All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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Le 17/07/2024 à 13:37, Moebius a écrit :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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

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Le 17/07/2024 à 15:42, FromTheRafters a écrit :

Moebius presented the following explanation :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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

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Am 17.07.2024 um 16:43 schrieb WM:

Le 17/07/2024 à 13:37, Moebius a écrit :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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?

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

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Am 17.07.2024 um 16:43 schrieb WM:

Le 17/07/2024 à 13:37, Moebius a écrit :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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.

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Le 17/07/2024 à 16:56, Moebius a écrit :

Am 17.07.2024 um 16:43 schrieb WM:

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

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Le 17/07/2024 à 17:21, Moebius a écrit :

Am 17.07.2024 um 17:08 schrieb WM:

There is a point where NUF is 0

Yes, for all points x <= 0.

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.

MEANING:

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

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Am 17.07.2024 um 17:08 schrieb WM:

There is a point where NUF is 0

Yes, for all points x <= 0.

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?

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?

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Am 17.07.2024 um 17:27 schrieb WM:

Le 17/07/2024 à 17:21, Moebius a écrit :

Am 17.07.2024 um 17:08 schrieb WM:

There is a point where NUF is 0

Yes, for all points x <= 0.

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}|

Hence:

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

<Nonsense deleted>

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Am 17.07.2024 um 17:27 schrieb WM:

Le 17/07/2024 à 17:21, Moebius a écrit :

Am 17.07.2024 um 17:08 schrieb WM:

There is a point where NUF is 0

Yes, for all points x <= 0.

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.

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>

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Am 17.07.2024 um 18:02 schrieb Moebius:

Am 17.07.2024 um 17:27 schrieb WM:

Le 17/07/2024 à 17:21, Moebius a écrit :

Am 17.07.2024 um 17:08 schrieb WM:

There is a point where NUF is 0

Yes, for all points x <= 0.

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.

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.

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>

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Am 17.07.2024 um 18:02 schrieb 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.

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Le 17/07/2024 à 18:32, Moebius a écrit :

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.

(Holy shit!)

for matheology. Indeed.

Regards, WM

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Am Wed, 17 Jul 2024 15:08:30 +0000 schrieb WM:

Le 17/07/2024 à 16:56, Moebius a écrit :

Am 17.07.2024 um 16:43 schrieb WM:

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.

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Le 17/07/2024 à 18:32, Moebius a écrit :

Am 17.07.2024 um 18:02 schrieb 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!)

We look at all points.

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

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Le 17/07/2024 à 19:01, joes a écrit :

Am Wed, 17 Jul 2024 15:08:30 +0000 schrieb WM:

Le 17/07/2024 à 16:56, Moebius a écrit :

Am 17.07.2024 um 16:43 schrieb WM:

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.

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

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Am 17.07.2024 um 18:58 schrieb WM:

Le 17/07/2024 à 18:32, Moebius a écrit :

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.

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Am 17.07.2024 um 19:16 schrieb WM:

Le 17/07/2024 à 18:32, Moebius a écrit :

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 .

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Le 17/07/2024 à 19:13, FromTheRafters a écrit :

WM presented the following explanation :

Le 17/07/2024 à 15:42, FromTheRafters a écrit :

Moebius presented the following explanation :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?] >>>>

This is true but difficult to understand.

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

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Am 17.07.2024 um 19:49 schrieb WM:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

WM presented the following explanation :

Le 17/07/2024 à 15:42, FromTheRafters a écrit :

Moebius presented the following explanation :

All unit fractions are separated. Therefore there is a first one >>>>>

Moebius> All integers are separated. Therefore there is a first one
[?]

This is true but difficult to understand.

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}.

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Le 17/07/2024 à 20:02, Moebius a écrit :

Am 17.07.2024 um 19:49 schrieb WM:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

WM presented the following explanation :

Le 17/07/2024 à 15:42, FromTheRafters a écrit :

Moebius presented the following explanation :

All unit fractions are separated. Therefore there is a first one >>>>>>

Moebius> All integers are separated. Therefore there is a first one >>>>>> [?]

This is true but difficult to understand.

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].

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

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

Regards, WM

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Am 17.07.2024 um 20:07 schrieb 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.

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Am Wed, 17 Jul 2024 17:17:54 +0000 schrieb WM:

Le 17/07/2024 à 19:01, joes a écrit :

Am Wed, 17 Jul 2024 15:08:30 +0000 schrieb WM:

Le 17/07/2024 à 16:56, Moebius a écrit :

Am 17.07.2024 um 16:43 schrieb WM:

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.

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Am 17.07.2024 um 21:48 schrieb 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

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Am 17.07.2024 um 22:37 schrieb joes:

Am Wed, 17 Jul 2024 17:17:54 +0000 schrieb 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).

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Le 17/07/2024 à 21:08, Moebius a écrit :

Am 17.07.2024 um 20:07 schrieb 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?

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

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Le 17/07/2024 à 22:37, joes a écrit :

Am Wed, 17 Jul 2024 17:17:54 +0000 schrieb WM:

Le 17/07/2024 à 19:01, joes a écrit :

Am Wed, 17 Jul 2024 15:08:30 +0000 schrieb WM:

Le 17/07/2024 à 16:56, Moebius a écrit :

Am 17.07.2024 um 16:43 schrieb WM:

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.

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

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Le 17/07/2024 à 21:48, "Chris M. Thomasson" a écrit :

On 7/17/2024 10:13 AM, FromTheRafters wrote:

WM presented the following explanation :

Le 17/07/2024 à 15:42, FromTheRafters a écrit :

Moebius presented the following explanation :

All unit fractions are separated. Therefore there is a first one >>>>>

Moebius> All integers are separated. Therefore there is a first one [?] >>>>>

This is true but difficult to understand.

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!

Don't claim. Explain how NUF increases.

We can say as the unit fractions get smaller and
smaller forevermore, that their limit is zero.

Irrelevant.

However, NO unit
fraction, no matter how small, ever equals zero. Period.

Of course not. Therefore NUF cannot change at zero.

Regards, WM

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Le 17/07/2024 à 22:24, Moebius a écrit :

Am 17.07.2024 um 21:48 schrieb 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.

Regards, WM

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Le 17/07/2024 à 21:50, "Chris M. Thomasson" a écrit :

On 7/17/2024 10:01 AM, joes wrote:

The distances between unit
fractions get infinitely small.

Right!

It remains finite in every case.

Regards, WM

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Le 17/07/2024 à 22:44, Moebius a écrit :

Am 17.07.2024 um 22:37 schrieb joes:

Am Wed, 17 Jul 2024 17:17:54 +0000 schrieb 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

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Le 17/07/2024 à 23:15, "Chris M. Thomasson" a écrit :

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

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Am 17.07.2024 um 23:17 schrieb WM:

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.

Das hingegen ist wieder mal unsyntaktischer Mist:

∀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.)

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Am 17.07.2024 um 23:17 schrieb WM:

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.

Das hingegen ist wieder mal unsyntaktischer Mist:

∀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.)

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Am 17.07.2024 um 23:37 schrieb 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.

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Am 17.07.2024 um 23:20 schrieb 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.

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On 7/17/24 10:43 AM, WM wrote:

Le 17/07/2024 à 13:37, Moebius a écrit :

All unit fractions are separated. Therefore there is a first one

Moebius> All integers are separated. Therefore there is a first one [?]

This is true but difficult to understand.

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.

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On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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

x⋅g∘f(y) = x⋅(y/√​̅y​̅²​̅+​̅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|

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Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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.)

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Am 18.07.2024 um 07:52 schrieb Moebius:

Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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} .

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Am 17.07.2024 um 23:20 schrieb WM:

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.

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Le 17/07/2024 à 23:37, "Chris M. Thomasson" a écrit :

On 7/17/2024 2:20 PM, WM wrote:

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

Correct.

Regards, WM

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Le 18/07/2024 à 00:16, Moebius a écrit :

Am 17.07.2024 um 23:20 schrieb 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

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Le 18/07/2024 à 00:08, Moebius a écrit :

Am 17.07.2024 um 23:17 schrieb WM:

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

Nein.

They must be subtracted from your claim

Man kann Zahlen ganz schlecht von Behauptungen "abziehen",

Aber von behaupteten Zahlen.

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

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Le 18/07/2024 à 04:15, Richard Damon a écrit :

On 7/17/24 10:43 AM, WM wrote:

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?

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.

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

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Le 18/07/2024 à 06:30, Jim Burns a écrit :

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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

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Le 18/07/2024 à 07:52, Moebius a écrit :

Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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

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Le 18/07/2024 à 07:55, "Chris M. Thomasson" a écrit :

On 7/17/2024 2:07 PM, WM wrote:

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

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Le 18/07/2024 à 08:00, Moebius a écrit :

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

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Le 18/07/2024 à 08:09, Moebius a écrit :

Am 17.07.2024 um 23:20 schrieb WM:

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

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Am 18.07.2024 um 15:22 schrieb WM:

Le 18/07/2024 à 00:16, Moebius a écrit :

Am 17.07.2024 um 23:20 schrieb WM:

sgn(x)

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!

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Am 18.07.2024 um 15:20 schrieb WM:

Le 18/07/2024 à 00:08, Moebius a écrit :

Am 17.07.2024 um 23:17 schrieb WM:

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.

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Am 18.07.2024 um 15:34 schrieb WM:

Le 18/07/2024 à 07:52, Moebius a écrit :

Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

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.

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Am 18.07.2024 um 15:39 schrieb WM:

Le 18/07/2024 à 08:00, Moebius a écrit :

         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.

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Am 18.07.2024 um 15:41 schrieb WM:

Le 18/07/2024 à 08:09, Moebius a écrit :

Am 17.07.2024 um 23:20 schrieb WM:

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!

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On 7/18/2024 1:52 AM, Moebius wrote:

Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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

x⋅g∘f(y) = x⋅(y/√​̅y​̅²​̅+​̅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.

|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|

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Le 18/07/2024 à 19:00, Jim Burns a écrit :

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

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Le 18/07/2024 à 17:43, Moebius a écrit :

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

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Le 18/07/2024 à 17:38, Moebius a écrit :

Am 18.07.2024 um 15:20 schrieb WM:

Le 18/07/2024 à 00:08, Moebius a écrit :

Am 17.07.2024 um 23:17 schrieb WM:

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

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On 7/18/2024 2:28 PM, WM wrote:

Le 18/07/2024 à 19:00, Jim Burns a écrit :

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.

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Am 18.07.2024 um 19:00 schrieb Jim Burns:

On 7/18/2024 1:52 AM, Moebius wrote:

Am 18.07.2024 um 06:30 schrieb Jim Burns:

On 7/17/2024 1:49 PM, WM wrote:

Le 17/07/2024 à 19:13, FromTheRafters a écrit :

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).

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XPost: de.sci.mathematik

Am 18.07.2024 um 22:34 schrieb joes:

Am Thu, 18 Jul 2024 18:24:28 +0000 schrieb WM:

Le 18/07/2024 à 17:38, Moebius a écrit :

Am 18.07.2024 um 15:20 schrieb WM:

Le 18/07/2024 à 00:08, Moebius a écrit :

Am 17.07.2024 um 23:17 schrieb WM:

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.

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XPost: de.sci.mathematik

Am Thu, 18 Jul 2024 18:24:28 +0000 schrieb WM:

Le 18/07/2024 à 17:38, Moebius a écrit :

Am 18.07.2024 um 15:20 schrieb WM:

Le 18/07/2024 à 00:08, Moebius a écrit :

Am 17.07.2024 um 23:17 schrieb WM:

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.

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.

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Le 18/07/2024 à 21:16, Jim Burns a écrit :

On 7/18/2024 2:28 PM, WM wrote:

Le 18/07/2024 à 19:00, Jim Burns a écrit :

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

--- SoupGate-Win32 v1.05
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Le 18/07/2024 à 17:44, Moebius a écrit :

         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

--- SoupGate-Win32 v1.05
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Le 18/07/2024 à 22:34, joes a écrit :

Am Thu, 18 Jul 2024 18:24:28 +0000 schrieb 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.

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

--- SoupGate-Win32 v1.05
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Am 18.07.2024 um 22:58 schrieb WM:

Le 18/07/2024 à 22:34, joes a écrit :

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!!!

--- SoupGate-Win32 v1.05
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Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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".

--- SoupGate-Win32 v1.05
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Am 18.07.2024 um 23:35 schrieb Moebius:

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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???

--- SoupGate-Win32 v1.05
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On 7/18/2024 4:55 PM, WM wrote:

Le 18/07/2024 à 21:16, Jim Burns a écrit :

On 7/18/2024 2:28 PM, WM wrote:

Le 18/07/2024 à 19:00, Jim Burns a écrit :

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

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,∞)

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.

--- SoupGate-Win32 v1.05
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Am Thu, 18 Jul 2024 21:12:02 +0000 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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.

--- SoupGate-Win32 v1.05
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Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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).

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

--- SoupGate-Win32 v1.05
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Le 19/07/2024 à 09:11, joes a écrit :

Am Thu, 18 Jul 2024 21:12:02 +0000 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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

--- SoupGate-Win32 v1.05
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Le 18/07/2024 à 23:42, Jim Burns a écrit :

On 7/18/2024 4:55 PM, WM wrote:

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.

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.

No unit fractions are smaller than (0,∞)

Right. That means Fritsches x are not all x.

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

--- SoupGate-Win32 v1.05
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Am 19.07.2024 um 15:53 schrieb WM:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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?!

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.

(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 .

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 7/19/2024 9:53 AM, WM wrote:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

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.

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.

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]

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Am 19.07.2024 um 20:52 schrieb Jim Burns:

On 7/19/2024 9:53 AM, WM wrote:

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

Your "abbreviating" is a quantifier shift.

Right.

Making it less visible doesn't make it prove anything.

Indeed!

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.

Making it less visible doesn't make it prove anything.

Indeed!

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

Proof by "obviously".

Right.

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

Indeed!

If wishes were horses, beggars would ride!

--- SoupGate-Win32 v1.05
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Le 19/07/2024 à 16:29, Moebius a écrit :

Am 19.07.2024 um 15:53 schrieb WM:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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)

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.

(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.

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

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 .

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

--- SoupGate-Win32 v1.05
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Le 19/07/2024 à 20:52, Jim Burns a écrit :

On 7/19/2024 9:53 AM, WM wrote:

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.

"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

--- SoupGate-Win32 v1.05
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Le 19/07/2024 à 20:55, "Chris M. Thomasson" a écrit :

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

--- SoupGate-Win32 v1.05
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Le 19/07/2024 à 21:59, Moebius a écrit :

Am 19.07.2024 um 20:52 schrieb Jim Burns:

Your "abbreviating" is a quantifier shift.

Right.

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.

Regards, WM

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 7/20/24 8:47 AM, WM wrote:

Le 19/07/2024 à 20:52, Jim Burns a écrit :

On 7/19/2024 9:53 AM, WM wrote:

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.

"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

But no value of x can BE that left hand side of (0, oo) because if such
an x did exist then x/2 would be outside the interval, but also positive.

The fact being that these systems are UNBOUNDED, and thus there does not
exist a number "next to" another number, but we just have an infinite
density of numbers.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

Le 19/07/2024 à 16:29, Moebius a écrit :

Am 19.07.2024 um 15:53 schrieb WM:

Le 18/07/2024 à 23:35, Moebius a écrit :

Am 18.07.2024 um 23:12 schrieb WM:

Le 18/07/2024 à 17:44, Moebius a écrit :

         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?!

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

Nobody is claiming that.

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.

(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.

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

It is not the same y that is smaller than every x.

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 .

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].

No rule saying a function has to be continuous.

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

--- SoupGate-Win32 v1.05
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Le 20/07/2024 à 15:31, joes a écrit :

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

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

Nobody is claiming that.

But it is true.

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

It is not the same y that is smaller than every x.

Irrelevant. Only few unit fractions lie right-hand side of the ℵo
claimed for every x > 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].

No rule saying a function has to be continuous.

This function has to increase by at most 1. Between [0, 1] and (0, 1] it
cannot increase at all.

Regards, WM

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Le 20/07/2024 à 15:31, joes a écrit :

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

...

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].

No rule saying a function has to be continuous.

The ONLY rule in WMaths is: "If Crank Wolfgang Mückenheim from
Hochschule Augburg doesn't like it it is false, if he likes it
is true."

This is not a big issue on Usenet, he's not alone with this way
of "thinking".

It is, though, a BIG scandal in Hochschule Augsburg because he is
TEACHING there, and the institution is hiding itself behind
"academic freedom".

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Am 20.07.2024 um 15:30 schrieb Richard Damon:

On 7/20/24 8:47 AM, WM wrote:

My theorem: X is left-hand side of every x > 0 <==> X is left-hand
side of (0, oo).

Of course he has forgotten to state the "domain" of his "theorem"
explicitely. But it's reasonable zu assume that "X" and "x" here are
used as as varibles for real numbers.

Find a counter example or accept it.

But no [...] x can BE that left hand side of (0, oo) because if such
an x did exist then x/2 would be outside the interval, but also positive.

0 can't be "left hand side of (0, oo)" (WM)?

Hint: WM uses non-standard terminology here. "X is left hand side of (0,
oo)" means: Ax e (0, oo): X < x. We might abbreviate this with X < (0, oo).

If we state his "X is left-hand side of every x > 0" in the usual formal/symbolic language (which is uses in the context of set theory) we
get: "Ax e IR, x > 0: X < x".

Now clearly: Ax(x e IR & x > 0 <-> x e (0, oo)). Hence his "theorem" is
just a simple tautology (based on a simple definition):

Def.: X < (0, oo) :<-> Ax e (0, oo): X < x

Now:

Ax e IR, x > 0: X < x <-> Ax e (0, oo): X < x

since by definition (of an intervall) Ax(x e IR & x > 0 <-> x e (0, oo).

Hence (with def. from above):

Ax e IR, x > 0: X < x <-> X < (0, oo) .

A trivial "result".

Though I'd prefere it to state it the following way:

AX e IR: (Ax e IR, x > 0: X < x <-> X < (0, oo)) .

_________________________________________

Now we may, say, consider the set {X e IR : X < (0, oo)}. Then we get:

{X e IR : X < (0, oo)} = (-oo, 0].

Using an additional "extension" of "<" we might even state this result as

(-oo, 0] < (0, oo)

Meaning: Ax,y e IR: x e (-oo, 0] & y e (0, oo) -> x < y.

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Am 20.07.2024 um 15:31 schrieb joes:

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

That means NUF(x) is constant over the whole interval (0, oo),

Exactly!

Hint@WM: Ax(x e (0, oo)): NUF(x) = aleph_0.

Das hast Du gut erkannt, Mückenheim!

NUF increases from 0 to infinity between [0, 1] and (0, 1].

No rule saying [...]

Hier nochmal die Frage an Dich: Was bedeutet "NUF increases from 0 to
infinity between [0, 1] and (0, 1]" insbesondere das "between [0, 1] and
(0, 1]" --- ICH weiß es nicht, kannst Du es mir erklären?

ich erwähne das hier, weil Du darüber hinweggehst, ALSO OB WM hier etwas "Sinnvolles" gesagt hätte, das einer sinnvollen/vernünftigen Antwort
bedarf. (M.E. hat er hier aber nur saudummen Scheißdreck dahergelabert.)

VERMUTLICH will er damit IRGENDWIE den Umstand zum Ausdruck bringen,
dass NUF(0) = 0 und Ax(x e (0, oo)): NUF(x) = aleph_0 ist. But...

Und ja: "No rule saying a function has to be continuous." Insbesondere
auch dann nicht, WENN sie es nicht ist. :-P

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Am 20.07.2024 um 17:58 schrieb Moebius:

Am 20.07.2024 um 15:31 schrieb joes:

Am Sat, 20 Jul 2024 12:27:11 +0000 schrieb WM:

That means NUF(x) is constant over the whole interval (0, oo),

Exactly!

Hint@WM: Ax(x e (0, oo)): NUF(x) = aleph_0.

Correction:

Ax e (0, oo): NUF(x) = aleph_0 .

Das hast Du gut erkannt, Mückenheim!

NUF increases from 0 to infinity between [0, 1] and (0, 1].

No rule saying [...]

Hier nochmal die Frage an Dich: Was bedeutet "NUF increases from 0 to infinity between [0, 1] and (0, 1]" insbesondere das "between [0, 1] and
(0, 1]" --- ICH weiß es nicht, kannst Du es mir erklären?

ich erwähne das hier, weil Du darüber hinweggehst, ALSO OB WM hier etwas "Sinnvolles" gesagt hätte, das einer sinnvollen/vernünftigen Antwort bedarf. (M.E. hat er hier aber nur saudummen Scheißdreck dahergelabert.)

VERMUTLICH will er damit IRGENDWIE den Umstand zum Ausdruck bringen,
dass NUF(0) = 0 und Ax(x e (0, oo)): NUF(x) = aleph_0 ist. But...

Und ja: "No rule saying a function has to be continuous." Insbesondere
auch dann nicht, WENN sie es nicht ist. :-P

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Am 20.07.2024 um 20:56 schrieb Chris M. Thomasson:

On 7/20/2024 5:52 AM, WM wrote:

Le 19/07/2024 à 20:55, "Chris M. Thomasson" a écrit :

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.

There is no smallest unit fraction. No unit fraction represents zero. Therefore there will always be a gap between any unit fraction and zero. Period. Done. You are a special type of thing... The "dark" thing... Too stupid for measure.

The Force is strong with him. Especially the the dark side of the Force.

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Am Sat, 20 Jul 2024 12:52:05 +0000 schrieb WM:

Le 19/07/2024 à 20:55, "Chris M. Thomasson" a écrit :

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).

That is not the claim.

Of course there is no unit fraction left-hand side of every x.

Aha!

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guarantedd that n+1 exists for every n.

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Am 20.07.2024 um 23:17 schrieb joes:

Am Sat, 20 Jul 2024 12:52:05 +0000 schrieb WM:

Of course there is no unit fraction left-hand side of every x.

I guess with "left-hand side of" he means "which is smaller than".
<faceplam>

Of course there is no unit fraction which is smaller than every x [e

IR, x > 0].

Aha!

But for every x e IR, x > 0 there is a unit fraction which is smaller
than x.

Der Spinner begreift einfach nicht, dass man

Ax Ey ...

von

Ey Ax ...

unterscheiden muss.

Mit Dummheit kann man das m. E. nicht erklären. Es liegt m. E. eine
profunde Denkstörung vor. Genauer gesagt ein Wahn.

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

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Le 21/07/2024 à 01:15, Moebius a écrit :

But for every x e IR, x > 0 there is a unit fraction which is smaller
than x.

Left-hand side of every x > 0 there are unit fractions <==> Left-hand side
of (0, oo) there are unit fractions.

Regards, WM

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Le 20/07/2024 à 23:17, joes a écrit :

Am Sat, 20 Jul 2024 12:52:05 +0000 schrieb WM:

My theorem: X is left-hand side of every x > 0 <==> X is left-hand side
of (0, oo).

That is not the claim.

No, that is a theorem.

Regards, WM

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