Gus Gassmann schrieb am Sonntag, 11. Juli 2021 um 18:24:57 UTC+2:
> On Sunday, 11 July 2021 at 08:32:55 UTC-3, WM wrote:

> > The unit fractions discerned and counted are well-ordered by this counting.
>
> Sure. If you assume there are only finitely many integers, then there are only finitely many integers.

It is clear that aleph_0 belong to the first bunch which is not counted and cannot be counted becaue it is dark. Therefore only a finite remainder is available for operations with individuals.

Regards, WM

Gus Gassmann schrieb am Sonntag, 11. Juli 2021 um 18:31:05 UTC+2:
> On Sunday, 11 July 2021 at 12:36:26 UTC-3, WM wrote:
> > You don't know and can't know about the order of aleph_0 of them.
> Of course you can know about the order. It is induced by the well-order on the natural numbers. Any unit fraction 1/n with n > 1 is smaller than 1. For any two unit fractions 1/n and 1/m you have either 1/n < 1/m or 1/n > 1/m or 1/n = 1/m, and in the last case you know that n = m.

But you cannot compare two numbers when you cannot address or choose them. Look at the unit fractions. A bunch of aleph_0 will be passed by the cursor without counting any element. Every counted element has aleph_0 not counted between itself and zero.

Regards, WM

Jim Burns schrieb am Sonntag, 11. Juli 2021 um 21:04:42 UTC+2:

> Consider the unit interval [0,1].
>
> In fact, consider it twice: [0,1]x[0,1], a square.
>
> We would like to know how to say that f: [0,1] --> [0,1] is a
> _continuous_ path from the square's lower left corner (0,0) to
> its upper right corner (1,1).
>
> What is a _discontinuity_ so that we can say to f, "Don't do that"?

Every definable point is surrounded by dark points - by infinitely many. This does not say don't do that. But it should be taken into account.

Regards, WM

On 7/11/2021 1:14 PM, WM wrote:
> William schrieb am Sonntag, 11. Juli 2021 um 19:10:42 UTC+2:
>> On Sunday, July 11, 2021 at 11:44:59 AM UTC-4, WM wrote:
>>
>>> The [elements of U_F] discerned by the cursor are well ordered:
>>
>> Add "discerned by the cursor" to the list of things that mean "can be written down. This is a finite set, trivially it is well ordered
>>
>> u_1, u_2, u_3, ..., but between u_1 and zero there are aleph_0 [elements of U_F]
>>
>> which are not well ordered by the natural order, and do not have first element under this ordering
>
> They cannot be ordered because they cannot be written down, distinguished from each other, discerned etc. etc. Of course they must be ordered such that every 1/n is between 1/(n-1) and 1/(n+1). And if n is even, then n+1 is odd. But it is impossible to find out which is which. But they exist and the cursor must have passed all of them.
>
> Regards, WM
>

premise is wrong.

The cursor passes ALL real numbers in the interval [0,1]

All are well ordered.

Just because you pick out a sequence, this fact does not change.

On 7/11/2021 2:58 PM, WM wrote:
> William schrieb am Sonntag, 11. Juli 2021 um 20:53:56 UTC+2:
>> On Sunday, July 11, 2021 at 2:14:54 PM UTC-4, WM wrote:
>>> William schrieb am Sonntag, 11. Juli 2021 um 19:10:42 UTC+2:
>>>> On Sunday, July 11, 2021 at 11:44:59 AM UTC-4, WM wrote:
>>>>
>>>>> The [elements of U_F] discerned by the cursor are well ordered:
>>>>
>>>> Add "discerned by the cursor" to the list of things that mean "can be written down. This is a finite set, trivially it is well ordered
>>>>
>>>> u_1, u_2, u_3, ..., but between u_1 and zero there are aleph_0 [elements of U_F]
>>>>
>>>> which are not well ordered by the natural order, and do not have first element under this ordering
>>> They cannot be ordered because they cannot be written down,
>> Piffle. U_F is, like any Peano set, a totally ordered set, despite the fact that most of the elements of U_F cannot be written down.
>
> and it cannot be checked whether they are ordered because they cannot be taken as individuals.
>
>> We know that for an two elements of U_F, x and y, x<y, x=y, or x>y. We know this if the two elements are with a fox. We know this if the two elements are in a box, We know this if the two elements are here or there. We know this if two element cannot be written down. So we know U_F is totally ordered.
>
> I said that every natural has neighbours. But the dark or not writable elements cannot be compared. How would you compare what cannot be written or chosen or addressed in any way?

wrong. the Cursor instantated, discovered, etc ALL real numbers in the
interval [0,1]

there are no dark numbers.

>>
>> It did not take you long to go from "I am right because the elements are well ordered" to "I am right because the elements cannot be ordered"
>
> You misunderstood. The elements counted by the cursor (or its pilot) are well-ordered. They have a first one.

there is no first one, we proved that there is no first one.

you dont use math much do you ?

The elements passed but not counted by the cursor cannot be ordered.
How would you compare two natural numbers which cannot be addressed in
any way?
>
> Regards, WM
>

On 7/11/2021 10:39 AM, WM wrote:
> Greg Cunt schrieb am Sonntag, 11. Juli 2021 um 15:58:43 UTC+2:
>> On Sunday, July 11, 2021 at 1:00:58 PM UTC+2, WM wrote:
>>>
>>> If the cursor meets a non-empty set, then it meets a first element.
>
>>> Proved by Zermelo: Every non-empty well-ordered set has a first element.
>> Look, you silly crank:
>>
>> The ordered set { ... 1/4 < 1/3 < 1/2 < 1/1} is neither well-ordered, nor does it have a first/smallest element.
>>
> But the set of unit fractions enumerated by the cursor u_1, u_2, u_3, ... is well-ordered.
>
> Regards, WM
>

bfd, so is the real line.

On 7/11/2021 8:57 AM, WM wrote:
> Sergio schrieb am Sonntag, 11. Juli 2021 um 15:55:22 UTC+2:
>> On 7/11/2021 5:53 AM, WM wrote:
>>> Dan Christensen schrieb am Samstag, 10. Juli 2021 um 17:26:52 UTC+2:
>>>> Not even you can deny (see above) that the point particle (your "cursor"), as it goes from the point x=0 to the point x=1, passes through or meets EVERY unit fraction in the interval.
>>>
>>> Of course, even the first one and the other infinitely many dark one, but the first discernible unit fraction has already infinitely many, i.e. almost all, unit fractions between itself and zero.
>> so you agree there is no first discernible unit fraction.
>
> There are many. Almost all.
>
> Regards, WM
>

you changed your mind ?

On Sunday, July 11, 2021 at 3:58:13 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 11. Juli 2021 um 20:53:56 UTC+2:
> > On Sunday, July 11, 2021 at 2:14:54 PM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 11. Juli 2021 um 19:10:42 UTC+2:
> > > > On Sunday, July 11, 2021 at 11:44:59 AM UTC-4, WM wrote:
> > > >
> > > > > The [elements of U_F] discerned by the cursor are well ordered:
> > > >
> > > > Add "discerned by the cursor" to the list of things that mean "can be written down. This is a finite set, trivially it is well ordered
> > > >
> > > > u_1, u_2, u_3, ..., but between u_1 and zero there are aleph_0 [elements of U_F]
> > > >
> > > > which are not well ordered by the natural order, and do not have first element under this ordering
> > > They cannot be ordered because they cannot be written down,
> > Piffle. U_F is, like any Peano set, a totally ordered set, despite the fact that most of the elements of U_F cannot be written down.
> and it cannot be checked whether they are ordered because they cannot be taken as individuals.

"Taken as individuals" is yet one more or your ways of saying "can be written down". So What? We do not need write the elements of U_F down to know that each element is ordered.

> > We know that for an two elements of U_F, x and y, x<y, x=y, or x>y. We know this if the two elements are with a fox. We know this if the two elements are in a box, We know this if the two elements are here or there. We know this if two element cannot be written down. So we know U_F is totally ordered.
> I said that every [element of U_F]l has neighbours.

Indeed, because every element of U_F is orderable.

> But the dark

Are totally irrelevant. No element of U_F is dark

--
William Hughes

On 7/11/2021 8:41 AM, WM wrote:
> Sergio schrieb am Sonntag, 11. Juli 2021 um 15:14:44 UTC+2:
>> On 7/11/2021 7:46 AM, WM wrote:
>>> Greg Cunt schrieb am Sonntag, 11. Juli 2021 um 00:36:30 UTC+2:
>
>>>> Hint: for all t e IR with 0 < t <= 1 this cardinality will be aleph_0.
>>>
>>> I fully agree. This proves that aleph_0 unit fractions exist and necessarily are passed by the cursor like all point in the unit interval but cannot be counted.
>> the Cursor illuminates, discovers, instantates ALL numbers on
>> the real line from [0,1]
>
> What is the first unit fraction that it discovers?
>
> Regards, WM
>

proof there is no first number;

assume the first is k

However k/2 is before k and is now first

contradiction

therefore k is not first number.

therefore there is no first number.

any of that sink in yet ?

On 7/11/2021 3:16 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 11. Juli 2021 um 18:31:05 UTC+2:
>> On Sunday, 11 July 2021 at 12:36:26 UTC-3, WM wrote:
>>> You don't know and can't know about the order of aleph_0 of them.
>> Of course you can know about the order. It is induced by the well-order on the natural numbers. Any unit fraction 1/n with n > 1 is smaller than 1. For any two unit fractions 1/n and 1/m you have either 1/n < 1/m or 1/n > 1/m or 1/n = 1/m, and in the last case you know that n = m.
>
> But you cannot compare two numbers when you cannot address or choose them. Look at the unit fractions. A bunch of aleph_0 will be passed by the cursor without counting any element.

Wrong. the Cursor passes ALL real numbers in the interval [0,1]

> Every counted element has aleph_0 not counted between itself and zero.

wrong.

>
> Regards, WM
>

On 7/11/2021 3:58 PM, WM wrote:
> William schrieb
> am Sonntag, 11. Juli 2021 um 20:53:56 UTC+2:
>> On Sunday, July 11, 2021 at 2:14:54 PM UTC-4,
>> WM wrote:

>>> They cannot be ordered because they cannot be written down,
>>
>> Piffle. U_F is, like any Peano set, a totally ordered set,
>> despite the fact that most of the elements of U_F cannot be
>> written down.
>
> and it cannot be checked whether they are ordered because
> they cannot be taken as individuals.

It can be reasoned from their description that they are totally
ordered.[1]

"Checking" whether they are ordered makes just as much sense as
"checking" whether triangles have three corners: none at all.

[1]
You (WM) may be interested to hear that being totally ordered
and being ordered mean different things.

Feel free to ask your colleagues at Hochschule Augsburg
for details.

On Sunday, 11 July 2021 at 17:13:10 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 11. Juli 2021 um 18:24:57 UTC+2:
> > On Sunday, 11 July 2021 at 08:32:55 UTC-3, WM wrote:
>
> > > The unit fractions discerned and counted are well-ordered by this counting.
> >
> > Sure. If you assume there are only finitely many integers, then there are only finitely many integers.
> It is clear that aleph_0 belong to the first bunch which is not counted and cannot be counted becaue it is dark. Therefore only a finite remainder is available for operations with individuals.

You can't even articulate which is the "first bunch" and which is the "remainder".

So your entire contribution is worthless. As I said "whoop-de-doodoo".

On Sunday, 11 July 2021 at 20:11:31 UTC-3, Jim Burns wrote:
> On 7/11/2021 3:58 PM, WM wrote:
> > William schrieb
> > am Sonntag, 11. Juli 2021 um 20:53:56 UTC+2:
> >> On Sunday, July 11, 2021 at 2:14:54 PM UTC-4,
> >> WM wrote:
>
> >>> They cannot be ordered because they cannot be written down,
> >>
> >> Piffle. U_F is, like any Peano set, a totally ordered set,
> >> despite the fact that most of the elements of U_F cannot be
> >> written down.
> >
> > and it cannot be checked whether they are ordered because
> > they cannot be taken as individuals.
> It can be reasoned from their description that they are totally
> ordered.[1]
>
> "Checking" whether they are ordered makes just as much sense as
> "checking" whether triangles have three corners: none at all.
>
> [1]
> You (WM) may be interested to hear that being totally ordered
> and being ordered mean different things.
>
> Feel free to ask your colleagues at Hochschule Augsburg
> for details.
Better he should go next door and ask at University of Augsburg. They have people there who know what they are talking about.

On 7/11/2021 10:36 AM, WM wrote:
> Greg Cunt schrieb am Sonntag, 11. Juli 2021 um 16:18:27 UTC+2:
>
>> Hint: All unit fractions are real numbers (i.e. elements in IR).
>
> Nevertheless, when the cursor enumerates the first discerned unit fraction u_1, already aleph_0 unit fractions appear suddenly between u_1 and zero. You don't know and can't know about the order of aleph_0 of them.
>
> Regards, WM
>

there is no first unit fraction, this has been proven.

It is clear you do not read nor understand Math Proofs, nor understand
Algebra.

So that disqualifies you from Math, and this is the reason you have so
much trouble with infinite series, and the rest of math.

On Sunday, July 11, 2021 at 4:07:49 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 11. Juli 2021 um 18:59:39 UTC+2:
> > On Sunday, July 11, 2021 at 11:32:27 AM UTC-4, WM wrote:
> > > William schrieb am Sonntag, 11. Juli 2021 um 17:19:35 UTC+2:
> > > > On Sunday, July 11, 2021 at 9:56:54 AM UTC-4, WM wrote:
> > >
> > > > For every element of U_F we know that it is not next to zero. For every "dark" element we do not know if it is "next" to zero. No element of U_F is "dark".
> > > But you agree that when the cursor moves from 0 to 1 aleph_0 [elements of U_F] appear suddenly as successors of the first defined U_F?
> >
> > The subset, S, of U_F of elements that are greater than the first U_F that can be written down has cardinality aleph_0.
> That is laudable statement. Unfortunately most set theorists will oppose it.
> > Every pair of elements, x and y of S are such that x<y, x=y, x>y. We know this, even thought we do not know which is true.
> Agreed.
> > 'We also know that S has no last element, e.g. for every g an element of S, there is a h(g) element of S such that g<h(g).
> But we cannot distinguish aleph_0 elements. They exist only as a bunch.

Piffle, U_F has cardinality aleph_0. Each element is distinguishable . One does not need to write an element down to distinguish it.

> >
> > > aleph_0 will always follow upon every [element of U_F that can be written down]. You will never know about the order of the last aleph_0 elements of U_F.
> >
> > We will never know everything, but we will know some things. What we know is enough to conclude that there is no last element among the last aleph_0 elements of U_F and that there is no "dark" element among the last aleph_0 elements of U_F
> Well, we can agree to say "that cannot be written down" instead of dark. In order to abbreviate this, say tcbwd-elements.

Nope. You use "dark" to mean a lot more than "cannot be written down". E..g. a "dark unit fraction" is 1/m where m is not the largest element of a FISON. Lots of elements of U_F cannot be written down. No element of U_F is 1/m where m is not the largest element of a FISON. How about a completely arbitrary name?Instead of "can be written down" we can talk about "penguin" numbers.

--
William Hughes

On 7/11/2021 3:13 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 11. Juli 2021 um 18:24:57 UTC+2:
>> On Sunday, 11 July 2021 at 08:32:55 UTC-3, WM wrote:
>
>>> The unit fractions discerned and counted are well-ordered by this counting.
>>
>> Sure. If you assume there are only finitely many integers, then there are only finitely many integers.
>
> It is clear that aleph_0 belong to the first bunch which is not counted and cannot be counted becaue it is dark. Therefore only a finite remainder is available for operations with individuals.
>
> Regards, WM
>

Ants of the First Bunch

Ants of the First Bunch Not Counted

Finite Remainder Ants

Ants Available for Operations With Individuals

Ants Available for Ant Operations With Individuals

Ants Available for Operations With Individual Ants

Unit Fractions Ants Discerned, Counted, and Well-Ordered

Clean Ants

Dirty Dark Ants

On Sunday, July 11, 2021 at 10:02:54 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Sonntag, 11. Juli 2021 um 20:30:34 UTC+2:
> >
> > the ARE already ordered, since {1/n : n e IN} is a totally ordered set.
> >
> What you call n concerns only natural numbers which <bla>

No, "n" is a variable whick "concerns" _all_ natural numbers; that is what "n e IN" in {1/n : n e IN} MEANS, you silly crank.

Hint: {n : n e IN} = IN .

Since ({1/n : n e IN}, <=) already *is* a totally ordered set there's

> > no need [for the cursur] "to order" them.
> >
> The [...] natural numbers which cannot be addressed cannot be compared. Therefore we believe that they are ordered, but we cannot verify it.

** Holy shit **

Look you silly dumbfuck: We can PROVE that the set ({1/n : n e IN}, <=) already *is* a totally ordered set. We do not have to "varify" or "check this fact" by considering each and every unit fraction "individually" (WM).

We already know, for example, that for any "two" unit fractions s,t in {1/n : n e IN} either s < t, t < s, or s = t.

It's getting worse with you, Mückenheim.

On Sunday, July 11, 2021 at 10:10:33 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Sonntag, 11. Juli 2021 um 17:56:48 UTC+2:
> >
> > The set {s : En e IN(s = 1/n) & 0 < s < u_1} is a totally ordered set.
> >
> Nevertheless you cannot [...] compare any one or two elements of this aleph_0 bunch.

Of course, we can, you silly crank. That's what "totally" refers to in "totally ordered set".

We may call any "two" elements x,y such that x <= y or y <= x "comparable".

Then the claim that M is a "totally ordered set" means 1) that M is a partially ordered set and 2) that any "two" elements in M are comparable.

On Monday, July 12, 2021 at 2:53:52 PM UTC+2, Greg Cunt wrote:
> On Sunday, July 11, 2021 at 10:02:54 PM UTC+2, WM wrote some nonsens:

> > The [...] natural numbers which cannot be addressed cannot be compared. (*)

Look, you silly dumbfuck, IN is a totally ordered set, even a well-ordered set.

We can easily prove by induction that for any "two" natural numbers n,m either n < m, m < n, or n = m.

So any two numbers in IN are "comparable".

HENCE, if (*) were true, we would get the conclusion:

_All_ natural numbers can be "adressesed" (whatever that means).

> > Therefore

Huh?!

> > we believe that they are ordered, but we cannot verify it.

Man, in math we "believe" what we can PROVE. Actually, we usually say/think that we KNOW that X, if we can PROVE that X.

We do not have to "varify" or "check" that the natural numbers are "totally orderd" by considering each and every natural number "individually" (WM).

Hint: We know that for all natural numbers 2*n = n + n, since we can PROVE that. We (at least most of us) do not have to "verify" or "check" that fact for each and every natural number "individually" (WM) before being able to "believe" it.

Of course, it seems that you'd prefer to "verify" that fact be considering each and every number "individually" (WM):

2 * 1 = 2 = 1 + 1, check!
2 * 2 = 4 = 2 + 2, check!
2 * 3 = 6 = 3 + 3, check!

Please go ahead, Mückenheim. And DON'T STOP!

On Monday, July 12, 2021 at 2:53:52 PM UTC+2, Greg Cunt wrote:
> On Sunday, July 11, 2021 at 10:02:54 PM UTC+2, WM wrote some nonsens:

> > The [...] natural numbers which cannot be addressed cannot be compared. (*)

Look, you silly dumbfuck, IN is a totally ordered set, even a well-ordered set.

We can easily prove by induction that for any "two" natural numbers n,m either n < m, m < n, or n = m.

So any two numbers in IN are "comparable".

HENCE, if (*) were true, we would get the conclusion:

_All_ natural numbers can be "addressed" (whatever that means).

> > Therefore

Huh?!

> > we believe that they are ordered, but we cannot verify it.

Man, in math we "believe" what we can PROVE. Actually, we usually say/think that we KNOW that X, if we can PROVE that X.

We do not have to "varify" or "check" that the natural numbers are "totally orderd" by considering each and every natural number "individually" (WM).

Hint: We know that for all natural numbers 2*n = n + n, since we can PROVE that. We (at least most of us) do not have to "verify" or "check" that fact for each and every natural number "individually" (WM) before being able to "believe" it.

Of course, it seems that you'd prefer to "verify" that fact be considering each and every number "individually" (WM):

2 * 1 = 2 = 1 + 1, check!
2 * 2 = 4 = 2 + 2, check!
2 * 3 = 6 = 3 + 3, check!

Please go ahead, Mückenheim. And DON'T STOP!

On 7/11/2021 5:06 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 11. Juli 2021 um 21:04:42 UTC+2:

>> Consider the unit interval [0,1].
>>
>> In fact, consider it twice: [0,1]x[0,1], a square.
>>
>> We would like to know how to say that f: [0,1] --> [0,1] is a
>> _continuous_ path from the square's lower left corner (0,0) to
>> its upper right corner (1,1).
>>
>> What is a _discontinuity_ so that we can say to f, "Don't do that"?
>
> Every definable point is surrounded by dark points
> - by infinitely many. This does not say don't do that.
> But it should be taken into account.

----
Fine.

I just want to say again, to Ross or whoever, that
what we know how to say is "Don't be discontinuous at x".
That involves a delta-epsilon kind of statement about
the path _near x_

A path can obey our command and still jump around if
the points x at which we tell it to not jump around leave
even one point out.

Thus, we need a delta-epsilon kind of statement about f,
and we need a point-between kind of statement about x.

The second part is easy to overlook or forget about.
It describes the real numbers, and the real numbers are
the water in which we fish swim. Forgotten or not,
we must have it.

----
My accounting for dark points consists of providing two boxes
box 1: "things to which we might be referring by x"
box 2: "things to which we are not referring by x"

Then I reason about the things to which we might be referring by x.

That's all. I don't put any dark points into either box.
I let you (WM) do that, if you wish. Or not do it.

I have not found what you've said about dark points to be
clear enough to decide which box they go in. That's just
how it is, no matter who's to blame.

And, since I'm using x to make statements about the things
we are interested in (for now, about lines and points) and
reasoning from those statements, _what I do_ doesn't change
if they're in box 1 or box 2, some in one and the rest in
the other, or if they pop back and forth with every
keystroke I make.

This is basically the reason I don't press harder for
clarity and consistency in which box the dark points go on.

On Monday, 12 July 2021 at 15:26:16 UTC+2, Greg Cunt wrote:
> On Monday, July 12, 2021 at 2:53:52 PM UTC+2, Greg Cunt wrote:
> > On Sunday, July 11, 2021 at 10:02:54 PM UTC+2, WM wrote some nonsens:
<snip>
> Man, in math we "believe" what we can PROVE.

There are so many even fundamental things that still haven't been proven, indeed for some it's not even proven if they are at all provable from the respective axioms.

> Actually, we usually say/think that we KNOW that X, if we can PROVE that X.

I concur with the rest of your objections to WM, but that's a misstatement in a fundamental way: we indeed have proof that in *every* theory of some specific (very) minimal complexity there are statements that are true but unprovable. Truth and provability must not be conflated: a Theorem by definition is *a true statement* (of some theory), not (i.e. not necessarily) a proven statement (of that theory). Indeed, talking of primality, another way to state that result is that in *every* correct system of at least that minimal complexity, that of provable statements is necessarily a strict subset of that of true statements.

Sure, I grant you that if I bring you a theorem with annex proof you could more easily entertain it than if I tried to justify it on plain logical or "meta-logical" grounds... yet truth and probability are no more the same thing than Newtonian physics is true beyond a certain very restricted sense.

> Please go ahead, Mückenheim. And DON'T STOP!

Nor will you apparently.

Julio

William schrieb am Montag, 12. Juli 2021 um 00:29:11 UTC+2:
> On Sunday, July 11, 2021 at 3:58:13 PM UTC-4, WM wrote:

> "Taken as individuals" is yet one more or your ways of saying "can be written down". So What? We do not need write the elements of U_F down to know that each element is ordered.

You believe it, but you cannot find any individual.

When the cursor rests at zero, all unit fractions are between itself and 1. When the cursor is at epsilon (however small you choose it), almost all unit fractions are between itself and zero. That means you cannot state anything about almost all unit fractions individually. You can say what properties they have to have, but you cannot check it. In particual you cannot check a bijection with other infinite sets.

> > I said that every [element of U_F]l has neighbours.
>
> Indeed, because every element of U_F is orderable.

No, almost all can only appear as an infinite bunch.
>
> > But the dark
>
> Are totally irrelevant. No element of U_F is dark

For almost all elements you cannot know that. In mathematics we do not believe properties which cannot be checked. That's matheology.

Regards, WM

Sergio schrieb am Montag, 12. Juli 2021 um 00:53:04 UTC+2:

> the Cursor passes ALL real numbers in the interval [0,1]

How can you know if there is not a first one?

> > Every counted element has aleph_0 not counted between itself and zero.
> wrong.

Show one unit fraction that has not infinitely many between itself and zero.

Regards, WM

Jim Burns schrieb am Montag, 12. Juli 2021 um 01:11:31 UTC+2:
> On 7/11/2021 3:58 PM, WM wrote:

> It can be reasoned from their description that they are totally
> ordered.

It can also be believed, but it cannot be checked.

When the cursor rests at zero, all unit fractions are between itself and 1. When the cursor is at epsilon (however small you choose it), almost all unit fractions are between itself and zero. That means you cannot state anything about almost all unit fractions individually. You can say what properties they have to have, but you cannot check it. In particual you cannot check a bijection with other infinite sets.
>
> "Checking" whether they are ordered makes just as much sense as
> "checking" whether triangles have three corners:

No. For a figure presented as a triangle it can be checked. For a mapping with |N presented as a bijection this cannot be checked.
>
Regards, WM