On 1/19/2024 5:19 AM, WM wrote:
> Le 18/01/2024 à 19:20, Jim Burns a écrit :

>> [...]
>
> In order to excorzise the bijective meaning
> I will henceforth use only ℵ,
> meaning "infinitely many".

¬(𝒫(ℕ) ⇉ ℕ)

| Assume otherwise.
| Assume 𝒫(ℕ) ⇉ ℕ
| Exists f: 𝒫(ℕ) ⇉ ℕ
| ∀S ∈ 𝒫(ℕ): ∃k ∈ ℕ: k = f(S)
| | However,
| consider D = {f(S) ∈ ℕ| S ∈ 𝒫(ℕ) ∧ f(S) ∉ S }
| f(D) ∈ D ∨ f(D) ∉ D
| | (i) f(D) ∈ D
| f(D) ∈ D ⇒ f(D) ∉ D
| Contradiction.
| | (ii) f(D) ∉ D
| f(D) ∉ D ⇒ f(D) ∈ D
| Contradiction.
| | Thus, contradiction.

Therefore,
¬(𝒫(ℕ) ⇉ ℕ)
and
|𝒫(ℕ)| > |ℕ|

> |ℕ| = ℵ, |ℚ| = ℵ, |ℝ| = ℵ,

|ℝ| > |ℚ| = |ℕ| ∉ ℕ

On 1/19/24 2:18 PM, WM wrote:
> Le 19/01/2024 à 19:09, Richard Damon a écrit :
>> On 1/19/24 12:07 PM, WM wrote:
>>> Le 19/01/2024 à 16:54, Richard Damon a écrit :
>>>> On 1/19/24 3:55 AM, WM wrote:
>>>>>>>
>>>>>>> No, it is bounded in length by the origin 0.
>>>>>>
>>>>>> Which isn't an element of the set,
>>>>>
>>>>> But NUF(x) is well defined.
>>>>
>>>> No, it isn't.
>>>
>>> NUF(x) is the number of unit fractions between 0 and x.
>>
>> Which is infinite for all x > 0.
>
> Not when mathematics is applied.
>>
>> And thus NUF(x) doesn't have a finite value answer for any x > 0, and
>> if you are saying you are working in the domain of finite values,
>> NUF(x) isn't actually defined for any x > 0, since its value isn't
>> defined to something in the domain of regard.
>>
>> Nothing in that definition allows it to have values other than 0 or
>> infinity, so claiming it has other values is just an error.
>>
>> To presume it has the value 1 somewhere, presumes that there exists a
>> smallest unit fraction, which is a false assumption.
>
> Not when mathematics is applied.

What mathematics?

BY YOUR DEFINITON, since there is an infinte number of unit fractions
below any finite x, NUF(x) has an infinite value for all finite x
greater than 0.

There is no point where NF(x) can be 1, as that implies that there is a
smallest unit fraction and thus a highest Natural Number, but ALL
Natural numbers, by definition, have a successor, so that couldn't have
been the smallest unit fraction.

>>
>>>>
>>>>> Most unit fractions are dark because most natural numbers are dark.
>>>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>> |ℕ \ {1, 2, 3, ...}| = 0
>>>>
>>>> But that doesn't define a set of just dark number.
>>>
>>> There is no set of only dark numbers. There is an infinite set ℕ, a
>>> finite part of which is visible, the complement is dark.
>>
>> But why is it only a finite part that is visible?
>
> Try to make more visible. Fail. Then you know it.

>>
>>>>
>>>> Your problem is you don't have a definiton of what "N_def" actually is,
>>>
>>> Every number that is defined individually is visible.
>>
>> So, all Natural Numbers are visible, as we can find the individual
>> definition of any of them.
>
> Try it.

I did. You just don't understand it.

>>>
>>>> You are implicitly assuming that there is a "highest" defined number,
>>>
>>> Not a constant number but only temporarily.
>>
>> So not actually existing.
>
> The number is actually existing like the biggest known prime number.

Nope, "Known" is different than "Existing".

>>
>> All you are doing is showing that there exist number bigger than HAVE
>> BEEN named, not bigger than CAN be named.
>
> Try to name all. Fail

Didn't say I could name ALL, I said I could name ANY.

Of course we can't name ALL of an unbounded set, as NAMING is a finite
function.

>>
>> This becomes a problem of what we have seen, not of what is, so not a
>> property of the numbers themselves, but of the observer.
>
> Exactly. Nevertheless no observer can see all numbers.

Doesn't matter if they have been actually obsevered, just that they be
observable.

You are just confusing Knowledge with Truth.

>
> Regards, WM
>
>

Le 19/01/2024 à 22:48, Richard Damon a écrit :
> On 1/19/24 2:18 PM, WM wrote:

>>> To presume it has the value 1 somewhere, presumes that there exists a
>>> smallest unit fraction, which is a false assumption.
>>
>> Not when mathematics is applied.
>
> What mathematics?

∀n ∈ ℕ: 1/n - 1/(n+1) > 0
>
> BY YOUR DEFINITON, since there is an infinte number of unit fractions
> below any finite x,

Wrong premise.

> NUF(x) has an infinite value for all finite x
> greater than 0.

Wrong result.
>
> There is no point where NF(x) can be 1, as that implies that there is a
> smallest unit fraction and thus a highest Natural Number, but ALL
> Natural numbers, by definition, have a successor,

This definition is obtained from definble numbers and erroneously
generatlized to all numbers.
>>>
>>> But why is it only a finite part that is visible?
>>
>> Try to make more visible. Fail. Then you know it.

>>> So, all Natural Numbers are visible, as we can find the individual
>>> definition of any of them.
>>
>> Try it.
>
> I did. You just don't understand it.

I don't accept your lies.
>
>>>>
>>>>> You are implicitly assuming that there is a "highest" defined number,
>>>>
>>>> Not a constant number but only temporarily.
>>>
>>> So not actually existing.
>>
>> The number is actually existing like the biggest known prime number.
>
> Nope, "Known" is different than "Existing".

All numbers are existing. Only few are visible.
>
>>>
>>> All you are doing is showing that there exist number bigger than HAVE
>>> BEEN named, not bigger than CAN be named.
>>
>> Try to name all. Fail
>
> Didn't say I could name ALL, I said I could name ANY.

Wrong. You can name only any which has ℵ successors. That is a big
difference.
>
> Of course we can't name ALL of an unbounded set, as NAMING is a finite
> function.

You must let almost all remain unnamed.

Regards, WM

Le 19/01/2024 à 20:53, Jim Burns a écrit :
> On 1/19/2024 5:19 AM, WM wrote:
>> Le 18/01/2024 à 19:20, Jim Burns a écrit :
>
>>> [...]
>>
>> In order to excorzise the bijective meaning
>> I will henceforth use only ℵ,
>> meaning "infinitely many".
>
> ¬(𝒫(ℕ) ⇉ ℕ)
>
> | Assume otherwise.
> | Assume 𝒫(ℕ) ⇉ ℕ
> | Exists f: 𝒫(ℕ) ⇉ ℕ
> | ∀S ∈ 𝒫(ℕ): ∃k ∈ ℕ: k = f(S)
> |
> | However,
> | consider D = {f(S) ∈ ℕ| S ∈ 𝒫(ℕ) ∧ f(S) ∉ S }
> | f(D) ∈ D ∨ f(D) ∉ D

This proof presumes that infinite bijections exist and that all subsets of
ℕ are definable. Both is wrong. Same with Cantor's diagonal. It assumes
that all natnumbers were definable. That is wrong. Therefore that is
wrong:

> |ℝ| > |ℚ|

Regards, WM

Fritz Feldhase schrieb am Freitag, 19. Januar 2024 um 21:40:34 UTC+1:

> After all, we already have a theory of (finite) and infinite cardinals in set
> theory.

But this theory requires that if not point of (0, 1] has less than ℵ
smaller unit fractions to its left-hand side, nevertheless there are not
ℵ unit fractions to the left-hand side of the interval (0, 1], i.e. a
theory which violates geometry.

Regards, WM

On 1/19/24 5:41 PM, WM wrote:
> Le 19/01/2024 à 22:48, Richard Damon a écrit :
>> On 1/19/24 2:18 PM, WM wrote:
>
>>>> To presume it has the value 1 somewhere, presumes that there exists
>>>> a smallest unit fraction, which is a false assumption.
>>>
>>> Not when mathematics is applied.
>>
>> What mathematics?
>
> ∀n ∈ ℕ: 1/n - 1/(n+1) > 0

Which says nothing about

>>
>> BY YOUR DEFINITON, since there is an infinte number of unit fractions
>> below any finite x,
>
> Wrong premise.

So, you lied? It was your premise. There ARE an infinite (aleph) number
of unit fractions below ANY unit fraction.

>
>> NUF(x) has an infinite value for all finite x greater than 0.
>
> Wrong result.

Where is it not?

>>
>> There is no point where NF(x) can be 1, as that implies that there is
>> a smallest unit fraction and thus a highest Natural Number, but ALL
>> Natural numbers, by definition, have a successor,
>
> This definition is obtained from definble numbers and erroneously
> generatlized to all numbers.

But what Natural Number or Unit Fraction isn't definable?

For that matter, what Rational Number isn't.

>>>>
>>>> But why is it only a finite part that is visible?
>>>
>>> Try to make more visible. Fail. Then you know it.
>
>>>> So, all Natural Numbers are visible, as we can find the individual
>>>> definition of any of them.
>>>
>>> Try it.
>>
>> I did. You just don't understand it.
>
> I don't accept your lies.

Your lose, since they weren't lies.

You are just rejecting the truth.

>>
>>>>>
>>>>>> You are implicitly assuming that there is a "highest" defined number,
>>>>>
>>>>> Not a constant number but only temporarily.
>>>>
>>>> So not actually existing.
>>>
>>> The number is actually existing like the biggest known prime number.
>>
>> Nope, "Known" is different than "Existing".
>
> All numbers are existing. Only few are visible.

Where do Natural Numbers stop becoming visible.

You can't actually define the set of Visible Numbers, which is needed
for your theory.

>>
>>>>
>>>> All you are doing is showing that there exist number bigger than
>>>> HAVE BEEN named, not bigger than CAN be named.
>>>
>>> Try to name all. Fail
>>
>> Didn't say I could name ALL, I said I could name ANY.
>
> Wrong. You can name only any which has ℵ successors. That is a big
> difference.

Which is all of them. You just don't understand the infinite.

>>
>> Of course we can't name ALL of an unbounded set, as NAMING is a finite
>> function.
>
> You must let almost all remain unnamed.
>

Nope.

> Regards, WM
>
>
>

On 1/19/24 5:46 PM, WM wrote:
> Le 19/01/2024 à 20:53, Jim Burns a écrit :
>> On 1/19/2024 5:19 AM, WM wrote:
>>> Le 18/01/2024 à 19:20, Jim Burns a écrit :
>>
>>>> [...]
>>>
>>> In order to excorzise the bijective meaning
>>> I will henceforth use only ℵ,
>>> meaning "infinitely many".
>>
>> ¬(𝒫(ℕ) ⇉ ℕ)
>>
>> | Assume otherwise.
>> | Assume 𝒫(ℕ) ⇉ ℕ
>> | Exists f: 𝒫(ℕ) ⇉ ℕ
>> | ∀S ∈ 𝒫(ℕ): ∃k ∈ ℕ: k = f(S)
>> |
>> | However,
>> | consider D = {f(S) ∈ ℕ| S ∈ 𝒫(ℕ) ∧ f(S) ∉ S }
>> | f(D) ∈ D  ∨  f(D) ∉  D
>
> This proof presumes that infinite bijections exist and that all subsets
> of ℕ are definable. Both is wrong. Same with Cantor's diagonal. It
> assumes that all natnumbers were definable. That is wrong. Therefore
> that is wrong:
>

So you think, but can not prove.

You are just wrong, because you think with logic insufficent to handle
unbounded sets.

>> |ℝ| > |ℚ|
>
> Regards, WM
>
>

Le 20/01/2024 à 00:18, Richard Damon a écrit :
> On 1/19/24 5:41 PM, WM wrote:
>
> But what Natural Number or Unit Fraction isn't definable?
>
The smallest ones. Those which, according to you, sit at infinitesimals
between 0 and (0, 1].
>>
>> Wrong. You can name only any which has ℵ successors. That is a big
>> difference.
>
> Which is all of them.

No, it is a small always finite part.
>
>>>
>>> Of course we can't name ALL of an unbounded set, as NAMING is a finite
>>> function.
>>
>> You must let almost all remain unnamed.
>
> Nope.

Liar.

Regards, WM

Le 20/01/2024 à 00:18, Richard Damon a écrit :
> On 1/19/24 5:46 PM, WM wrote:

>> This proof presumes that infinite bijections exist and that all subsets
>> of ℕ are definable. Both is wrong. Same with Cantor's diagonal. It
>> assumes that all natnumbers were definable. That is wrong. Therefore
>> that is wrong:
>>
> So you think, but can not prove.

Simple. The diagonal number has no last digit. Therefore it is undefined.

Regards, WM

On 1/20/24 6:08 AM, WM wrote:
> Le 20/01/2024 à 00:18, Richard Damon a écrit :
>> On 1/19/24 5:46 PM, WM wrote:
>
>>> This proof presumes that infinite bijections exist and that all
>>> subsets of ℕ are definable. Both is wrong. Same with Cantor's
>>> diagonal. It assumes that all natnumbers were definable. That is
>>> wrong. Therefore that is wrong:
>>>
>> So you think, but can not prove.
>
> Simple. The diagonal number has no last digit. Therefore it is undefined.
>
> Regards, WM
>
>
>

Are you jumping to the Reals now? The proof that they are not countable.

On 1/20/24 6:05 AM, WM wrote:
> Le 20/01/2024 à 00:18, Richard Damon a écrit :
>> On 1/19/24 5:41 PM, WM wrote:
>>
>> But what Natural Number or Unit Fraction isn't definable?
>>
> The smallest ones. Those which, according to you, sit at infinitesimals
> between 0 and (0, 1].
>>>
>>> Wrong. You can name only any which has ℵ successors. That is a big
>>> difference.
>>
>> Which is all of them.
>
> No, it is a small always finite part.

Nope.

Which ones, NAME THEM OR THEIR SET, don't

>>
>>>>
>>>> Of course we can't name ALL of an unbounded set, as NAMING is a
>>>> finite function.
>>>
>>> You must let almost all remain unnamed.
>>
>> Nope.
>
> Liar.

You described yourself.

>
> Regards, WM
>
>
>
>

Le 20/01/2024 à 13:29, Richard Damon a écrit :
> On 1/20/24 6:05 AM, WM wrote:
>> Le 20/01/2024 à 00:18, Richard Damon a écrit :
>>> On 1/19/24 5:41 PM, WM wrote:
>>>
>>> But what Natural Number or Unit Fraction isn't definable?
>>>
>> The smallest ones. Those which, according to you, sit at infinitesimals
>> between 0 and (0, 1].

Do you agree?
>>>>
>>>> Wrong. You can name only any which has ℵ successors. That is a big
>>>> difference.
>>>
>>> Which is all of them.
>>
>> No, it is a small always finite part.
>
> Nope.
>
> Which ones, NAME THEM OR THEIR SET, don't

Those which, according to you, sit at infinitesimals between 0 and (0, 1].

>>>> You must let almost all remain unnamed.
>>>
>>> Nope.

Name one of them.

Regards, WM

Le 20/01/2024 à 13:29, Richard Damon a écrit :
> On 1/20/24 6:08 AM, WM wrote:

>>> So you think, but can not prove.
>>
>> Simple. The diagonal number has no last digit. Therefore it is undefined.
>>
> Are you jumping to the Reals now?

I prove that the diagonal number is not defined.

Regards, WM

On 1/21/24 2:58 AM, WM wrote:
> Le 20/01/2024 à 13:29, Richard Damon a écrit :
>> On 1/20/24 6:05 AM, WM wrote:
>>> Le 20/01/2024 à 00:18, Richard Damon a écrit :
>>>> On 1/19/24 5:41 PM, WM wrote:
>>>>
>>>> But what Natural Number or Unit Fraction isn't definable?
>>>>
>>> The smallest ones. Those which, according to you, sit at
>>> infinitesimals between 0 and (0, 1].
>
> Do you agree?

Yes, it seems that what you think of as your dark numbers (of unit
fractions) are actually the infinitesimals, but they don't have the
non-definable property of your dark numbers. They are dark to you only
because you don't understand them.

>>>>>
>>>>> Wrong. You can name only any which has ℵ successors. That is a big
>>>>> difference.
>>>>
>>>> Which is all of them.
>>>
>>> No, it is a small always finite part.
>>
>> Nope.
>>
>> Which ones, NAME THEM OR THEIR SET, don't
>
> Those which, according to you, sit at infinitesimals between 0 and (0, 1].

In other words, you are admitting that you "dark numbers" are just the
transfinite numbers, the infintesimals between 0 and (0, and the
infinite number beyond the finites.

This negates your claim that they are actually parts of the Natural
Numbers or the Unit Fractions.

You are just admitting that your darkness is well known numbers that you
just don't how to handle in your limite logic system.

>
>>>>> You must let almost all remain unnamed.
>>>>
>>>> Nope.
>
> Name one of them.
Since you can't define what set you consider Visible, I can't tell what
you don't think are visible.

I could say 5, but you are probably able to count that high.

I could say a googleplexplex, but maybe you have heard of that and
consider it visible.

And, since your putitive definition of visible isn't actually a formal
definition, anything I say you can just use to redefine your visible
set, showing you don't actually have a definition.

>
> Regards, WM
>
>

On 1/21/24 2:59 AM, WM wrote:
> Le 20/01/2024 à 13:29, Richard Damon a écrit :
>> On 1/20/24 6:08 AM, WM wrote:
>
>>>> So you think, but can not prove.
>>>
>>> Simple. The diagonal number has no last digit. Therefore it is
>>> undefined.
>>>
>> Are you jumping to the Reals now?
>
> I prove that the diagonal number is not defined.
>
> Regards, WM
>
>

So, you think that 0.333.... isn't defined?

You need to get your definitions straight.

And the Reals need better logic than you have, since you can't even
handle the Natural Numbers.

Le 21/01/2024 à 14:01, Richard Damon a écrit :
> On 1/21/24 2:58 AM, WM wrote:
>
> Yes, it seems that what you think of as your dark numbers (of unit
> fractions) are actually the infinitesimals, but they don't have the
> non-definable property of your dark numbers.

They cannot be distinguished.

> They are dark to you only

Can you distinguish them?

>> Those which, according to you, sit at infinitesimals between 0 and (0, 1].
>
> In other words, you are admitting that you "dark numbers" are just the
> transfinite numbers, the infintesimals between 0 and (0, and the
> infinite number beyond the finites.

No. Unit fractions are not infinitesimals. But you raised this topic
because you cannot distinguish the smallest ℵ unit fractions.

Regards, WM

On Sunday, January 21, 2024 at 8:52:22 AM UTC-8, WM wrote:
> Le 21/01/2024 à 14:01, Richard Damon a écrit :
> > On 1/21/24 2:58 AM, WM wrote:
> >
> > Yes, it seems that what you think of as your dark numbers (of unit
> > fractions) are actually the infinitesimals, but they don't have the
> > non-definable property of your dark numbers.
> They cannot be distinguished.
> > They are dark to you only
> Can you distinguish them?
> >> Those which, according to you, sit at infinitesimals between 0 and (0, 1].
> >
> > In other words, you are admitting that you "dark numbers" are just the
> > transfinite numbers, the infintesimals between 0 and (0, and the
> > infinite number beyond the finites.
> No. Unit fractions are not infinitesimals. But you raised this topic
> because you cannot distinguish the smallest ℵ unit fractions.
>
> Regards, WM

Language

Language: words for all the things.

Rules

Rules: in logic, physics, mathematics, and science, you don't get to make them.

Furthermore, it's suggested MW's made nothing but a mess of them.
Driving reactionary rejection, it's never made much a theorem,
though I did point out how sweep applies to the infinite balanced binary tree,
and otherwise offered clear words and rules to solve his mathematical problems,
but, the bot's either unwilling or incapable, except to be what it is.

A waste of our time

On 1/21/24 11:52 AM, WM wrote:
> Le 21/01/2024 à 14:01, Richard Damon a écrit :
>> On 1/21/24 2:58 AM, WM wrote:
>>
>> Yes, it seems that what you think of as your dark numbers (of unit
>> fractions) are actually the infinitesimals, but they don't have the
>> non-definable property of your dark numbers.
>
> They cannot be distinguished.

But in the infinitesimals, the can be.

Depending on which system you choose to use, they may be called delta or
epsilon (or something else).

>
>> They are dark to you only
>
> Can you distinguish them?

I just did.

Admittedly, you need to adjust your definition of NUF(x) to allow
NUF(delta) == 1, but if you don't, then you just have to accept that
NUF(x), not taking into account the infintesimals, just jumps from 0 to
ℵ0 as it passes over the GAP caused by them.

>
>>> Those which, according to you, sit at infinitesimals between 0 and
>>> (0, 1].
>>
>> In other words, you are admitting that you "dark numbers" are just the
>> transfinite numbers, the infintesimals between 0 and (0, and the
>> infinite number beyond the finites.
>
> No. Unit fractions are not infinitesimals. But you raised this topic
> because you cannot distinguish the smallest ℵ unit fractions.
> Regards, WM
>
>
>

But you just admitted that the area that NUF(x) went from 0 to infinity
was the infinitesimals. WHen asked where those numbers where, you said
"those which, according to you, sit at infinitesimals between 0 and
(0,1]" and then you try to say that the are not infinitesimals.

Which are they?

Why can you not see that we CAN distinguish ALL the unit fractions, as
we can distinguish ALL Natural Numbers. Every natural number has a
"name" being effectively its definition. The number n being n
applications of the successor function to 0. Since all natural numbers
are finite, all numbers have a finite name they can be given.

We don't get a non-finite number until we actually try to count all of
the number, to use the complete set, then we get to ℵ0, an infinity,
which is bigger than all finite numbers.

Note, to say we can not distinguish the "smallest" ℵ unit fractions, you
are saying we can not distinguish ANY of them, as there are only ℵ unit
fractions.

You have to deal with the fact that ℵ has different mathematics than the
finite numbers, and that ALL unbounded subsets of the Natural Numbers or
Unit Fractions all have the "same" number of members, that is ℵ0 (which
you are mistakenly calling ℵ)

On Sunday, January 21, 2024 at 9:52:29 AM UTC-8, Richard Damon wrote:
> On 1/21/24 11:52 AM, WM wrote:
> > Le 21/01/2024 à 14:01, Richard Damon a écrit :
> >> On 1/21/24 2:58 AM, WM wrote:
> >>
> >> Yes, it seems that what you think of as your dark numbers (of unit
> >> fractions) are actually the infinitesimals, but they don't have the
> >> non-definable property of your dark numbers.
> >
> > They cannot be distinguished.
> But in the infinitesimals, the can be.
>
> Depending on which system you choose to use, they may be called delta or
> epsilon (or something else).
> >
> >> They are dark to you only
> >
> > Can you distinguish them?
> I just did.
>
> Admittedly, you need to adjust your definition of NUF(x) to allow
> NUF(delta) == 1, but if you don't, then you just have to accept that
> NUF(x), not taking into account the infintesimals, just jumps from 0 to
> ℵ0 as it passes over the GAP caused by them.
> >
> >>> Those which, according to you, sit at infinitesimals between 0 and
> >>> (0, 1].
> >>
> >> In other words, you are admitting that you "dark numbers" are just the
> >> transfinite numbers, the infintesimals between 0 and (0, and the
> >> infinite number beyond the finites.
> >
> > No. Unit fractions are not infinitesimals. But you raised this topic
> > because you cannot distinguish the smallest ℵ unit fractions.
> > Regards, WM
> >
> >
> >
> But you just admitted that the area that NUF(x) went from 0 to infinity
> was the infinitesimals. WHen asked where those numbers where, you said
> "those which, according to you, sit at infinitesimals between 0 and
> (0,1]" and then you try to say that the are not infinitesimals.
>
> Which are they?
>
> Why can you not see that we CAN distinguish ALL the unit fractions, as
> we can distinguish ALL Natural Numbers. Every natural number has a
> "name" being effectively its definition. The number n being n
> applications of the successor function to 0. Since all natural numbers
> are finite, all numbers have a finite name they can be given.
>
> We don't get a non-finite number until we actually try to count all of
> the number, to use the complete set, then we get to ℵ0, an infinity,
> which is bigger than all finite numbers.
>
> Note, to say we can not distinguish the "smallest" ℵ unit fractions, you
> are saying we can not distinguish ANY of them, as there are only ℵ unit
> fractions.
>
> You have to deal with the fact that ℵ has different mathematics than the
> finite numbers, and that ALL unbounded subsets of the Natural Numbers or
> Unit Fractions all have the "same" number of members, that is ℵ0 (which
> you are mistakenly calling ℵ)

Cardinals are counts not numbers, and while having transitive equality and
inequality among themselves, have no business in formulas with non-cardinal
numbers, at all.

Then, there's a usual naive extensionality between finite cardinals and natural integers,
and finite ordinals, but otherwise naively putting them together is wrong.

So, cardinals in formulas among other numbers is usually nonsense.

Yes, I know there's a usual trivial assignment of initial ordinals to cardinals,
and it's sloppy, lazy, risky, and not type-safe.

Infinity's even better!

On 1/21/2024 5:03 AM, Richard Damon wrote:
> On 1/21/24 2:59 AM, WM wrote:
>> Le 20/01/2024 à 13:29, Richard Damon a écrit :
>>> On 1/20/24 6:08 AM, WM wrote:
>>
>>>>> So you think, but can not prove.
>>>>
>>>> Simple. The diagonal number has no last digit. Therefore it is
>>>> undefined.
>>>>
>>> Are you jumping to the Reals now?
>>
>> I prove that the diagonal number is not defined.
>>
>> Regards, WM
>>
>>
>
> So, you think that 0.333.... isn't defined?

..(3) is how base 10 handles 1 divided by 3.

>
>
> You need to get your definitions straight.
>
> And the Reals need better logic than you have, since you can't even
> handle the Natural Numbers.

:^)

On 1/21/24 3:30 PM, Chris M. Thomasson wrote:
> On 1/21/2024 5:03 AM, Richard Damon wrote:
>> On 1/21/24 2:59 AM, WM wrote:
>>> Le 20/01/2024 à 13:29, Richard Damon a écrit :
>>>> On 1/20/24 6:08 AM, WM wrote:
>>>
>>>>>> So you think, but can not prove.
>>>>>
>>>>> Simple. The diagonal number has no last digit. Therefore it is
>>>>> undefined.
>>>>>
>>>> Are you jumping to the Reals now?
>>>
>>> I prove that the diagonal number is not defined.
>>>
>>> Regards, WM
>>>
>>>
>>
>> So, you think that 0.333.... isn't defined?
>
> .(3) is how base 10 handles 1 divided by 3.
>

There are various notations used, but it still has no last digit, as
that is just a notation for repeating.

he also says that the square root of 2 isn't defined, or pi.

>
>>
>>
>> You need to get your definitions straight.
>>
>> And the Reals need better logic than you have, since you can't even
>> handle the Natural Numbers.
>
> :^)

On Sunday, January 21, 2024 at 12:45:33 PM UTC-8, Richard Damon wrote:
> On 1/21/24 3:30 PM, Chris M. Thomasson wrote:
> > On 1/21/2024 5:03 AM, Richard Damon wrote:
> >> On 1/21/24 2:59 AM, WM wrote:
> >>> Le 20/01/2024 à 13:29, Richard Damon a écrit :
> >>>> On 1/20/24 6:08 AM, WM wrote:
> >>>
> >>>>>> So you think, but can not prove.
> >>>>>
> >>>>> Simple. The diagonal number has no last digit. Therefore it is
> >>>>> undefined.
> >>>>>
> >>>> Are you jumping to the Reals now?
> >>>
> >>> I prove that the diagonal number is not defined.
> >>>
> >>> Regards, WM
> >>>
> >>>
> >>
> >> So, you think that 0.333.... isn't defined?
> >
> > .(3) is how base 10 handles 1 divided by 3.
> >
> There are various notations used, but it still has no last digit, as
> that is just a notation for repeating.
>
> he also says that the square root of 2 isn't defined, or pi.
> >
> >>
> >>
> >> You need to get your definitions straight.
> >>
> >> And the Reals need better logic than you have, since you can't even
> >> handle the Natural Numbers.
> >
> > :^)

You mean the quantity that squared equals 2, and
the quantity that is the ratio of a circle's circumference to diameter?

The Real Numbers or Reelle Zahlen have their properties if we don't even exist.
It's called Platonism and is the premier theory in history, including today..

Reelle: like Crelle's.

I have at least three formal continuous domains, so, they make one thing.
It's a theory of real numbers. I even just found them laying around.

On 1/21/2024 12:45 PM, Richard Damon wrote:
> On 1/21/24 3:30 PM, Chris M. Thomasson wrote:
>> On 1/21/2024 5:03 AM, Richard Damon wrote:
>>> On 1/21/24 2:59 AM, WM wrote:
>>>> Le 20/01/2024 à 13:29, Richard Damon a écrit :
>>>>> On 1/20/24 6:08 AM, WM wrote:
>>>>
>>>>>>> So you think, but can not prove.
>>>>>>
>>>>>> Simple. The diagonal number has no last digit. Therefore it is
>>>>>> undefined.
>>>>>>
>>>>> Are you jumping to the Reals now?
>>>>
>>>> I prove that the diagonal number is not defined.
>>>>
>>>> Regards, WM
>>>>
>>>>
>>>
>>> So, you think that 0.333.... isn't defined?
>>
>> .(3) is how base 10 handles 1 divided by 3.
>>
>
> There are various notations used, but it still has no last digit, as
> that is just a notation for repeating.

1 divided by 3

3 does not go into 1

3 goes into 10 3 times, 3 * 3

3 does not go into (10 - 9) aka, 3 does not go into 1.

We just hit a cycle at one third.

Fair enough?

;^)

>
> he also says that the square root of 2 isn't defined, or pi.
>
>>
>>>
>>>
>>> You need to get your definitions straight.
>>>
>>> And the Reals need better logic than you have, since you can't even
>>> handle the Natural Numbers.
>>
>> :^)
>

Le 21/01/2024 à 14:03, Richard Damon a écrit :
> On 1/21/24 2:59 AM, WM wrote:

>> I prove that the diagonal number is not defined.
>
> So, you think that 0.333.... isn't defined?

Why that? "0.333..." is a formula determining every digit uniquely.
Cantor's arbitrary list and its diagonal are undefined.

Regards, WM

Le 21/01/2024 à 18:52, Richard Damon a écrit :
> On 1/21/24 11:52 AM, WM wrote:
>>
>> Can you distinguish them?
>
> I just did.
>
> Admittedly, you need to adjust your definition of NUF(x) to allow
> NUF(delta) == 1, but if you don't, then you just have to accept that
> NUF(x), not taking into account the infintesimals, just jumps from 0 to
> ℵ0 as it passes over the GAP caused by them.

Fine. Say NUF(delta) == 1. But you cannot express delta in real numbers.

Regards, WM