On 9/4/2021 5:00 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 4. September 2021 um 22:00:31 UTC+2:
>> On 9/4/2021 8:43 AM, WM wrote:

>>> This chain cannot have an empty intersection unless
>>> one endsegment is empty.
>>
>> It is sufficient for an empty intersection of a collection
>> that each element be not-in at least one set in the collection.
>
> Yes, but
> If there is no empty endsegment or no last endsegment,

If there is no empty endsegment or no last endsegment,
it might still be that, for each element, an end segment exists
which that element is not-in.
In that case, we would have an empty intersection.

And so it happens that the intersection of all FISONable-natural
end segments is empty.
Ak e N: k ~e E(k+1)

> If there is no empty endsegment or no last endsegment,
> then there will remain natural nunbers in the intersection.
> If these cannot be detected, they are dark. What you call
> "each" fails to cover the dark numbers.

What I call "each" _succeeds_ at NOT covering your dark numbers.

We describe what we intend to reason about, and reason from
their description.
We describe the things-we-count-with, and reason from their
description.
Dark numbers are NOT things-we-count-with, so, if "each"
covered dark numbers, we would have failed at describing
things-we-count-with.

On Saturday, September 4, 2021 at 5:52:51 PM UTC-3, WM wrote:
> William schrieb am Samstag, 4. September 2021 um 16:12:27 UTC+2:
> > On Saturday, September 4, 2021 at 9:46:32 AM UTC-3, WM wrote:
> > > William schrieb am Samstag, 4. September 2021 um 02:26:43 UTC+2:
> > > > On Friday, September 3, 2021 at 5:38:46 PM UTC-3, WM wrote:
> > >
> > > > > A dark number cannot be identified by any means. It cannot be handled as a single individual.
> > > > However, a number that does not have the property "Can be written down" still has the property "can be handled a a single individual'.
> > > How would you handle it?
> > Even if n cannot be written down it is still an element of |N_F and still has any property shared by all elements of |N_F. One of these properties is that n "can be distinguished".
> Even if n cannot be written down or distinguished in any way from others,

Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.

--
William Hughes

William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:

> Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.

How do you distinguish it from another number that cannot be written down?

Regards, WM

William schrieb am Sonntag, 5. September 2021 um 01:51:37 UTC+2:
> On Saturday, September 4, 2021 at 5:54:16 PM UTC-3, WM wrote:

> > > > > Because of inclusion monotony every endsegment is a subset of its predecessors. This chain cannot have an empty intersection unless one endsegment is empty.
> > > or if there is no last endsegment.
> > Then there will remain infinitely many natural numbers beyond every existing endsegment.
> There are no elements of |N_F beyond any endsegment.
>
If there is no last one, then no last one can be defined. Then there are infinitely many beyond every defined endsegment. Because when you define endsegments individually, then you define necessarily a last one. How long you will try ever.

Regards, WM

On 9/5/2021 5:38 AM, WM wrote:
> William schrieb
> am Sonntag, 5. September 2021 um 01:51:37 UTC+2:
>> On Saturday, September 4, 2021 at 5:54:16 PM UTC-3,
>> WM wrote:

>>> Then there will remain infinitely many natural numbers
>>> beyond every existing endsegment.
>>
>> There are no elements of |N_F beyond any endsegment.
>
> If there is no last one, then no last one can be defined.

Each of them is not-last.
If it's defined, it's not-last.
If it's not defined, it's not-last.

> Then there are infinitely many beyond every defined endsegment.

Each of them can be counted to, in principle.
If it's defined, it can be counted to, in principle.
If it's not defined, it can be counted to, in principle.

> Because when you define endsegments individually,
> then you define necessarily a last one.

You're equivocating on "last".
The last defined and the (non-existent) last in N_F
are different.

If I ask you to point to "the green one",
you will want to know the green one of _what_
It's the same with "the last one".

> How long you will try ever.

"However long you will try."
Your English is usually very good,
except when you are trying to shake off pursuit.

What does it imply when you are trying to keep people
from understanding you?

If even you realize how bad your arguments are,
perhaps you could just change your mind.
People do that. It's not a tragedy.

Jim Burns schrieb am Sonntag, 5. September 2021 um 13:38:07 UTC+2:
> On 9/5/2021 5:38 AM, WM wrote:
> > William schrieb
> > am Sonntag, 5. September 2021 um 01:51:37 UTC+2:
> >> On Saturday, September 4, 2021 at 5:54:16 PM UTC-3,
> >> WM wrote:
>
> >>> Then there will remain infinitely many natural numbers
> >>> beyond every existing endsegment.
> >>
> >> There are no elements of |N_F beyond any endsegment.
> >
> > If there is no last one, then no last one can be defined.
> Each of them is not-last.
> If it's defined, it's not-last.
> If it's not defined, it's not-last.
> > Then there are infinitely many beyond every defined endsegment.
> Each of them can be counted to, in principle.

No, each counted one is not the last.
> If it's defined, it can be counted to, in principle.
> If it's not defined, it can be counted to, in principle.

All that can be counted have aleph_0 successors, most of which cannot be coubted, because then there were no successors remaining.

> > Because when you define endsegments individually,
> > then you define necessarily a last one.
> You're equivocating on "last".
> The last defined and the (non-existent) last in N_F
> are different.

The last defined one is the last definable one, not a fixed one though.
>
> If I ask you to point to "the green one",
> you will want to know the green one of _what_
> It's the same with "the last one".

No. The last one defined by you is the last one defined by you.
The last one defined by anyone is the last one definable.

> > How long you will try ever.
> "However long you will try."
> Your English is usually very good,
> except when you are trying to shake off pursuit.

Thank you.
>
> What does it imply when you are trying to keep people
> from understanding you?

It was simply a mistake.
>
> If even you realize how bad your arguments are,
> perhaps you could just change your mind.
> People do that. It's not a tragedy.

If you exhaust the whole set collectively, then nothing remains.
If you try to exhaust the set individually, then aleph_0 elements remain.
This means that you can remove more collectively than individually. Why? Do you know a better answer than me?

Regads, WM

Jim Burns schrieb am Sonntag, 5. September 2021 um 01:26:11 UTC+2:
> On 9/4/2021 5:00 PM, WM wrote:
> > Jim Burns schrieb
> > am Samstag, 4. September 2021 um 22:00:31 UTC+2:
> >> On 9/4/2021 8:43 AM, WM wrote:
>
> >>> This chain cannot have an empty intersection unless
> >>> one endsegment is empty.
> >>
> >> It is sufficient for an empty intersection of a collection
> >> that each element be not-in at least one set in the collection.
> >
> > Yes, but
> > If there is no empty endsegment or no last endsegment,
>
> If there is no empty endsegment or no last endsegment,
> it might still be that, for each element, an end segment exists
> which that element is not-in.
> In that case, we would have an empty intersection.

No, the endsegment consists of elements, even of infinitely many, as you say,. so there remain infinitely many in every case.
>
> And so it happens that the intersection of all FISONable-natural
> end segments is empty.
> Ak e N: k ~e E(k+1)

It does not, because every k has aleph_0 successors.

Regards, WM

William schrieb am Sonntag, 5. September 2021 um 01:47:16 UTC+2:
> On Saturday, September 4, 2021 at 5:55:32 PM UTC-3, WM wrote:

> > The step to zero is the last step.
> Piffle. No step "reaches 0".

But collectively you can do all steps.

> If you want to talk about something (e.g. the cursor) "reaching 0" use a method that is suitable for infinite sets, not a method that can only succeed if there is a last step (lots of steps exist, no step that exists has the property that it is a last step).
>
If you exhaust or step the whole set collectively, then nothing remains.
If you try to exhaust or step the set individually, then aleph_0 elements remain.
This means that you can remove or step more collectively than individually. Why? Do you know a better answer than me?

Regads, WM

On 9/5/2021 4:34 AM, WM wrote:
> William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:
>
>> Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.
>
> How do you distinguish it from another number that cannot be written down?
>
> Regards, WM
>

How distinguished ?

Distinguished Ants that Cannot Be Written Down

Indistinguishable Ants that have been Written Down.

On Sunday, September 5, 2021 at 3:22:31 PM UTC+2, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 01:26:11 UTC+2:
> >
> > If there is no empty endsegment [and] no last endsegment,
> > it might still be that, for each element, an end segment exists
> > which that element is not-in.

Indeed! If n is an arbitrary natural number, then it is not element in the endsegment E(n+1).

In other words,

> > Ak e N: k ~e E(k+1)
> >
> > In that case, we would have an empty intersection.
> >
> No

Yes. BY DEFINITION OF ***INTERSECTION*** YOU SILLY ASSHOLE!

> the endsegment <nonsense deleted>

> > And so it happens that the intersection of all [...] end segments is empty.
> >
> It does not, because every k has aleph_0 successors.

Huh?! Non sequitur, dumbo.

"Ak e N: k ~e E(k+1)" IMPLIES that INTERSECTION_(k e N) E(k) is empty.

On Sunday, September 5, 2021 at 6:34:37 AM UTC-3, WM wrote:
> William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:
>
> > Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.
> How do you distinguish it from another number that cannot be written down?

Why would one have to write down elements of |N_F is order to distinguish them?

--
William Hughes

On Sunday, 5 September 2021 at 10:22:31 UTC-3, WM wrote:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 01:26:11 UTC+2:
> > And so it happens that the intersection of all FISONable-natural
> > end segments is empty.
> > Ak e N: k ~e E(k+1)
> It does not, because every k has aleph_0 successors.

Proving once again that Muckenheim is too stupid to understand the intersection operator, is utterly and totally resistant to any kind of learning, and is *WAY* out of his depth in discussing the matters in this thread. Mucke, go home!

On 9/5/2021 9:22 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 5. September 2021 um 01:26:11 UTC+2:
>> On 9/4/2021 5:00 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Samstag, 4. September 2021 um 22:00:31 UTC+2:
>>>> On 9/4/2021 8:43 AM, WM wrote:

>>>>> This chain cannot have an empty intersection unless
>>>>> one endsegment is empty.
>>>>
>>>> It is sufficient for an empty intersection of a collection
>>>> that each element be not-in at least one set in the collection.
>>>
>>> Yes, but
>>> If there is no empty endsegment or no last endsegment,
>>
>> If there is no empty endsegment or no last endsegment,
>> it might still be that, for each element, an end segment exists
>> which that element is not-in.
>> In that case, we would have an empty intersection.
>
> No, the endsegment consists of elements, even of infinitely many,
> as you say,. so there remain infinitely many in every case.

The intersection of a collection of sets contains only such
elements as are in each set in the collection.
And you can't say "no" to this. It's what an intersection *IS*

>> And so it happens that the intersection of all
>> FISONable-natural end segments is empty.
>> Ak e N: k ~e E(k+1)
>
> It does not, because every k has aleph_0 successors.

For each successor of k, there is an end segment it is not-in.
For each successor of k, it is not in all end segments.
For each successor of k, it is not in the intersection of
end segments.

On 9/5/2021 4:38 AM, WM wrote:
> William schrieb am Sonntag, 5. September 2021 um 01:51:37 UTC+2:
>> On Saturday, September 4, 2021 at 5:54:16 PM UTC-3, WM wrote:
>
>>>>>> Because of inclusion monotony every endsegment is a subset of its predecessors. This chain cannot have an empty intersection unless one endsegment is empty.
>>>> or if there is no last endsegment.
>>> Then there will remain infinitely many natural numbers beyond every existing endsegment.
>> There are no elements of |N_F beyond any endsegment.
>>
> If there is no last one, then no last one can be defined. Then there are infinitely many beyond every defined endsegment. Because when you define endsegments individually, then you define necessarily a last one. How long you will try ever.

there are no last ones,
it is Easy proof in Math to show there is no "last" one.
If you understood the math,
you would not be down this wrong path.

>
> Regards, WM
>

On 9/5/2021 9:07 AM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 5. September 2021 um 13:38:07 UTC+2:
>> On 9/5/2021 5:38 AM, WM wrote:
>>> William schrieb
>>> am Sonntag, 5. September 2021 um 01:51:37 UTC+2:
>>>> On Saturday, September 4, 2021 at 5:54:16 PM UTC-3,
>>>> WM wrote:

>>>>> Then there will remain infinitely many natural numbers
>>>>> beyond every existing endsegment.
>>>>
>>>> There are no elements of |N_F beyond any endsegment.
>>>
>>> If there is no last one, then no last one can be defined.
>>
>> Each of them is not-last.
>> If it's defined, it's not-last.
>> If it's not defined, it's not-last.
>>
>>> Then there are infinitely many beyond every defined endsegment.
>>
>> Each of them can be counted to, in principle.
>
> No, each counted one is not the last.

We can reason about each one countable to by describing k
as countable to, in principle.
We might say something like a steppable {0,...,k} exists
in which, for adjacent i,j, j = i+1.

This is true of k, as long as k refers to one that is
countable to, in principle.
We can reason from _this_ fact about k, even if we do not have
_other_ facts we might like to have about k.
Such as whether k is even, or if it is odd.
Such as the unique identity of k.
Even though we don't know _those_ facts, we know that steppable
{0,...,k} exists in which, for adjacent i,j, j = i+1.
And we can reason from _that_ fact about k.

When we describe k as having steppable {0,...,k} exist
in which, for adjacent i,j, j = i+1,
we EXCLUDE the possibility of k NOT having steppable
{0,...,k} exist in which, for adjacent i;j, j = i+1.

If what we mean by "natural number" is that it is
countable to, in principle (something we do mean), then
all natural numbers are countable to, in principle, and
anything NOT, even in principle, countable to is NOT
a natural number.

That is as inevitable as a right triangle having
three sides, and for much the same reason.

>> If it's defined, it can be counted to, in principle.
>> If it's not defined, it can be counted to, in principle.

Defining and refraining from defining do not affect the
existence of {0,...,k} or its relevant properties.

William schrieb am Sonntag, 5. September 2021 um 16:27:42 UTC+2:
> On Sunday, September 5, 2021 at 6:34:37 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:
> >
> > > Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.
> > How do you distinguish it from another number that cannot be written down?
> Why would one have to write down elements of |N_F is order to distinguish them?

How do you distinguish them?

Regards, WM

William schrieb am Sonntag, 5. September 2021 um 16:34:44 UTC+2:
> On Sunday, September 5, 2021 at 10:27:29 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 5. September 2021 um 01:47:16 UTC+2:
> > > On Saturday, September 4, 2021 at 5:55:32 PM UTC-3, WM wrote:
> >
> > > > The step to zero is the last step.
> > > Piffle. No step "reaches 0".
> > But collectively you can do all steps.
> So What? No step that you do "reaches 0". The fact that doing "all steps" "reaches 0" does not mean that one of the steps "reaches 0"

That is true because all steps collectively cover the dark ones. Why else should there be a difference?

Regards, WM

William schrieb am Sonntag, 5. September 2021 um 16:42:22 UTC+2:
> On Sunday, September 5, 2021 at 6:38:27 AM UTC-3, WM wrote:
> > when you define endsegments individually, then you define necessarily a last one.
> Nope.
>
Simple fact. At ever time you have defined finitely many whilke infinitely many are undefined.

Regards, WM

Jim Burns schrieb am Sonntag, 5. September 2021 um 20:53:35 UTC+2:
> On 9/5/2021 9:22 AM, WM wrote:

> > It does not, because every k has aleph_0 successors.
> For each successor of k, there is an end segment it is not-in.

As long as the endsegments are infinite, they contain infinitely many natnumbers and therefore have not an empty intersection. This is independent of how many endsegments there are.

Regards, WM

On Sunday, September 5, 2021 at 5:22:35 PM UTC-3, WM wrote:
> William schrieb am Sonntag, 5. September 2021 um 16:27:42 UTC+2:
> > On Sunday, September 5, 2021 at 6:34:37 AM UTC-3, WM wrote:
> > > William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:
> > >
> > > > Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.
> > > How do you distinguish it from another number that cannot be written down?
> > Why would one have to write down elements of |N_F is order to distinguish them?
> How do you distinguish them?
>
By the fact that they are different.

--
William Hughes

On Sunday, September 5, 2021 at 10:28:14 PM UTC+2, WM wrote:

> As long as the endsegments are infinite,

All endsegments are infinite BY DEFINITION, you silly crank.

> they contain infinitely many natnumbers

Ideed. :-)

> and therefore have not an empty intersection.

Non sequitur.

Hint: The intersection of each and every infinite set of endsegments is empty.

> This is independent of how many endsegments there are.

Nope. This depends ONLY on the number of endsements. (*sigh*)

On Sunday, 5 September 2021 at 20:24:15 UTC-3, Greg Cunt wrote:
> On Sunday, September 5, 2021 at 10:28:14 PM UTC+2, WM wrote:
>
> > As long as the endsegments are infinite,
> All endsegments are infinite BY DEFINITION, you silly crank.
You note, of course, how he surreptitiously snuck in his version of what is going on, namely a stepwise process ("As long as...")

I can't really imagine that he truly believes that there will come a spot where the end segments fail to be infinite, since any end segment E(n) is the set {n, n+1, n+2, n+3, ...} or {m : m = n - 1 + k for some k in IN}, since he has acknowledged the cardinality of IN is aleph_0. Then again, he believes a million different things every day before breakfast, and every single one of them is wrong.
> > they contain infinitely many natnumbers
> Ideed. :-)
> > and therefore have not an empty intersection.
> Non sequitur.
>
> Hint: The intersection of each and every infinite set of endsegments is empty.
> > This is independent of how many endsegments there are.
> Nope. This depends ONLY on the number of endsements. (*sigh*)

söndag 5 september 2021 kl. 22:28:14 UTC+2 skrev WM:
> Jim Burns schrieb am Sonntag, 5. September 2021 um 20:53:35 UTC+2:
> > On 9/5/2021 9:22 AM, WM wrote:
>
> > > It does not, because every k has aleph_0 successors.
> > For each successor of k, there is an end segment it is not-in.
> As long as the endsegments are infinite, they contain infinitely many natnumbers and therefore have not an empty intersection. This is independent of how many endsegments there are.
>
> Regards, WM
incorrect. This is only true for a FINITE collection of endsegments.

William schrieb am Sonntag, 5. September 2021 um 23:57:26 UTC+2:
> On Sunday, September 5, 2021 at 5:25:03 PM UTC-3, WM wrote:
> > William schrieb am Sonntag, 5. September 2021 um 16:42:22 UTC+2:
> > > On Sunday, September 5, 2021 at 6:38:27 AM UTC-3, WM wrote:
> > > > when you define endsegments individually, then you define necessarily a last one.
> > > Nope.
> > >
> > Simple fact. At every time you have defined finitely many while infinitely many are undefined.

> So what? If there are still infinitely many you do not have a last one.
But you have a last defined one. And all are infinite. And all definable endsegments are the intersections of endsegments which are possible for definable endsegments..

Regards, WM

William schrieb am Sonntag, 5. September 2021 um 23:40:55 UTC+2:
> On Sunday, September 5, 2021 at 5:22:35 PM UTC-3, WM wrote:
> > William schrieb am Sonntag, 5. September 2021 um 16:27:42 UTC+2:
> > > On Sunday, September 5, 2021 at 6:34:37 AM UTC-3, WM wrote:
> > > > William schrieb am Sonntag, 5. September 2021 um 01:56:55 UTC+2:
> > > >
> > > > > Piffle. The fact that n cannot be written down does not mean n cannot be distinguished.
> > > > How do you distinguish it from another number that cannot be written down?
> > > Why would one have to write down elements of |N_F is order to distinguish them?
> > How do you distinguish them?
> >
> By the fact that they are different.
>
They are nothing unless you can address them.

Regards, WM