Some results about unit fractions

[WM]

NUF(x), the Number of Unit Fractions between 0 and x, increases in (0, 1] from 0 to ℵo.
According to
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 (*)
never two or more unit fractions sit at the same place.

My results:
(1) I conclude from the increase from 0 to ℵo that there is an increase from 0 to ℵo.
(2) I conclude from (*) that never more than one unit fraction will add to NUF(x) at a point x (and a short distance afterwards).
(3) I conclude from the fact that this cannot be followed between 0 and ℵo that this cannot be followed between 0 and ℵo.

Too bold conclusions?

Regards, WM

[Julio Di Egidio]

On Wednesday, 12 April 2023 at 16:06:37 UTC+2, WM wrote:

> NUF(x), the Number of Unit Fractions between 0 and x, increases in (0, 1] from 0 to ℵo.

Wrong, that would hold on [0,1].

Julio

[WM]

In 0 there is no unit fraction. So there cannot be any more increase in [0, 1] than in (0, 1] .

Regards, WM

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
> 1] from 0 to ℵo.

No. NUF(x) is constant on (0,1].

--
Ben.

[Fritz Feldhase]

On Wednesday, April 12, 2023 at 4:06:37 PM UTC+2, WM wrote:

> NUF(x), the Number of Unit Fractions between 0 and x, increases in (0, 1] from 0 to ℵo.

No, NUF is constant on (0, 1].

Hint: Ax,x' e (0, 1]: NUF(x) = NUF(x').

Actually, Ax e (0, 1]: NUF(x) = ℵo.

[Fritz Feldhase]

Nonsense. Hint: NUF(0) = 0, while NUF(x) = ℵo for all x in (0, 1]. (This means that NUF is _constant_ on (0,1], no "increase" at all there.)

Though I would prefer to refer to this phenomenon as a /jump/ at 0 (from 0 to ℵo).

The term "increase" seems to be misleading to me in this context (though this point/question may be debatable).

[Fritz Feldhase]

On Wednesday, April 12, 2023 at 4:06:37 PM UTC+2, WM wrote:

> My results:
> (1) I conclude from [A that A].

> (2) I conclude from (*) that never more than one unit fraction will add to NUF(x) at a point x (and a short distance afterwards).

> (3) I conclude from [B that B].

Yeah, (1) and (3) are fascinating results, but (2) is utter nonsense, sorry.

[WM]

Wrong. Impossible because I conclude from (*) that never more than one unit fraction will add to NUF(x) at a point x (and a short distance afterwards).

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 (*)

That is dictated by mathematics.

Regards, WM

[WM]

It is the only non-trivial result. Only if you disregard mathematics

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

you can disregard this result.

Regards, WM

[Fritz Feldhase]

On Wednesday, April 12, 2023 at 6:39:26 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 12. April 2023 um 17:30:36 UTC+2:
> > On Wednesday, April 12, 2023 at 4:06:37 PM UTC+2, WM wrote:
> > >
> > > My results:
> > > (1) I conclude from [A that A].
> > > (2) I conclude from (*) that never more than one unit fraction will add to NUF(x) at a point x (and a short distance afterwards).
> > > (3) I conclude from [B that B].
> > >
> > Yeah, (1) and (3) are fascinating results, but (2) is utter nonsense, sorry.
> >
> It is the only non-trivial result.

Ok. So utter nonsense is your only non-trivial "result". I see.

[Fritz Feldhase]

On Wednesday, April 12, 2023 at 6:37:51 PM UTC+2, WM wrote:
> Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
> >
> > NUF(x) is constant on (0,1].
> >
> Wrong.

No, not wrong, but correct: NUF *is* constant on (0, 1].

Hint: Ax,x' e (0, 1]: NUF(x) = NUF(x').

Actually, Ax e (0, 1]: NUF(x) = ℵo, you silly crank.

> Impossible because I conclude from (*) <bla bla bla>

Yes, Mückenheim, we already know:

"[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

[Ben Bacarisse]

WM <askas...@gmail.com> writes:

> Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>> > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
>> > 1] from 0 to ℵo.
>> No. NUF(x) is constant on (0,1].
>
> Wrong.

NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
interval (0,1]. It's hard to see anything you can be more obviously
wrong about.

--
Ben.

[Jim Burns]

On 4/12/2023 10:06 AM, WM wrote:

> NUF(x), the Number of Unit Fractions
> between 0 and x,
> increases in (0, 1] from 0 to ℵo.

> ∀n ∈ ℕ:
> 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> (*)

∀n ∈ ℕ:
1/n > 1/(n+1)
(*)

> (2) I conclude from (*) that
> never more than one unit fraction
> will add to NUF(x) at a point x
> (and a short distance afterwards).

From (*)
the unit fractions are 1×1 1-ended.

1×1 1-ended sets are all the same size
(same cardinality).

UF = ⟨ ... 1/2 1/1 ⟩

Split ⟨ ... 1/2 1/1 ⟩
⟨ ... 1/i⁺⁺ ⟩ ⟨ 1/i ... 1/1 ⟩
are 1×1 1-ended and 1×1 2-ended
⟨ ... 1/i⁺⁺ ⟩ is the same size as
⟨ ... 1/2 1/1 ⟩

Split ⟨ ... 1/i⁺⁺ ⟩
⟨ ... 1/j⁺⁺ ⟩ ⟨ 1/j⁺⁺ ... 1/i⁺⁺ ⟩
are 1×1 1-ended and 1×1 2-ended
⟨ ... 1/j⁺⁺ ⟩ is the same size as
⟨ ... 1/1⁺⁺ ⟩ and ⟨ ... 1/2 1/1 ⟩

Split ⟨ ... 1/j⁺⁺ ⟩
⟨ ... 1/k⁺⁺ ⟩ ⟨ 1/k ... 1/j⁺⁺ ⟩
are 1×1 1-ended and 1×1 2-ended
⟨ ... 1/k⁺⁺ ⟩ is the same size as
⟨ ... 1/j⁺⁺ ⟩ and ⟨ ... 1/i⁺⁺ ⟩
and ⟨ ... 1/2 1/1 ⟩

Ad infinitum.

Infinity is not
a reallyreallyreallyreallyreallyreally large
natural number.

[WM]

Jim Burns schrieb am Mittwoch, 12. April 2023 um 21:55:38 UTC+2:
> On 4/12/2023 10:06 AM, WM wrote:
>
> > NUF(x), the Number of Unit Fractions
> > between 0 and x,
> > increases in (0, 1] from 0 to ℵo.
> > ∀n ∈ ℕ:
> > 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > (*)
> ∀n ∈ ℕ:
> 1/n > 1/(n+1)

Therefore NUF(x) starts with 1.

Regards, WM

[WM]

Matheology may say so. Mathematics denies it because never more than one unit fractions is added at a point:
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
Let alone ℵo unit fractions.

Why do you deny this equation?

Regards, WM

[WM]

Fritz Feldhase schrieb am Mittwoch, 12. April 2023 um 18:58:54 UTC+2:
> On Wednesday, April 12, 2023 at 6:37:51 PM UTC+2, WM wrote:
> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
> > >
> > > NUF(x) is constant on (0,1].
> > >
> > Wrong.
> No, not wrong, but correct: NUF *is* constant on (0, 1].

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 forbids the addition of more than one unit fraction in one point. Let alone of ℵo unit fractions.

But even if you were right, the ℵo unit fractions starting at 0 are dark.

Regards, WM

[WM]

Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 21:43:13 UTC+2:
> WM <askas...@gmail.com> writes:
>
> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> >> Unendlichen" at Hochschule Augsburg.)
> >> > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
> >> > 1] from 0 to ℵo.
> >> No. NUF(x) is constant on (0,1].
> >
> > Wrong.
> NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
> interval (0,1].

Even if you were right, the first ℵo unit fractions starting at 0 and remaining until the first definable unit fraction are dark. You cannot subdivide them individually to less than ℵo.

Regards, WM

[Python]

This is quite of a desesperate answer, Professor Crank Wolfgang
Mückenheim, from Hochschule Augsburg.

[Jim Burns]

No.
∀n ∈ ℕ:
If x > 1/n
then x > 1/n > 1/(n+1)

If NUF(x) > 0
then NUF(x) > 1

NUF(x) ≠ 1

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 21:43:13 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> >> Unendlichen" at Hochschule Augsburg.)
>> >> > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
>> >> > 1] from 0 to ℵo.
>> >> No. NUF(x) is constant on (0,1].
>> >
>> > Wrong.
>> NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
>> interval (0,1]. It's hard to see anything you can be more obviously
>> wrong about.
>
> Matheology may say so.

What's that? Mathematics says that NUF(x) = ℵo for x > 0.

> Mathematics denies it because never more than one unit fractions is
> added at a point

Mathematics is not done by waffling like this.

> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

From this trivial fact you would have to show that there are x1 and x2 >
0 with NUF(x1) =/= NUF(x2). Since you can't, you should just admit that
your denial that NUF(x) is constant on (0,1] is wrong.

The step change happens in WMaths too. UF(x) (the set of unit fractions
between 0 and x) is empty when x = 0 and not finite for all x > 0. How
is this possible when "never more than one unit fractions is added at a
point"?

--
Ben.

[Julio Di Egidio]

On Wednesday, 12 April 2023 at 16:49:14 UTC+2, WM wrote:
> Julio Di Egidio schrieb am Mittwoch, 12. April 2023 um 16:42:56 UTC+2:
> > On Wednesday, 12 April 2023 at 16:06:37 UTC+2, WM wrote:
> >
> > > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0, 1] from 0 to ℵo.
> >
> > Wrong, that would hold on [0,1].
>
> In 0 there is no unit fraction.

"No unit fraction" = "(a Number of) zero (i.e. 0) Unit Fractions".

> So there cannot be any more increase in [0, 1] than in (0, 1] .

LOL, yesterday I have had to explain the concept of
non-strict vs strict inequality to a secondary school
pupil: you wouldn't pass that test...

Julio

[WM]

Ben Bacarisse schrieb am Donnerstag, 13. April 2023 um 02:44:18 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
>
> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 21:43:13 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >>
> >> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
> >> >> WM <askas...@gmail.com> writes:
> >> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> >> >> Unendlichen" at Hochschule Augsburg.)
> >> >> > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
> >> >> > 1] from 0 to ℵo.
> >> >> No. NUF(x) is constant on (0,1].
> >> >
> >> > Wrong.
> >> NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
> >> interval (0,1]. It's hard to see anything you can be more obviously
> >> wrong about.
> >
> > Matheology may say so.
> What's that? Mathematics says that NUF(x) = ℵo for x > 0.

Yes, for every definable x, usually called eps, this is true. But mathematics says also that all unit fractions sit at different places: ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 and can appear in NUF only one by one.
Therefore there is only one way out of the apparent contradiction: Dark numbers.

> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> From this trivial fact you would have to show that there are x1 and x2 >
> 0 with NUF(x1) =/= NUF(x2).

Every unit fraction increases the number NUF. If the number starts with ℵo then you contradict the mathematical fact that ℵo unit fractions cannot start at the same x.

> Since you can't, you should just admit that
> your denial that NUF(x) is constant on (0,1] is wrong.

If it is constant for every x > 0, then ℵo unit fractions sit at one point. Contradition.

> The step change happens in WMaths too. UF(x) (the set of unit fractions
> between 0 and x) is empty when x = 0 and not finite for all x > 0. How
> is this possible when "never more than one unit fractions is added at a
> point"?

It is not possible. It is clearly wrong when all x are in question. Therefore it is enforcing the existence of numbers x which cannot be chosen as eps - dark numbers x.

Regards, WM

[WM]

Julio Di Egidio schrieb am Donnerstag, 13. April 2023 um 08:56:00 UTC+2:
> On Wednesday, 12 April 2023 at 16:49:14 UTC+2, WM wrote:

> > So there cannot be any more increase in [0, 1] than in (0, 1] .
> LOL, yesterday I have had to explain

That does not change the fact. NUF(0) = 0. NUF(1) = ℵo, NUF(eps) = ℵo for every eps > 0.

Regards, WM

[WM]

Jim Burns schrieb am Donnerstag, 13. April 2023 um 00:32:21 UTC+2:

> If NUF(x) > 0
> then NUF(x) > 1
>
> NUF(x) ≠ 1

This is true for all definable x. But ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 shows that never two unit fractions sit at the same place. Therefore they cannot be included in NUF(x) simultaneously. A first one must start. Therefore there are undefinable x.

It is simple to see: Every eps that you can choose contains almost all unit fractions.

Regards, WM

[WM]

Python schrieb am Mittwoch, 12. April 2023 um 23:44:30 UTC+2:
> Crank Wolfgang Mückenheim, aka WM wrote:
> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 21:43:13 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >>
> >>> Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
> >>>> WM <askas...@gmail.com> writes:
> >>>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> >>>> Unendlichen" at Hochschule Augsburg.)
> >>>>> NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
> >>>>> 1] from 0 to ℵo.
> >>>> No. NUF(x) is constant on (0,1].
> >>>
> >>> Wrong.
> >> NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
> >> interval (0,1].
> >
> > Even if you were right, the first ℵo unit fractions starting at 0 and remaining until the first definable unit fraction are dark. You cannot subdivide them individually to less than ℵo.
> >

> This is quite of a desesperate answer,

What an answer? It is certainly desperate for set theory.

Regards, WM

[Fritz Feldhase]

On Thursday, April 13, 2023 at 6:44:47 PM UTC+2, WM wrote:

> Every unit fraction increases [...] NUF.

No, it doesn't (necessarily), you silly crank.

Hint: For each and every x: If card(S) = ℵo, then card(S u {x}) = ℵo. No increase.

Holy shit!

Wie dumm kann man eigentlich sein, Mückenheim?

[Jim Burns]

On 4/13/2023 12:48 PM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 13. April 2023 um 00:32:21 UTC+2:

>> ∀n ∈ ℕ:
>> If x > 1/n
>> then x > 1/n > 1/(n+1)
>>

>> If NUF(x) > 0
>> then NUF(x) > 1
>>
>> NUF(x) ≠ 1
>
> This is true for all definable x.

It is true for all x which are
either x ≥ 1/n > 1/n⁺⁺ for some 1/n
or not x ≥ 1/n > 1/n⁺⁺ for some 1/n

> But
> ∀n ∈ ℕ:
> 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> shows that never two unit fractions sit
> at the same place.
> Therefore they cannot be
> included in NUF(x) simultaneously.
> A first one must start.

No.
"A first one must start" _in a finite list_
However,
the unit fractions are not a finite list.

Each unit fraction u,
whether u is dark or visible,
is preceded by another unit fraction v
1/v = 1/u + 1
0 < v < u

Each unit fraction u,
whether u is dark or visible,
is _not first_

A finite list is 1×1 2-ended.
The unit fractions are 1×1 1-ended.
They are not a finite list.

> Therefore there are undefinable x.

The description is of a unit fraction,
whether visible or dark.

The not-first-possibly-false claims
augmenting the description refer to
a unit fraction, whether visible or dark.

Let's suppose being-undefinable
to be relevant to these facts equally true
of the definable and the undefinable.

It seems as though you've stopped
_just before_ you let us in on the
secret of _how it's relevant_

How is it relevant?

> It is simple to see:
> Every eps that you can choose
> contains almost all unit fractions.

That does not have the consequences
to which you feel entitled.

All 1×1 1-ended sets are the same size
(same cardinality, can match elements).

Split the unit fractions at eps > 0
m := ⌈1/eps⌉

⟨ ... 1/m ⟩ ⟨ 1/m⁻⁻ ... 1/1 ⟩

⟨ ... 1/m ⟩ is 1×1 1-ended.
It has the same cardinality as
1×1 1-ended UF = ⟨ ... 1/1 ⟩

----
Consider the two splits
⟨ ... 1/m⁺⁺ 1/m ⟩ ⟨ 1/m⁻⁻ ... 1/1 ⟩
⟨ ... 1/m⁺⁺ ⟩ ⟨ 1/m 1/m⁻⁻ ... 1/1 ⟩
1×1 1-ended, 1×1 2-ended
1×1 1-ended, 1×1 2-ended

|⟨ ... 1/m ⟩| =
|⟨ ... 1/m⁺⁺ ⟩|

|⟨ 1/m⁻⁻ ... 1/1 ⟩| <
|⟨ 1/m ... 1/1 ⟩|

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Donnerstag, 13. April 2023 um 02:44:18 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>
>> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 21:43:13 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >>
>> >> > Ben Bacarisse schrieb am Mittwoch, 12. April 2023 um 17:11:27 UTC+2:
>> >> >> WM <askas...@gmail.com> writes:
>> >> >> > NUF(x), the Number of Unit Fractions between 0 and x, increases in (0,
>> >> >> > 1] from 0 to ℵo.
>> >> >> No. NUF(x) is constant on (0,1].
>> >> >
>> >> > Wrong.
>> >> NUF(x) = ℵo for x > 0. NUF(x) has the same value over the whole
>> >> interval (0,1]. It's hard to see anything you can be more obviously
>> >> wrong about.
>> >
>> > Matheology may say so.
>> What's that? Mathematics says that NUF(x) = ℵo for x > 0.
>
> Yes, for every definable x, usually called eps, this is true.

Speculation about something in (0,1] for which NUF(x) =/= ℵo can be
ruled out because we know what kind of thing in is (0,1] and not one of
those things has a finite set of unit fractions between 0 and it. We
know this by proving it.

If you were a mathematician, you would formally characterise everything
in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀. I don't think you
can even do the first part of that. You certainly can't do the second
because it's not true.

> But mathematics says also that all unit fractions sit at different
> places: ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

Yes.

> and can appear in NUF only one by one.

No. Infinitely many of them are in (0,x) for every x > 0. They don't
appear or disappear, they don't come or go. You have invented a
metaphorical view of them "appearing" which has confused you and which,
sadly, you will probably use to try to trick your students.

> Therefore there is only one way out of the apparent contradiction:
> Dark numbers.

We can define the basic properties possessed of every thing in (0,1]
(see the axioms of the reals). Nothing else is in (0,1]. From those
basic properties we can prove x in (0,1] implies NUF(x) = ℵo. That's
how mathematics works. There isn't anything else in (0,1]. No unit
fractions appear anywhere. Infinite unit fractions either are or are
not in (0,x).

>> The step change happens in WMaths too. UF(x) (the set of unit fractions
>> between 0 and x) is empty when x = 0 and not finite for all x > 0. How
>> is this possible when "never more than one unit fractions is added at a
>> point"?
>
> It is not possible.

So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n. Amazing. Can you
show me such an x or is too dark to see? Or are you relying on a
failure of logic in WMaths where ~∃x in S: P(x) does not imply ∀x in S:
P(x)? You've had to rely on that nonsense in the past.

--
Ben.

[Fritz Feldhase]

On Friday, April 14, 2023 at 1:00:34 AM UTC+2, Ben Bacarisse wrote:

> WM <askas...@gmail.com> writes:
> >
> > But mathematics says also that all unit fractions sit at different
> > places: ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> >
> Yes.
> >
> > and can appear in NUF only one by one.
> >
> No. Infinitely many of them are in (0,x) for every x > 0. They don't
> appear or disappear, they don't come or go. You have invented a
> metaphorical view of them "appearing" which has confused you and which,
> sadly, you will probably use to try to trick your students.

Sadly, indeed!

Actually, a very good comment by you. Imho he invented a metaphorical view of set theory (or math) in general. Numbers are visible or dark, they can be found (or not), etc. etc.

> > Therefore there is only one way out of the apparent contradiction:
> > Dark numbers.

*sigh*

"[WM’s] conclusions are based on the sloppiness of his notions, his inability of giving precise definitions, his fundamental misunderstanding of elementary mathematical concepts, and sometimes, as the late Dik Winter remarked [...], on nothing at all."

> We can define the basic properties possessed of every thing in (0,1]
> (see the axioms of the reals). Nothing else is in (0,1]. From those
> basic properties we can prove x in (0,1] implies NUF(x) = ℵo. That's
> how mathematics works. There isn't anything else in (0,1]. No unit
> fractions appear anywhere. Infinite unit fractions either are or are
> not in (0,x).

Did you intend to write "Infinitely many unit fractions either are or are not in (0,x)"?

> So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n. Amazing. Can you
> show me such an x or is too dark to see? Or are you relying on a
> failure of logic in WMaths where ~∃x in S: P(x) does not imply ∀x in S:
> P(x)? You've had to rely on that nonsense in the past.

Did you intend to write "where ~∃x in S: ~P(x) does not imply ∀x in S: P(x)" (or: "where ~∃x in S: P(x) does not imply ∀x in S: ~P(x)")?

[WM]

Jim Burns schrieb am Freitag, 14. April 2023 um 00:51:03 UTC+2:
> On 4/13/2023 12:48 PM, WM wrote:

> > But
> > ∀n ∈ ℕ:
> > 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > shows that never two unit fractions sit
> > at the same place.
> > Therefore they cannot be
> > included in NUF(x) simultaneously.
> > A first one must start.
> No.
> "A first one must start" _in a finite list_
> However,
> the unit fractions are not a finite list.

The real numbers in (0, 1] are all finite. All unit fractions are finite and sit at points of the real axis.

>
> Each unit fraction u,
> whether u is dark or visible,
> is preceded by another unit fraction v

No, that is only valid for visible unit fractions. Dark unit fractions have no discernible order.
It cannot be valid, because there is 0 as an end.

> Each unit fraction u,
> whether u is dark or visible,
> is _not first_

There must be a first because at 0 there is none and afterwards unit fractions come only as singles.

Regards, WM

[Julio Di Egidio]

On Friday, 14 April 2023 at 16:15:04 UTC+2, WM wrote:
> Jim Burns schrieb am Freitag, 14. April 2023 um 00:51:03 UTC+2:

> > the unit fractions are not a finite list.
>
> The real numbers in (0, 1] are all finite.

LOL. Liar.

> There must be a first because at 0 there is none and
> afterwards unit fractions come only as singles.

Back to square one: there is no first *unit fraction*
as long as there is no last natural number.

What's a sequence, WM?

Liar and spammer.

Julio

[WM]

Ben Bacarisse schrieb am Freitag, 14. April 2023 um 01:00:34 UTC+2:

> WM <askas...@gmail.com> writes:

> >> What's that? Mathematics says that NUF(x) = ℵo for x > 0.
> >
> > Yes, for every definable x, usually called eps, this is true.
> Speculation about something in (0,1] for which NUF(x) =/= ℵo can be
> ruled out

No. ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 holds for all n. Every 1/n is a single point in (0, 1].
Therefore there is a first one. If you disregard mathematics, you should be exorcized from teaching institutions.

> If you were a mathematician, you would formally characterise everything
> in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.

Simple. If ℵ₀ unit fractions are smaller than all x in (0,1], then then you cannot distinguish them. If two of them, p and q, are distinct, then you have p > q > 0 and p is not in the set of ℵ₀ unit fractions smaller than every x > 0.

> > But mathematics says also that all unit fractions sit at different
> > places: ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> Yes.
> > and can appear in NUF only one by one.
> No.

Yes, that is the result of appearing one by one.

> Infinitely many of them are in (0,x) for every x > 0.

How do you distinguish them? By size? Then they sit at different positions x. If you could distinguish all, then you would get a last one. But that is impossible. ℵo cannot be distinguished, whether or not you accept mathematics all unit fractions at distances 1/(n(n+1)) or whether you believe in mathemaology where ℵo appear before every x > 0.

> They don't
> appear or disappear, they don't come or go.

They appear with increasing x.

> You have invented a
> metaphorical view

I have invented a mathematical function NUF(x).

> > Therefore there is only one way out of the apparent contradiction:
> > Dark numbers.
> We can define the basic properties possessed of every thing in (0,1]
> (see the axioms of the reals). Nothing else is in (0,1]. From those
> basic properties we can prove x in (0,1] implies NUF(x) = ℵo. That's
> how mathematics works.

But you cannot distinguish ℵo unit fractios which sit before every x in (0,1].

> Infinite unit fractions either are or are
> not in (0,x).

Nonsense. The function NUF(x) starts with 0 at 0 and increases where every unit fraction is met at a single point.

> >> The step change happens in WMaths too. UF(x) (the set of unit fractions
> >> between 0 and x) is empty when x = 0 and not finite for all x > 0. How
> >> is this possible when "never more than one unit fractions is added at a
> >> point"?
> >
> > It is not possible.

> So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n.

Yes, it is dictated by the universal quantifier in ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
Nut the first points are dark.

> Amazing. Can you
> show me such an x or is too dark to see?

I can show you mathematics if you were unable to see it yet:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

All unit fractions have distances. They come as singles.
But even your coarse vison needs undistinguishable unit fractions, namely those ℵo sitting before (0, 1]. You see there ℵo, don't you? Otherwise the function NUF(x) would not be constant over (0, 1]. Can you distingusih them by size?

Will you really sacrifice basic mathematics to maintain the nonsense of set theory that meanwhile every third-class pupil can refute???

Regards, WM

[WM]

Julio Di Egidio schrieb am Freitag, 14. April 2023 um 16:27:29 UTC+2:
> On Friday, 14 April 2023 at 16:15:04 UTC+2, WM wrote:
> > Jim Burns schrieb am Freitag, 14. April 2023 um 00:51:03 UTC+2:
>
> > > the unit fractions are not a finite list.
> >
> > The real numbers in (0, 1] are all finite.
> LOL.

What is an infinite one?

> > There must be a first because at 0 there is none and
> > afterwards unit fractions come only as singles.
> Back to square one: there is no first *unit fraction*
> as long as there is no last natural number.

Right. But there is a first unit fraction, because they come as singles with distances:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

Regards, WM

[Julio Di Egidio]

On Friday, 14 April 2023 at 16:39:53 UTC+2, WM wrote:
> Julio Di Egidio schrieb am Freitag, 14. April 2023 um 16:27:29 UTC+2:
> > On Friday, 14 April 2023 at 16:15:04 UTC+2, WM wrote:

> > > There must be a first because at 0 there is none and
> > > afterwards unit fractions come only as singles.
> >
> > Back to square one: there is no first *unit fraction*
> > as long as there is no last natural number.
>
> Right. But there is a first unit fraction,

WRONG, there is NO first *unit fraction*.

Liar and spammer. Ad nauseam.
The only mistery is how the flies find
your stupid bullshit worth debunking
and CANNOT EVEN MANAGE THAT MUCH!

Go fuck yourself/yourselves.

*Plonk*

Julio

[WM]

Julio Di Egidio schrieb am Freitag, 14. April 2023 um 17:36:44 UTC+2:
> On Friday, 14 April 2023 at 16:39:53 UTC+2, WM wrote:
> > Julio Di Egidio schrieb am Freitag, 14. April 2023 um 16:27:29 UTC+2:
> > > On Friday, 14 April 2023 at 16:15:04 UTC+2, WM wrote:
>
> > > > There must be a first because at 0 there is none and
> > > > afterwards unit fractions come only as singles.
> > >
> > > Back to square one: there is no first *unit fraction*
> > > as long as there is no last natural number.
> >
> > Right. But there is a first unit fraction,
> WRONG, there is NO first *unit fraction*.

In part you are right. We cannot find the first one. It is dark. Fact however is that by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

we can prove that all unit fractions sit at isolated places on the real axis. Can you comprehend that? For all 1/n: 1/n sits at its own place. There appear ℵo after the origin at zero. Since never more than one appears at a point, there is a first one. To claim that ℵo appear simultaneously shows outrageous stupidity or being stultified by set theory.

Regards, WM

[Julio Di Egidio]

On Friday, 14 April 2023 at 18:38:05 UTC+2, WM wrote:
> Julio Di Egidio schrieb am Freitag, 14. April 2023 um 17:36:44 UTC+2:
> > On Friday, 14 April 2023 at 16:39:53 UTC+2, WM wrote:
> > > Julio Di Egidio schrieb am Freitag, 14. April 2023 um 16:27:29 UTC+2:
> > > > On Friday, 14 April 2023 at 16:15:04 UTC+2, WM wrote:
> >
> > > > > There must be a first because at 0 there is none and
> > > > > afterwards unit fractions come only as singles.
> > > >
> > > > Back to square one: there is no first *unit fraction*
> > > > as long as there is no last natural number.
> > >
> > > Right. But there is a first unit fraction,
> >
> > WRONG, there is NO first *unit fraction*.
>
> In part you are right. We cannot find the first one. It is dark.

There is NOTHING DARK about it,
you piece of lying shit, either you take
the limit of that sequence or you don't.

Stop lying.

Julio

[WM]

Julio Di Egidio schrieb am Freitag, 14. April 2023 um 18:45:18 UTC+2:
> On Friday, 14 April 2023 at 18:38:05 UTC+2, WM wrote:

> > In part you are right. We cannot find the first one. It is dark.

> either you take
> the limit of that sequence or you don't.

I don't. The limit is only useful when ignoring details. I can only repeat for those intelligent enough to understand my advanced result:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

shows that never two unit fractions sit at the same place. Therefore they have distances. The limit of these distances is zero, but no unit fractions has zero distance from its neighbour or from the origin. NUF(x) can only increase one by one. That is mathematics. I cannot change it. Neither can you.

Regards, WM

[Julio Di Egidio]

On Friday, 14 April 2023 at 18:56:36 UTC+2, WM wrote:
> Julio Di Egidio schrieb am Freitag, 14. April 2023 um 18:45:18 UTC+2:
> > On Friday, 14 April 2023 at 18:38:05 UTC+2, WM wrote:
> >
> > > In part you are right. We cannot find the first one. It is dark.
> >
> > either you take the limit of that sequence or you don't.
>
> I don't.

Then STFU.

Julio

[Ben Bacarisse]

Fritz Feldhase <franz.fri...@gmail.com> writes:

> On Friday, April 14, 2023 at 1:00:34 AM UTC+2, Ben Bacarisse wrote:

>> We can define the basic properties possessed of every thing in (0,1]
>> (see the axioms of the reals). Nothing else is in (0,1]. From those
>> basic properties we can prove x in (0,1] implies NUF(x) = ℵo. That's
>> how mathematics works. There isn't anything else in (0,1]. No unit
>> fractions appear anywhere. Infinite unit fractions either are or are
>> not in (0,x).
>
> Did you intend to write "Infinitely many unit fractions either are or
> are not in (0,x)"?

Yes, thanks.

>> So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n. Amazing. Can you
>> show me such an x or is too dark to see? Or are you relying on a
>> failure of logic in WMaths where ~∃x in S: P(x) does not imply ∀x in S:
>> P(x)? You've had to rely on that nonsense in the past.
>
> Did you intend to write "where ~∃x in S: ~P(x) does not imply ∀x in S:
> P(x)" (or: "where ~∃x in S: P(x) does not imply ∀x in S: ~P(x)")?

Indeed I did. I must stop posting late at night. In fact, I must stop
posting altogether!

Because one of WM's word games involves two meaning for "all" I once
(many years ago now) had an exchange with him where I used only ∃x and
~∃x to avoid making any claims about "all" this or that. Because he got
stuck, WM had to eventually deny the standard relationship between ∃ and
∀. Having denied it, he then walked that back saying it applied in some
cases but, of course, only he knew which cases!

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Freitag, 14. April 2023 um 01:00:34 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> >> What's that? Mathematics says that NUF(x) = ℵo for x > 0.
>> >
>> > Yes, for every definable x, usually called eps, this is true.
>> Speculation about something in (0,1] for which NUF(x) =/= ℵo can be
>> ruled out
>
> No. ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 holds for all n. Every 1/n
> is a single point in (0, 1]. Therefore there is a first one.

You have been spoiled by your position. You get to tell your general
studies students anything you like, they write it down, and then they
churn it out back to in the exam. You have grown used to students who
accept proof by assertion -- "Therefore there is a first one".

Walk up the road to the university and see if it works there. You need
only those two lines, though "first" needs to have the context
explained. Do let us know how it goes.

On the other hand, it is trivial to prove (you know, really prove) that
~∃x in (0,1] with NUF(x) =/= ℵo.

> If you disregard mathematics, you should be exorcized from teaching
> institutions.

True.

>> If you were a mathematician, you would formally characterise everything
>> in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.
>
> Simple.

Ah. Good...

> If ℵ₀ unit fractions are smaller than all x in (0,1],

But no unit fractions are smaller than all x in (0,1]. The only real
that is smaller than all x in (0,1] is zero and, as you know, that is
not a unit fraction.

So, as expected, you did not characterise everything in (0,1] and prove
∃x in (0,1] with NUF(e) =/= ℵ₀. You can't.

> then then you
> cannot distinguish them. If two of them, p and q, are distinct, then
> you have p > q > 0 and p is not in the set of ℵ₀ unit fractions
> smaller than every x > 0.

No unit fraction is smaller than every x > 0. Do your general studies
students really fall for this stuff? I suppose they must. They have no
intellectual armour against it, and there is no institutional protection
for them either.

>> So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n.
>
> Yes,

That's a great result. If you could prove it (you know, really prove
it) it would be only the third theorem that distinguishes WMaths from
proper mathematics. The other two (also unproven) are the existence of
sets P and E with E in P and P \ {E} = P, and the existence of a bounded
monotonic sequence of rationals that does not converge in R.

That last one (the Bacarisse-Mueckenheim theorem) was so long ago, I'd
nearly forgotten it, but I seem to have a knack for finding WMaths
theorems for you.

>> Amazing. Can you
>> show me such an x or is too dark to see?
>
> I can show you mathematics if you were unable to see it yet:
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> All unit fractions have distances. They come as singles.

But, no, it seems you can't show such an x. Do you even have some bound
on x? It is less the 1/2? Is it less than 1/10^100? It is really,
really, really small? It would be shame if all you knew is that WMaths
predicts it, but no idea where it was (other than being in (0,1] of
course).

Why waste your time here? Get on with these proofs or at least
establish some bounds for this mysterious x in (0,1] with |UF(x)| in N.
Surely that's a better use of your time?

--
Ben.

[Fritz Feldhase]

On Friday, April 14, 2023 at 10:27:58 PM UTC+2, Ben Bacarisse wrote:

> Indeed I did. I must stop posting late at night. In fact, I must stop
> posting altogether!

C'mon, man, your postings are like pearls in shit, you know.

> Because one of WM's word games involves two meaning for "all" [...]

Right. [...]

Keep up the good work, man!

[Fritz Feldhase]

On Saturday, April 15, 2023 at 2:17:05 AM UTC+2, Ben Bacarisse wrote:
> WM <askas...@gmail.com> writes:
> >
> > If [one] disregard[s] mathematics, [one] should be exorcized from teaching
> > institutions.
> >
> True.

Yeah. Not so in germany, though.

Hint: Germany has changed to an intellectual, political and cultural shit hole.

[Julio Di Egidio]

On Saturday, 15 April 2023 at 04:25:47 UTC+2, Fritz Feldhase wrote:
> On Friday, April 14, 2023 at 10:27:58 PM UTC+2, Ben Bacarisse wrote:
>
> > Indeed I did. I must stop posting late at night. In fact, I must stop
> > posting altogether!
>
> C'mon, man, your postings are like pearls in shit, you know.

Two pieces of polluting nazi-retarded shit complimenting each other.

ESAD, both of you.

Julio

[WM]

Ben Bacarisse schrieb am Freitag, 14. April 2023 um 22:27:58 UTC+2:

> Because one of WM's word games involves two meaning for "all"

Instead of words use mathematics. Simply answer the question: Does

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

enforce the function NUF(x), the number of unit fractions between 0 and x, to increase from NUF(0) = 0 in steps of height 1? Or does NUF(x) increase in a step of height ℵo?

Regards, WM

[WM]

Ben Bacarisse schrieb am Samstag, 15. April 2023 um 02:17:05 UTC+2:

> WM <askas...@gmail.com> writes:

> > I can show you mathematics if you were unable to see it yet:
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > All unit fractions have distances. They come as singles.
> But, no, it seems you can't show such an x.

Of course not. All unit fractions which start the game are dark - like all natural numbers beyond those definable
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
and like all finite endsegments.

> Do you even have some bound
> on x?

There is not a bound on the definable natural numbers. With n also n^n^n is definable. That makes things so difficult to teach.

Nevertheless I asked the deceisive question already in the other post: Does the function NUF(x) increase one by one in steps of height 1 or does it explode from 0 to ℵo suddenly. And if so, where does this happen?

And if it happens at a singl point, how do you distinguish the unit fractions with same 1/n = x.

Regards, WM

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Samstag, 15. April 2023 um 02:17:05 UTC+2:
>> WM <askas...@gmail.com> writes:

... Oh, I see you cut your mistake. It would show some courage
had you left it and agreed that it was a mistake. It's OK to be wrong
(I am often wrong -- check my post just yesterday) but it's shameful to
be intellectually dishonest.

For the record, your argument started with a false premise:

You: "If ℵ₀ unit fractions are smaller than all x in (0,1], ..."

Me: "But no unit fractions are smaller than all x in (0,1]. The only real

that is smaller than all x in (0,1] is zero and, as you know, that is
not a unit fraction."

You went on to repeat the error:

You: ".. then then you cannot distinguish them. If two of them, p and

q, are distinct, then you have p > q > 0 and p is not in the set of
ℵ₀ unit fractions smaller than every x > 0."

Me: "No unit fraction is smaller than every x > 0."

>> > I can show you mathematics if you were unable to see it yet:
>> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
>> > All unit fractions have distances. They come as singles.
>> But, no, it seems you can't show such an x.
>
> Of course not. All unit fractions which start the game are dark

More dishonesty. This part was about WMaths. You cut from this reply
my asking:

Me: "So in WMaths ∃x in (0,1] and n in N with |UF(x)| = n."

To which you replied: "Yes" -- an extraordinary claim. That
extraordinary claim was the context of my saying that you can't show
such an x.

So does WMaths now have dark numbers or have you abandoned all pretence
of intellectual honesty, switching away from WMaths when the questions
get hard?

To summarise (unless you want to retract C), these are the three great
(as yet unproven) theorems WMaths:

A. There are sets P and E such that E ∈ P and P \ {E} = P.

B. There is a bounded monotonic sequence of rationals that does not
converge to a limit in R.

C. ∃x in (0,1] with |UF(x)| ∈ N. (NF(x) = {1/n < x with n ∈ N}.)

And I see no word of thanks for my helping to find these. I can't help
with the proofs, of course, because you told me that you can't yet
define set membership, equality and difference in WMaths. Apparently
the definitions in your textbook are simplifications for beginners.

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

Why do you keep removing the attribution? Don't you want your students
to find these threads?

> Ben Bacarisse schrieb am Freitag, 14. April 2023 um 22:27:58 UTC+2:
>
>> Because one of WM's word games involves two meaning for "all"
>
> Instead of words use mathematics.

I wish you would.

> Simply answer the question: Does
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> enforce the function NUF(x), the number of unit fractions between 0
> and x, to increase from NUF(0) = 0 in steps of height 1?

No.

> Or does NUF(x) increase in a step of height ℵo?

No. ℵo is not a "step height". Instead of bad metaphors, use
mathematics:

NUF(x) = 0 iff x in (-oo, 1]
NUM(x) = ℵ₀ iff x in (0, 1]

No need for rhetorical words, just mathematics. You will call it what
you like, but you /must/ avoid simply stating the facts as above because
the mathematical facts do not permit you to play your word games.

If you want to reclaim some integrity, ask someone to prove that NUF is
constant in (0,1] (or do it yourself if you think you are up to it) and
then come back and admit you should not have said

"NUF(x), the Number of Unit Fractions between 0 and x, increases in
(0,1] from 0 to ℵo."

It can't "increase" in that interval because it's constant. It can't go
"from 0" in (0,1] because it is not zero at any point in (0,1]. You've
re-worded that nonsense now because you know that it was wrong, but an
honest person would admit that before trying a new game.

--
Ben.

[WM]

Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:28:11 UTC+2:

> WM <askas...@gmail.com> writes:

> For the record, your argument started with a false premise:
>
> You: "If ℵ₀ unit fractions are smaller than all x in (0,1], ..."
>
> Me: "But no unit fractions are smaller than all x in (0,1].

Then not ℵ₀ unit fractions are smaller than all x in (0,1]. Fine. Then NUF(x) = ℵ₀ only for an x > 0.

> You: ".. then you cannot distinguish them. If two of them, p and

> q, are distinct, then you have p > q > 0 and p is not in the set of
> ℵ₀ unit fractions smaller than every x > 0."
> Me: "No unit fraction is smaller than every x > 0."

Fine. I got the contrary impression from your arguing. But then some sit at x > 0. That is also what mathematics proves:

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

all are there sitting at different points of the real line.
Alas we cannot find any real point where unit fractions sit as long as there are not yet infinitely many. That proves dark unit fractions. Or do you want again to change your position?

Regards, WM

[WM]

Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:
> WM <askas...@gmail.com> writes:

> > Instead of words use mathematics.

> > Simply answer the question: Does
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > enforce the function NUF(x), the number of unit fractions between 0
> > and x, to increase from NUF(0) = 0 in steps of height 1?
> No.

Then you refuse or ignore mathematics.

> NUF(x) = 0 iff x in (-oo, 0]

> NUM(x) = ℵ₀ iff x in (0, 1]

Then you refuse or ignore mathematics.

> If you want to reclaim some integrity, ask someone to prove that NUF is
> constant in (0,1]

That means ℵo unit fractions are between 0 and (0, 1].

> you should not have said
> "NUF(x), the Number of Unit Fractions between 0 and x, increases in
> (0,1] from 0 to ℵo."

That is your claim because never a finite number of unit fractions is realized. Mathematics proves: Every unit fraction sits at a real point but all are sitting at different points. Therefore the distances 1/(n(n+1)) lie on the real axis.

> It can't "increase" in that interval because it's constant.

Then it must increase before, because NUF(0) = 0. The difference between 0 and ℵo is an infinite increase.

> It can't go
> "from 0" in (0,1] because it is not zero at any point in (0,1].

But it is at most 1 before it is 2.

> You've
> re-worded that nonsense now because you know that it was wrong, but an
> honest person would admit that before trying a new game.

You have answered the question above already. Therefore the twaddle above is useless.

> NUM(x) = ℵ₀ iff x in (0, 1]

The sum (not the limit!) of all distances 1/(n(n+1)) is 1. All lie on the real axis in (0, 1]. Your statement is incompatible with mathematics.

Regards, WM

[Fritz Feldhase]

On Sunday, April 16, 2023 at 2:29:34 AM UTC+2, Ben Bacarisse wrote:
>
> NUF(x) = 0 iff x in (-oo, 1]
> NUM(x) = ℵ₀ iff x in (0, 1]

| NUF(x) = 0 iff x in (-oo, 0]
| NUF(x) = ℵ₀ iff x in (0, 1]

Good post though, as usual. But lost on WM.

[WM]

Fritz Feldhase schrieb am Sonntag, 16. April 2023 um 15:02:50 UTC+2:

> | NUF(x) = 0 iff x in (-oo, 0]
> | NUF(x) = ℵ₀ iff x in (0, 1]
>
> Good post though, as usual. But lost on WM.

Because he faithfully adheres to mathematics.
Mathematics proves that all unit fractions exist at *different* points of the real axis:
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
This implies an increase over more than the difference between 0 and (0, 1].
Only who disregards mathematics can ignore this result.

Regards, WM

[Ben Bacarisse]

Rats! (But thanks) Actually I /intended/ to write

NUF(x) = 0 iff x in (-oo, 0]

NUF(x) = ℵ₀ iff x in (0, oo]

so that NUF is defined on R. So that's 3 typos in two lines.

--
Ben.

[Fritz Feldhase]

On Sunday, April 16, 2023 at 5:50:17 PM UTC+2, Ben Bacarisse wrote:

I'm sorry, but I'd say (in this case) it should read

| NUF(x) = ℵ₀ iff x in (0, oo).

instead of

> NUF(x) = ℵ₀ iff x in (0, oo].

:-P

[Ben Bacarisse]

Oh dear!! You know what? I started with "NUF: R -> {0, ℵ₀}" and
"NUF(x) = 0 iff x <= 0, ℵ₀ otherwise" but I wanted to use intervals
since the rest of the discussion was about intervals; so I re-wrote it.
How hard can it be to split R into (-oo, 0] U (0, oo)? Too hard for me!

Now, how many slip-ups are there in /this/ post?

(To be fair, WM does not typically use typos to score points, so they
rarely derail the discussion.)

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:

>> If you want to reclaim some integrity, ask someone to prove that NUF is
>> constant in (0,1]
>
> That means ℵo unit fractions are between 0 and (0, 1].

No. NUF is provably constant in (0,1], so if that fact entailed the
nonsense that there are any (much less ℵ₀) unit fractions between 0 and
(0,1] you would have a proof of the inconsistency of (most of)
mathematics. That would be the paper of the century, and you'd win
medals an honours beyond imagining. Yet you prefer to post in this
forgotten corner of the Internet. I find that ... telling.

>> you should not have said
>> "NUF(x), the Number of Unit Fractions between 0 and x, increases in
>> (0,1] from 0 to ℵo."
>
> That is your claim

That's dishonest. It was the very first claim in the first post in this
thread. It is a literal quote from a head post you made.

You should not have said it because it is patently wrong. Mathematics
(what you call set theory) shows that NUF(x) is never zero in the
interval (0,1] so saying that it increases "from 0" in (0,1] is wrong.

It is also wrong because NUF(x) is constant in (0,1] so saying that it
increases in (0,1] is absurd.

>> NUM(x) = ℵ₀ iff x in (0, 1]

(aside: NUF is probably defined on R so that "iff" should be "if")

> The sum (not the limit!) of all distances 1/(n(n+1)) is 1. All lie on
> the real axis in (0, 1]. Your statement is incompatible with
> mathematics.

If you could prove that, you would be famous, but none of the
mathematical facts are incompatible with each other. They are, sadly
for you, incompatible with your initial statement in this thread.
That's why it's wrong.

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:28:11 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> For the record, your argument started with a false premise:
>>
>> You: "If ℵ₀ unit fractions are smaller than all x in (0,1], ..."
>>
>> Me: "But no unit fractions are smaller than all x in (0,1].
>
> Then not ℵ₀ unit fractions are smaller than all x in (0,1]. Fine. Then
> NUF(x) = ℵ₀ only for an x > 0.
>
>> You: ".. then you cannot distinguish them. If two of them, p and
>> q, are distinct, then you have p > q > 0 and p is not in the set of
>> ℵ₀ unit fractions smaller than every x > 0."
>> Me: "No unit fraction is smaller than every x > 0."
>
> Fine. I got the contrary impression from your arguing.

So here's the hole you got yourself into and are trying to pretend I
didn't say what I said:

Me: Mathematics says that NUF(x) = ℵo for x > 0.

You: Yes, for every definable x, usually called eps, this is true.

Me: Speculation about something in (0,1] for which NUF(x) =/= ℵo can be
ruled out

You: No.

Me: If you were a mathematician, you would formally characterise everything

in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.

You: Simple.

It is at this point you started an argument with a false premise. So
start over and try again to do this "simple" thing: characterise

everything in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.

I'll wait... Meanwhile, mathematics characterises everything in (0,1]
with axioms and proves that ~∃x in (0,1] with NUF(e) =/= ℵ₀ so if you
can do what you said was "simple" you will have proved conventional
mathematics inconsistent. That must be worth a few minutes typing.

> But then some sit at x > 0.
> That is also what mathematics proves:
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> all are there sitting at different points of the real line.

Yes. You cling to this trivial fact like a drowning man to an oar. But
you can't get from it to "so therefore there is a first" nor to "so
NUF(x) 'increases' by 1" nor to anything that contradicts the equally
trivial fact that NUF(x) is two-valued.

> Alas we cannot find any real point where unit fractions sit as long as
> there are not yet infinitely many.

Use mathematics not words. Your current metaphor soup includes "sit"
(equals?) "as long as" (nothing changes in my scenario) "not yet" (time
has nothing to do with it.

> That proves dark unit fractions.

You can't even say what a dark number is much less prove anything about
one. Meanwhile, in mathematics, we characterise the set (0,1] (see the
axioms of the reals) and prove that for any x in (0,1], NUF(x) = ℵ₀.
It's a simple proof. There is nothing else in (0,1] other than the
reals for which NUF is ℵ₀. This is why you will fail to do what you
said was "simple" (see above).

> Or do you want again to change your position?

Please cite the two positions of mine that you think are different.
Yes, I know, you never comply with requests to back up these claims,
but I feel I must ask in case you have some evidence.

Now let's address your other embarrassing claim: the third of the great
unproven (and never to be proven) "theorems" of WMaths:

∃x in (0,1] with |UF(x)| ∈ N. (NF(x) = {1/n < x with n ∈ N}.)

You say you can't show such a x. You say you can't put any bounds on
such an x. But neither can you prove that such as x exists.
Nonetheless you make the claim. Do you want again to change your
position? I need to know if I can cite it along with the other junk
"theorems" of WMaths.

--
Ben.

[WM]

Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:38 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:
>
> >> If you want to reclaim some integrity, ask someone to prove that NUF is
> >> constant in (0,1]
> >
> > That means ℵo unit fractions are between 0 and (0, 1].
> No. NUF is provably constant in (0,1],

Then there must be ℵ₀ unit fractions before the interval. That means they cannot be discerned. They are dark.

> so if that fact entailed the
> nonsense that there are any (much less ℵ₀) unit fractions between 0 and
> (0,1] you would have a proof of the inconsistency of (most of)
> mathematics.

Mathematics has nothing in common with set theoty.

> That would be the paper of the century, and you'd win
> medals an honours beyond imagining. Yet you prefer to post in this
> forgotten corner of the Internet. I find that ... telling.

The reason is simple. Matheologians don't want to be unmasked as charlatans.

> Mathematics
> (what you call set theory) shows that NUF(x) is never zero in the
> interval (0,1] so saying that it increases "from 0" in (0,1] is wrong.

In increases in the interval [0, 1] from 0 to ℵ₀ . Since 0 is not unit fraction, it increases in the interval (0, 1] from 0 to ℵ₀.

>
> It is also wrong because NUF(x) is constant in (0,1] so saying that it
> increases in (0,1] is absurd.

If there are ℵ₀ unit fractions in the interval, then there are at least two unit fractions. They have a distance. Therefore one of them does not contribute to NUF(x) = ℵ₀ over the complete interval. This arguments however holds for all unit fractions.

> >> NUM(x) = ℵ₀ iff x in (0, 1]
> (aside: NUF is probably defined on R so that "iff" should be "if")
> > The sum (not the limit!) of all distances 1/(n(n+1)) is 1. All lie on
> > the real axis in (0, 1]. Your statement is incompatible with
> > mathematics.
> If you could prove that

I did. ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0. Therefore at most one count of a unit fraction could be present over the complete interval (0, 1].

You claim NUM(x) = ℵ₀ over the whole interval (0, 1]. Therefore at least one unit fraction must be counted. The next one can only be counted after the distance 1/(n(n+1)) > 0.

Regards, WM

[WM]

Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:42 UTC+2:

> WM <askas...@gmail.com> writes:

> It is at this point you started an argument with a false premise.

You said: Speculation about something in (0,1] for which NUF(x) =/= ℵo can be ruled out

I said: If ℵ₀ unit fractions are smaller than all x in (0,1], then you cannot distinguish them.

But if you now want to withdraw this statement, then NUF(x) is not constant over the whole interval.

> So
> start over and try again to do this "simple" thing: characterise
> everything in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.

Impossible, because these things are dark. But the existence is proven by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

If there is a unit fraction at the beginning of the interval, then the next one has the distance 1/(n(n+1)) and sits inside nthe interval.

> I'll wait... Meanwhile, mathematics characterises everything in (0,1]
> with axioms and proves that ~∃x in (0,1] with NUF(e) =/= ℵ₀ so if you
> can do what you said was "simple" you will have proved conventional
> mathematics inconsistent. That must be worth a few minutes typing.

If there are NUF = ℵ₀ unit fractions over the whole interval, then you cannot distinguish them by different x. They are dark.

But mathematics says, if there are ℵ₀ unit fractions ate the beginning of the interval, then there is at least one unit fraction. But if there is any unit fraction at the beginning of the interval, then by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

the next one has the distance 1/(n(n+1)) and sits inside the interval.

> > But then some sit at x > 0.
> > That is also what mathematics proves:
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > all are there sitting at different points of the real line.
> Yes. You cling to this trivial fact like a drowning man to an oar. But
> you can't get from it to "so therefore there is a first"

Why should I? It is clear that if there are many at the beginning of the interval, then there is at least one at the beginning of the interval. Is you mathematics unable to undertstand this simple estimation?

> nor to "so
> NUF(x) 'increases' by 1"

What about SUF(x), the Set of Unit Fractions between 0 and x? Is it constant too?

Regards, WM

[Fritz Feldhase]

On Monday, April 17, 2023 at 9:23:44 PM UTC+2, WM wrote:

> What about SUF(x), the Set of Unit Fractions between 0 and x? Is it constant too?

No, it's not constant on (0, 1].

Hint: SUF(x) = {u e U : u <= x} with U = {1/n : n e IN}.

Hence 1/1 e SUF(1) and, say, 1/1 !e SUF(1/2).

But card SUF(1) = card SUF(1/2) = aleph_0.

Actually, Ax e (0, 1]: card SUF(x) = aleph_0.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:42 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> It is at this point you started an argument with a false premise.
>
> You said: Speculation about something in (0,1] for which NUF(x) =/= ℵo
> can be ruled out

And I explained why.

> I said: If ℵ₀ unit fractions are smaller than all x in (0,1], then
> you cannot distinguish them.

If there is a largest prime, I can't identify it. If there are real
solutions to x^2 = -1 then I cannot distinguish them.

You know that there are no unit fractions smaller than all x in (0,1]
and I know that there are no unit fractions smaller than all x in (0,1].
Everyone knows that there are no unit fractions smaller than all x in
(0,1]. What is it you think you are showing by stating that I can't
"distinguish" between non-existent unit fractions?

(This is a word game I'll play because although "distinguish" is one of
your magic, undefined "get out of jail free" words, I can agree to any
statement about the members of an empty set!)

> But if you now want to withdraw this statement, then NUF(x) is not
> constant over the whole interval.

Specifically, NUF(x) = ℵ₀ for every x in (0,1] (the set of reals >0 and
<=1). This is trivial to prove. For any real x in (0,1], let k =
floor(1/x) + 1 so 1/k is in (0,x). If 1/k is in (0,x) so are 1/(k+1),
1/(k+2), ... and therefore

f(1/k+n) = n (for n in N)

is a bijection between N and the unit fractions less than x. Hence
NUF(x) = ℵ₀.

>> So
>> start over and try again to do this "simple" thing: characterise
>> everything in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.
>
> Impossible,

So you are withdrawing the claim that it's simple to do what I asked? I
presume you find the "characterise everything in (0,1]" as the
impossible bit because that would give the whole game away. You'd have
to admit that everything in (0,1] is a real to which the above argument
applies.

> because these things are dark. But the existence is proven by

But if you can't do both parts: (a) characterise everything in (0,1] and
(b) prove ∃x in (0,1] with NUF(e) =/= ℵ₀, you have nothing. If you
don't do (a) there might be magic goblins in (0,1]. There might be
complex numbers in (0,1]. Without (a) and (b) you are just waffling.

> What about SUF(x), the Set of Unit Fractions between 0 and x? Is it
> constant too?

Obviously not. Why would you even ask? Are you planning to change the
subject because you've got stuck again?

--
Ben.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:38 UTC+2:
>> WM <askas...@gmail.com> writes:
>> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:
>>
>> >> If you want to reclaim some integrity, ask someone to prove that NUF is
>> >> constant in (0,1]
>> >
>> > That means ℵo unit fractions are between 0 and (0, 1].
>> No. NUF is provably constant in (0,1],
>
> Then there must be ℵ₀ unit fractions before the interval.

Are you seriously suggesting that there are unit fractions <= 0? Not
only does it seem like you are suggesting their existence, you are
suggesting that somehow they, magically, affect the cardinality of a set
they are not members of!

I think you are now just saying anything, no matter how daft.

The facts: NUF is a two-valued function, zero for x <= 0 and ℵ₀
otherwise. Functions with a discontinuity like this are nothing special
and sets so dense in the reals that the "number of them" in (0,x] is
infinite for all x > 0 are also common. All your arguments apply to
these sets as well. The positive rationals are all "at distinct points"
and have non-zero distances from each other yet NPR(x) = ℵ₀ for x > 0.
The positive irrationals are all "at distinct points" and have non-zero
distances from each other yet NPI(x) = c for x > 0...

> That means they cannot be discerned. They are dark.

They are not even unit fractions, so any description of them can be
taken to be vacuously true: the unit fractions that come before (0,1]
are all green. Yes! The reciprocals of the unit fractions that come
before (0,1] are all even primes. Yes!

>> That would be the paper of the century, and you'd win
>> medals an honours beyond imagining. Yet you prefer to post in this
>> forgotten corner of the Internet. I find that ... telling.
>
> The reason is simple. Matheologians don't want to be unmasked as
> charlatans.

Ah, you subscribe to the world-wide conspiracy theory. Is this why you
can only teach this stuff in an optional course in a college with no
mathematics program? I ask because, although I am a part of the
world-wide conspiracy, I don't always read the newsletters.

>> Mathematics
>> (what you call set theory) shows that NUF(x) is never zero in the
>> interval (0,1] so saying that it increases "from 0" in (0,1] is wrong.
>
> In increases in the interval [0, 1] from 0 to ℵ₀.

Yes (though the metaphor is a bad one).

> Since 0 is not unit fraction, it increases in the interval (0, 1] from
> 0 to ℵ₀.

No function can increase "from 0" in an interval if it never has the
value 0 in that interval! This is just a misuse of language. You find
the unbounded density of unit fractions near zero to have properties that
run counter to your intuition, but "how odd" does not constitute a proof.

The error above is simply one of words. There's no mathematical issue
here unless you claim that

∃x in (0,1] with NUF(x) = 0?

If so, please do say that. As you tell me: "use mathematics not words".
If you don't claim that, we are in agreement about the facts.

>> It is also wrong because NUF(x) is constant in (0,1] so saying that it
>> increases in (0,1] is absurd.
>
> If there are ℵ₀ unit fractions in the interval, then there are at
> least two unit fractions. They have a distance. Therefore one of them
> does not contribute to NUF(x) = ℵ₀ over the complete interval.

Use mathematics, not words:

Let UF(x) = { 1/n | n in N and 1/n < x }.
∀u1, u2 in UF(1) with u1 =/= u2
∃x in (0,1] with u1 in UF(x) and u2 not in UF(x)
or u2 in UF(x) and u1 not in UF(x)

> This arguments however holds for all unit fractions.

Word game alert: attempted exploitation of double meaning of "all"! Use
mathematics not words. The above holds for all pairs of unit fractions.
If you mean something else, write it property with the quantifiers in
the right order.

NUF(x) is constant in (0,1]. You know what you need to show to prove
otherwise, and you just can't do it.

--
Ben.

[JVR]

On Monday, April 17, 2023 at 9:23:44 PM UTC+2, WM wrote:

And may the Lord bless you and keep you and have pity on you and
disabuse you of the notion that you can do math.

[Julio Di Egidio]

On Tuesday, 18 April 2023 at 13:32:36 UTC+2, JVR wrote:

> And may the Lord bless you and keep you and have pity on you and
> disabuse you of the notion that you can do math.

May the Lord stick a pole our your asses, you
pieces of abusive retarded spamming nazi-shit!!

I have been searching for my seminal posts from the past,
from the re-foundation of mathematics to my declaration
of self-sovereignty at some point in alt.philosophy.

IT'S ALL GONE!! Somebody has been systematically
deleting it all, guess who...

You mortally stupid retarded insane filthy cunts,
I'll call you by your real names till you get livid!!!

-LV

[WM]

Ben Bacarisse schrieb am Dienstag, 18. April 2023 um 03:49:53 UTC+2:

> WM <askas...@gmail.com> writes:

> > You said: Speculation about something in (0,1] for which NUF(x) =/= ℵo
> > can be ruled out

> You know that there are no unit fractions smaller than all x in (0,1]
> and I know that there are no unit fractions smaller than all x in (0,1].
> Everyone knows that there are no unit fractions smaller than all x in
> (0,1].

Then your above statement is wrong. "Something" in (0,1] can be every x > 0.

> > But if you now want to withdraw this statement, then NUF(x) is not
> > constant over the whole interval.
> Specifically, NUF(x) = ℵ₀ for every x in (0,1] (the set of reals >0 and
> <=1). This is trivial to prove. For any real x in (0,1]

Any real x = eps that you can prove something for has many unit fractions in (0, eps). But every real x that is existing has not many unit fractions in (0, x). That is the difference between visible and dark.

> >> So
> >> start over and try again to do this "simple" thing: characterise
> >> everything in (0,1] and prove ∃x in (0,1] with NUF(e) =/= ℵ₀.
> >
> > Impossible,
> So you are withdrawing the claim that it's simple to do what I asked?

Why should I? I use dark numbers, not only your rough epsilontics. For every eps, there are infinitely many smaller distances 1/(n(n+1)) between unit fractions. But if you believe it or not, these distances are real numbers although you cannot define every of them. This is a mathematical proof:

Your claim is NUF(x) is infinite for all x in (0, 1] which you can define. If however this were true for all distances 1/(n(n+1)) then you'd get this contradiction: If there are many unit fractions at the beginning of the interval, then there is also at least one unit fraction 1/n among them. Then by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

the next one, 1/(n+1) sits inside the interval, not at the beginning. But this is true for each of these infinitely many unit fractions. Therefore only one can sit at the beginning before every x. Your claim is true for all eps that you can define, not for all distances between unit fractions.

Regards, WM

[WM]

Ben Bacarisse schrieb am Dienstag, 18. April 2023 um 03:50:18 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
>
> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:38 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:
> >>
> >> >> If you want to reclaim some integrity, ask someone to prove that NUF is
> >> >> constant in (0,1]
> >> >
> >> > That means ℵo unit fractions are between 0 and (0, 1].
> >> No. NUF is provably constant in (0,1],
> >
> > Then there must be ℵ₀ unit fractions before the interval.
> Are you seriously suggesting that there are unit fractions <= 0?

No, you are if you really mean every x in (0, 1]. But I know that you mean only those which cover many distances of unit fractions.

> Functions with a discontinuity like this are nothing special
> and sets so dense in the reals that the "number of them" in (0,x] is
> infinite for all x > 0 are also common.

This idea is wrong. The unit fractions have distances. They are smaller than all the epsilons you can choose.

> The above holds for all pairs of unit fractions.

Therefore not both can sit before every x > 0.

> If you mean something else, write it property with the quantifiers in
> the right order.

Here no quantifier magic applies. The real line is constant. What is true for every x is true for all x.

>
> NUF(x) is constant in (0,1]. You know what you need to show to prove
> otherwise, and you just can't do it.

I did: If there are many unit fractions at the beginning of the interval, then there is also at least one unit fraction 1/n among them. Then by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

the next one, 1/(n+1) sits inside the interval, not at the beginning before every x > 0. But this is true for each of these infinitely many unit fractions. Therefore only one can sit at the beginning before every x > 0.

Regards, WM

[Julio Di Egidio]

On Tuesday, 18 April 2023 at 13:59:00 UTC+2, WM wrote:

> Therefore only one can sit at the beginning

May the Lord stick a pole up your asses, you

[Julio Di Egidio]

On Tuesday, 18 April 2023 at 13:59:00 UTC+2, WM wrote:

> Ben Bacarisse schrieb am Dienstag, 18. April 2023 um 03:49:53 UTC+2:

> Therefore only one can sit

[Mostowski Collapse]

Maybe you had an X-expire header in your posts?
Watch out what tools and what news provider you are using.

LoL

[JVR]

Here we go round the prickly pear
Prickly pear prickly pear
Here we go round the prickly pear
At five o'clock in the morning.

Between the idea
And the reality
Between the motion
And the act
Falls the Shadow
For Thine is the Kingdom

Between the conception
And the creation
Between the emotion
And the response
Falls the Shadow
Life is very long
--- T.S. Eliot

[Ben Bacarisse]

WM <askas...@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Dienstag, 18. April 2023 um 03:50:18 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 23:30:38 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >> > Ben Bacarisse schrieb am Sonntag, 16. April 2023 um 02:29:34 UTC+2:
>> >>
>> >> >> If you want to reclaim some integrity, ask someone to prove that NUF is
>> >> >> constant in (0,1]
>> >> >
>> >> > That means ℵo unit fractions are between 0 and (0, 1].
>> >> No. NUF is provably constant in (0,1],
>> >
>> > Then there must be ℵ₀ unit fractions before the interval.
>> Are you seriously suggesting that there are unit fractions <= 0?
>
> No, you are if you really mean every x in (0, 1].

So you claim (without proof) that NUF is provably constant in (0,1]
implies that there are unit fractions <= 0. You are wrong. The facts
are that

(a) NUF(x) = ℵ₀ for x > 0, and
(b) ~∃n in N with 1/n <= 0.

These are entirely consistent with each other.

> But I know that you mean only those which cover many distances of unit
> fractions.

And chance you could use maths rather than words? No.

>> Functions with a discontinuity like this are nothing special
>> and sets so dense in the reals that the "number of them" in (0,x] is
>> infinite for all x > 0 are also common.
>
> This idea is wrong. The unit fractions have distances. They are
> smaller than all the epsilons you can choose.

Any chance of writing that in mathematics?

>> The above holds for all pairs of unit fractions.
>
> Therefore not both can sit before every x > 0.

Indeed. Not even one can set before every x > 0. You really must start
to write these silly claims in maths. I don't think you know what the
words mean.

>> If you mean something else, write it property with the quantifiers in
>> the right order.
>
> Here no quantifier magic applies. The real line is constant. What is
> true for every x is true for all x.

You are playing quantifier games but either you don't know what the
words you use mean or you are doing it deliberately. There are unit
fractions less than every x > 0 but there are no unit fractions less
that all x > 0. In English, you can't switch every and all without
playing games.

I am sure this is deliberate on your part. Even you could write these
things symbolically, but your argument would evaporate when written
without the word games.

>> NUF(x) is constant in (0,1]. You know what you need to show to prove
>> otherwise, and you just can't do it.
>
> I did:

No you waffled. You need to (a) characterise all the elements in (0,1]
and (b) prove that there is a x in (0,1] with NUF(x) =/= ℵ₀. You can't
do (a) because that would give the game away.

> If there are many unit fractions at the beginning of the interval,
> then there is also at least one unit fraction 1/n among them.

Waffle. What is "at the beginning" of an open interval?

There's no need for this dance about if there are many there is one.
There is a formula for the largest unit fraction less the x.

> Then by
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 the next one, 1/(n+1) sits
> inside the interval,

Yes. For any x in (0,1] 1/n where n = floor(x) + 1 is the largest unit
fraction less that x. The infinite sequence of fractions 1/n, 1/(n+1),
1/(n+2), ... are all in the interval (0,x). That's why NUF(x) = ℵ₀.

> not at the beginning before every x > 0.

Waffle. You just don't have the right mental picture of the open set
and of the set of unit fractions. Only 0 is before every x > 0.
Nothing is "at the beginning" of (0,1] for any reasonable guess about
what the beginning means. You could try mathematics.

> But this
> is true for each of these infinitely many unit fractions. Therefore
> only one can sit at the beginning before every x > 0.

Nothing in (0,1] sits before every x > 0. No unit fraction is the
"first" (i.e. closest to the beginning).

--
Ben.

[JVR]

Here we go around the prickly pear
Prickly pear prickly pear
Here we go round the prickly pear ....

[Ben Bacarisse]

JVR <jrenne...@googlemail.com> writes:

> Here we go around the prickly pear
> Prickly pear prickly pear
> Here we go round the prickly pear ....

Yes. I think I've done enough circles on this one!

--
Ben.

[WM]

Ben Bacarisse schrieb am Freitag, 21. April 2023 um 04:39:14 UTC+2:


The facts
> are that
>
> (a) NUF(x) = ℵ₀ for x > 0, and
> (b) ~∃n in N with 1/n <= 0.
>
> These are entirely consistent with each other.

But not with
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

ℵ₀ unit fractions have ℵ₀ internal positive distances. They cannot all sit at a point. Let the sum of ℵ₀ distances be D. Then
NUF(x) = ℵ₀ for x > D, but not for x < D.

Regards, WM

[JVR]

Is there some magic number of repetitions, after which you expect
that people will no longer see that the daily dose of nonsense
that you post is nonsense?

[WM]

JVR schrieb am Freitag, 21. April 2023 um 17:34:54 UTC+2:
> On Friday, April 21, 2023 at 3:42:37 PM UTC+2, WM wrote:
> > Ben Bacarisse schrieb am Freitag, 21. April 2023 um 04:39:14 UTC+2:
> >
> >
> > The facts
> > > are that
> > >
> > > (a) NUF(x) = ℵ₀ for x > 0, and
> > > (b) ~∃n in N with 1/n <= 0.
> > >
> > > These are entirely consistent with each other.
> > But not with
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> >
> > ℵ₀ unit fractions have ℵ₀ internal positive distances. They cannot all sit at a point. Let the sum of ℵ₀ distances be D. Then
> > NUF(x) = ℵ₀ for x > D, but not for x < D.

> Is there some magic number of repetitions, after which you expect
> that people will no longer see that the daily dose of nonsense
> that you post is nonsense?

It is impossible to have
NUF(x) = ℵ for all x > 0
because
NUF(x) > 1 for all x > 0
is excluded by

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

What don't you understand?

Regards, WM

[JVR]

“When I use a word,’ Humpty Dumpty said in rather a scornful
tone, ‘it means just what I choose it to mean — neither more nor less.’

What I don't understand is why you keep repeating this nonsense
10 times every day.

Do you know what the conjunction 'because' means in English?

[Jim Burns]

On 4/21/2023 3:52 PM, WM wrote:

> It is impossible to have
> NUF(x) = ℵ for all x > 0
> because
> NUF(x) > 1 for all x > 0
> is excluded by
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

∀n ∈ ℕ: 1/n > 1/(n+1)

∀n ∈ ℕ:
x > 1/n ⟹
x > 1/n > 1/(n+1)

NUF(x) > 0 ⟹
NUF(x) > 1

> What don't you understand?

Define ℝ such that
a jumping function is not all-continuous.

Define
f(x) := [NUF(x) > 0]

f() jumps,
thus f() is not all-continuous.

If
x > β ∧ NUF(x) = 0
then
f() is continuous at β

If
y < β ∧ NUF(x) > 0
then
f() is continuous at β

f() jumps,
thus f() is not all-continuous
thus a point β exists such that
x > β ⟹ NUF(x) > 0
y < β ⟹ NUF(x) = 0


If
NUF(2β) > 0
then
2β > 1/m₂ᵦ
(2β)/4 > (1/m₂ᵦ)/4 [!]
β/2 > 1/(4m₂ᵦ)
NUF(β/2) > 0

lemma.
If NUF(2β) > 0
then NUF(β/2) > 0


If β > 0
then
2β > β
NUF(2β) > 0
(and by lemma)
NUF(β/2) > 0

However,
if β > 0
then
β/2 < β
NUF(β/2) = 0
Contradiction.

Therefore, ¬(β > 0)

x > 0 ≥ β ⟹ NUF(x) > 0

[WM]

JVR schrieb am Freitag, 21. April 2023 um 22:08:15 UTC+2:
> On Friday, April 21, 2023 at 9:52:04 PM UTC+2, WM wrote:

> > NUF(x) > 1 for all x > 0
> > is excluded by
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> > What don't you understand?
>

> What I don't understand is why you keep repeating this nonsense
> 10 times every day.

You don't understand that every pair of unit fractions is separated by a finite interval? Or don't you unerstand that this interval subtracted from (0, 1] diminishes the latter? Or is it only that you don't understand why you have accepted set theory?

Regards, WM

[Gus Gassmann]

On Friday, 21 April 2023 at 17:08:15 UTC-3, JVR wrote:
> On Friday, April 21, 2023 at 9:52:04 PM UTC+2, WM wrote:

[...]

> > It is impossible to have
> > NUF(x) = ℵ for all x > 0
> > because
> > NUF(x) > 1 for all x > 0
> > is excluded by
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> > What don't you understand?
>

> What I don't understand is why you keep repeating this nonsense
> 10 times every day.
>

At least he now covers the quantifier switcheroo so he can't even see it any more. Don't know if that is progress or progressive dementia. (My money is on the latter.)

[WM]

Jim Burns schrieb am Freitag, 21. April 2023 um 23:12:43 UTC+2:
> On 4/21/2023 3:52 PM, WM wrote:
>
> > It is impossible to have
> > NUF(x) = ℵ for all x > 0
> > because
> > NUF(x) > 1 for all x > 0
> > is excluded by
> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

So if one unit fraction would sit at zero, then the next one would sit already insidet of (0, 1]. Only NUF(x) > 0 for all x > 0 can be correct.

>
> ∀n ∈ ℕ: 1/n > 1/(n+1)
>
> ∀n ∈ ℕ:
> x > 1/n ⟹
> x > 1/n > 1/(n+1)

That's true for visible numbers. Dark numbers have no discernible order.

Regards, WM

[WM]

Gus Gassmann schrieb am Samstag, 22. April 2023 um 15:48:51 UTC+2:

> At least he now covers the quantifier switcheroo so he can't even see it any more.

If the universal quantifier in

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

deserves its name, then there are all unit fractions and all their internal distances. If one unit fraction sits before all elements of (0, 1] then no other does so.

NUF(x) = ℵo for all x ∈ (0, 1] is a bad mistake.

Regards, WM

[Jim Burns]

On 4/22/2023 9:49 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 21. April 2023 um 23:12:43 UTC+2:
>> On 4/21/2023 3:52 PM, WM wrote:

>>> It is impossible to have
>>> NUF(x) = ℵ for all x > 0
>>> because
>>> NUF(x) > 1 for all x > 0
>>> is excluded by
>>> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
>>
>> ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
>
> So
> if one unit fraction would sit at zero,

A unit fraction which sits at zero
not-exists.

u is a unit fraction :⇔
∃n ∈ ℕ⁺: n*u = 1

However,
¬∃n ∈ ℕ⁺: n*0 = 1
¬(0 is a unit fraction)

> So
> if one unit fraction would sit at zero,
> then the next one would sit already
> insidet of (0, 1].

We don't intentionally reason about
not-existing things,
not beyond learning that they not-exist.

The conclusions reached in that way
are useless.
Yes,
there are derivations, following all
the rules, that the next one
sits inside (0,1]
However,
because the zero unit fraction not-exists
there are also derivations that the next one
sits inside (blue,green] and sits inside
(flying rainbow sparkle pony}.

We spend considerable effort on
questions of existing v. not-existing.
That effort is not wasted if it saves us
from unintentionally relying upon such
useless derivation from the not-existing.

However,
once we've established that it's not-existing,
the only thing left to do is to
find something else to talk about.

> So
> if one unit fraction would sit at zero,
> then the next one would sit already
> insidet of (0, 1].
> Only NUF(x) > 0 for all x > 0
> can be correct.

Yes, but
not because of a zero unit fraction.

| A jumping function of ℝ without
| a point of discontinuity
| not-exists.

| A FISON without another FISON after it
| not-exists.

A point-between β exists between
points x above some unit fraction and
points y not above some unit fraction.

Positive point-between β not-exists.

| For any positive β
| 2β > β
| a unit fraction 1/m₂ᵦ < 2β exists
| thus
| a unit fraction 1/(4m₂ᵦ) < β/2 exists
|
| However.
| β/2 < β
| a unit fraction 1/(4m₂ᵦ) < β/2 not-exists
| Contradiction.

β exists because of what ℝ is
0 ≥ β because 1/(4m₂ᵦ) < β/2
and what ℕ is

> Only NUF(x) > 0 for all x > 0
> can be correct.

Yes,
because

x > 0 ≥ β ⟹ NUF(x) > 0

because
β > 0 ⟹
1/(4m₂ᵦ) < β/2 ∧ ¬(1/(4m₂ᵦ) < β/2)

[Jim Burns]

On 4/22/2023 2:56 PM, WM wrote:
> Gus Gassmann schrieb am Samstag, 22. April 2023 um 15:48:51 UTC+2:

>> At least he now covers the quantifier switcheroo so he can't even see it any more.
>
> If the universal quantifier in
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> deserves its name, then there are all unit fractions and all their internal distances.


> If one unit fraction sits before
> all elements of (0, 1]
> then no other does so.

A unit fraction before all elements of (0,1]
not-exists.

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

> NUF(x) = ℵo for all x ∈ (0, 1]

Define R such that
b exists such that
{NUF(y)=0} ᵋ≤ β ≤ᵋ {NUF(z)>0}

If 1/n ∈ (0,2x]
then 1/(4n) ∈ (0,x/2]

lemma.
If 2x ∈ {NUF(z)>0}
then x/2 ∈ {NUF(z)>0}

If
β > 0

then
2β > β

2β ∈ {NUF(z)>0}
(and by lemma)
β/2 ∈ {NUF(z)>0}

and
β/2 < β
β/2 ∉ {NUF(z)>0}

β > 0 ⟹ β/2 ∈∉ {NUF(z)>0}
0 ≥ β

x > 0 ≥ β ⟹ x ∈ {NUF(z)>0}

> NUF(x) = ℵo for all x ∈ (0, 1]
> is a bad mistake.

Say why it's a mistake.
The more explicit that you are,
the better chance that you have of
spotting an unjustified assumption.

[Fritz Feldhase]

On Saturday, April 22, 2023 at 8:56:12 PM UTC+2, WM wrote:

> If one unit fraction sits before all elements of (0, 1] then <bla>

NO unit fraction "sits before all elements of (0, 1]".

Hint: For each and every unit fraction u there is an x in (0, 1] such that x < u.

[Fritz Feldhase]

Proof: If u is a unit fraction then 1/(1/u + 1) is a real number in (0, 1] which is smaller than u.

[JVR]

No Mücke, what I don't understand is why you keep repeating the same nonsense
10 times every day.
That would be a mystery even if what you keep saying made sense.

[WM]

Jim Burns schrieb am Samstag, 22. April 2023 um 21:24:47 UTC+2:
> On 4/22/2023 9:49 AM, WM wrote:
> > Jim Burns schrieb am Freitag,
> > 21. April 2023 um 23:12:43 UTC+2:
> >> On 4/21/2023 3:52 PM, WM wrote:
>
> >>> It is impossible to have
> >>> NUF(x) = ℵ for all x > 0
> >>> because
> >>> NUF(x) > 1 for all x > 0
> >>> is excluded by
> >>> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
> >>
> >> ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.
> >
> > So
> > if one unit fraction would sit at zero,
> A unit fraction which sits at zero
> not-exists.

It is a worst-case analysis. If none sits at 0, then all reasults are shifted twowards the right-hand side.

>
> > if one unit fraction would sit at zero,
> > then the next one would sit already
> > insidet of (0, 1].
> We don't intentionally reason about
> not-existing things,
> not beyond learning that they not-exist.
>
> The conclusions reached in that way
> are useless.

Estimations like this are often used in mathematics.

> Yes,
> there are derivations, following all
> the rules, that the next one
> sits inside (0,1]
> However,
> because the zero unit fraction not-exists

Simply define 0 = 1/ω.

> > So
> > if one unit fraction would sit at zero,
> > then the next one would sit already
> > insidet of (0, 1].
> > Only NUF(x) > 0 for all x > 0
> > can be correct.
> Yes, but
> not because of a zero unit fraction.

Agreed.

Regards, WM

[WM]

You already have agreed:
************************************

> if one unit fraction would sit at zero,
> then the next one would sit already
> insidet of (0, 1].
> Only NUF(x) > 0 for all x > 0
> can be correct.

JB: Yes, but ...
*************************************

> The more explicit that you are,
> the better chance that you have of
> spotting an unjustified assumption.

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 shows that never unit fractions share a point. Only one can sit at the edge of (0, 1]. The next one is inside. Therefore NUF(x) = ℵ for all x ∈ (0, 1] is impossible.

Regards, WM

[WM]

JVR schrieb am Sonntag, 23. April 2023 um 00:30:55 UTC+2:
> On Saturday, April 22, 2023 at 3:45:34 PM UTC+2, WM wrote:

> > > > NUF(x) > 1 for all x > 0
> > > > is excluded by
> > > > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.

> > You don't understand that every pair of unit fractions is separated by a finite interval? Or don't you unerstand that this interval subtracted from (0, 1] diminishes the latter? Or is it only that you don't understand why you have accepted set theory?
> >

> No

Then try to learn it. Perhaps you will understand more afterwards.

Regards, WM

[WM]

Fritz Feldhase schrieb am Sonntag, 23. April 2023 um 00:07:54 UTC+2:
> On Sunday, April 23, 2023 at 12:03:43 AM UTC+2, Fritz Feldhase wrote:
> > On Saturday, April 22, 2023 at 8:56:12 PM UTC+2, WM wrote:
> > >
> > > If one unit fraction sits before all elements of (0, 1] then <bla>
> > >
> > NO unit fraction "sits before all elements of (0, 1]"

You have claimed more: NUF(x) = ℵ for x ∈ (0, 1].

> >
> > Hint: For each and every unit fraction u there is an x in (0, 1] such that x < u.
> Proof: If u is a unit fraction then 1/(1/u + 1) is a real number in (0, 1] which is smaller than u.

That does not prove your claim NUF(x) = ℵ for x ∈ (0, 1].

Regards, WM

[Jim Burns]

On 4/23/2023 9:30 AM, WM wrote:
> Jim Burns schrieb am Samstag,
> 22. April 2023 um 22:36:26 UTC+2:
>> On 4/22/2023 2:56 PM, WM wrote:

>>> NUF(x) = ℵo for all x ∈ (0, 1]
>>> is a bad mistake.
>>
>> Say why it's a mistake.
>
> You already have agreed:
> ************************************
>> if one unit fraction would sit at zero,
>> then the next one would sit already
>> insidet of (0, 1].
>> Only NUF(x) > 0 for all x > 0
>> can be correct.
>
> JB: Yes, but ...

Both of these are correct:

| Only NUF(x) > 0 for all x > 0

| NUF(x) = ℵo for all x ∈ (0, 1]

A positive lower-bound of unit-fractions
not-exists.

Positive β not-exists between lower-bounds
and not-lower-bounds
because there can't be both
2β > 1/n and ¬(β/2 > 1/(4n))

Thus,
x > 0 ≥ β is a not-bound of unit fractions
x > 0 ⟹ x > 1/n ⟹ NUF(x) > 0


The unit fractions are 1×1 1-ended.

u is a unit fraction ⟺
∃n ∈ ℕ⁺: u*n = 1

ℕ⁺ is 1×1 1-ended and
the unit fractions have the order of ℕ⁺
but reversed.


Infinity is not
a reallyreallyreallyreallyreallyreally large
finite cardinal.

A particular example of this not-being
is how each 1×1 1-ended sequence is
the same size (matchable to each other)

_That includes sub-sequences_
⟨ a₁ ... ⟩ is 1×1 1-ended
Split it.
⟨ a₁ ... a₂₉₉₇₉₂₄₅₈ ⟩ is 1×1 2-ended
⟨ a₂₉₉₇₉₂₄₅₉ ... ⟩ is 1×1 1-ended

⟨ a₂₉₉₇₉₂₄₅₉ ... ⟩ is matchable to
⟨ a₁ ... ⟩

For example, use this matching:
| a₁:a₂₉₉₇₉₂₄₅₉ are matched
| if aᵢ:aⱼ are matched,
| then aᵢ₊₁:aⱼ₊₁ are matched.

⟨ a₁ ... ⟩ and ⟨ a₂₉₉₇₉₂₄₅₉ ... ⟩ are 1×1
if some aₖ is not matched
there is a _first_ aᵢ₊₁ not matched
but with aᵢ:aⱼ matched.

That contradicts the matching, above.
A first unmatched aᵢ₊₁ not-exists.
An unmatched aₖ not-exists.
⟨ a₁ ... ⟩ can match
⟨ a₂₉₉₇₉₂₄₅₉ ... ⟩

|⟨ a₂₉₉₇₉₂₄₅₉ ... ⟩| =
|⟨ a₁ ... ⟩| = |ℕ⁺| = ℵ₀


x > 0 ⟹ NUF(x) = ℵ₀

x > 0 ⟹ NUF(x) > 0 ⟹ x > 1/n
⟨ 1/1 ... 1/n⁻⁻ ⟩ is 1×1 2-ended
⟨ 1/n ... ⟩ is 1×1 1-ended

|⟨ 1/n ... ⟩| = ℵ₀ = NUF(x)


tl;dr
x > 0 -> NUF(x) > 0
x > 0 -> NUF(x) = ℵ₀

> *************************************
>> The more explicit that you are,
>> the better chance that you have of
>> spotting an unjustified assumption.
>
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> shows that
> never unit fractions share a point.

Yes.
It shows that unit fractions have
the 1×1 1-ended order of ℕ⁺, reversed.

> Only one can sit at the edge of (0, 1].

Please be explicit about why _at least one_
element of (0,1] sits at the edge of (0,1]

I know, you said "only one".
But _what you use_ is "at least one".
Why at least one?

[WM]

Jim Burns schrieb am Sonntag, 23. April 2023 um 21:23:27 UTC+2:
> On 4/23/2023 9:30 AM, WM wrote:
> > Jim Burns schrieb am Samstag,
> > 22. April 2023 um 22:36:26 UTC+2:
> >> On 4/22/2023 2:56 PM, WM wrote:
>
> >>> NUF(x) = ℵo for all x ∈ (0, 1]
> >>> is a bad mistake.
> >>
> >> Say why it's a mistake.
> >
> > You already have agreed:
> > ************************************
> >> if one unit fraction would sit at zero,
> >> then the next one would sit already
> >> insidet of (0, 1].
> >> Only NUF(x) > 0 for all x > 0
> >> can be correct.
> >
> > JB: Yes, but ...
> Both of these are correct:

No, you agreed to "only".

> | Only NUF(x) > 0 for all x > 0

> | NUF(x) = ℵo for all x ∈ (0, 1]

That would deny the distances of every pair of unit fractions.

> x > 0 ⟹ NUF(x) = ℵ₀

∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0

There are ℵ distances 1/(n(n+1)).
Only one can be next to zero. At its end is the scond unit fraction, but it is not befoer (0, 1].

> x > 0 -> NUF(x) > 0
> x > 0 -> NUF(x) = ℵ₀

That is true for all x you can choose.
It proves that ℵ₀ unit fractions cannot be distinguished by their places x. They are dark.

> > ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> > shows that
> > never unit fractions share a point.
> Yes.

How should all be less than every x > 0?

> > Only one can sit at the edge of (0, 1].
> Please be explicit about why _at least one_
> element of (0,1] sits at the edge of (0,1]

Because the distance to the next one places this beyond the edge.

>
> I know, you said "only one".
> But _what you use_ is "at least one".
> Why at least one?

Not even at least one. Between the unit fraction next to 0 and 0 other real numbers can exist.

> > The next one is inside.
> > Therefore
> > NUF(x) = ℵ for all x ∈ (0, 1]
> > is impossible.

Too hard to comprehend for you?

Regards, WM

[Gus Gassmann]

On Saturday, 22 April 2023 at 15:56:12 UTC-3, WM wrote:
> Gus Gassmann schrieb am Samstag, 22. April 2023 um 15:48:51 UTC+2:
>
> > At least he now covers the quantifier switcheroo so he can't even see it any more.
> If the universal quantifier in
> ∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0
> deserves its name, then there are all unit fractions and all their internal distances.
> If one unit fraction sits before all elements of (0, 1]

Yes, well, if. Too bad there is no such unit fraction.

[Jim Burns]

On 4/23/2023 4:43 PM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 23. April 2023 um 21:23:27 UTC+2:
>> On 4/23/2023 9:30 AM, WM wrote:
>>> Jim Burns schrieb am Samstag,
>>> 22. April 2023 um 22:36:26 UTC+2:
>>>> On 4/22/2023 2:56 PM, WM wrote:

>>>>> NUF(x) = ℵo for all x ∈ (0, 1]
>>>>> is a bad mistake.
>>>>
>>>> Say why it's a mistake.
>>>
>>> You already have agreed:
>>> ************************************
>>>> if one unit fraction would sit at zero,
>>>> then the next one would sit already
>>>> insidet of (0, 1].
>>>> Only NUF(x) > 0 for all x > 0
>>>> can be correct.
>>>
>>> JB: Yes, but ...
>> Both of these are correct:
>
> No, you agreed to "only".

ℵ₀ > 0

It would appear that
you have misunderstood me.

x > 0 ⟹ NUF(x) = ℵ₀

x ≤ 0 ⟹ NUF(x) = 0
x > 0 xor x ≤ 0

>> | Only NUF(x) > 0 for all x > 0
>
>> | NUF(x) = ℵo for all x ∈ (0, 1]
>
> That would deny the distances of
> every pair of unit fractions.

No.
That denies that
unit fraction 1/n exists without
unit fraction 1/(4n)
That denies that
a jumping function exists without
a point of discontinuity.
That denies that
a 1×1 1-ended sequence exists which
hasn't cardinality ℵ₀

>> x > 0 -> NUF(x) > 0
>> x > 0 -> NUF(x) = ℵ₀
>
> That is true for all x you can choose.

That is true for all x which are
between-points of splits of Q

[WM]

There are finite distances, larger than points, between all unit fractions. One of them is next to zero. It is impossible that two or more are next to zero. It is impossible, in finished infinty, that nothig is next to zero.

Regards, WM