Greg Cunt schrieb am Montag, 5. Juli 2021 um 16:32:24 UTC+2:
> On Monday, July 5, 2021 at 4:19:15 PM UTC+2, WM wrote:
>
> > Then the cursor would [have] aleph_0 unit fractions between itself and zero.
>
> Indeed! Till it reaches zero.

How could it? Try to think logically! It is impossible that aleph_0 unit fractions can be passed simultaneously.

Regards, WM

Greg Cunt schrieb am Montag, 5. Juli 2021 um 17:01:54 UTC+2:
> On Monday, July 5, 2021 at 4:19:15 PM UTC+2, WM wrote:
> > I only use linearity.
>
> Note that we use IR to model "space" and "time" in physics.

Here is no mention of |R. We have a sequence of unit fractions. Never two occupy the same point. Therefore never two or more can be passed at the same point by the cursor.
>
> Hence t can take any value in [0, 1].

There is no mention of t.
>
> With v = -1 (the const. speed of the cursor)

There is no constnt speed.

> So the path of the cursor consists of "uncountably" many points "in space" (i.e. [0, 1]).

THere is no mention of uncountability. It is shown to be nonsense.
>
> Since there is no last/smallest unit fraction in {1/n : n e IN}, the cursor can't pass "a last" unit fraction.

It passes every unit fraction that is there. Of those it passes a last one.

Regards, WM

William schrieb am Montag, 5. Juli 2021 um 18:10:57 UTC+2:
> On Monday, July 5, 2021 at 10:23:33 AM UTC-4, WM wrote:
> > William schrieb am Montag, 5. Juli 2021 um 01:57:24 UTC+2:
> > > Note that the sequence (t_1, t_2, t_3,...) (for every m element of |N_F, t_m is the time at which the cursor passes 1/m ) is a sequence without last element. Stepwise processes are not appropriate.
> > But it is appropriate to state that all *existing*
> The set of elements of U_F that can be written down has a last element, so a stepwise processes can be applied do this set. Yes a question we are not asking can be dealt with by a stepwise process. Recall that U_F does not have a last element and no element of U_F is "dark"

The cursor passes only unit fractions which are there. Of those it passes a last one because it cannot pass two or more or even infinitely many at the same point.

Regards, WM

Sergio schrieb am Montag, 5. Juli 2021 um 19:25:33 UTC+2:
> On 7/5/2021 9:23 AM, WM wrote:
> > William schrieb am Montag, 5. Juli 2021 um 01:57:24 UTC+2:
> >> Note that the sequence (t_1, t_2, t_3,...) (for every m element of |N_F, t_m is the time at which the cursor passes 1/m ) is a sequence without last element. Stepwise processes are not appropriate.
> >
> > But it is appropriate to state that all *existing* unit fractions are passed at different positions. Therefore if all have been passed, then a last one must have been passed.
> wrong, there are no "last one"
> > For not existing unit fractions nothing is claimed.
> wrong, the cursor has identified ALL numbers in the interval.

It passes all existing numbers. It cannot identify them because then a last existing one would be identified.
> >
> > By the way, time does not play a role. The cursor has only to satisfy one condition, namely to arrive at 0.
> that makes no sense.

I am of different opinion.

Regards, WM

On Monday, 5 July 2021 at 15:35:13 UTC-3, WM wrote:
> Sergio schrieb am Montag, 5. Juli 2021 um 19:25:33 UTC+2:
> > On 7/5/2021 9:23 AM, WM wrote:
> > > William schrieb am Montag, 5. Juli 2021 um 01:57:24 UTC+2:
> > >> Note that the sequence (t_1, t_2, t_3,...) (for every m element of |N_F, t_m is the time at which the cursor passes 1/m ) is a sequence without last element. Stepwise processes are not appropriate.
> > >
> > > But it is appropriate to state that all *existing* unit fractions are passed at different positions. Therefore if all have been passed, then a last one must have been passed.
> > wrong, there are no "last one"
> > > For not existing unit fractions nothing is claimed.
> > wrong, the cursor has identified ALL numbers in the interval.
> It passes all existing numbers. It cannot identify them because then a last existing one would be identified.
> > >
> > > By the way, time does not play a role. The cursor has only to satisfy one condition, namely to arrive at 0.
> > that makes no sense.
> I am of different opinion.

Evidently.

Your difference of opinion is noted but is utterly irrelevant. Anyone who thinks the cursor can only reach zero by passing only finitely many unit fractions in order to get there "because it cannot pass two or more or even infinitely many at the same point" has lost *ALL* ability to reason along with any credibility (but that was lost years ago, anyway.).

On Monday, July 5, 2021 at 2:33:08 PM UTC-4, WM wrote:
> William schrieb am Montag, 5. Juli 2021 um 18:10:57 UTC+2:
> > On Monday, July 5, 2021 at 10:23:33 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 5. Juli 2021 um 01:57:24 UTC+2:
> > > > Note that the sequence (t_1, t_2, t_3,...) (for every m element of |N_F, t_m is the time at which the cursor passes 1/m ) is a sequence without last element. Stepwise processes are not appropriate.
> > > But it is appropriate to state that all *existing*
> > The set of elements of U_F that can be written down has a last element, so a stepwise processes can be applied do this set. Yes a question we are not asking can be dealt with by a stepwise process. Recall that U_F does not have a last element and no element of U_F is "dark"
> The cursor passes only unit fractions which are there. Of those it passes a last one because it cannot pass two or more or even infinitely many at the same point.

There is no need for the cursor to pass a last point, or to pass more than one point at any one time for the cursor to pass every point in U_F.

--
William Hughes

William schrieb am Montag, 5. Juli 2021 um 20:11:49 UTC+2:
> On Monday, July 5, 2021 at 1:56:31 PM UTC-4, WM wrote:
> > William schrieb am Montag, 5. Juli 2021 um 19:40:19 UTC+2:
> > > > On 7/5/2021 9:23 AM, WM wrote:
> > >
> > > > > For not existing unit fractions nothing is claimed.
> > > However, the elements of U_F that cannot be written down form a set with cardinality aleph_0
> > This assertion is disproved by the fact the cursor passes dark elements..
> Nope, The question of putative "dark" elements passed by the cursor has nothing to do with U_F which does not contain "dark" elements. U_F is a Peano set and like any Peano set has cardinality aleph_0 (And like any Peano set it contains "non existing" elements, elements that can not be wriitten down)

But they can be passed by the cursor. As long as every passed unit fraction has aleph_0 successors, the cursor cannot reach zero (because aleph_0 unit fractions are between it and zero).

Regards, WM

Gus Gassmann schrieb am Montag, 5. Juli 2021 um 20:44:09 UTC+2:
> On Monday, 5 July 2021 at 15:35:13 UTC-3, WM wrote:

es not play a role. The cursor has only to satisfy one condition, namely to arrive at 0.
> > > that makes no sense.
> > I am of different opinion.
> Evidently.
>
> Your difference of opinion is noted but is utterly irrelevant.

I am of different opinion.

> Anyone who thinks the cursor can only reach zero by passing only finitely many unit fractions in order to get there "because it cannot pass two or more or even infinitely many at the same point" has lost *ALL* ability to reason along with any credibility

in matheology? That is easy to bear.

The cursor cannot reach zero as long as unit fractions are between itself an zero. Not even one is allowed. That is a matter of fact. Note that for ever n 0 < 1/n < cursor position. Also passing aleph_0 unit fractions simultaneously is impossible. Note that 1/n < 1/m or vice versa for different unit fractions.

Regards, WM

William schrieb am Montag, 5. Juli 2021 um 20:50:27 UTC+2:
> On Monday, July 5, 2021 at 2:33:08 PM UTC-4, WM wrote:

> > The cursor passes only unit fractions which are there. Of those it passes a last one because it cannot pass two or more or even infinitely many at the same point.
> There is no need for the cursor to pass a last point, or to pass more than one point at any one time for the cursor to pass every point in U_F.

I would also think so. An potentielly infinite collection has no last point.. But the claim that every unit fraction has aleph_0 successors can be excluded. The cursor cannot reach zero before it has passed all unit fractions - one by one. Note that for any different unit fractions 1/n < 1/m or vice versa.

Regards, WM

On Monday, July 5, 2021 at 3:06:17 PM UTC-4, WM wrote:
> William schrieb am Montag, 5. Juli 2021 um 20:11:49 UTC+2:
> > On Monday, July 5, 2021 at 1:56:31 PM UTC-4, WM wrote:
> > > William schrieb am Montag, 5. Juli 2021 um 19:40:19 UTC+2:
> > > > > On 7/5/2021 9:23 AM, WM wrote:
> > > >
> > > > > > For not existing unit fractions nothing is claimed.
> > > > However, the elements of U_F that cannot be written down form a set with cardinality aleph_0
> > > This assertion is disproved by the fact the cursor passes dark elements.
> > Nope, The question of putative "dark" elements passed by the cursor has nothing to do with U_F which does not contain "dark" elements. U_F is a Peano set and like any Peano set has cardinality aleph_0 (And like any Peano set it contains "non existing" elements, elements that can not be wriitten down)
> But they can be passed by the cursor.

Indeed,

>As long as every passed unit fraction has aleph_0 successors, the cursor cannot reach zero

"As long" is a period of time [0,1) (Have the cursor start at time 0 and travel at a speed of 1). It is not "forever". The cursor cannot reach 0 for any time t<1. The cursor can read 0 for any time t>=1.

-William Hughes

On Monday, July 5, 2021 at 8:30:50 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Montag, 5. Juli 2021 um 17:01:54 UTC+2:
> >
> > Note that we use IR to model "space" and "time" in physics.
> >
> Here is no mention of IR.

Look you psychotic crank: You wrote:

"The cursor runs from 1 to 0."

You didn't explicitly mention "space" and IR here. But IR is a rather natural choice for describing the movement of some "entity", you silly dumbfuck. Especially when the "start point" is given as 1 and the "end point" as 0.

"Every passed accessible unit fraction has infinitely many smaller unit fractions as successors. When the cursor passes 0 all unit fractions have been passed."

This can be described as I told you, you silly asshole. We don't want a jumping "cursor."

> We have a sequence of unit fractions.

No, we don't. Hint: 0 is not a unit fraction, you fucking asshole full of shit!

Now a MOVEMENT is something whcih can be described with an equation for the path of the moving entity. This description (formula) usually uses (time) t as a parameter.

Here:

> > t can take any value in [0, 1].
> >
> There is no mention of t.

It IS _implicitely_, you silly asshole!

You wrote: "When the cursor passes 0 all unit fractions have been passed."

WHEN [...] PASSES 0 then [...] HAVE BEEN PASSED.

What's the matter with you? Lost your last marbles?

> > With v = -1 (the const. speed of the cursor)
> >
> There is no constant speed.

Right. It needn't be constant. But things are simplified this way. No need to obscure simple states of affairs, dumbo.

So *I assume* a constant speed of the "cursor".

> > the path of the cursor consists of "uncountably" many points "in space" (i.e. [0, 1]).
> >
> There is no mention of uncountability.

This can be PROVEN, you silly dumbfuck.

Didn't you read my post or are you on drugs?

> > Since there is no last/smallest unit fraction in {1/n : n e IN}, the cursor can't pass "a last" unit fraction.
> >
> It passes every unit fraction that is there.

"That is there" is a nonsensical formulation, psycho.

But you may claim, "it passes" every unit fraction in [0, 1] or any unit fraction in {1/n : n e IN}.

William schrieb am Montag, 5. Juli 2021 um 21:26:59 UTC+2:
> On Monday, July 5, 2021 at 3:06:17 PM UTC-4, WM wrote:

> >As long as every passed unit fraction has aleph_0 successors, the cursor cannot reach zero
> "As long" is a period of time [0,1)

No, it means "given that". No time is involved.

> (Have the cursor start at time 0 and travel at a speed of 1). It is not "forever".

That proves that not always unit fractions are between the curser and zero. When the cursor has arrived, then there is no unit fraction between it and zero. All are passed. But never two or more are passed at the same position.

> The cursor cannot reach 0 for any time t<1. The cursor can read 0 for any time t>=1.

We should forget the time. Fact is: When the cursor reaches 0 no unit fraction is between itself and 0. And fact is further that the cursor never passes aleph_0 unit fractions at the same point.

Regards, WM

On Monday, July 5, 2021 at 3:18:32 PM UTC-4, WM wrote:
> William schrieb am Montag, 5. Juli 2021 um 20:50:27 UTC+2:
> > On Monday, July 5, 2021 at 2:33:08 PM UTC-4, WM wrote:
>
> > > The cursor passes only unit fractions which are there. Of those it passes a last one because it cannot pass two or more or even infinitely many at the same point.
> > There is no need for the cursor to pass a last point, or to pass more than one point at any one time for the cursor to pass every point in U_F.
> I would also think so. An potentielly infinite collection has no last point.

As does the set U_F which (unlike a "potentially infinite collection") does not change.

> the claim that every unit fraction has aleph_0 successors can be excluded. The cursor cannot reach zero before it has passed all unit >
> fractions - one by one.

Correct, The cursor can and does pass each element of U_F one by one. What the cursor does not do is pass a last element of U_F. The fact that it passes each element of U_F one by one does not mean it passes as last element of U_F.

--
William Hughes

Greg Cunt schrieb am Montag, 5. Juli 2021 um 21:35:48 UTC+2:
> On Monday, July 5, 2021 at 8:30:50 PM UTC+2, WM wrote:

> "The cursor runs from 1 to 0."
>
> You didn't explicitly mention "space" and IR here. But IR is a rather natural choice for describing the movement of some "entity",

There are to many dark points. We need only the unit fractions and if you like the distances between them.
>
> "Every passed accessible unit fraction has infinitely many smaller unit fractions as successors. When the cursor passes 0 all unit fractions have been passed."

> We don't want a jumping "cursor."

Therefore we cannot agree that all unit fractions have aleph_0 successors.
>
> Here:
> > > t can take any value in [0, 1].

You can take any accessible value.
> > >
> > There is no mention of t.
> It IS _implicitely

No, the cursor need not even move with constant velocity.
>
> You wrote: "When the cursor passes 0 all unit fractions have been passed."
>
> WHEN [...] PASSES 0 then [...] HAVE BEEN PASSED.

Yes. No velocity given.
>
> So *I assume* a constant speed of the "cursor".

It is not prohibited. If you like time in mathematics.

> > There is no mention of uncountability.
>
> This can be PROVEN,

No longer. Note that dark numbers cannot be put in order.
>
> > It passes every unit fraction that is there.
> "That is there" is a nonsensical formulation

It is important because too many matheologians believ in unit fractions which are no there-
>
> But you may claim, "it passes" every unit fraction in [0, 1] or any unit fraction in {1/n : n e IN}.

Including the last one, which however is dark and therefore cannot be put in any order and cannot be determined as the last one.

Regards, WM

On Monday, July 5, 2021 at 9:39:29 PM UTC+2, WM wrote:
> William schrieb am Montag, 5. Juli 2021 um 21:26:59 UTC+2:
> > On Monday, July 5, 2021 at 3:06:17 PM UTC-4, WM wrote:
> >
> > "As long" is a period of time [0,1)
> >
> No [bla bla]. No time is involved.

LOOK, YOU FUCKIN LIAR, in de.sci.mathematik YOU WROTE:

"Der Cursor schafft die Strecke in sagen wir mal einer Sekunde." (Wolfgang Mückenheim)

[ "The cursor traverses the path in, say, a second." ] (Wolfgang Mückenheim)

> > Have the cursor start at time 0 and travel at a speed of 1). It is not "forever".
> > The cursor cannot reach 0 for any time t < 1. The cursor can reach 0 for any time t>=1.
> >
> We should forget the time.

No, we shouldn't, idiot

Hint: "The cursor traverses the path in, say, a second." (Wolfgang Mückenheim)

On Monday, July 5, 2021 at 3:39:29 PM UTC-4, WM wrote:
> William schrieb am Montag, 5. Juli 2021 um 21:26:59 UTC+2:
> > On Monday, July 5, 2021 at 3:06:17 PM UTC-4, WM wrote:
>
> > >As long as every passed unit fraction has aleph_0 successors, the cursor cannot reach zero
> > "As long" is a period of time [0,1)
> No, it means "given that". No time is involved.
> > (Have the cursor start at time 0 and travel at a speed of 1). It is not "forever".
> That proves that not always unit fractions are between the curser and zero. When the cursor has arrived, then there is no unit fraction between it and zero. All are passed. But never two or more are passed at the same position.

Correct, So What?

> > The cursor cannot reach 0 for any time t<1. The cursor can read 0 for any time t>=1.
> We should forget the time Fact is: When the cursor reaches 0 no unit fraction is between itself and 0. And fact is further that the cursor never passes aleph_0 unit fractions at the same point.

Correct. So what?

--

William Hughes

On 7/4/2021 9:08 AM, WM wrote:

> The cursor runs from 1 to 0. Every passed accessible unit fraction
> has infinitely many smaller unit fractions as successors.

Every movement of the cursor passes infinitely-many points.
Perhaps some of them are your unit fractions. Perhaps not.

> When the cursor passes 0 all unit fractions have been passed.

Agreed. All are all.

> None remains, not even the infinitely many following upon
> every accessible unit fraction.

Any point x before 0 and after each fraction 1/k corresponds to
a point 1/x after each (finite) natural k. Archimedes would say
that no such 1/x exist in the line, but that each point must be
able to be encompassed by finitely-many unit-interval extensions
of the line.

No 1/x, no x, no dark points.

> They are not accessible. They are dark.

Your argument is
< If we describe movement with the 1/k, it is incorrect.
< Therefore, dark points exist.

But you don't say anything about what your dark points are like.
The only use I see for dark points is that they give you (WM)
something to say to impertinent students who expect you to
know something about the material you're lecturing on.

----
A better description of these points is
(i)
definable points passed in every movement
("definable" might only be "finite-length decimals"), and
(ii)
single-component points existing between every cut of the
line, in order to prevent continuous (by definition) paths
from teleporting.

( Anywhere a single-component point is missing, a path can
( be continuous before and continuous after, but still have
( the before and after arbitrarily far apart, hence, teleporting.
( ( In order to ban teleporting, include all single-component
( points. This gives Dedekind completeness and the real numbers.

On Monday, July 5, 2021 at 9:46:46 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Montag, 5. Juli 2021 um 21:35:48 UTC+2:
> >
> > IR is a rather natural choice for describing the movement of some "entity",
> >
> There are to many dark points.

That's why we use IR. In IR there are only so called "real numbers" (no dark matter)!

Maybe you have learned in Analysis I how to deal with them, but maybe you didn't.

Now:

> > > > t can take any value in [0, 1].
> > > >
> > > There is no mention of t.

LOOK, YOU FUCKIN LIAR, in de.sci.mathematik YOU WROTE:

"Der Cursor schafft die Strecke in sagen wir mal einer Sekunde." (Wolfgang Mückenheim)

[ "The cursor traverses the path in, say, a second." ] (Wolfgang Mückenheim)

"t" is a common choice in physics for a variable referring to "time".

> > No velocity given.

I already told you that for simplicity we may choose v = -1.

Then we get the simple equation

x(t) = 1 + v * t (with t e [0, 1])

for the path of the "cursor".

Hint:

Maybe you have learned in school:

V = S / T (average speed)

Here S = -1 ("The cursor runs from 1 to 0." [WM])

T = 1 ("The cursor traverses the path in, say, a second." [WM])

Hence V = -1.

So v = -1 (const.) seems to be a good choice.

Hence

> > *I assume* a constant speed of the "cursor".

Things are simplified this way. No need to obscure simple states of affairs, dumbo.

On Monday, 5 July 2021 at 16:14:49 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 5. Juli 2021 um 20:44:09 UTC+2:
> > On Monday, 5 July 2021 at 15:35:13 UTC-3, WM wrote:
>
> es not play a role. The cursor has only to satisfy one condition, namely to arrive at 0.
> > > > that makes no sense.
> > > I am of different opinion.
> > Evidently.
> >
> > Your difference of opinion is noted but is utterly irrelevant.
>
> I am of different opinion.
>
> > Anyone who thinks the cursor can only reach zero by passing only finitely many unit fractions in order to get there "because it cannot pass two or more or even infinitely many at the same point" has lost *ALL* ability to reason along with any credibility
> in matheology? That is easy to bear.
>
> The cursor cannot reach zero as long as unit fractions are between itself an zero. Not even one is allowed. That is a matter of fact. Note that for ever n 0 < 1/n < cursor position. Also passing aleph_0 unit fractions simultaneously is impossible. Note that 1/n < 1/m or vice versa for different unit fractions.

As I said: Your ability to reason is now officially zero. As such, it doesn't interest me any longer, and neither does this discussion.

On Monday, July 5, 2021 at 9:46:46 PM UTC+2, WM wrote:
> >
> > But you may claim, "it passes" every unit fraction in [0, 1] or any unit fraction in {1/n e IR : n e IN}.
> >
> Including the last one, which <bla bla>

There is no "last" unit fraction in {1/n e IR : n e IN} and hence not in [0, 1] (since there is no unit fraction in [0, 1] which is not in {1/n e IR : n e IN}).

Since the cursor "passes" exactly all the points in [0, 1]
(remember its path is describe by the equation x(t) = 1 - t (with t e [0, 1])
it passes all unit fractions. Hence there is no "last" unit fraction it passes.

On Monday, July 5, 2021 at 9:14:49 PM UTC+2, WM wrote:

> The cursor cannot reach zero as long as unit fractions are between itself an zero.

Well..."cannot reach ... as long as ..." seems to be an "inappropriate" choice of words here.

Anyway, the cursor reaches 0 at t = 1 (if we assume v = -1).

So what does this mean?

This means: There are (always) infinitely many unit fractions between 0 and the position of the cursor x(t) for all t e [0, 1). And there are NO unit fractions (anymore) between 0 and the position of the cursor x(t) for t = 1 (since x(1) = 0).

EOD

Greg Cunt schrieb am Montag, 5. Juli 2021 um 22:36:22 UTC+2:
> On Monday, July 5, 2021 at 9:14:49 PM UTC+2, WM wrote:
>
> > The cursor cannot reach zero as long as unit fractions are between itself an zero.
> Well..."cannot reach ... as long as ..." seems to be an "inappropriate" choice of words here.
>
> Anyway, the cursor reaches 0 at t = 1 (if we assume v = -1).
>
> So what does this mean?

It means that even the last unit fractions has been passed.
>
> This means: There are (always) infinitely many unit fractions between 0 and the position of the cursor x(t) for all t e [0, 1).

Of course for every definable position or time you are right. But the cursor does not pass infinitely many unit fractions at the same position or at the same time. Can you understand this?

> And there are NO unit fractions (anymore) between 0 and the position of the cursor x(t) for t = 1 (since x(1) = 0).
>
Nonsense. The cursor does not pass two or more at the same position or time.

Regards, WM

Gus Gassmann schrieb am Montag, 5. Juli 2021 um 22:25:23 UTC+2:
> On Monday, 5 July 2021 at 16:14:49 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 5. Juli 2021 um 20:44:09 UTC+2:
> > > On Monday, 5 July 2021 at 15:35:13 UTC-3, WM wrote:
> >
> > es not play a role. The cursor has only to satisfy one condition, namely to arrive at 0.
> > > > > that makes no sense.
> > > > I am of different opinion.
> > > Evidently.
> > >
> > > Your difference of opinion is noted but is utterly irrelevant.
> >
> > I am of different opinion.
> >
> > > Anyone who thinks the cursor can only reach zero by passing only finitely many unit fractions in order to get there "because it cannot pass two or more or even infinitely many at the same point" has lost *ALL* ability to reason along with any credibility
> > in matheology? That is easy to bear.
> >
> > The cursor cannot reach zero as long as unit fractions are between itself an zero. Not even one is allowed. That is a matter of fact. Note that for ever n 0 < 1/n < cursor position. Also passing aleph_0 unit fractions simultaneously is impossible. Note that 1/n < 1/m or vice versa for different unit fractions.
> As I said: Your ability to reason is now officially zero.

At zero rematkable hings happen.

> As such, it doesn't interest me any longer, and neither does this discussion.

You suspect that you will no longer be able to adhere to the nonsense of set theory? Note that never two unit fractions reside at the same position, let alone aleph_0.

Regards, WM

WM formulated on Monday :
> William schrieb am Montag, 5. Juli 2021 um 21:26:59 UTC+2:
>> On Monday, July 5, 2021 at 3:06:17 PM UTC-4, WM wrote:
>
>>> As long as every passed unit fraction has aleph_0 successors, the cursor
>>> cannot reach zero
>> "As long" is a period of time [0,1)
>
> No, it means "given that". No time is involved.
>
>> (Have the cursor start at time 0 and travel at a speed of 1). It is not
>> "forever".
>
> That proves that not always unit fractions are between the curser and zero.
> When the cursor has arrived, then there is no unit fraction between it and
> zero. All are passed. But never two or more are passed at the same position.
>
>> The cursor cannot reach 0 for any time t<1. The cursor can read 0 for any
>> time t>=1.
>
> We should forget the time.

Then what does 'arrive' and 'passing' even mean? Time is implicit in
your adopting a step by step method.

On Monday, July 5, 2021 at 11:25:43 PM UTC+2, WM wrote:
> >
> > Anyway, the cursor reaches 0 at t = 1 (if we assume v = -1, and x(0) = 1).
> >
> > So what does this mean?
> >
> It means that [all] unit fractions has been passed [when the cursor is at x = 0].

Exactly!

Moreover,

> > this means: There are (always) infinitely many unit fractions between 0 and the position of the cursor x(t) for all t e [0, 1).
> >
> Of course for every [...] position [x e (0, 1] or time [t e [0, 1)] you are right.

Sure. Moreover (!) there are (always) only finitel many unit fractions between 1 and the position of the cursor x(t) for all t e [0, 1).

> But the cursor does not pass infinitely many unit fractions at the same position or at the same time. Can you understand this?

Sure.

At a certain position/time (x, t) the cursor "passes" a (exactly one) unit fraction, i. e. x = x(t) = 1/n for some n e IN, or none, for example for a time t where x = x(t) = 2/3; or for x = x(t) with t = 1.

> > And there are NO unit fractions (anymore) between 0 and the position of the cursor x(t) for t = 1 (since x(1) = 0).
> >
> Nonsense.

Nope. A simple fact.

Hint: there is no unit fraction between 0 and 0. :-)

> The cursor does not pass two or more at the same position or time.

Did anyone claim that?

Hint: For any x, x e IR, x > 0, there are uncountably many real numbers between x and 0. For x = 0 there is no number "between" x and 0. (Between here means ... < r < ...)