be true:

Knowing that it is impossible to subtract individually definable natural numbers from ℕ such that ℕ gets exhausted but that it is possible to subtract collectively natural numbers such that ℕ gets exhausted, is enough to see the difference between definable and undefinable natnumbers.

Regards, WM

zelos...@gmail.com schrieb am Montag, 20. Dezember 2021 um 06:53:48 UTC+1:
> fredag 17 december 2021 kl. 23:10:59 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 13:27:08 UTC+1:
> > > fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:
> >
> > >
> > > But with that, congrats, you have L=N :)
> > >
> > > because n ∉ E(n+1) is true for ALL n
> > What contain the non-empty endsegments?
> > > >But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
> > > Who says any endsegment is empty?
> > What do non-empty endsegments contain when all n are gone?

> what do you mean "when all n are gone"?

Your proof is based on the idea that by
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
no natnumber is remaining in the intersection. That means all are deleted. But this applies as well to the endsegments.

Regards, WM

Chris M. Thomasson schrieb am Montag, 20. Dezember 2021 um 09:48:17 UTC+1:
> On 12/19/2021 1:52 AM, WM wrote:
>
> It goes on forever. There is no end.

There is nothing beyond omega. And all natnumbers before omega can be exhausted in bijections if set theory is correct.

Regards, WM

On 12/20/2021 1:17 PM, WM wrote:
> William schrieb am Sonntag, 19. Dezember 2021 um 16:48:45 UTC+1:
>> On Sunday, December 19, 2021 at 5:51:32 AM UTC-4, WM wrote:
>>> William schrieb am Samstag, 18. Dezember 2021 um 16:49:01 UTC+1:
>>>> On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
>>>>> William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1:
>>>> l
>>>>>> Let the state of the process at step k be T(k), where k is a natural number. T(k) is the set produced by applying the operator to the set (E(1),E(2), ...,E(k))
>>>>>>
>>>>>> There is no reason why S should be have any connection to the T(k). Knowing that the T(k) share a common property does not mean that S must share this property.
>>>>> If all numbers were definable and all endsegments were infinite, they would not have lost all their numbers.
>>>> What happens to "they", the T(k), does not allow you to determine S.
>>> I agree.
>>> But all endegments
>> Knowing what happens at finite steps does not allow you to say anything about S.
>
> In mathematics that is the only way to say anything about S. Matheology is different. But even in matheology the following should be true:
>
> Knowing that it is impossible to subtract individually definable natural numbers from ℕ such that ℕ gets exhausted but that it is possible to subtract collectively natural numbers such that ℕ gets exhausted, is enough to see the difference between definable and undefinable natnumbers.
>
> Regards, WM

you have confused yourself, really good this time.

WM submitted this idea :
> William schrieb am Sonntag, 19. Dezember 2021 um 16:48:45 UTC+1:
>> On Sunday, December 19, 2021 at 5:51:32 AM UTC-4, WM wrote:
>>> William schrieb am Samstag, 18. Dezember 2021 um 16:49:01 UTC+1:
>>>> On Saturday, December 18, 2021 at 11:29:54 AM UTC-4, WM wrote:
>>>>> William schrieb am Samstag, 18. Dezember 2021 um 15:05:18 UTC+1: l
>>>>>> Let the state of the process at step k be T(k), where k is a natural
>>>>>> number. T(k) is the set produced by applying the operator to the set
>>>>>> (E(1),E(2), ...,E(k))
>>>>>>
>>>>>> There is no reason why S should be have any connection to the T(k).
>>>>>> Knowing that the T(k) share a common property does not mean that S must
>>>>>> share this property.
>>>>> If all numbers were definable and all endsegments were infinite, they
>>>>> would not have lost all their numbers.
>>>> What happens to "they", the T(k), does not allow you to determine S.
>>> I agree.
>>> But all endegments
>> Knowing what happens at finite steps does not allow you to say anything
>> about S.
>
> In mathematics that is the only way to say anything about S.

Stop Lying.

On 12/20/2021 1:28 PM, WM wrote:
> Chris M. Thomasson schrieb am Montag, 20. Dezember 2021 um 09:48:17 UTC+1:
>> On 12/19/2021 1:52 AM, WM wrote:
>>
>> It goes on forever. There is no end.
>
> There is nothing beyond omega. And all natnumbers before omega can be exhausted in bijections if set theory is correct.
>
> Regards, WM

you make mistake;

"exhausted" is not a term used in Math.

On 12/20/2021 1:20 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 20. Dezember 2021 um 06:53:48 UTC+1:
>> fredag 17 december 2021 kl. 23:10:59 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 13:27:08 UTC+1:
>>>> fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:
>>>
>>>>
>>>> But with that, congrats, you have L=N :)
>>>>
>>>> because n ∉ E(n+1) is true for ALL n
>>> What contain the non-empty endsegments?
>>>>> But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
>>>> Who says any endsegment is empty?
>>> What do non-empty endsegments contain when all n are gone?
>
>> what do you mean "when all n are gone"?
>
> Your proof is based on the idea that by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}

that equation only shows the relation of one endsegment to the adjacent endsegment, and has nothing to do with the subject.

> no natnumber is remaining in the intersection.

why do you treat endsegments as variables ? They are fixed sets.

> That means all are deleted.

"remaining" and "deleted" are not terms used in Math.

> But this applies as well to the endsegments.

no.

>
> Regards, WM

On Monday, December 20, 2021 at 3:17:37 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 19. Dezember 2021 um 16:48:45 UTC+1:
But all endegments
> > Knowing what happens at finite steps does not allow you to say anything about S.
> In mathematics that is the only way to say anything about S.

Nonsense. The intersection operator, I, is well defined. R=I(Q) where Q is a set of sets.
R is defined to be the set of elements that is are every element of Q. Let S=I(E(1), E(2), E(3),...).
From the facts that every E(i) is a set of natural numbers and for every natural number, m, there is
an element of (E(1), E(2), E(3),...) that does note contain m we can conclude that S is empty.
(note there is no need to be able to write m down).

The properties of the T(k) =E(k), in particular their cardinalities, do not allow you to say anything about S.

>Matheology is different. But even in matheology the following should be true:
>
> Knowing that it is impossible to subtract individually definable natural numbers from ℕ such that ℕ gets exhausted
but that it is possible to subtract collectively natural numbers such that ℕ gets exhausted,

"subtract individually" means subtract a finite set, and "subtract collectively" means subtract an infinite set. Since every natural number (even numbers you cannot write down) is an element of a finite set and an infinite set, this triviality does allow you to split the natural numbers into two parts.

--
William Hughes