Le 23/12/2023 à 14:37, Richard Damon a écrit :
> On 12/23/23 7:40 AM, WM wrote:

>> Of course. And since the leap down from omega never hits a dark number,
>> their existence is proven by the fact that after leaping down you can
>> run upwards through all dark numbers collectively by "and so on" or by
>> "...".
>
> But there is no "leap down". What is the leap down from 0 (in the
> Naturals).

There is a leap from omega to 17.

> You never reach "up" from the Ordinals to the Transfinite,

The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...

>>> If 0 isn't a unit fraction, it can't be the lower bound of unit
>>> fractions.
>>
>> Then find another one.
>
> It doesn't exist.

You are so wrong that I presume you never have studied mathematics.
>
> There is no requirement that a bound exist, read your article on infimum:
>
> In mathematics, the infimum (abbreviated inf; plural infima) of a subset
> S of a partially ordered set P is the greatest element in P that is less
> than or equal to each element of S, *if such an element exists.*
>
> So, there need not be a bound, in fact, that is why they are called
> "unbounded" sets.

No, there is no bound that is an element of S. The unit fractions
have a supremum that is a unit fraction, namely 1 = 1/1, and an infimum
that is not a unit fraction, namely 0.

Regards, WM

On 12/24/23 3:37 AM, WM wrote:
> Le 23/12/2023 à 14:37, Richard Damon a écrit :
>> On 12/23/23 7:40 AM, WM wrote:
>
>>> Of course. And since the leap down from omega never hits a dark
>>> number, their existence is proven by the fact that after leaping down
>>> you can run upwards through all dark numbers collectively by "and so
>>> on" or by "...".
>>
>> But there is no "leap down". What is the leap down from 0 (in the
>> Naturals).
>
> There is a leap from omega to 17.

Really, a ONE STEP leap?

>
>> You never reach "up" from the Ordinals to the Transfinite,
>
> The sequence has been given by Cantor:
> 1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...

Please show a reference where you can actually "Step up" from one of the
.... to the following multiple of omega (or down).

>
>>>> If 0 isn't a unit fraction, it can't be the lower bound of unit
>>>> fractions.
>>>
>>> Then find another one.
>>
>> It doesn't exist.
>
> You are so wrong that I presume you never have studied mathematics.

No, YOU are wrong, as the set doesn't have a lower bound (in the set).
To claim one, you need to show how it exists.

BOUNDS, as I am using them are the maximum or minimum if the set, which
must be a member of the set.

The fact that the infimum is NOT a member of the set says it is not a
minimum of the set, so there is no "smallest unit faction", so no finite
point where NUF(x) == 1.

>>
>> There is no requirement that a bound exist, read your article on infimum:
>>
>> In mathematics, the infimum (abbreviated inf; plural infima) of a subset
>> S of a partially ordered set P is the greatest element in P that is
>> less than or equal to each element of S, *if such an element exists.*
>>
>> So, there need not be a bound, in fact, that is why they are called
>> "unbounded" sets.
>
> No, there is no bound that is an element of S. The unit fractions have a
> supremum that is a unit fraction, namely 1 = 1/1, and an infimum that is
> not a unit fraction, namely 0.

And since the infimum isn't is the Unit fractions, there is no smallest
unit fraction, and thus no finite point where NUF(x) == 1.

>
> Regards, WM
>
>

Le 24/12/2023 à 18:04, Richard Damon a écrit :
> On 12/24/23 3:37 AM, WM wrote:
>> Le 23/12/2023 à 14:37, Richard Damon a écrit :
>>> On 12/23/23 7:40 AM, WM wrote:
>>
>>>> Of course. And since the leap down from omega never hits a dark
>>>> number, their existence is proven by the fact that after leaping down
>>>> you can run upwards through all dark numbers collectively by "and so
>>>> on" or by "...".
>>>
>>> But there is no "leap down". What is the leap down from 0 (in the
>>> Naturals).
>>
>> There is a leap from omega to 17.
>
> Really, a ONE STEP leap?

Really. But we can go every number of finite steps, for instance
ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
At every term we can reverse the direction and can go upwards ℵo steps.
>
>>> You never reach "up" from the Ordinals to the Transfinite,
>>
>> The sequence has been given by Cantor:
>> 1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
>
> Please show a reference where you can actually "Step up" from one of the
> ... to the following multiple of omega (or down).

If not, then not all natural numbers can be applied in bijections.
>
> BOUNDS, as I am using them are the maximum or minimum

That is not correct.

>> No, there is no bound that is an element of S. The unit fractions have a
>> supremum that is a unit fraction, namely 1 = 1/1, and an infimum that is
>> not a unit fraction, namely 0.
>
> And since the infimum isn't is the Unit fractions, there is no smallest
> unit fraction, and thus no finite point where NUF(x) == 1.

But there is a real number, namely 0, that is not reached. Hence when
going upwards from 0, the NUF(x) increases. But it cannot increase by more
than 1 before it is constat over the gap following every unit fraction
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

Do you disagree with this mathematical formula?

Regards, WM

On 12/25/23 2:22 PM, WM wrote:
> Le 24/12/2023 à 18:04, Richard Damon a écrit :
>> On 12/24/23 3:37 AM, WM wrote:
>>> Le 23/12/2023 à 14:37, Richard Damon a écrit :
>>>> On 12/23/23 7:40 AM, WM wrote:
>>>
>>>>> Of course. And since the leap down from omega never hits a dark
>>>>> number, their existence is proven by the fact that after leaping
>>>>> down you can run upwards through all dark numbers collectively by
>>>>> "and so on" or by "...".
>>>>
>>>> But there is no "leap down". What is the leap down from 0 (in the
>>>> Naturals).
>>>
>>> There is a leap from omega to 17.
>>
>> Really, a ONE STEP leap?
>
> Really. But we can go every number of finite steps, for instance
> ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
> At every term we can reverse the direction and can go upwards ℵo steps.

Why is the step down from ω 2023? Why not 100000, or 1?

There is no unique value that works, so the operation isn't actually
defined.

>>
>>>> You never reach "up" from the Ordinals to the Transfinite,
>>>
>>> The sequence has been given by Cantor:
>>> 1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
>>
>> Please show a reference where you can actually "Step up" from one of
>> the ... to the following multiple of omega (or down).
>
> If not, then not all natural numbers can be applied in bijections.

Why not?

The bijection rules are based only on the set 1, 2, 3, ... and don't
need the omega stuff.

>>
>> BOUNDS, as I am using them are the maximum or minimum
>
> That is not correct.

Why not?

That is the key diffence between the minimum and the infimum, and
unbounded sets will be missing either minimum or maximum, even if they
have a infimum.

Since there is no minimum unit fraction, there is no lower unit
fraction, so you can't start counting at that which doesn't exist.

>
>>> No, there is no bound that is an element of S. The unit fractions
>>> have a supremum that is a unit fraction, namely 1 = 1/1, and an
>>> infimum that is not a unit fraction, namely 0.
>>
>> And since the infimum isn't is the Unit fractions, there is no
>> smallest unit fraction, and thus no finite point where NUF(x) == 1.
>
> But there is a real number, namely 0, that is not reached. Hence when
> going upwards from 0, the NUF(x) increases. But it cannot increase by
> more than 1 before it is constat over the gap following every unit
> fraction ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.

But it CAN'T jump from 0 to 1 at a unit fraction, as there will always
be a unit fraction smaller than were you want it to go to 1. Thus your
definition is broken.

So, you theory seems to be based on the existance of Magical Unicorns,
which just don't exist.

Maybe your NUF is "dark".

Note, to actually try to define it for finite resultant values, you need
to allow the domain to include the right sort of infinitesimals, where
perhaps NUF(x) can be 1 for x = delta, the smallest infinitesimal. then
to 2 at 2*delta, the next infinitesimal.

Of course, that breaks your claims, as the infinitesimals are usable
individually, so not "dark"

>
> Do you disagree with this mathematical formula?

No, only your conclusion off of it.

For NUF to increase from 0 to 1 at a finite value, that value must be
the minimum unit fraction, which doesn't exist.

Therefore your definition of NUF is defective.

>
> Regards, WM
>
>

Le 25/12/2023 à 20:41, Richard Damon a écrit :
> On 12/25/23 2:22 PM, WM wrote:
>> Le 24/12/2023 à 18:04, Richard Damon a écrit :
>>> On 12/24/23 3:37 AM, WM wrote:
>>>> Le 23/12/2023 à 14:37, Richard Damon a écrit :
>>>>> On 12/23/23 7:40 AM, WM wrote:
>>>>
>>>>>> Of course. And since the leap down from omega never hits a dark
>>>>>> number, their existence is proven by the fact that after leaping
>>>>>> down you can run upwards through all dark numbers collectively by
>>>>>> "and so on" or by "...".
>>>>>
>>>>> But there is no "leap down". What is the leap down from 0 (in the
>>>>> Naturals).
>>>>
>>>> There is a leap from omega to 17.
>>>
>>> Really, a ONE STEP leap?
>>
>> Really. But we can go every number of finite steps, for instance
>> ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
>> At every term we can reverse the direction and can go upwards ℵo steps.
>
> Why is the step down from ω 2023? Why not 100000, or 1?

All that is possible. All these sequences end after a finite number of
steps. But in reverse diraction there are infinitely many step possible.
>
> There is no unique value that works, so the operation isn't actually
> defined.

You can define whateverr you like. The sequence will be finite.
>
>>>
>>>>> You never reach "up" from the Ordinals to the Transfinite,
>>>>
>>>> The sequence has been given by Cantor:
>>>> 1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
>>>
>>> Please show a reference where you can actually "Step up" from one of
>>> the ... to the following multiple of omega (or down).
>>
>> If not, then not all natural numbers can be applied in bijections.
>
> Why not?

Because all implies that none is missing.

> The bijection rules are based only on the set 1, 2, 3, ... and don't
> need the omega stuff.
>
>>>
>>> BOUNDS, as I am using them are the maximum or minimum
>>
>> That is not correct.
>
> Why not?

If you understand German, you may consult W. Mückenheim: "Mathematik für
die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
https://www.degruyter.com/document/doi/10.1515/9783110377347/html
Ask Wiki or a math teacher if you don't believe me.

Regards, WM

On 12/25/23 5:40 PM, WM wrote:
> Le 25/12/2023 à 20:41, Richard Damon a écrit :
>> On 12/25/23 2:22 PM, WM wrote:
>>> Le 24/12/2023 à 18:04, Richard Damon a écrit :
>>>> On 12/24/23 3:37 AM, WM wrote:
>>>>> Le 23/12/2023 à 14:37, Richard Damon a écrit :
>>>>>> On 12/23/23 7:40 AM, WM wrote:
>>>>>
>>>>>>> Of course. And since the leap down from omega never hits a dark
>>>>>>> number, their existence is proven by the fact that after leaping
>>>>>>> down you can run upwards through all dark numbers collectively by
>>>>>>> "and so on" or by "...".
>>>>>>
>>>>>> But there is no "leap down". What is the leap down from 0 (in the
>>>>>> Naturals).
>>>>>
>>>>> There is a leap from omega to 17.
>>>>
>>>> Really, a ONE STEP leap?
>>>
>>> Really. But we can go every number of finite steps, for instance
>>> ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
>>> At every term we can reverse the direction and can go upwards ℵo steps.
>>
>> Why is the step down from ω 2023? Why not 100000, or 1?
>
> All that is possible. All these sequences end after a finite number of
> steps. But in reverse diraction there are infinitely many step possible.

So, your mathematics is inconsistant. If step_down(x) can have multiple
answers, you system is just not consistant.

>>
>> There is no unique value that works, so the operation isn't actually
>> defined.
>
> You can define whateverr you like. The sequence will be finite.

So, the infinite number has a finite number of predecessors. Sounds a
bit inconsistant to me.

>>
>>>>
>>>>>> You never reach "up" from the Ordinals to the Transfinite,
>>>>>
>>>>> The sequence has been given by Cantor:
>>>>> 1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
>>>>
>>>> Please show a reference where you can actually "Step up" from one of
>>>> the ... to the following multiple of omega (or down).
>>>
>>> If not, then not all natural numbers can be applied in bijections.
>>
>> Why not?
>
> Because all implies that none is missing.

Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...

>
>> The bijection rules are based only on the set 1, 2, 3, ... and don't
>> need the omega stuff.
>>
>>>>
>>>> BOUNDS, as I am using them are the maximum or minimum
>>>
>>> That is not correct.
>>
>> Why not?
>
> If you understand German, you may consult W. Mückenheim: "Mathematik für
> die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
> https://www.degruyter.com/document/doi/10.1515/9783110377347/html
> Ask Wiki or a math teacher if you don't believe me.

But you can't claim yourself as an athority to prove yourself.

You are just proving your "logic" is defective.

>
> Regards, WM

Le 26/12/2023 à 01:31, Richard Damon a écrit :
> On 12/25/23 5:40 PM, WM wrote:

>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>
>> All that is possible. All these sequences end after a finite number of
>> steps. But in reverse diraction there are infinitely many step possible.
>
> So, your mathematics is inconsistant.

This is set theory.
>
>>>
>>> There is no unique value that works, so the operation isn't actually
>>> defined.
>>
>> You can define whatever you like. The sequence will be finite.
>
> So, the infinite number has a finite number of predecessors. Sounds a
> bit inconsistant to me.

Without dark numbers it is inconsistent.

>> Because all implies that none is missing.
>
> Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...

as well as the unit fractions which *never* sit together at one and the
same place. There is no discontinuity near zero.

>>>>>
>>>>> BOUNDS, as I am using them are the maximum or minimum
>>>>
>>>> That is not correct.
>>>
>>> Why not?
>>
>> If you understand German, you may consult W. Mückenheim: "Mathematik für
>> die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
>> https://www.degruyter.com/document/doi/10.1515/9783110377347/html
>> Ask Wiki or a math teacher if you don't believe me.
>
> But you can't claim yourself as an athority to prove yourself.

Og course I am an authority in this question. If you don't believe, then
ask someone else. No mathematician will share your personal theory.
>
> You are just proving your "logic" is defective.

Chuckle.

Regards, WM

On 12/26/23 5:17 AM, WM wrote:
> Le 26/12/2023 à 01:31, Richard Damon a écrit :
>> On 12/25/23 5:40 PM, WM wrote:
>
>>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>>
>>> All that is possible. All these sequences end after a finite number
>>> of steps. But in reverse diraction there are infinitely many step
>>> possible.
>>
>> So, your mathematics is inconsistant.
>
> This is set theory.
>>

It seems your "Set Theory" is defective.

>>>>
>>>> There is no unique value that works, so the operation isn't actually
>>>> defined.
>>>
>>> You can define whatever you like. The sequence will be finite.
>>
>> So, the infinite number has a finite number of predecessors. Sounds a
>> bit inconsistant to me.
>
> Without dark numbers it is inconsistent.

Nope, WITH dark numbers it is inconsistent.

Dark numbers persume that there exists a n that is the highest Natuaral
Number that is "usable indivitually", and ignores that n+1 meett

>>> Because all implies that none is missing.
>>
>> Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
>
> as well as the unit fractions which *never* sit together at one and the
> same place. There is no discontinuity near zero.

No one said they "sit together", just that they get unboundedly packed
together.

Also, your step from 0 to some unit fraction is NEVER to the "first"
unit fraction, so your "finite length of a step" doesn't apply.
>
>>>>>>
>>>>>> BOUNDS, as I am using them are the maximum or minimum
>>>>>
>>>>> That is not correct.
>>>>
>>>> Why not?
>>>
>>> If you understand German, you may consult W. Mückenheim: "Mathematik
>>> für die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
>>> https://www.degruyter.com/document/doi/10.1515/9783110377347/html
>>> Ask Wiki or a math teacher if you don't believe me.
>>
>> But you can't claim yourself as an athority to prove yourself.
>
> Og course I am an authority in this question. If you don't believe, then
> ask someone else. No mathematician will share your personal theory.

Then show an AcTUAL PROOF that isn't based on errors.

You point out that BETWEEN two unit fractions there needs to be a finite
space, and then try to claim that this must apply between a point that
isn't a unit fraction.

That you think it make sense shows you are NOT an actual authority on
this question, but an IDIOT.

You make the same kind of errors the idiot Peter Olcott does, you think
you can ignore when people point out your errors, but that just shows
that your logic is flawed, and you understand this, but can't face it,
so you just repeat your refuted statements, showing the world that you
are just a liar.

>>
>> You are just proving your "logic" is defective.
>
> Chuckle.

Yep, laugh while the world laughs at you.

>
> Regards, WM
>
>

Le 26/12/2023 à 13:44, Richard Damon a écrit :
> On 12/26/23 5:17 AM, WM wrote:
>> Le 26/12/2023 à 01:31, Richard Damon a écrit :
>>> On 12/25/23 5:40 PM, WM wrote:
>>
>>>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>>>
>>>> All that is possible. All these sequences end after a finite number
>>>> of steps. But in reverse diraction there are infinitely many step
>>>> possible.
>>>
>>> So, your mathematics is inconsistant.
>>
>> This is set theory.
>>>
>
> It seems your "Set Theory" is defective.

It is common set theory. Obviously you don't know set theory as little as
mathematics.

>>>
>>> Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
>>
>> as well as the unit fractions which *never* sit together at one and the
>> same place. There is no discontinuity near zero.
>
> No one said they "sit together", just that they get unboundedly packed
> together.

If they do not sit together at the sma epoint, then one of them is the
first leading to an increase of NUF. There is no third alternative.
>
> Also, your step from 0 to some unit fraction is NEVER to the "first"
> unit fraction,

Either one is the first or many are the first. There is no alternative.

>> Og course I am an authority in this question. If you don't believe, then
>> ask someone else. No mathematician will share your personal theory.
>
> Then show an AcTUAL PROOF that isn't based on errors.

The infimum is by definition the greatest lower bound. There is no proof
for that definition. But there is a proof that 0 is the infimum of the set
of unit fractions because every larger x can be undercut.
>
> You point out that BETWEEN two unit fractions there needs to be a finite
> space, and then try to claim that this must apply between a point that
> isn't a unit fraction.

No, it must apply every unit fraction. Therefroe there can't be two
without a gap.

Regards, WM

On 12/26/23 1:14 PM, WM wrote:
> Le 26/12/2023 à 13:44, Richard Damon a écrit :
>> On 12/26/23 5:17 AM, WM wrote:
>>> Le 26/12/2023 à 01:31, Richard Damon a écrit :
>>>> On 12/25/23 5:40 PM, WM wrote:
>>>
>>>>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>>>>
>>>>> All that is possible. All these sequences end after a finite number
>>>>> of steps. But in reverse diraction there are infinitely many step
>>>>> possible.
>>>>
>>>> So, your mathematics is inconsistant.
>>>
>>> This is set theory.
>>>>
>>
>> It seems your "Set Theory" is defective.
>
> It is common set theory. Obviously you don't know set theory as little
> as mathematics.

I don't see it in any set theory that I know of. What statement in set
theory defines a step down from omega to be some (or ANY) finite number.

The set thory that defines the Natural Numbers, the finite numbers 1, 2,
3, 4, ... (continuing counting forever) make them all "finite" numbers,
all individually finitely definable and thus individually finitely
nameable (admittedly with unbounded length and value). Thus NO element
of this meets your definition of "Dark" of not being individually
usable/definable/namable.

The set theory that does this NEVER gets to "omega" as a number, so
never provides a way to "step up" to it, or to step down from it.

A SEPARATE set theory, extends the definition of numbers to what what
Cantor called the Transifinte Numbers gives us omega and its relatives,
but still doesn't define a finite "step" either up to, or down from
omega (or a multiple of omega) to a natural number or a lower multiple
of omega. Such a "step" would be a Transfinite step, not a finite step.

It seems your "Common Set Thoery" that you are relying on is also
"Dark", not able to be named or used "individually", but only as some
indefinite clump, which makes it not actually part of logic.

Try to prove me wrong, and give some ACTUAL set theory that shows otherwise.

>
>>>>
>>>> Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
>>>
>>> as well as the unit fractions which *never* sit together at one and
>>> the same place. There is no discontinuity near zero.
>>
>> No one said they "sit together", just that they get unboundedly packed
>> together.
>
> If they do not sit together at the sma epoint, then one of them is the
> first leading to an increase of NUF. There is no third alternative.

Nope. They just get increasingly tight together with out bound.

There is no "first"

You can not "step" from 0 up to a "first" finite number (unit fraction,
rational, or real) as that step just isn't defined.

The "Third Alternative" that you apparently can't comprehend is that
they are infinitely dense at this point and don't have a smallest.

It isn't that we can't name it, it just doesn't exist.

>>
>> Also, your step from 0 to some unit fraction is NEVER to the "first"
>> unit fraction,
>
> Either one is the first or many are the first. There is no alternative.

There are NO first. Why does there need to be? This is just something
that comes out of unbounded sets. If your logic can't handle it, it
can't handle the Natural Numbers, and thus can't get up to the Unit
Fractions, Rational or Reals.

In fact, it is PROVABLE that there is no first, that comes out of basic
set theory. This doesn't say the first is "dark", it says there is no first.

Your "Dark" just can't exist if you limit your self to the Unit
Fractions and Natural Numbers, as ALL such number are PROVABLE to be
individually namable and usable, by their method of construction. So
your logic is just proven to be incorrect.

>
>>> Og course I am an authority in this question. If you don't believe,
>>> then ask someone else. No mathematician will share your personal theory.
>>
>> Then show an AcTUAL PROOF that isn't based on errors.
>
> The infimum is by definition the greatest lower bound. There is no proof
> for that definition. But there is a proof that 0 is the infimum of the
> set of unit fractions because every larger x can be undercut.

Yes, but 0 isn't a member of the set, and thus doesn't provide a
definition of a "first".

>>
>> You point out that BETWEEN two unit fractions there needs to be a
>> finite space, and then try to claim that this must apply between a
>> point that isn't a unit fraction.
>
> No, it must apply every unit fraction. Therefroe there can't be two
> without a gap.

Right, and who said there were.

The problem is that 0 isn't a unit fraction, so there doesn't need to be
a finite gap between it and the begining of the unit
fractions/rationals/real.

Your step property only applies BETWEEN unit fractions, and thus not
between 0 and something past it.

>
> Regards, WM
>

Le 27/12/2023 à 03:29, Richard Damon a écrit :
> On 12/26/23 1:14 PM, WM wrote:
>>>>
>>>>>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>>>>>
>>>>>> All that is possible. All these sequences end after a finite number
>>>>>> of steps. But in reverse diraction there are infinitely many step
>>>>>> possible.
>>>>>
>>>>> So, your mathematics is inconsistant.
>>>>
>>>> This is set theory.
>>>>>
>>>
>>> It seems your "Set Theory" is defective.
>>
>> It is common set theory. Obviously you don't know set theory as little
>> as mathematics.
>
> I don't see it in any set theory that I know of. What statement in set
> theory defines a step down from omega to be some (or ANY) finite number.

https://en.wikipedia.org/wiki/Axiom_of_regularity
https://en.wikipedia.org/wiki/Goodstein%27s_theorem
The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take the
hereditary base bn representation of G(m)(n), and replace each occurrence
of the base bn with the first infinite ordinal number ω.

Regards, WM

On 12/27/23 3:47 AM, WM wrote:
> Le 27/12/2023 à 03:29, Richard Damon a écrit :
>> On 12/26/23 1:14 PM, WM wrote:
>>>>>
>>>>>>>> Why is the step down from ω 2023? Why not 100000, or 1?
>>>>>>>
>>>>>>> All that is possible. All these sequences end after a finite
>>>>>>> number of steps. But in reverse diraction there are infinitely
>>>>>>> many step possible.
>>>>>>
>>>>>> So, your mathematics is inconsistant.
>>>>>
>>>>> This is set theory.
>>>>>>
>>>>
>>>> It seems your "Set Theory" is defective.
>>>
>>> It is common set theory. Obviously you don't know set theory as
>>> little as mathematics.
>>
>> I don't see it in any set theory that I know of. What statement in set
>> theory defines a step down from omega to be some (or ANY) finite number.
>
> https://en.wikipedia.org/wiki/Axiom_of_regularity
> https://en.wikipedia.org/wiki/Goodstein%27s_theorem
> The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take the
> hereditary base bn representation of G(m)(n), and replace each
> occurrence of the base bn with the first infinite ordinal number ω.
>
> Regards, WM

Which is against the laws of numbers, since omega isn't a natural
number, and those terms you are plugging it into are supposed to be
terms that are Natural Numbers.

In other words, you are just admitting your theory is based on lies.

Le 27/12/2023 à 14:37, Richard Damon a écrit :
> On 12/27/23 3:47 AM, WM wrote:

>>> I don't see it in any set theory that I know of. What statement in set
>>> theory defines a step down from omega to be some (or ANY) finite number.
>>
>> https://en.wikipedia.org/wiki/Axiom_of_regularity
>> https://en.wikipedia.org/wiki/Goodstein%27s_theorem
>> The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take the
>> hereditary base bn representation of G(m)(n), and replace each
>> occurrence of the base bn with the first infinite ordinal number ω.
>>
> Which is against the laws of numbers,

If you are unable to read and understand the given references, then you
are unable to discuss this topic.
EOD.

Regards, WM

On 12/27/23 5:19 PM, WM wrote:
> Le 27/12/2023 à 14:37, Richard Damon a écrit :
>> On 12/27/23 3:47 AM, WM wrote:
>
>>>> I don't see it in any set theory that I know of. What statement in
>>>> set theory defines a step down from omega to be some (or ANY) finite
>>>> number.
>>>
>>> https://en.wikipedia.org/wiki/Axiom_of_regularity
>>> https://en.wikipedia.org/wiki/Goodstein%27s_theorem
>>> The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take
>>> the hereditary base bn representation of G(m)(n), and replace each
>>> occurrence of the base bn with the first infinite ordinal number ω.
>>>
>> Which is against the laws of numbers,
>
> If you are unable to read and understand the given references, then you
> are unable to discuss this topic.
> EOD.
>
> Regards, WM
>

So, you are admitting that is worthless talking to YOU, since you can't
follow the reference you use.

Since each of these G(m)(n) are Natural Numbers, and you state you
change those to omega, which isn't a Natural Number.