There is a first/smallest integer (in Mückenland)

Post by Moebius
WM> All unit fractions are separated. Therefore there is a first one
Moebius> All integers are separated. Therefore there is a first one [?]
WM> This is true but difficult to understand.

Perhaps, with professional counseling, they could get back together.

2024-07-17 14:41:37 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one point
although all unit fractions are separated by finite distances?

Regards, WM

FromTheRafters

2024-07-17 17:13:09 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one point
although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

2024-07-17 17:49:59 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one point
although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

Og course it jumps, but what is the maximum size of a jump?

Regards, WM

Moebius

2024-07-17 18:02:08 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

Of course it jumps, but what is the maximum size of a jump?

The jump "at" 0 is THE ONLY jump here, Mückenheim: For x <= 0 NUF(x) = 0
and for ALL x > 0 NUF(x) = aleph_0. So the answer is: aleph_0.

Hint: img(NUF) = {0, aleph_0}.

2024-07-17 18:07:31 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

Of course it jumps, but what is the maximum size of a jump?

The jump "at" 0 is THE ONLY jump here,

No, ℵo finite intervals do not fit between [0, 1] and (0, 1].

Post by Moebius
Hint: img(NUF) = {0, aleph_0}.

It is highly deplorable how acquired "knowledge" can paralyse the brain.

Regards, WM

Moebius

2024-07-17 19:08:29 UTC

Post by WM
ℵo finite intervals do not fit between [0, 1] and (0, 1].

Was immer das auch heißen soll, Mückenheim.

Aber für jedes x > 0 passen ℵo Stammbrüche zwischen 0 und x.

2024-07-17 21:01:32 UTC

Post by WM
ℵo finite intervals do not fit between [0, 1] and (0, 1].

Was immer das auch heißen soll,

Are you unable to understand simple sentences?

Post by Moebius
Aber für jedes x > 0 passen ℵo Stammbrüche zwischen 0 und x.

No. Only for every x > 0 that is in a distance of at least ℵo finite
intervals from 0.

Regards, WM

Jim Burns

2024-07-18 04:30:01 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Og course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

∀ᴿx≤0: NPR(x) = |(0,x)| = |{}| = 0
∀ᴿx>0: NPR(x) = |(0,x)| = |ℝ|

f(y) = y/√̅y̅²̅+̅1
f: ℝ → (-1,1): 1.to.1

g(y) = (y+1)/2
g: (-1,1) → (0,1): 1.to.1

g∘f(y) = (y/√̅y̅²̅+̅1+1)/2
g∘f: ℝ → (0,1): 1.to.1

x⋅g∘f(y) = x⋅(y/√̅y̅²̅+̅1+1)/2
x⋅g∘f: ℝ → (0,x): 1.to.1

|ℝ| ≤ |(0,x)|

Also
ℝ ⊇ |(0,x)|
|ℝ| ≥ |(0,x)|
|ℝ| = |(0,x)|

∀ᴿx>0: NPR(x) = |(0,x)| = |ℝ|

Moebius

2024-07-18 05:52:26 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Nope. |IN| is "the maximum size of a jump", namely "the size" of the
jump "at" 0. (Hint: NUF has only one jump, namely "at" 0.)

Moebius

2024-07-18 06:00:07 UTC

Post by FromTheRafters
Sure, it jumps because of your stepwise function.

Of course it jumps, but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Nope. |IN| (aleph_0) is "the maximum size of a jump", namely "the size" of the
jump "at" 0. (Hint: NUF has only one jump, namely "at" 0.)

Hint: NUF(x) := |{u e {1/n : n e IN} : u <= x}| (x e IR) .

Hence

NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 ,

as well as

img(NUF) = {0, aleph_0} .

2024-07-18 13:39:35 UTC

Post by Moebius
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 ,

Don't claim such obviously wrong stuff. Try to understand how the function
increases.
Hint: Use ∀n ∈ ℕ: 1/n - 1/(n+1) > 0.

Regards, WM

Moebius

2024-07-18 15:44:39 UTC

NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 ,

Don't claim <bla>

Mückenheim, Du bist selbst zum Scheißen zu blöde.

2024-07-18 21:12:02 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means there are ℵo unit fractions smaller than (0, oo) because this
interval contains all x > 0 and nothing else.

Regards, WM

Moebius

2024-07-18 21:35:34 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means

that for each and every x e IR, x > 0 there are aleph_0 unit fractions
which are <= x.

Ich glaube, das hat man Dir jetzt schon so um die 500- bis 1000-mal
erklärt, Mückenheim.

Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

Mückenheim, Du hast wirklich einen schweren Sprung in der Schüssel.
Dafür kannst Du natürlich nichts. Bedenklich ist, dass Dir die Leute in
Deinem Umfeld dabei offenbar keine Hilfe sind und das einfach
ignorieren. (Ich denke dabei vor allem an die Leute mit denen Du im
Kontext der "Technischen Hochschule Augsburg" interagierst. Es ist eine
Schande!)

Bitte geh doch einfach mal zu einem guten Psychiater und lass Dich
"durchchecken".

Moebius

2024-07-18 21:51:18 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means

that for each and every x e IR, x > 0 there are aleph_0 unit fractions
which are <= x.

Es kann doch nicht so schwer zu verstehen sein, Mückenheim!

Wenn x eine reelle Zahl > 0 ist, dann sind die (abzählbar) unendlich
vielen Stammbrüche

1/ceil(1/x), 1/ceil(1/x + 1), 1/ceil(1/x + 2), 1/ceil(1/x + 3), ...

(allesamt) kleiner-gleich x.

WAS GENAU verstehst Du daran nicht, Mückenheim???

2024-07-19 13:53:15 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means

that for each and every x e IR, x > 0 there are aleph_0 unit fractions
which are <= x.

Then we can abbreviate each and every x > 0 by (0, oo).

Post by Moebius
Ich glaube, das hat man Dir jetzt schon so um die 500- bis 1000-mal
erklärt, Mückenheim.
Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

That is not claimed by quantifier shifting but proven by the fact that (0,
oo) contains nothing but all x > 0. Obviously it contains more than your
x. What could that be?

Regards, WM

Moebius

2024-07-19 14:29:24 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 , (*)

That means

that for each and every x e IR, x > 0 there are aleph_0 unit fractions
which are <= x.

Then we can abbreviate each and every x > 0 by (0, oo).

Huh?!

Post by Moebius
Ich glaube, das hat man Dir jetzt schon so um die 500- bis 1000-mal
erklärt, Mückenheim.
Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

That is not claimed by quantifier shifting but <bla>

Wie auch immer, Mückenheim. Du scheinst aber zu blöde zu sein, den
Unterschied zu verstehen.

Post by WM
(0, oo) contains nothing but all x > 0.

Ja, mückenheim, das Intervall (die Menge) (0, oo) enthällt genau alle
reellen Zahlen > 0.

In Zeichen: x e (0, oo) <-> x e IR & x > 0.

Das bedeutet (ins Unreine gesprochen), dass man in Formeln/Aussagen den
Ausdruck "x e IR & x > 0" durch den Ausdruck "x e (0, oo)" ersetzen kann
und umgekehrt.

Man kann also z. B. (*) auch so schreiben:

NUF(x) = aleph_0 for all x e (0, oo) .

Rein formal solle man aber die Formeln SO schreiben:

for all x e IR, x > 0: NUF(x) = aleph_0 ,
bzw.
for all x e (0, oo): NUF(x) = aleph_0 .

2024-07-20 12:27:11 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 , (*)

That means

that for each and every x e IR, x > 0 there are aleph_0 unit fractions
which are <= x.

Then we can abbreviate each and every x > 0 by (0, oo).

Huh?!

Is in (0, oo) any real number which is not positive?
Is in (0, oo) any positive realnumber missing?

If something is left of all x > 0 then it is left of the interval (0, oo)

Post by Moebius
Ich glaube, das hat man Dir jetzt schon so um die 500- bis 1000-mal
erklärt,

No explanations of your quantifier nonsensebut a counter example please.

Post by WM
(0, oo) contains nothing but all x > 0.

One of these is the result left-hand side of every x > 0 <==> left-hand
side of (0, oo).

Post by Moebius
NUF(x) = aleph_0 for all x e (0, oo) .
for all x e IR, x > 0: NUF(x) = aleph_0 ,
bzw.
for all x e (0, oo): NUF(x) = aleph_0 .

That means NUF(x) is constant over the whole interval (0, oo), i.e., a
false expression because it follows that NUF increases from 0 to infinity
between [0, 1] and (0, 1].

Regards, WM

Jim Burns

2024-07-19 18:52:26 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means

that for each and every x e IR, x > 0
there are aleph_0 unit fractions which are <= x.

Then we can abbreviate each and every x > 0 by (0, oo).

Your "abbreviating" is a quantifier shift.
Making it less visible doesn't make it prove anything.

"Abbreviate" rock.paper.scissors and
you get wrong answers.
That method is unreliable.

----
For each A ⊆ ⅟ℕ∩(0,1]: ∃u ∈ A ᴬ≤ u ⇐ A ≠ {}
For each u ∈ ⅟ℕ∩(0,1]: ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
For each v ∈ ⅟ℕ∩(0,1]: ⅟(-1+⅟v) = min.⅟ℕ∩(v,1] ⇐ v < 1

The order of all unit fractions ⅟ℕ∩(0,1] is
well.ordered with step.down non.max.step.up
(max = 1)

There are ℵ₀.many unit.fractions in (0,1]

⎛ For each A ⊆ ⅟ℕ∩(0,x]: ∃u ∈ A ᴬ≤ u ⇐ A ≠ {}
⎜ because
⎜ for each A ⊆ ⅟ℕ∩(0,1]: ∃u ∈ A ᴬ≤ u ⇐ A ≠ {}
⎜ and
⎝ A ⊆ ⅟ℕ∩(0,x] ⊆ ⅟ℕ∩(0,1]

⎛ For each u ∈ ⅟ℕ∩(0,x]: ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
⎜ because
⎜ for each u ∈ ⅟ℕ∩(0,1]: ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
⎜ and
⎝ u ∈ ⅟ℕ∩(0,x] ⊆ ⅟ℕ∩(0,1]

⎛ For each v ∈ ⅟ℕ∩(0,x]: ⅟(-1+⅟v) = min.⅟ℕ∩(v,x] ⇐ v < ⅟⌈⅟x⌉
⎜ because
⎜ for each v ∈ ⅟ℕ∩(0,1]: ⅟(-1+⅟v) = min.⅟ℕ∩(v,1] ⇐ v < 1
⎜ and,
⎜ for v < ⅟⌈⅟x⌉:
⎜ ⅟ℕ∩(v,x] /= {}
⎝ min.⅟ℕ∩(v,x] = min.⅟ℕ∩(v,1]

For each A ⊆ ⅟ℕ∩(0,x]: ∃u ∈ A ᴬ≤ u ⇐ A ≠ {}
For each u ∈ ⅟ℕ∩(0,x]: ⅟(1+⅟u) = max.⅟ℕ∩(0,u)
For each v ∈ ⅟ℕ∩(0,x]: ⅟(-1+⅟v) = min.⅟ℕ∩(v,x] ⇐ v < ⅟⌈⅟x⌉

The order of all unit fractions ⅟ℕ∩(0,x] is
well.ordered with step.down non.max.step.up
(max = ⅟⌈⅟x⌉)

There are ℵ₀.many unit.fractions in (0,x]

Infinite and humongous are different.

Post by Moebius
Ich glaube, das hat man Dir jetzt
schon so um die 500- bis 1000-mal erklärt, Mückenheim.
Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

That is not claimed by quantifier shifting

Your "abbreviating" is a quantifier shift.
Making it less visible doesn't make it prove anything.

"Abbreviate" rock.paper.scissors and
you get wrong answers.
That method is unreliable.

Post by WM
but proven by the fact that
(0, oo) contains nothing but all x > 0.
Obviously it contains more than your x.
What could that be?

Proof by "obviously".
You depended upon the unreliable quantifier shift
and wishing upon a star ("obviously").

⅟ℕ∩(0,x] doesn't contain more than
max in each nonempty A
a step.down for each unit.fraction
a step.up for each non.max unit.fraction

That is sufficient for ℵ₀.many in ⅟ℕ∩(0,x]

Moebius

2024-07-19 19:59:11 UTC

Post by WM
Then we can abbreviate each and every x > 0 by (0, oo).

Your "abbreviating" is a quantifier shift.

Right.

Post by Jim Burns
Making it less visible doesn't make it prove anything.

Indeed!

Post by Moebius
Hint: Ax > 0: E^aleph_0 u: ... <=/=> E^aleph_0 u: Ax > 0: ...

That is not claimed by quantifier shifting

Your "abbreviating" is a quantifier shift.

Right.

Post by Jim Burns
Making it less visible doesn't make it prove anything.

Indeed!

Post by WM
[...] proven by the fact that
(0, oo) contains nothing but all x > 0.
Obviously it contains more than your x.

Proof by "obviously".

Right.

Post by Jim Burns
You depended upon the unreliable quantifier shift
and wishing upon a star ("obviously").

Indeed!

If wishes were horses, beggars would ride!

2024-07-20 12:47:50 UTC

Post by WM
Then we can abbreviate each and every x > 0 by (0, oo).

Your "abbreviating" is a quantifier shift.
Making it less visible doesn't make it prove anything.

My theorem: X is left-hand side of every x > 0 <==> X is left-hand side of
(0, oo).
Find a counter example or accept it.
If you believe it can be interpreted as a quantifier shift, then note that
not every quantifier shift produces a wrong result. Example: see above.

Post by Jim Burns
"Abbreviate" rock.paper.scissors and
you get wrong answers.
That method is unreliable.

Don't waffle, do what is usual in mathematics: Show a counter example for
my theorem.

Regards, WM

joes

2024-07-19 07:11:26 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means there are ℵo unit fractions smaller than (0, oo) because this
interval contains all x > 0 and nothing else.

It makes no sense to talk about something smaller than an interval.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.

2024-07-19 13:55:58 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means there are ℵo unit fractions smaller than (0, oo) because this
interval contains all x > 0 and nothing else.

It makes no sense to talk about something smaller than an interval.

It makes sense, but if you dislike this expression, use left-hand side of
every x > 0 and of all x > 0, abbreviated by (0, oo).

Regards, WM

Chris M. Thomasson

2024-07-19 18:55:35 UTC

NUF(x) = aleph_0 for all x e IR, x > 0 ,

That means there are ℵo unit fractions smaller than (0, oo) because this
interval contains all x > 0 and nothing else.

It makes no sense to talk about something smaller than an interval.

It makes sense, but if you dislike this expression, use left-hand side
of every x > 0 and of all x > 0, abbreviated by (0, oo).

There will always be a gap between any unit fraction and zero. There is
no unit fraction that can represent 0. 0/1 is not a unit fraction.

2024-07-20 12:52:05 UTC

Post by Chris M. Thomasson
There will always be a gap between any unit fraction and zero.

My theorem: X is left-hand side of every x > 0 <==> X is left-hand side of
(0, oo).
Find a counter example or accept it.
Of course there is no unit fraction left-hand side of every x.

Regards, WM

2024-07-18 13:34:04 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Nope. |IN| is "the maximum size of a jump", namely "the size" of the
jump "at" 0. (Hint: NUF has only one jump, namely "at" 0.)

At 0 there are no unit fractions. Therefore you are wrong.

Regards, WM

Moebius

2024-07-18 15:43:30 UTC

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

|IN| is "the maximum size of a jump", namely "the size" of the
jump "at" 0. (Hint: NUF has only one jump, namely "at" 0.)

At 0

Ja, Mückenheim, man sagt ja auch nur, dass der Sprung "bei 0" ist, Du
hirnloser Affe.

Gemeint ist: NUF(0) = 0 und NUF(x) = aleph_0 für x > 0.

2024-07-18 18:31:45 UTC

Post by Moebius
Gemeint ist: NUF(0) = 0 und NUF(x) = aleph_0 für x > 0.

Every x > 0 without exception means every point of (0, oo). No point has
less smaller unit fractions. That means the interval (0, oo) has
infinitely many smaller unit fractions. That is wrong.

Regards, WM

Jim Burns

2024-07-18 17:00:28 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Nope.

You are answering a different question,
| What is the size of the jump of NUF(x) at 0?

I had read "a jump" as a reference more generic than that.
-- However, in WM's most recent post, it seems that
your reading is more correct than mine.

Paraphrasing, WM seemed to ask
| How can a jump at one point be by
| more than one point? Anywhere. Any jump.

Paraphrasing you
| Well, it _is_ more.
| <same proof again>

Paraphrasing me
| Here is how.
| A jump _can_ be by
| much more than it _is_ in NUF(x) at 0
| <different conceivably.understood proof>

|ℝ| is the maximum size of a jump.

∀ᴿx≤0: NPR(x) = |(0,x)| = |{}| = 0
∀ᴿx>0: NPR(x) = |(0,x)| = |ℝ|

f(y) = y/√̅y̅²̅+̅1
f: ℝ → (-1,1): 1.to.1

g(y) = (y+1)/2
g: (-1,1) → (0,1): 1.to.1

g∘f(y) = (y/√̅y̅²̅+̅1+1)/2
g∘f: ℝ → (0,1): 1.to.1

x⋅g∘f(y) = x⋅(y/√̅y̅²̅+̅1+1)/2
x⋅g∘f: ℝ → (0,x): 1.to.1

|ℝ| ≤ |(0,x)|

Also
ℝ ⊇ |(0,x)|
|ℝ| ≥ |(0,x)|
|ℝ| = |(0,x)|

∀ᴿx>0: NPR(x) = |(0,x)| = |ℝ|

Jumps "at" a point are between
nearby points.
WM admitted that much in a recent post,
but changed what "change" means to him.

Post by Moebius
|IN| is "the maximum size of a jump",
namely "the size" of the jump "at" 0.
(Hint: NUF has only one jump, namely "at" 0.)

I continue to accept the proofs that
∀ᴿx>0: NUF(x) = |⅟ℕ∩(0,x)| = |⅟ℕ|

2024-07-18 18:28:26 UTC

Post by Jim Burns
Jumps "at" a point are between
nearby points.
WM admitted that much in a recent post,
but changed what "change" means to him.

A claim for all x > 0 is a claim for all points of the interval (0, oo).
The claim "between 0 and every point of (0, oo) NUF(x) = ℵo" implies the
existence of ℵo unit fractions between 0 and (0, oo) and is false.

Regards, WM

Jim Burns

2024-07-18 19:16:32 UTC

Post by Jim Burns
Jumps "at" a point are between
nearby points.
WM admitted that much in a recent post,
but changed what "change" means to him.

A claim for all x > 0 is
a claim for all points of the interval (0, oo).
The claim
"between 0 and every point of (0, oo) NUF(x) = ℵo"
implies the existence of
ℵo unit fractions between 0 and (0, oo)
and is false.

The unit.fractions between 0 and different points x
are not the same unit.fractions.

For each x > 0
the unit fractions in (0,x] are
step.down non.max.step.up well.ordered.by.>

That order is the same type as ℕ
step.up non.min.step.down well.ordered.by.<

Anything with the order.type of ℕ holds ℵ₀.many.

For each x > 0
max.(⅟ℕ∩(0,x]) = ⅟⌈⅟x⌉

The sets ⅟ℕ∩(0,x] are different.
The order.type of each ⅟ℕ∩(0,x] is the same.
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
Each ⅟ℕ∩(0,x] holds ℵ₀.many.

Infinite does not mean humongous.

2024-07-18 20:55:57 UTC

Post by Jim Burns
Jumps "at" a point are between
nearby points.
WM admitted that much in a recent post,
but changed what "change" means to him.

A claim for all x > 0 is
a claim for all points of the interval (0, oo).
The claim
"between 0 and every point of (0, oo) NUF(x) = ℵo"
implies the existence of
ℵo unit fractions between 0 and (0, oo)
and is false.

The unit.fractions between 0 and different points x
are not the same unit.fractions.

Not claimed that they are the same. But if for all points x > 0 there are
ℵo smaller unit fractions, then for the interval (0, oo) there are ℵo
smaller unit fractions. Or is there a point in (0, oo) which is not an x >
0?

Regards, WM

Jim Burns

2024-07-18 21:42:42 UTC

Post by Jim Burns
Jumps "at" a point are between
nearby points.
WM admitted that much in a recent post,
but changed what "change" means to him.

A claim for all x > 0 is
a claim for all points of the interval (0, oo).
The claim
"between 0 and every point of (0, oo) NUF(x) = ℵo"
implies the existence of
ℵo unit fractions between 0 and (0, oo)
and is false.

The unit.fractions between 0 and different points x
are not the same unit.fractions.

Not claimed that they are the same.

For different points x₁ x₂
the unit.fractions ⅟ℕ∩(0,x₁] and ⅟ℕ∩(0,x₂] are
step.down non.⅟⌈⅟x₁⌉.step.up well.ordered.by.>
and
step.down non.⅟⌈⅟x₂⌉.step.up well.ordered.by.>

⅟ℕ∩(0,x₁] and ⅟ℕ∩(0,x₂] are order.isomorphic to ℕ
step.up non.0.step.down well.ordered.by.<

|⅟ℕ∩(0,x₁]| = |⅟ℕ∩(0,x₂]| = |ℕ| = ℵ₀

Post by WM
But if
for all points x > 0
there are ℵo smaller unit fractions,
then
for the interval (0, oo)
there are ℵo smaller unit fractions.

No.
For each point x > 0
there are
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
unit.fractions in (0,x] which,
because
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
are ℵ₀.many.

No unit fractions are smaller than (0,∞)

Post by WM
Or is there a point in (0, oo) which
is not an x > 0?

No, but
there is no point x in (0,∞) such that
⅟⌈⅟x⌉ isn't = max.(⅟ℕ∩(0,x])
and
unit.fractions in ⅟ℕ∩(0,x] aren't
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
and
unit.fractions in ⅟ℕ∩(0,x] aren't ℵ₀.many.

2024-07-19 14:03:48 UTC

Post by WM
But if
for all points x > 0
there are ℵo smaller unit fractions,
then
for the interval (0, oo)
there are ℵo smaller unit fractions.

No.

Like Bob. No reason to continue.

Post by Jim Burns
For each point x > 0
there are
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
unit.fractions in (0,x] which,
because
step.down non.⅟⌈⅟x⌉.step.up well.ordered.by.>
are ℵ₀.many.

Of all unit fractions left of any x > 0, ℵ₀ are the same, finitely
many are different.

Post by Jim Burns
No unit fractions are smaller than (0,∞)

Right. That means Fritsches x are not all x.

Post by WM
Or is there a point in (0, oo) which
is not an x > 0?

No, but
there is no point x in (0,∞) such that

irrelevant. All x > 0 are there and nothing else.

Regards, WM

Moebius

2024-07-18 20:34:52 UTC

Post by FromTheRafters
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Nope.

You are answering [the] question,
| What is the size of the jump of NUF(x) at 0?
I had read "a jump" as a reference more generic than that.

Yes, sometimes it's hard to guess what he's babbling about.

-- However, in WM's most recent post, it seems that
your reading is more correct than mine.

May be.

Paraphrasing, WM seemed to ask
| How can a jump at one point be by
| more than one point? Anywhere. Any jump.

I'd say ... "can be higher than 1".

Paraphrasing you
| Well, it _is_ more.
| <same proof again>

Sure.

[...]

Jumps "at" a point are between
nearby points.

Sort of. :-P

WM admitted that much in a recent post,
but changed what "change" means to him.

Yeah, WM sometimes adjust to the replies he gets.

A =/= A (depending on time: A (at t = t_1) may differ from A (at t =
t_2, if t_1 =/= t_2).

2024-07-18 13:31:33 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Then |ℝ| unit fractions must occupy that point where it jumps. Alas
there are not |ℝ| unit fractions at all.

Regards, WM

Chris M. Thomasson

2024-07-18 18:53:09 UTC

Post by FromTheRafters
Sure,
it jumps because of your stepwise function.

Of course it jumps,
but what is the maximum size of a jump?

|ℝ| is the maximum size of a jump.

Then |ℝ| unit fractions must occupy that point where it jumps. Alas
there are not |ℝ| unit fractions at all.

All unit fractions are real numbers.

Not all real numbers are unit fractions.

Chris M. Thomasson

2024-07-17 19:48:33 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

Real line going to the right:

(0)->(1/1)

There is no so-called smallest unit fraction that magically sits next to
zero! Damn it WM! We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero. However, NO unit
fraction, no matter how small, ever equals zero. Period.

Moebius

2024-07-17 20:24:22 UTC

Post by Chris M. Thomasson
There is no so-called smallest unit fraction that magically sits next to
zero! Damn it WM! We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero. However, NO unit
fraction, no matter how small, ever equals zero. Period.

Well, that's the point of view of "classical mathematics".

See: https://en.wikipedia.org/wiki/Classical_mathematics

2024-07-17 21:10:30 UTC

Well, that's the point of view of "classical mathematics".

Therefore NUF cannot change at 0.

Regards, WM

Chris M. Thomasson

2024-07-17 21:27:17 UTC

Post by Chris M. Thomasson
There is no so-called smallest unit fraction that magically sits next
to zero! Damn it WM! We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero. However, NO unit
fraction, no matter how small, ever equals zero. Period.

Well, that's the point of view of "classical mathematics".

Therefore NUF cannot change at 0.

The number of unit fractions is infinite in nature for there are
infinitely many, indeed.

There is no unit fraction that can represent zero for 0/1 is NOT a unit
fraction.

Chris M. Thomasson

2024-07-18 19:05:29 UTC

Post by Chris M. Thomasson
There is no so-called smallest unit fraction that magically sits next
to zero! Damn it WM! We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero. However, NO unit
fraction, no matter how small, ever equals zero. Period.

Well, that's the point of view of "classical mathematics".
See: https://en.wikipedia.org/wiki/Classical_mathematics

Indeed. :^)

2024-07-17 21:07:44 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

(0)->(1/1)
There is no so-called smallest unit fraction that magically sits next to
zero! Damn it WM!

Don't claim. Explain how NUF increases.

Post by Chris M. Thomasson
We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero.

Irrelevant.

Post by Chris M. Thomasson
However, NO unit
fraction, no matter how small, ever equals zero. Period.

Of course not. Therefore NUF cannot change at zero.

Regards, WM

Chris M. Thomasson

2024-07-18 05:55:18 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

Sure, it jumps because of your stepwise function.

(0)->(1/1)
There is no so-called smallest unit fraction that magically sits next
to zero! Damn it WM!

Don't claim. Explain how NUF increases.

Post by Chris M. Thomasson
We can say as the unit fractions get smaller and smaller forevermore,
that their limit is zero.

Irrelevant.

Post by Chris M. Thomasson
However, NO unit fraction, no matter how small, ever equals zero. Period.

Of course not. Therefore NUF cannot change at zero.

If there are no unit fractions then the number of unit fractions is
zero. ;^) Anyway, there is no unit fraction that equals zero. Therefore
there cannot be a smallest unit fraction. Behold, 0/1 is not a unit
fraction! There is no unit fraction that can represent zero. Get over
it. However, the largest unit fraction, is good ol' 1/1. :^)

2024-07-18 13:37:04 UTC

Post by Chris M. Thomasson
However, NO unit fraction, no matter how small, ever equals zero. Period.

Of course not. Therefore NUF cannot change at zero.

If there are no unit fractions then the number of unit fractions is
zero.

Of course. NUF(0) = 0 and the increase follows at x > 0. Th question is
how large the first increase is.

Regards, WM

Chris M. Thomasson

2024-07-18 19:00:08 UTC

Post by Chris M. Thomasson
However, NO unit fraction, no matter how small, ever equals zero. Period.

Of course not. Therefore NUF cannot change at zero.

If there are no unit fractions then the number of unit fractions is zero.

Of course. NUF(0) = 0 and the increase follows at x > 0.

The first increase, humm... What about 1/1 - 0?
A smaller increase, what about 1/2 - 0?

Smaller and smaller all the way down, yet no unit fraction ever equals
zero. There is no smallest one. See?

[...]

Chris M. Thomasson

2024-07-17 19:44:01 UTC

Perhaps, with professional counseling,

you could explain how NUF(x) can increase from 0 to many more in one
point although all unit fractions are separated by finite distances?

0/1 is not a unit fraction!

2024-07-17 14:43:09 UTC

Can you explain how NUF(x) can increase from 0 to many more in one point x
although all unit fractions are separated by finite distances of
uncountably many x each?

Regards, WM

Moebius

2024-07-17 14:54:31 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0] in [any] point
x [>] although all unit fractions are separated by finite distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No first
one.)

Not that hard, is it?

Relevance concerning your nonsensical claim "This is true but difficult
to understand"? - None.

Moebius

2024-07-17 14:56:08 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0] [at any] point
x [> 0] although all unit fractions are separated by finite distances [...]

2024-07-17 15:08:30 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0] [at any] point
x [> 0] although all unit fractions are separated by finite distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No first
one.)
Not that hard, is it?

Thema verfehlt. The question is: How does NUF(x) increase from 0 to more?
You know? There is a point where NUF is 0 and then it increases. How?

Regards, WM

Moebius

2024-07-17 15:21:16 UTC

Post by WM
There is a point where NUF is 0

Yes, for all points x <= 0.

Post by WM
and then it increases.

No, it does not "increase", it _jumps_ "at" x = 0.

MEANING:

NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0

Not that hard, is it?

Post by WM
How?

Huh?! Ask it, it might tell you!

A reasonable question might be: "Why?"

Because for each and every x e IR, x > 0 there are countably-infinitely
many unit fractions which are <= x (hint: No first one) AND for each and
every x e IR, x <= 0 there is no unit fractions which is <= x.

Not that hard, is it?

2024-07-17 15:27:02 UTC

Post by WM
There is a point where NUF is 0

Yes, for all points x <= 0.

Post by WM
and then it increases.

No, it does not "increase", it _jumps_ "at" x = 0.

Nonsense. Jumping by X at a point x requires X unit fractions at that
point x. That is forbidden by mathematics.

Post by Moebius
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0

For all x > 0 that can be named. Obviously that are not the x where NUF
increases first.

Regards, WM

Moebius

2024-07-17 15:59:33 UTC

Post by WM
There is a point where NUF is 0

Yes, for all points x <= 0.

Post by WM
and then it increases.

No, it does not "increase", it _jumps_ "at" x = 0.

Jumping by [aleph_0] at a point x requires [aleph_0] unit fractions at that
point x.

It seems that you have forgotten your OWN definition of NUF, Mückenheim.

No, it does not "require aleph_0 unit fractions at that point x" but
aleph_0 unit fractions THAT ARE <= x. (Holy shit!)

Hint: NFU(x) := |{u e {1/n : n e IN} : u <= x}|

Post by Moebius
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0

Moebius

2024-07-17 16:02:17 UTC

Post by WM
There is a point where NUF is 0

Yes, for all points x <= 0.

Post by WM
and then it increases.

No, it does not "increase", it _jumps_ "at" x = 0.

MEANING:

NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 .

[NFU(x) = aleph_0] at a point x requires [aleph_0] unit fractions at that
point x.

Moebius

2024-07-17 16:17:42 UTC

Post by WM
There is a point where NUF is 0

Yes, for all points x <= 0.

Post by WM
and then it increases.

No, it does not "increase", it _jumps_ "at" x = 0.

Viell. Hast Du ja nur Probleme mit dem Begriff "Sprungstelle"?

Hier wird ein schönes Beispiel geliefert:

"Ein einfaches Beispiel für eine Sprungstelle unendliche Höhe
liefert die Funktion

f(x) := 1/x, x e IR\{0}
f(x) := 0, x = 0

bei x0 = 0."

Quelle: https://www.spektrum.de/lexikon/mathematik/unstetigkeit/10568

In our case we have (mutatis mutandis):

NUF(x) := aleph_0, x e IR+
NUF(x) := 0, x = 0.

Post by Moebius
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 .

[NFU(x) = aleph_0] at a point x requires [aleph_0] unit fractions at
that point x.

It seems that you have forgotten your OWN definition of NUF, Mückenheim.
No, it does not "require aleph_0 unit fractions at that point x" but
aleph_0 unit fractions THAT ARE <= x. (Holy shit!)
Hint: NFU(x) := |{u e {1/n : n e IN} : u <= x}|
<Nonsense deleted>

Moebius

2024-07-17 16:32:08 UTC

Post by Moebius
It seems that you have forgotten your OWN definition of NUF, Mückenheim.
No, it does not "require aleph_0 unit fractions at that point x" but
aleph_0 unit fractions THAT ARE <= x. (Holy shit!)
Hint: NFU(x) := |{u e {1/n : n e IN} : u <= x}|

This means NFU(x) = a iff the (cardinal)number of the set of unit
fractions which are <= x is a.

2024-07-17 16:58:57 UTC

Post by Moebius
No, it does not "require aleph_0 unit fractions at that point x" but
aleph_0 unit fractions THAT ARE <= x.

All points =< x are in the preimage of NUF. Not only those after
infinitely many finite intervals.

Post by Moebius
(Holy shit!)

for matheology. Indeed.

Regards, WM

Moebius

2024-07-17 17:30:23 UTC

Post by Moebius
No, it does not "require aleph_0 unit fractions at that point x" but
aleph_0 unit fractions THAT ARE <= x.

All points [in IR+] are in the [domain] of NUF. Not only those after
infinitely many finite intervals.

*lol* There ARE ONLY such x in IR+ you silly idiot! :-)

Hint: For each and every x e IR, x > 0 there are infinitely many unit
fractions u such that u <= x.

2024-07-17 17:16:07 UTC

We look at all points.

Post by Moebius
Hint: NFU(x) := |{u e {1/n : n e IN} : u <= x}|

This means NFU(x) = a iff the (cardinal)number of the set of unit
fractions which are <= x is a.

NUF increases at poits x, but never by more than 1 at one point..

Regards, WM

Moebius

2024-07-17 17:26:23 UTC

Post by Moebius
Hint: NFU(x) := |{u e {1/n : n e IN} : u <= x}|

This means NFU(x) = a iff the (cardinal)number of the set of unit
fractions which are <= x is a.

NUF <bla>

Yes, hence:

NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 .

joes

2024-07-17 17:01:07 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function. There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.

2024-07-17 17:17:54 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function.

No, ℵo finite intervals do not fit between [0, 1] and (0, 1]. The sign
function fits.

Post by joes
There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

They remain finite in every case.

Regards, WM

Chris M. Thomasson

2024-07-17 19:51:21 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function.

No, ℵo finite intervals do not fit between [0, 1] and (0, 1]. The sign
function fits.

Post by joes
There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

They remain finite in every case.

There are infinitely many of them, and none of them equals zero... 0/1
is not a unit fraction! Damn it. :^)

joes

2024-07-17 20:37:01 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function.

No, ℵo finite intervals do not fit between [0, 1] and (0, 1]. The sign
function fits.

Where do you get this requirement from?
Consider the sign function times infinity.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.

Moebius

2024-07-17 20:44:25 UTC

Post by WM
ℵo finite intervals do not fit between [0, 1] and (0, 1].

Where do you get this requirement from?

Warum bezeichnest Du so einen saudummen Scheißdreck als "requirement"?

Please tell us what "requirement" you are talking about (using
mathematical terminology/language/symbols).

2024-07-17 21:17:17 UTC

Post by WM
ℵo finite intervals do not fit between [0, 1] and (0, 1].

Where do you get this requirement from?

Please tell us what "requirement" you are talking about (using
mathematical terminology/language/symbols).

ℵo unit fractions occupy at least ℵo points and ℵo finite intervals.
Therefore NUF(x) cannot change from 0 to ℵo without passing an interval
of 2^ℵo points. They must be subtracted from your claim
∀ x > 0: NUF(x) = ℵo.
Corrected
∀ x > 2^ℵo positive points: NUF(x) = ℵo.

Regards, WM

Moebius

2024-07-17 21:50:14 UTC

ℵo unit fractions occupy [...] ℵo points and [leads to] ℵo finite intervals.

Indeed! Das hast Du gut bemerkt, Mückenheim!

Therefore NUF(x) cannot change from 0 to ℵo without [x] passing [...] 2^ℵo points.

Das ist ZIEMLICH blumiges Gerede. Aber was Du vielleicht meinst, ist,
dass zwischen 0 und x (für jedes x e IR, x > 0) "2^ℵo points" liegen.
Ja, das ist in der Tat so. (->Cantor/Mengenlehre)

Es gilt also für alle x e IR, x > 0: |(0, x]| = 2^ℵo.

They must be subtracted from your claim

Man kann Zahlen ganz schlecht von Behauptungen "abziehen", Mückenheim
(außer vielleicht in der Irrenanstalt in Mückenhausen).

Man muss das wohl leider wieder einmal als "saudummen Scheißdreck"
bezeichnen.

Hinweis: Die korrekte Aussage ist nach wie vor: ∀x > 0: NUF(x) = ℵo.

∀x > 2^ℵo positive points: NUF(x) = ℵo.

Also nochmal:

∀ x > 0: E^(2^ℵo) y: y < x.

Daher kann man die "Pseudobedingung" "> 2^ℵo positive points" weglassen.
"x > 0" drückt GENAU das aus, was WIR sagen wollen. (Du natürlich nicht,
da Du ja nur saudummen Scheißdreck daherreden kannst.)

Moebius

2024-07-17 22:08:54 UTC

ℵo unit fractions occupy [...] ℵo points and [leads to] ℵo finite intervals.

Indeed! Das hast Du gut bemerkt, Mückenheim!

Therefore NUF(x) cannot change from 0 to ℵo without [x] passing [...] 2^ℵo points.

They must be subtracted from your claim

∀x > 2^ℵo positive points: NUF(x) = ℵo.

Also nochmal:

∀ x > 0: E^(2^ℵo) y: 0 < y < x.

Daher kann man die "Pseudobedingung" "> 2^ℵo positive points" weglassen.
"x > 0" drückt GENAU das aus, was WIR sagen wollen. (Du natürlich nicht,
da Du ja nur saudummen Scheißdreck daherreden kannst.)

2024-07-18 13:20:43 UTC

Post by Moebius
Es gilt also für alle x e IR, x > 0: |(0, x)| = 2^ℵo.

Nein.

Post by WM
They must be subtracted from your claim

Man kann Zahlen ganz schlecht von Behauptungen "abziehen",

Aber von behaupteten Zahlen.

Post by Moebius
Hinweis: Die korrekte Aussage ist nach wie vor: ∀x > 0: NUF(x) = ℵo.

Explain how NUF(x) increases! There are two alternatives: Exists NUF(x) =
1 or not. If not, then basic mathematics is false.

Regards, WM

Moebius

2024-07-18 15:38:12 UTC

Post by Moebius
Es gilt also für alle x e IR, x > 0: |(0, x)| = 2^ℵo.

Nein.

Ja, doch.

Hast Du das nicht gewusst? Jedes noch so kleine reelle Intervall (das
nicht zu einem Punkt ausgeartet ist), hat die Mächtigkeit von ganz IR,
also 2^ℵo. Das hat schon Cantor bewiesen.

2024-07-18 18:24:28 UTC

Post by Moebius
Es gilt also für alle x e IR, x > 0: |(0, x)| = 2^ℵo.

Nein.

Ja, doch.
Hast Du das nicht gewusst? Jedes noch so kleine reelle Intervall (das
nicht zu einem Punkt ausgeartet ist), hat die Mächtigkeit von ganz IR,
also 2^ℵo. Das hat schon Cantor bewiesen.

Das ist ein Indiz dafür, dass diese Theorie falsch ist, hier aber
irrelevant.

A claim for all x > 0 is a claim for all points of the interval (0, oo).
The claim "between 0 and every point of (0, oo) NUF(x) = ℵo" implies the
existence of ℵo unit fractions betwee 0 and (0, oo) and is false.

Regards, WM

joes

2024-07-18 20:34:51 UTC

Post by Moebius
Es gilt also für alle x e IR, x > 0: |(0, x)| = 2^ℵo.

Hast Du das nicht gewusst? Jedes noch so kleine reelle Intervall (das
nicht zu einem Punkt ausgeartet ist), hat die Mächtigkeit von ganz IR,
also 2^ℵo. Das hat schon Cantor bewiesen.

Das ist ein Indiz dafür, dass diese Theorie falsch ist, hier aber
irrelevant.

Ich finde die Theorie sehr richtig.

Post by WM
A claim for all x > 0 is a claim for all points of the interval (0, oo).
The claim "between 0 and every point of (0, oo) NUF(x) = ℵo" implies the
existence of ℵo unit fractions betwee 0 and (0, oo) and is false.

How do you come up with with the "between 0 and (0, oo)"? The interval
(0, 0) is empty anyway.

--
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.

Moebius

2024-07-18 20:49:01 UTC

Post by Moebius
Es gilt also für alle x e IR, x > 0: |(0, x)| = 2^ℵo.

Hast Du das nicht gewusst? Jedes noch so kleine reelle Intervall (das
nicht zu einem Punkt ausgeartet ist), hat die Mächtigkeit von ganz IR,
also 2^ℵo. Das hat schon Cantor bewiesen.

Das ist ein Indiz dafür, dass diese Theorie falsch ist, hier aber
irrelevant.

Ich finde die Theorie sehr richtig.

For what it's worth: Ich auch. :-P

At least "rather helpful", as most mathematicians, I'd say.

Of course, WM is *not* a mathematician, but a fucking asshole full of shit.

2024-07-18 20:58:54 UTC

How do you come up with with the "between 0 and (0, oo)"? The interval
(0, 0) is empty anyway.

A claim of ℵo smaller unit fractions for all x > 0 is a claim of ℵo
smaller unit for (0, oo) because the interval contains all x > 0 and
nothing else. But at x = 0 there is no unit fraction.

Regards, WM

Moebius

2024-07-18 21:02:26 UTC

Post by joes
How do you come up with with the "between 0 and (0, oo)"? The interval
(0, 0) is empty anyway.

A claim of <bla>

<facepalm>

Mückenheim, Du bist selbst zum Scheißen zu blöde.

Geh endlich mal zu einem Psychiater! DO IT!!!

2024-07-17 21:05:38 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function.

No, ℵo finite intervals do not fit between [0, 1] and (0, 1]. The sign
function fits.

Where do you get this requirement from?

ℵo unit fractions occupy ℵo finite intervals.

Post by joes
Consider the sign function times infinity.

The sign function can change from point 0 to all points of the interval
(0, oo) without exception. NUF needs some of these points to acquire ℵo
unit fractions.

Regards, WM

Chris M. Thomasson

2024-07-17 19:50:24 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function. There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

Right!

2024-07-17 21:08:40 UTC

Post by joes
The distances between unit
fractions get infinitely small.

Right!

It remains finite in every case.

Regards, WM

Chris M. Thomasson

2024-07-17 21:15:27 UTC

Post by joes
The distances between unit
fractions get infinitely small.

Right!

It remains finite in every case.

Do you think you can count forever, and right before you die you write
down the number? Ahh, that is the smallest unit fraction for sure...
Then you are in the afterlife counting again? lol. Anyway, there is no
smallest unit fraction just like there is no largest natural number.

Smallest and largest in your mind seems to mean something akin to "has
this number been computed, seen by a single person, anywhere in the
universe as a whole, ect...?" All others are simply, dark. ;^) There is
no dark number that is the smallest unit fraction... Got it?

Is that better, WM?

2024-07-17 21:20:46 UTC

Post by Chris M. Thomasson
Is that better, WM?

No. The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

Regards, WM

Chris M. Thomasson

2024-07-17 21:36:06 UTC

Post by Chris M. Thomasson
Is that better, WM?

No. The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

0/1 is NOT a unit fraction. 1/1 is, 1/2 is, ect... on and on...

Chris M. Thomasson

2024-07-17 21:37:47 UTC

Post by Chris M. Thomasson
Is that better, WM?

No. The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

There is ALWAYS a gap between ANY unit fraction and zero. This is
because any unit fraction does not equal zero.

The limit is just what it tends to, not the actual results of the
individual iterates, so to speak.

Moebius

2024-07-17 22:46:15 UTC

Post by Chris M. Thomasson
There is ALWAYS a gap between ANY unit fraction and zero. This is
because any unit fraction does not equal zero.

Indeed!

Moreover, between any unit fractions an zero lies another unit fraction.

If u is a unit fraction, then u' = 1/(1/u + 1) is a unit fraction such
that 0 < u' < u.

Except in Mückenland (i.e. the ward in Mückenhausen) of course.

2024-07-18 13:13:15 UTC

Post by WM
The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

There is ALWAYS a gap between ANY unit fraction and zero.

So it is. Therefore there is no change before the end of the first gap.

Post by Chris M. Thomasson
The limit is just what it tends to, not the actual results of the
individual iterates, so to speak.

Correct.

Regards, WM

Chris M. Thomasson

2024-07-18 18:48:18 UTC

Post by WM
The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

There is ALWAYS a gap between ANY unit fraction and zero.

So it is. Therefore there is no change before the end of the first gap.

The first gap? You mean between 0 and 1/1? There must be a gap because
no unit fraction can ever equal zero...

Post by Chris M. Thomasson
The limit is just what it tends to, not the actual results of the
individual iterates, so to speak.

Correct.

[...]

Moebius

2024-07-17 22:16:49 UTC

Post by WM
The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

How does sgn(x) increase from 0 to more? There is a point where sgn(x)
is 0 [namely x = 0] and then it increases. How?

Ja, Mückenheim, Deine "Fragen" sind ähnlich hirnrissig (hirntot) wie
Dein Geschwafel im Allgemeinen.

Du kannst wirklich nur noch saudummen Scheißdreck daherreden.

2024-07-18 13:22:38 UTC

Post by WM
The question is: How does NUF(x) increase from 0 to more? There is a
point where NUF is 0 and then it increases. How?

How does sgn(x) increase from 0 to more? There is a point where sgn(x)
is 0 [namely x = 0] and then it increases. How?

It increases from 0 at 0 to 1 at (0, oo). This is impossible for unit
fractions.

Regards, WM

Moebius

2024-07-18 15:40:57 UTC

sgn(x)

Post by WM
It increases from 0 at 0 to 1 at (0, oo).

Nonsense. "at" ("bei") x = 0, Du dummer Spinner. Lern doch wenigstens
mal DIE GRUNDLAGEN der Matematik!

Moebius

2024-07-18 06:09:42 UTC

The question is: How does NUF(x) [jump] from 0 to more? There is a
point [namely x = 0] where NUF(x) is 0 and then it [jumps]. How?

Hint: NUF(x) := |{u e {1/n : n e IN} : u <= x}| (x e IR) .

Hence
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 .

That's "how", Du dummer Spinner.

2024-07-18 13:41:32 UTC

The question is: How does NUF(x) [jump] from 0 to more? There is a
point [namely x = 0] where NUF(x) is 0 and then it [jumps]. How?

Hint: NUF(x) := |{u e {1/n : n e IN} : u <= x}| (x e IR) .
Hence
NUF(x) = 0 for all x e IR, x <= 0
and
NUF(x) = aleph_0 for all x e IR, x > 0 .
That's "how"

Not if all unit fractions have distances. Do you despise basic
mathematics?

Regards, WM

Moebius

2024-07-18 15:45:33 UTC

Hint:   NUF(x) := |{u e {1/n : n e IN} : u <= x}| (x e IR) .
Hence
         NUF(x) = 0   for all x e IR, x <= 0
and
         NUF(x) = aleph_0   for all x e IR, x > 0 .
That's "how"

Not if <bla>

Mückenheim, geh doch endlich mal zum Psychiater!

Chris M. Thomasson

2024-07-17 21:11:55 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function. There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

Right!

Wrt sign, there is no least positive number. Take any positive number
and multiply it by -1. Therefore there is no least negative number. The
negatives and the positives get infinitely small, tending towards zero...

The abs of the number tending towards zero is interesting. Always a
positive distance towards zero... :^)

Chris M. Thomasson

2024-07-18 05:48:03 UTC

Can you explain how NUF(x) can [jump] from 0 [at x = 0] to [aleph_0]
[at any]
point x [> 0] although all unit fractions are separated by finite
distances [...]

Yes, of course: For each and every x e IR, x > 0 there are
countably-infinitely many unit fractions which are <= x. (Hint: No
first one.)

Thema verfehlt. The question is: How does NUF(x) increase from 0 to
more? There is a point where NUF is 0 and then it increases. How?

The same as the sign function. There simply is no such "point", as
there is no least positive number. The distances between unit
fractions get infinitely small.

Right!

A monkey wrench is so called signed zero. Where positive zero is the
least positive number? Fun twists that are besides the point. lol. ;^)

Post by Chris M. Thomasson
The abs of the number tending towards zero is interesting. Always a
positive distance towards zero... :^)

Chris M. Thomasson

2024-07-17 19:40:25 UTC

Can you explain how NUF(x) can increase from 0 to many more in one point
x although all unit fractions are separated by finite distances of
uncountably many x each?

Are you trying to mix and match things here, on purpose? The granularity
of the unit fractions is hyper coarse compared to the reals...

[...]

Richard Damon

2024-07-18 02:15:18 UTC

Can you explain how NUF(x) can increase from 0 to many more in one point
x although all unit fractions are separated by finite distances of
uncountably many x each?
Regards, WM

Because is isn't a properly defined function.

It has no finite value for any finite value of x > 0.

Thus, one need no figure how it gets from 0 to any other number, since
it doesn't.

2024-07-18 13:28:13 UTC

Post by Richard Damon

Post by WM
Can you explain how NUF(x) can increase from 0 to many more in one point
x although all unit fractions are separated by finite distances of
uncountably many x each?

Because is isn't a properly defined function.

What do you miss in its definition?

Post by Richard Damon
It has no finite value for any finite value of x > 0.

Then basic mathematics is false. ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 shows
that every unit fraction occupies its own point.

Post by Richard Damon
Thus, one need no figure how it gets from 0 to any other number, since
it doesn't.

Does it go from 0 to ℵo? How can that happen?

Regards, WM

Chris M. Thomasson

2024-07-18 18:46:59 UTC