On Wednesday, June 9, 2021 at 12:26:54 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 9. Juni 2021 um 17:09:16 UTC+2:
> > On Wednesday, June 9, 2021 at 10:08:40 AM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 8. Juni 2021 um 23:27:04 UTC+2:
> > >

> > Achilles will never find a unit fraction that is not followed by another unit fraction. He has to pass unit fraction 1, then unit fraction 2, then unit fraction 3 and so on. There are an endless number of unit fractions to pass. Achilles has to do an infinite number of things to pass every unit fraction.
> But he cannot do this with visible natnumbers because they are glued to the origin by FISONs. And there is no infinite FISON.

Yes like every natural number (and unlike your misleadingly named "dark natural numbers") every element of |N (the set you call the definable natural numbers) is the largest element in a FISON. There is, of course, no infinite FISON. There is, of course, an infinite number of FISONS. So the set F of the FISONs is an infinite set, comprised of finite elements. Finite but unbounded only makes sense for things that can change

There is a trivial bijection between F and |N. By the definition of cardinality the two sets have the same cardinality. And |N (the set you call the definable natural numbers) is a Peano set with cardinality aleph_0

--
William Hughes

On 6/9/2021 11:23 AM, WM wrote:
> Greg Cunt schrieb
> am Dienstag, 8. Juni 2021 um 21:15:18 UTC+2:

>> Switching to the (all) reals for a change:
>> There is no point x > 0 _from where_ he "enters" 0.
>
> There is no determinable point from where the cursor enters 0.
> But if there was no point, then the continuum would not be
> a continuum.

No, the opposite of that.
If there *IS* a point from where the cursor enters 0,
then the continuum would not be a continuum.

a point from where the cursor enters 0
== a point with no other points between it and 0
== a point in a different _component_ of the continuum from 0

If there is more than one _component_ to the continuum,
a function can be _definitionally_ continuous (delta, epsilon, etc)
within the same component and still possibly have big _gaps_
between components.

"Possibly having big gaps" is not _what we mean_ by
"continuous function".
Therefore, "more than one component" is not _what we mean_
by "continuum".

Therefore, if there *IS* a point (even a dark point) from which
to enter 0, then the continuum would not be the continuum.

William schrieb am Mittwoch, 9. Juni 2021 um 19:16:47 UTC+2:
> On Wednesday, June 9, 2021 at 12:26:54 PM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 9. Juni 2021 um 17:09:16 UTC+2:
> > > On Wednesday, June 9, 2021 at 10:08:40 AM UTC-3, WM wrote:
> > > > William schrieb am Dienstag, 8. Juni 2021 um 23:27:04 UTC+2:
> > > >
>
> > > Achilles will never find a unit fraction that is not followed by another unit fraction. He has to pass unit fraction 1, then unit fraction 2, then unit fraction 3 and so on. There are an endless number of unit fractions to pass. Achilles has to do an infinite number of things to pass every unit fraction.
> > But he cannot do this with visible natnumbers because they are glued to the origin by FISONs. And there is no infinite FISON.
> Yes like every natural number (and unlike your misleadingly named "dark natural numbers") every element of |N (the set you call the definable natural numbers) is the largest element in a FISON. There is, of course, no infinite FISON. There is, of course, an infinite number of FISONS.

Here you use potential infinity! An actual infinity of chains would require at least one infinite chain by the pigeon hole principle.

In order to distinguish 3 chains you need one chain with at least 3 elements (longer chains are possible):
o oo
ooo

In order to distinguish n chains you need one chain with at least n elements (longer chains are possible).
In order to distinguish more than any finite number of chains you need one chain with at least more than any finite number of elements. AS this is impossible for FISONs, you cannot gather aleph_0 FISONs.

> So the set F of the FISONs is an infinite set, comprised of finite elements.

Wrong. Up to every finite chain the set is finite.

> Finite but unbounded only makes sense for things that can change.

The set of defined natnunbers can change.
>
> There is a trivial bijection between F and |N.

That is an unfounded claim. Bijections can only exist between defnied elements.

> By the definition of cardinality the two sets have the same cardinality. And |N (the set you call the definable natural numbers) is a Peano set with cardinality aleph_0

That is traditional but it is nonsense nevertheless.

Why do you think that induction fails when infinitely many steps are done? As long as FISONs are concerned, it is impossible to leave their set by induction. Induction proves that this is impossible.

Regards, WM

On 6/9/2021 11:08 AM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 9. Juni 2021 um 17:56:07 UTC+2:
>> WM explained on 6/9/2021 :
>>> William schrieb am Mittwoch, 9. Juni 2021 um 17:09:16 UTC+2:
>>>> On Wednesday, June 9, 2021 at 10:08:40 AM UTC-3, WM wrote:
>>>>> William schrieb am Dienstag, 8. Juni 2021 um 23:27:04 UTC+2:
>>>>>
>>>>>>> When will he find a unit fration that is not followed by another unit
>>>>>>> fraction but by zero?
>>>>>> Never.
>>>>> Never means never
>>>>
>>>> Indeed. Achilles will never find a unit fraction that is not followed by
>>>> another unit fraction. He has to pass unit fraction 1, then unit fraction 2,
>>>> then unit fraction 3 and so on. There are an endless number of unit
>>>> fractions to pass. Achilles has to do an infinite number of things to pass
>>>> every unit fraction.
>>> But he cannot do this with visible natnumbers because they are glued to the
>>> origin by FISONs. And there is no infinite FISON. Can you comprehend this
>>> simple proof? There is no infinite FISON. Only dark natnumbers can supply the
>>> required infinity.
>> It doesn't need to, infinity is not in the set,
>
> The set of natnumbers is said to be infinite. Its cardinality is aleph_0. Of course visible natnumbers cannot accomplish this because they are all in finite sets (FISONs). aleph_0 however is larger than every finite set. Therefore the set of visible natnumbers is not infinite. It has cardinality less than aleph_0.

that is because you stopped, at your last visible natural number, you
make mistake stopping.

>
>> omega is the successor of all finite natnumbers.
>
> But the set cannot become infinite by adding only one element omega. There must be infinitely many dark elements in that set.

no, only if you stop.

the set is infinite, regardless of your trivial "defined", "visable" etc.

>
> Regards, WM
>
>

On 6/9/2021 1:49 PM, WM wrote:
> William schrieb am Mittwoch, 9. Juni 2021 um 19:16:47 UTC+2:
>
>
> Why do you think that induction fails when infinitely many steps are done?

it does not fail, that is how limits are determined.

you are still misleading yourself in circles, go read real math books.

> Regards, WM
>

On 6/9/2021 11:31 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 9. Juni 2021 um 16:00:25 UTC+2:

>> This is why I call the rational numbers bright points.
>> If anything can be uncontroversial, it seems like they should be.
>> I could easily keep calling them "rationals", but I would
>> like to emphasize where WM and I _agree_ The existence of
>> the rationals seems to be one such place.
>
> Every defined point is visible or bright if you like.
> This includes pi.

So I understand.
Maybe I could say _at least_ the rationals are bright.

Taking the rationals and requiring
| No teleporting while continuous
includes pi, every defined point, and uncountably-many others.

If we follow Ockham, anything _beyond_ the rationals and
the anti-teleporting points are without necessity when it comes
to describing the line -- and, thus, without existence in the line.

This is how it seems to me:
Archimedes: At least the rational points exist.
Dedekind: At least points exist which make one component.
Ockham: No more than those points exist.

If we follow Archimedes, Dedekind, and Ockham in describing
the line, any dark points left out of that description are
not in the line. As such, they cannot be counter-examples
to a claim about points in the line.

> But there are dark naturals and rationals and others.

Which are dark and which are not doesn't seem to be a
property of the naturals or rationals themselves.

An argument concerning darkness or brightness --
if darkness and brightness are NOT properties of naturals
or rationals or points in the line
-- will not be able to proceed from ONLY properties of
naturals or rationals or points in the line.

----
Suppose the frames are at
1, 1/2, 1/3, 1/4, 1/5, ..., 0
Then Achilles _enters_ (from a frame with no frame between)
1/2, 1/3, 1/4, 1/5, ... but does not enter frame 0

That's a particular choice of where the frames are.
If we had chosen instead for frames to be at
1/3+1/2, 1/4+1/2, 1/5+1/2, 1/6+1/2, ... , 1/2
then Achilles would not _enter_ the frame at 1/2.

Even though Achilles _enters_ the frame at 1/2 in the first case,
and does not _enter_the frame at 1/2 in the second case,
his motion is the same in both cases.

Or, with still different frames, where-Achilles-passes-the-tortoise
could be a finite frame, and he would _enter_ that frame.
1, 1/2, 1/3, 0

On 6/9/2021 11:22 AM, WM wrote:
> Greg Cunt schrieb
> am Dienstag, 8. Juni 2021 um 22:53:01 UTC+2:

>> At t = t_0 Achilles reaches 0.
>> But "induction" (mentioned by you) is not
>> in contradiction with this fact.
>
> Induction dictates that when he is at a finite index
> he will reach thze net finite index.

"reach" definition 1

Presumably, thze net == the next

> Induction is in contradiction with the idea that he
> reaches omega after infinitely many finite natnumbers.

"reach" definition 2

"He always reaches[1] a finite index"
does not contradict
"He reaches[2] omega".

> Every natnumber is the end of a FISON and therefore it is
> not possible to reach omega if only visible natnumbers exist.

Every natnumber is the end of a FISON.
It is possible to reason about all the things that are
the end of a FISON. Start with "k is the end of a FISON".

We can reason about infinitely-many extensions of a
finite path by one step. Start with
"k is the end of a FISON that k+1 is not in".

If you choose to poetically describe infinitely-many
next steps as "arriving at omega", that's up to you.
Saying that won't make the naturals be finitely-many
or make a last natural _of any kind_ exist one step
from omega.

And we are back to the set of FISONs is finite, I feel young again.

> An actual infinity of chains would require at least one infinite chain by the pigeon hole principle.

or instead of an infinite chain a set of chains without longest chain would also work.

>
> In order to distinguish 3 chains you need one chain with at least 3 elements (longer chains are possible):
> o
> oo
> ooo
>
> In order to distinguish n chains you need one chain with at least n elements (longer chains are possible).
> In order to distinguish more than any finite number of chains you need

a set of chains without longest chain (all of the chains can be finite of course).

--
William Hughes

On 6/9/2021 11:26 AM, WM wrote:

> Only dark natnumbers can supply the required infinity.

We can reason about infinitely-many instances of supplying
one more natural number. Start with
"A completable linear crowd 0,...,k exists which k+1 is not in".

If k is described this way, k is not omega.

However, there are other descriptions which also describe omega.
Start with
"Any non-empty collection of these things (ordinals)
contains a first element."

This is equivalent to _transfinite_ induction.

| If, forall x,
| ( forall z < x, P(z) implies P(x) )
| then, forall y,
| P(y)

If omega exists, _transfinite_ induction includes it.

Jim Burns schrieb am Mittwoch, 9. Juni 2021 um 20:28:40 UTC+2:
> On 6/9/2021 11:23 AM, WM wrote:
> > Greg Cunt schrieb
> > am Dienstag, 8. Juni 2021 um 21:15:18 UTC+2:
>
> >> Switching to the (all) reals for a change:
> >> There is no point x > 0 _from where_ he "enters" 0.
> >
> > There is no determinable point from where the cursor enters 0.
> > But if there was no point, then the continuum would not be
> > a continuum.
> No, the opposite of that.
> If there *IS* a point from where the cursor enters 0,
> then the continuum would not be a continuum.

It is obvious that the curser never touches anything else than points.

But since this is hard to understand, I have devised a simpler example without continuity: Achilles will pass all finite frames and from there will enter frame omega. It is silly to claim that his path has any interruption such that there is not finite frame immediately before he occupies frame omega.

Set theorists are blinded but everyone else will recognize that this claim is wrong, and it is silly to sacrifice the "continuity" of the path in order to maintain a wrong understanding of infinity. omega has immediate predecessors. Otherwise it would sit elsewhere. But they are dark. For Achilles this means: There is a last finite frame and there is not a last finite frame because dark frames have no conceivable order.

> a point from where the cursor enters 0

> == a point with no other points between it and 0
> == a point in a different _component_ of the continuum from 0

Since the points next to zero is dark, their configuration cannot be determined whether there is one or are many points next to zero.
>
> Therefore, if there *IS* a point (even a dark point) from which
> to enter 0, then the continuum would not be the continuum.

It is impossible to find out what is next to zero. But certainly there is something. The only alternative (in potential infinity) would be nothing.

Regards, WM

Jim Burns schrieb am Mittwoch, 9. Juni 2021 um 21:14:37 UTC+2:
> On 6/9/2021 11:31 AM, WM wrote:
> > Jim Burns schrieb
> > am Mittwoch, 9. Juni 2021 um 16:00:25 UTC+2:
>
> >> This is why I call the rational numbers bright points.
> >> If anything can be uncontroversial, it seems like they should be.
> >> I could easily keep calling them "rationals", but I would
> >> like to emphasize where WM and I _agree_ The existence of
> >> the rationals seems to be one such place.
> >
> > Every defined point is visible or bright if you like.
> > This includes pi.
> So I understand.
> Maybe I could say _at least_ the rationals are bright.

Not all! Only finitely many can be defined. That's the same as with natural numbers. Never more than finitely many will be defined.

> If we follow Archimedes, Dedekind, and Ockham in describing
> the line, any dark points left out of that description are
> not in the line.

The cursor from 1 to 0 does pass all unit fractions. None is missing and there is no gap between them and zero. It is silly to claim that the cursor gets to zero from a not existing unit fraction. However, the cursor cannot get to zero from a visible unit fractions. Therefore there are dark unit fractions necessary.

> > But there are dark naturals and rationals and others.
> Which are dark and which are not doesn't seem to be a
> property of the naturals or rationals themselves.

What we know is that all bright elements are a finite set and the dark elements are an infinite set which will under all circumstances remain infinite..
>
> An argument concerning darkness or brightness --
> if darkness and brightness are NOT properties of naturals
> or rationals or points in the line

The points next to zero or any other defined point are dark.

You say that they don't exist, but that is wrong because that would mean gaps.

> -- will not be able to proceed from ONLY properties of
> naturals or rationals or points in the line.
>
> ----
> Suppose the frames are at
> 1, 1/2, 1/3, 1/4, 1/5, ..., 0
> Then Achilles _enters_ (from a frame with no frame between)
> 1/2, 1/3, 1/4, 1/5, ... but does not enter frame 0
> That's a particular choice of where the frames are.
> If we had chosen instead for frames to be at
> 1/3+1/2, 1/4+1/2, 1/5+1/2, 1/6+1/2, ... , 1/2
> then Achilles would not _enter_ the frame at 1/2.

Every definabvle point on his path can be treated like the omega frame. But it would complicate the matter.

Regards, WM

What are mathematicians doing with their set?
That is just man's order arrangement of quantity element.
What order are you creating?
It is not a mathematical quantity operation...

Mitchell Raemsch

William schrieb am Mittwoch, 9. Juni 2021 um 22:24:35 UTC+2:
> And we are back to the set of FISONs is finite, I feel young again.
> > An actual infinity of chains would require at least one infinite chain by the pigeon hole principle.
> or instead of an infinite chain a set of chains without longest chain would also work.
> >
> > In order to distinguish 3 chains you need one chain with at least 3 elements (longer chains are possible):
> > o
> > oo
> > ooo
> >
> > In order to distinguish n chains you need one chain with at least n elements (longer chains are possible).
> > In order to distinguish more than any finite number of chains you need
> a set of chains without longest chain (all of the chains can be finite of course).

That is silly magic. aleph_0 is a number larger than any finite number. From aleph_0 natnumbers you can subtract every FISON without exhausting the set. What you claim may help in potential infinity.

But that becomes much clearer with Achilles and his frames, or the cursor moving from 1 to 0. There a not existing last frame or unit fraction does not help. Achilles' path has no gaps and t5he curser enters zero from an existing unit fraction, not from a not existing unit fraction. The "not existing last" is no remedy.

Regards, WM