Le 21/02/2024 à 13:32, Richard Damon a écrit :
> On 2/21/24 3:42 AM, WM wrote:

>> If all are there, timeless and static, then one of them is the first.

> You don't understand the properties of UNBOUNDED sets.

Sets between 0 and 1 have bounds.

Regards, WM

Le 21/02/2024 à 13:32, Richard Damon a écrit :
> On 2/21/24 3:52 AM, WM wrote:

>> Then take the first one existing there.
>
> There isn't one, and you are just proving your ignornacd.

There is a first one in a static chain of points 1/n with gaps between
them. To deny this means falling victim to nonsense. Matheology.

Regards, WM

On 2/22/24 7:01 AM, WM wrote:
> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>> On 2/21/24 3:42 AM, WM wrote:
>
>>> If all are there, timeless and static, then one of them is the first.
>
>> You don't understand the properties of UNBOUNDED sets.
>
> Sets between 0 and 1 have bounds.
>
> Regards, WM
>
>
>

Only if "Between" is INCLUSIVE, as the bounds are 0 and 1.

If the set EXCLUDES one or both of the bounds, it doesn't have it in
anymore.

The fact that 0 is not > 0 means there is no lowest possible unit
fraction / rational number / real number > 0.

Do you really think that the set can both exclude the bounding value and
also contain it?

You are just admitting that your logic is broken.

On 2/22/24 7:04 AM, WM wrote:
> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>> On 2/21/24 3:52 AM, WM wrote:
>
>>> Then take the first one existing there.
>>
>> There isn't one, and you are just proving your ignornacd.
>
> There is a first one in a static chain of points 1/n with gaps between
> them. To deny this means falling victim to nonsense. Matheology.
>
> Regards, WM

Nope.

YOU have fallen victim to your lies and nonsense;

If the is, then NAME IT or explain how it can be. (Not that your system
says it must be, that just shows your system is broken)

Either your logic system doesn't actually HAVE the Natural Numbers in it
or you are lying.

After all if some 1/n was actually the smallest, then that says that n
must be the highest natural number, but the definition of the Natural
Numbers says that the include the successor to all Natural Numbers, and
every number has a successor, so n+1 must be a Natural Number.

By your "Static" rule, it can't be that it comes into being when we look
at it, as that isn't "Stati".

So, you are just proven to be stupid and a liar.

Le 21/02/2024 à 20:26, Jim Burns a écrit :

> No positive point is next to zero,

If all are there and timeless, then there is a first one. But it is more
obvious that the chain of unit fractiond must have a first one, whenever
there is a unit fraction at all.

>> Note that
>> if reciprocals are existing on the real axis and
>> if all points are timeless,
>> then there is a point next to zero.
>
> Elaborate.

Nothing to elaborate.
>
> Do you reject
> all skipping.functions being discontinuous.somewhere?

Functions measuring elements are not discontinuous for more than 1 when
the elements have point between them.
>
> Do you reject
> only both or neither ⅟m ⅟(4⋅m) being
> final.ordinal.reciprocals findable by
> geometric procedure?

Dark numbers cannot be found.

Regards, WM

Le 22/02/2024 à 13:17, Richard Damon a écrit :
> On 2/22/24 7:01 AM, WM wrote:
>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>> On 2/21/24 3:42 AM, WM wrote:
>>
>>>> If all are there, timeless and static, then one of them is the first.
>>
>>> You don't understand the properties of UNBOUNDED sets.
>>
>> Sets between 0 and 1 have bounds.
>
> Only if "Between" is INCLUSIVE, as the bounds are 0 and 1.
>
> If the set EXCLUDES one or both of the bounds, it doesn't have it in
> anymore.

The bounds are there, if not at 0 then before. Note: linearity. No magical
leaps.

Regards, WM

Le 22/02/2024 à 13:17, Richard Damon a écrit :
> On 2/22/24 7:04 AM, WM wrote:
>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>> On 2/21/24 3:52 AM, WM wrote:
>>
>>>> Then take the first one existing there.
>>>
>>> There isn't one, and you are just proving your ignornacd.
>>
>> There is a first one in a static chain of points 1/n with gaps between
>> them. To deny this means falling victim to nonsense. Matheology.
>>
> If the is, then NAME IT or explain how it can be.

I did.

> (Not that your system
> says it must be, that just shows your system is broken)

It is simply mathematics and logic: By logic there must be a start of
NUF(x), by mathematics the start can only be 1. ∀n ∈ ℕ: 1/n -
1/(n+1) = d_n > 0.
>
> After all if some 1/n was actually the smallest, then that says that n
> must be the highest natural number, but the definition of the Natural
> Numbers says that the include the successor to all Natural Numbers, and
> every number has a successor, so n+1 must be a Natural Number.

The definition of unit fractions says that all have gaps and there is no
point where more than 1 sit.
>
> By your "Static" rule, it can't be that it comes into being when we look
> at it, as that isn't "Stati".
>
We cannot look at dark numbers.

Regards, WM

on 2/22/2024, WM supposed :
> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>> On 2/21/24 3:42 AM, WM wrote:
>
>>> If all are there, timeless and static, then one of them is the first.
>
>> You don't understand the properties of UNBOUNDED sets.
>
> Sets between 0 and 1 have bounds.

Which sets?

On 2/22/2024 7:01 AM, WM wrote:
> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>> On 2/21/24 3:42 AM, WM wrote:

>>> If all are there, timeless and static,
>>> then one of them is the first.
>
>> You don't understand
>> the properties of UNBOUNDED sets.
>
> Sets between 0 and 1 have bounds.

Consider a finite set B

In each
transitive.trichotomous.order on B
each non.empty subset S of B holds
two extrema (minimum and maximum) of S

Each
transitive.trichotomous.order on B
has
the each.subset.two.extrema property.

More at
https://en.wikipedia.org/wiki/Finite_set

----
Consider ⟨1,…,n⟩

∀S ᙾ⁰⊆ ⟨1,…,n⟩: S holds two extrema of S

The standard
transitive.trichotomous.order on ⟨1,…,n⟩
has
the each.subset.two.extrema property.

If any
transitive.trichotomous.order on ⟨1,…,n⟩
has
the each.subset.two.extrema property,
then each
transitive.trichotomous.order on ⟨1,…,n⟩
has
the each.subset.two.extrema property,
provably.

Each
transitive.trichotomous.order on ⟨1,…,n⟩
has
the each.subset.two.extrema property.

⟨1,…,n⟩ is finite

----
Consider the union ⋃ₙ⟨⟨1,…,n⟩⟩
Standardly ℕ = ⋃ₙ⟨⟨1,…,n⟩⟩

In the standard
transitive.trichotomous.order on ℕ
ℕ holds one extremum

The standard
transitive.trichotomous.order on ℕ
doesn't have
the each.subset.two.extrema property.

Each
transitive.trichotomous.order on ℕ
doesn't have
the each.subset.two.extrema property,
provably.

ℕ is not finite.

----
Consider ℕ∪{ω}
Standardly ℕ ᣔ< ω

In the standard
transitive.trichotomous.order on ℕ∪{ω}
ℕ∪{ω} holds two extrema, however,
ℕ ⊆ ℕ∪{ω} holds one extremum

The standard
transitive.trichotomous.order on ℕ∪{ω}
doesn't have
the each.subset.two.extrema property.

Each
transitive.trichotomous.order on ℕ∪{ω}
doesn't have
the each.subset.two.extrema property,
provably.

ℕ∪{ω} is not finite.

----
Consider [0,1] ⊆ ℝ

In the standard
transitive.trichotomous.order on [0,1]
[0,1] holds two extrema, however,
[0,1) ⊆ [0,1] holds one extremum

The standard
transitive.trichotomous.order on [0,1]
doesn't have
the each.subset.two.extrema property.

Each
transitive.trichotomous.order on [0,1]
doesn't have
the each.subset.two.extrema property,
provably.

[0,1] is not finite.

On 2/22/2024 8:06 AM, WM wrote:
> Le 21/02/2024 à 20:26, Jim Burns a écrit :

>> No positive point is next to zero,
>
> If all are there and timeless,
> then there is a first one.

All points are timelessly described as
final.ordinals ℕ and
differences.of.ratios ℚ of final.ordinals and
least.upper.bounds ℝ of bounded.non.empty.sets of
differences.of.ratios of final.ordinals

Each positive point is preceded by
a positive point,
and no positive precedes each positive.

There isn't a first one.

> But it is more obvious that
> the chain of unit fractiond must have a first one,
> whenever there is a unit fraction at all.

Each is preceded by some.
None precedes all.

>>> Note that
>>> if reciprocals are existing on the real axis and
>>> if all points are timeless,
>>> then there is a point next to zero.
>>
>> Elaborate.
>
> Nothing to elaborate.

Doesn't that worry you?

If you opened the hood of your auto,
and no engine was in there,
wouldn't you get a glimmer of a sense that,
perhaps, things were not as they should be?

>> Do you reject
>> all skipping.functions being discontinuous.somewhere?
>
> Functions measuring elements are not
> discontinuous for more than 1
> when the elements have point between them.

A point β exists between
lower.bounds of final.ordinal.reciprocals and
not.lower.bounds of final.ordinal.reciprocals.

>> Do you reject
>> only both or neither ⅟m ⅟(4⋅m) being
>> final.ordinal.reciprocals findable by
>> geometric procedure?
>
> Dark numbers cannot be found.

Positive β requires
final.ordinal.reciprocal ⅟(4⋅m) < β/2 with
β/2 a lower bound of final.ordinal.reciprocals
Note the contradiction.

β = 0 requires that darkᵂᴹ numbers not.exist.

On 2/22/2024 5:06 AM, WM wrote:
> Le 21/02/2024 à 20:26, Jim Burns a écrit :
>
>> No positive point is next to zero,
>
> If all are there and timeless, then there is a first one. But it is more
> obvious that the chain of unit fractiond must have a first one, whenever
> there is a unit fraction at all.
>
>>> Note that
>>> if reciprocals are existing on the real axis and
>>> if all points are timeless,
>>> then there is a point next to zero.
>>
>> Elaborate.
>
> Nothing to elaborate.
>>
>> Do you reject
>> all skipping.functions being discontinuous.somewhere?
>
> Functions measuring elements are not discontinuous for more than 1 when
> the elements have point between them.
>>
>> Do you reject
>> only both or neither ⅟m ⅟(4⋅m) being
>> final.ordinal.reciprocals findable by
>> geometric procedure?
>
> Dark numbers cannot be found.

Ummm. Well, WM just might be able to find a number covered in shit right
up is own hind end?

On 2/21/2024 12:52 AM, WM wrote:
> Le 21/02/2024 à 03:17, Richard Damon a écrit :
>> On 2/20/24 4:59 PM, WM wrote:
>
>>> We cannot use everything that exists on the real line, because among
>>> them there is the smallest unit fraction, at least the smallest unit
>>> fraction that exists on the real line. Where else should it be? This
>>> existence is static. You seem to deny it. If we could point to it, we
>>> caught the smallest unit fraction. But we cannot point to it although
>>> it must be there. That proves: It is dark.
>>
>> No, there ISN'T a "Smallest Unit Fraction" as has been shown.
>>
>> Since ALL Unit Fractions "exist" in the mathematical sense,.
>
> Then take the first one existing there.
[...]

Are you a dingbat, or something else, even worse?

On 2/22/2024 6:11 AM, WM wrote:
> Le 22/02/2024 à 13:17, Richard Damon a écrit :
>> On 2/22/24 7:04 AM, WM wrote:
>>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>>> On 2/21/24 3:52 AM, WM wrote:
>>>
>>>>> Then take the first one existing there.
>>>>
>>>> There isn't one, and you are just proving your ignornacd.
>>>
>>> There is a first one in a static chain of points 1/n with gaps
>>> between them. To deny this means falling victim to nonsense. Matheology.
>>>
>> If the is, then NAME IT or explain how it can be.
>
> I did.
>
>> (Not that your system says it must be, that just shows your system is
>> broken)
>
> It is simply mathematics and logic: By logic there must be a start of
> NUF(x), by mathematics the start can only be 1. ∀n ∈ ℕ: 1/n - 1/(n+1) =
> d_n > 0.
>>
>> After all if some 1/n was actually the smallest, then that says that n
>> must be the highest natural number, but the definition of the Natural
>> Numbers says that the include the successor to all Natural Numbers,
>> and every number has a successor, so n+1 must be a Natural Number.
>
> The definition of unit fractions says that all have gaps and there is no
> point where more than 1 sit.
>>
>> By your "Static" rule, it can't be that it comes into being when we
>> look at it, as that isn't "Stati".
>>
> We cannot look at dark numbers.

I think you plug up toilets with your dark numbers everyday. Ask your
plumbers about it...

On 2/22/2024 4:25 PM, Chris M. Thomasson wrote:
> On 2/22/2024 5:06 AM, WM wrote:
>> Le 21/02/2024 à 20:26, Jim Burns a écrit :
>>
>>> No positive point is next to zero,
>>
>> If all are there and timeless, then there is a first one. But it is
>> more obvious that the chain of unit fractiond must have a first one,
>> whenever there is a unit fraction at all.
>>
>>>> Note that
>>>> if reciprocals are existing on the real axis and
>>>> if all points are timeless,
>>>> then there is a point next to zero.
>>>
>>> Elaborate.
>>
>> Nothing to elaborate.
>>>
>>> Do you reject
>>> all skipping.functions being discontinuous.somewhere?
>>
>> Functions measuring elements are not discontinuous for more than 1
>> when the elements have point between them.
>>>
>>> Do you reject
>>> only both or neither ⅟m ⅟(4⋅m) being
>>> final.ordinal.reciprocals findable by
>>> geometric procedure?
>>
>> Dark numbers cannot be found.
>
> Ummm. Well, WM just might be able to find a number covered in shit right
> up is own hind end?
>

Then we can all say, okay, that plus one? ;^)

Le 22/02/2024 à 15:39, FromTheRafters a écrit :
> on 2/22/2024, WM supposed :

>> Sets between 0 and 1 have bounds.
>
> Which sets?

All sets with elements x which obey 0 < x < 1.

Regards, WM

Le 22/02/2024 à 15:54, Jim Burns a écrit :

> ℕ is not finite.
>
Ye4s, but what does this mean? The visible part has no final element n,
because n+1 and 2n and n^n^n can be found for every n. The complete set
however is complete such that no element can be added. Based on Cantor's
Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
Unendlichseiendem = kategorematice infinitum there is a fixed number of
numbers, although we cannot count or determine it better than by |ℕ|.

> [0,1] is not finite.

It has bounds. The numer of points is fixed.

Regards, WM

WM was thinking very hard :
> Le 22/02/2024 à 15:39, FromTheRafters a écrit :
>> on 2/22/2024, WM supposed :
>
>>> Sets between 0 and 1 have bounds.
>>
>> Which sets?
>
> All sets with elements x which obey 0 < x < 1.

Then your interval was unbounded (0,1) and your 'x' is strictly between
zero and one.

Le 22/02/2024 à 19:12, Jim Burns a écrit :
> On 2/22/2024 8:06 AM, WM wrote:
>> Le 21/02/2024 à 20:26, Jim Burns a écrit :
>
>>> No positive point is next to zero,
>>
>> If all are there and timeless,
>> then there is a first one.

> Each positive point is preceded by
> a positive point,
> and no positive precedes each positive.

Your claim requires to consider the predecessors. "Timeless" means that
every point exists independent of a predecessor.
> There isn't a first one.

NUF(x) together with 1/n =/= 1/(n+1) proves the first one.
>
>> But it is more obvious that
>> the chain of unit fractiond must have a first one,
>> whenever there is a unit fraction at all.
>
> Each is preceded by some.
> None precedes all.

You violate 1/n =/= 1/(n+1) and timeless existence.
>
>>>> Note that
>>>> if reciprocals are existing on the real axis and
>>>> if all points are timeless,
>>>> then there is a point next to zero.
>>>
>>> Elaborate.
>>
>> Nothing to elaborate.
>
> Doesn't that worry you?

Why. It is obvious. Timeless existence in linear order proves a first one.
>
> If you opened the hood of your auto,
> and no engine was in there,
> wouldn't you get a glimmer of a sense that,
> perhaps, things were not as they should be?

Failed analogy.
>
>>> Do you reject
>>> all skipping.functions being discontinuous.somewhere?
>>
>> Functions measuring elements are not
>> discontinuous for more than 1
>> when the elements have point between them.
>
> A point β exists between
> lower.bounds of final.ordinal.reciprocals and
> not.lower.bounds of final.ordinal.reciprocals.

You violate 1/n =/= 1/(n+1) and timeless existence.

Regards, WM

Le 23/02/2024 à 09:54, FromTheRafters a écrit :
> WM was thinking very hard :
>> Le 22/02/2024 à 15:39, FromTheRafters a écrit :
>>> on 2/22/2024, WM supposed :
>>
>>>> Sets between 0 and 1 have bounds.
>>>
>>> Which sets?
>>
>> All sets with elements x which obey 0 < x < 1.
>
> Then your interval was unbounded (0,1) and your 'x' is strictly between
> zero and one.

The border is either 0 an 1 or between 0 and 1. Real timeless complete
existence requires fixed smallest and largest points.
Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
Unendlichseiendem = kategorematice infinitum.

Regards, WM

On 2/22/24 9:05 AM, WM wrote:
> Le 22/02/2024 à 13:17, Richard Damon a écrit :
>> On 2/22/24 7:01 AM, WM wrote:
>>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>>> On 2/21/24 3:42 AM, WM wrote:
>>>
>>>>> If all are there, timeless and static, then one of them is the first.
>>>
>>>> You don't understand the properties of UNBOUNDED sets.
>>>
>>> Sets between 0 and 1 have bounds.
>>
>> Only if "Between" is INCLUSIVE, as the bounds are 0 and 1.
>>
>> If the set EXCLUDES one or both of the bounds, it doesn't have it in
>> anymore.
>
> The bounds are there, if not at 0 then before. Note: linearity. No
> magical leaps.
>
> Regards, WM

Right, so there isn't a number just "next to" 0, as you had to "leap"
over all the infinite numbers between 0 and that number.

The Lower Bound for (0, 1] is 0, which is out of the set, so there is no
lower bound IN the set to use, so there is not "lowest" value in the set.

You just don't understand how unbounded sets work.

Your math is just exploded in contradictions because it can't hold that
much safely.

On 2/22/24 9:11 AM, WM wrote:
> Le 22/02/2024 à 13:17, Richard Damon a écrit :
>> On 2/22/24 7:04 AM, WM wrote:
>>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>>> On 2/21/24 3:52 AM, WM wrote:
>>>
>>>>> Then take the first one existing there.
>>>>
>>>> There isn't one, and you are just proving your ignornacd.
>>>
>>> There is a first one in a static chain of points 1/n with gaps
>>> between them. To deny this means falling victim to nonsense. Matheology.
>>>
>> If the is, then NAME IT or explain how it can be.
>
> I did.

Nope, you just say you assume that there must be a smallest.

But if there is, why isn't the one smaller to that in it to, making it
not the smallest?

>
>> (Not that your system says it must be, that just shows your system is
>> broken)
>
> It is simply mathematics and logic: By logic there must be a start of
> NUF(x), by mathematics the start can only be 1. ∀n ∈ ℕ: 1/n - 1/(n+1) =
> d_n > 0.

Nope

By your logic, I have shown that the aquare root of two is Rational.

You err in assuming somethng that doesn't exist.

Your "Matheolgy" is based on Magic Faeries that can make the impossible
happen.

You claim to not want to use "Matheologies", but then you do, but you
use ones not up to the task you ask of them.

You ignore the natural and obvious facts of the numbers because you
can't face the limitations of your logic.

>>
>> After all if some 1/n was actually the smallest, then that says that n
>> must be the highest natural number, but the definition of the Natural
>> Numbers says that the include the successor to all Natural Numbers,
>> and every number has a successor, so n+1 must be a Natural Number.
>
> The definition of unit fractions says that all have gaps and there is no
> point where more than 1 sit.

Right, but the gap is smaller than the number itself, so there is alway
room for more between any of them and 0, so there is not first.

You just are to dim to see that the gaps shrink faster than the numbers
themselves so you get more and more "room" to put the smaller and
smaller values in.

>>
>> By your "Static" rule, it can't be that it comes into being when we
>> look at it, as that isn't "Stati".
>>
> We cannot look at dark numbers.

Because they don't exist.

>
> Regards, WM
>
>

On 2/23/24 4:04 AM, WM wrote:
> Le 23/02/2024 à 09:54, FromTheRafters a écrit :
>> WM was thinking very hard :
>>> Le 22/02/2024 à 15:39, FromTheRafters a écrit :
>>>> on 2/22/2024, WM supposed :
>>>
>>>>> Sets between 0 and 1 have bounds.
>>>>
>>>> Which sets?
>>>
>>> All sets with elements x which obey 0 < x < 1.
>>
>> Then your interval was unbounded (0,1) and your 'x' is strictly
>> between zero and one.
>
> The border is either 0 an 1 or between 0 and 1. Real timeless complete
> existence requires fixed smallest and largest points.
> Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
> Unendlichseiendem = kategorematice infinitum.
>
>
> Regards, WM

No, it doesn't

Remember, you have an INFINITE sized set (even if all the values are
finite) so your rule doesn't apply.

Don't understand you german, but you must be wrong.

It seem your problem is you just don't understand how infinity works.

And yes, if be "Real" you mean existing in the real physical world, it
can't actually have even Natural Numbers, since the physical world is
finite, and thus can't hold the infinite, so appeals to it just show the
limitation of trying to argue about Mathematics with a tool belt of
inadequacy tools.

On 2/23/2024 3:42 AM, WM wrote:
> Le 22/02/2024 à 15:54, Jim Burns a écrit :

>> Consider a finite set B
>>
>> In each
>> transitive.trichotomous.order on B
>> each non.empty subset S of B holds
>> two extrema (minimum and maximum) of S
>>
>> Each
>> transitive.trichotomous.order on B
>> has
>> the each.subset.two.extrema property.
>>
>> More at
>> https://en.wikipedia.org/wiki/Finite_set

>> ℕ is not finite.
>
> Ye4s, but what does this mean?

It means that
there is
transitive.trichotomous.order on ℕ
according to which there is
a non.empty subset of ℕ
which
does not hold two extrema.

> The visible part has no final element n,

Right.
Thus, ℕ is not finite.

Even in ℕᵂᴹ with your darkᵂᴹ numbers
there is the visibleᵂᴹ standard order
according to which there is
there is the visibleᵂᴹ number subset
which,
as you (WM) have noted,
does not hold two extrema.

The insertion of darkᵂᴹ numbers
cannot turn an infinite set finite,
_according to what "finite" means_

> because n+1 and 2n and n^n^n can be found
> for every n.
> The complete set however
> is complete such that
> no element can be added.
> Based on Cantor's
> Eigentlichunendlichem =
> Transfinitum =
> Vollendetunendlichem =
> Unendlichseiendem =
> kategorematice infinitum
> there is a fixed number of numbers,
> although we cannot count or determine it
> better than by |ℕ|.

When we "count" the elements of S
we determine the last final.ordinal[1]
which fits[2] in S, or, equivalently,
the predecessor of
the first final ordinal which doesn't fit in S

Flocks of sheep and pockets of pebbles have
their first.not.fitting.final.ordinal.

The set N of final.ordinals does not have
its first.not.fitting.final.ordinal.

It not.exists.
It doesn't exist.in.the.darkᵂᴹ

[2]
fits
1.to.1 map exists

[1]
final
another one doesn't fit

>> [0,1] is not finite.
>
> It has bounds.
> The numer of points is fixed.

[0,1] ⊇ [0,1) which has one extremum.

On 02/23/2024 05:11 AM, Jim Burns wrote:
> On 2/23/2024 3:42 AM, WM wrote:
>> Le 22/02/2024 à 15:54, Jim Burns a écrit :
>
>>> Consider a finite set B
>>>
>>> In each
>>> transitive.trichotomous.order on B
>>> each non.empty subset S of B holds
>>> two extrema (minimum and maximum) of S
>>>
>>> Each
>>> transitive.trichotomous.order on B
>>> has
>>> the each.subset.two.extrema property.
>>>
>>> More at
>>> https://en.wikipedia.org/wiki/Finite_set
>
>>> ℕ is not finite.
>>
>> Ye4s, but what does this mean?
>
> It means that
> there is
> transitive.trichotomous.order on ℕ
> according to which there is
> a non.empty subset of ℕ
> which
> does not hold two extrema.
>
>> The visible part has no final element n,
>
> Right.
> Thus, ℕ is not finite.
>
> Even in ℕᵂᴹ with your darkᵂᴹ numbers
> there is the visibleᵂᴹ standard order
> according to which there is
> there is the visibleᵂᴹ number subset
> which,
> as you (WM) have noted,
> does not hold two extrema.
>
> The insertion of darkᵂᴹ numbers
> cannot turn an infinite set finite,
> _according to what "finite" means_
>
>> because n+1 and 2n and n^n^n can be found
>> for every n.
>> The complete set however
>> is complete such that
>> no element can be added.
>> Based on Cantor's Eigentlichunendlichem =
>> Transfinitum =
>> Vollendetunendlichem = Unendlichseiendem =
>> kategorematice infinitum
>> there is a fixed number of numbers,
>> although we cannot count or determine it
>> better than by |ℕ|.
>
> When we "count" the elements of S
> we determine the last final.ordinal[1]
> which fits[2] in S, or, equivalently,
> the predecessor of
> the first final ordinal which doesn't fit in S
>
> Flocks of sheep and pockets of pebbles have
> their first.not.fitting.final.ordinal.
>
> The set N of final.ordinals does not have
> its first.not.fitting.final.ordinal.
>
> It not.exists.
> It doesn't exist.in.the.darkᵂᴹ
>
> [2]
> fits
> 1.to.1 map exists
>
> [1]
> final
> another one doesn't fit
>
>>> [0,1] is not finite.
>>
>> It has bounds.
>> The numer of points is fixed.
>
> [0,1] ⊇ [0,1) which has one extremum.
>
>

It's pretty simply,

n/d: the function the continuum limit as d -> oo

and

n/d: the equivalence classes of ratios, d =/= 0

that

one's a function the range a continuous domain [0,1]
the other a model of the non-negative rationals.

If you might condense _all_ of WM/MW's "axioms"
since he ever wrote to sci.math or sci.logic, and
catalog their reasonings and lack thereof, and
then relate it to ratios of integers either apiece
or in the continuum limit, then it seems
the entire bit could be much reduced,
instead of perpetuating the freak-out.

So, for these mutual notions

n/d: the continuum limit of the function as d -> oo

and

n/d: the equivalence classes of ratios, with d =/= 0

then these are very well-understood numbers
and their forms have very familiar properties
and they stop disagreeing with each other
and stop disagreeing with them.

On 2/23/2024 4:23 AM, Richard Damon wrote:
> On 2/22/24 9:05 AM, WM wrote:
>> Le 22/02/2024 à 13:17, Richard Damon a écrit :
>>> On 2/22/24 7:01 AM, WM wrote:
>>>> Le 21/02/2024 à 13:32, Richard Damon a écrit :
>>>>> On 2/21/24 3:42 AM, WM wrote:
>>>>
>>>>>> If all are there, timeless and static, then one of them is the first.
>>>>
>>>>> You don't understand the properties of UNBOUNDED sets.
>>>>
>>>> Sets between 0 and 1 have bounds.
>>>
>>> Only if "Between" is INCLUSIVE, as the bounds are 0 and 1.
>>>
>>> If the set EXCLUDES one or both of the bounds, it doesn't have it in
>>> anymore.
>>
>> The bounds are there, if not at 0 then before. Note: linearity. No
>> magical leaps.
>>
>> Regards, WM
>
> Right, so there isn't a number just "next to" 0, as you had to "leap"
> over all the infinite numbers between 0 and that number.
>
> The Lower Bound for (0, 1] is 0, which is out of the set, so there is no
> lower bound IN the set to use, so there is not "lowest" value in the set.
>
> You just don't understand how unbounded sets work.

We have tried to explain this to him... His mind seems to be locked in.

>
> Your math is just exploded in contradictions because it can't hold that
> much safely.