On Friday, September 3, 2021 at 7:22:58 PM UTC+2, WM wrote:
> Jim Burns schrieb am Freitag, 3. September 2021 um 02:40:50 UTC+2:
> >
> > Every end segment [...] is an end segment.
> >
> But it is not necessarily infinite.

Shut up, you sillly asshole full of shit!

Each and every end segment is INFINITE by definition (and the usual properties of IN).

Hint: An e IN: card({m e IN: m >= n} = aleph_0.

Jim Burns schrieb am Freitag, 3. September 2021 um 23:48:51 UTC+2:
> On 9/3/2021 4:47 PM, WM wrote:
> > Jim Burns schrieb

> >> | In short,
> >> | if INT(Ends) is NOT empty,
> >> | then Ends is NOT the collection of all end segments of
> >> | FISONable naturals.
> >
> > The intersection cannot be empty without an empty endsegment
> > because of inclusion monotony.
> Why?
>
Because of inclusion monotony every endsegment is a subset of its predecessors. This chain cannot have an empty intersection unless one endsegment is empty.
> ----
> What is the first claim you object to?
>
> (i)
> If INT(Ends) is NOT empty,
> then a FISONable natural m is in INT(Ends).

No. "Not empty" means containing a natnural number but not necessarily a FISONable one.

Regards, WM

William schrieb am Samstag, 4. September 2021 um 02:19:47 UTC+2:
> On Friday, September 3, 2021 at 5:41:24 PM UTC-3, WM wrote:
> > Then you could remove all real numbers individually such that nothing remains. Then necessarily a last one would be removed.
> Niope, You can remove all without removing a last

This is only possible collectively. It is not possible one by one or individually.

Regards, WM

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?

Regards, WM

On Saturday, 4 September 2021 at 09:43:50 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.
Wrong. You clearly have no understanding of intersections, end segments, or infinity. I truly pity your students for having to sit through your drivel year after year.

On Saturday, September 4, 2021 at 2:48:28 PM UTC+2, Gus Gassmann wrote:
> On Saturday, 4 September 2021 at 09:43:50 UTC-3, WM wrote:

Not so fast, Gus. HERE WM is right:

> > Because of [strict] inclusion monotony every endsegment is a [proper] subset of its predecessors.

True.

Picking a certain endsegment we may consider the "chain" consisting of this endsegment and all its predecessors.

> > This chain [i.e. a "chain" of FINITELY MANY endsegements] cannot have an empty intersection unless one endsegment is empty.

Indeed! (And since no endsegment is empty all those "chains" will not have an empty intersection .)

But WM's original claim was:

| The intersection [of the set of all endsegments] cannot be empty without an empty endsegment
| because of inclusion monotony.

And that's simply WRONG. (Note that the set of all endsegments is infinite, while each and every of those "chains" mentioned by Mückenheim is finite.)

> You clearly have no understanding of intersections, end segments, or infinity. I truly pity your students for having to sit through your drivel year after year.

Yeah.

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".

--
William Hughes

On Saturday, 4 September 2021 at 10:34:05 UTC-3, Greg Cunt wrote:
> On Saturday, September 4, 2021 at 2:48:28 PM UTC+2, Gus Gassmann wrote:
> > On Saturday, 4 September 2021 at 09:43:50 UTC-3, WM wrote:
> Not so fast, Gus. HERE WM is right:
>
> > > Because of [strict] inclusion monotony every endsegment is a [proper] subset of its predecessors.
>
> True.
>
> Picking a certain endsegment we may consider the "chain" consisting of this endsegment and all its predecessors.
>
> > > This chain [i.e. a "chain" of FINITELY MANY endsegements] cannot have an empty intersection unless one endsegment is empty.
>
> Indeed! (And since no endsegment is empty all those "chains" will not have an empty intersection .)
>
> But WM's original claim was:
>
> | The intersection [of the set of all endsegments] cannot be empty without an empty endsegment
> | because of inclusion monotony.
>
> And that's simply WRONG. (Note that the set of all endsegments is infinite, while each and every of those "chains" mentioned by Mückenheim is finite.)
> > You clearly have no understanding of intersections, end segments, or infinity. I truly pity your students for having to sit through your drivel year after year.
> Yeah.

Alright. Point taken. However, I severely doubt that WM acknowledges that there is a difference between the two statements of his that you quoted, nor that he will make the distinction in the future. And he certainly did not add the crucial parenthetical remarks you provided.

On Saturday, September 4, 2021 at 4:59:21 PM UTC+2, Gus Gassmann wrote:

> Alright. Point taken. However, I severely doubt that WM acknowledges that there is a difference between the two statements of his that you quoted, nor that he will make the distinction in the future. And he certainly did not add the crucial parenthetical remarks you provided.

Yes, yes and yes.

He loves to use ambigous formulations. It's a well known strategy if you are not interested in truth but "debate".

https://www.logicallyfallacious.com/logicalfallacies/Ambiguity-Fallacy

On Saturday, September 4, 2021 at 4:59:21 PM UTC+2, Gus Gassmann wrote:

> Alright. Point taken. However, I severely doubt that WM acknowledges that there is a difference between the two statements of his that you quoted, nor that he will make the distinction in the future. And he certainly did not add the crucial parenthetical remarks you provided.

Yes, yes and yes.

He loves to use ambiguous formulations. It's a well known strategy if you are not interested in truth but "debate".

https://www.logicallyfallacious.com/logicalfallacies/Ambiguity-Fallacy

On 9/4/2021 7:45 AM, WM wrote:
> William schrieb am Samstag, 4. September 2021 um 02:19:47 UTC+2:
>> On Friday, September 3, 2021 at 5:41:24 PM UTC-3, WM wrote:
>>> Then you could remove all real numbers individually such that nothing remains. Then necessarily a last one would be removed.
>> Niope, You can remove all without removing a last
>
> This is only possible collectively. It is not possible one by one or individually.
>
> Regards, WM
>

gad, so there is no knowledge of infinity, nor limits, nor equations with WM, no wonder he is shepherded around by his imaginary Darkies, WM has no
mathematical structures to rely on.

On 9/4/2021 7:48 AM, Gus Gassmann wrote:
> On Saturday, 4 September 2021 at 09:43:50 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.
>
> Wrong. You clearly have no understanding of intersections, end segments, or infinity. I truly pity your students for having to sit through your drivel year after year.
>

agree, WM is doing so badly on very simple stuff. Teacher ? NFW. I would walk out of class in the first 15 min. and never go back.

SO, what value is he creating here ?

Oh yea, his vast contributions to the ANT LIST 6.0 !!

On 9/4/2021 8:43 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 3. September 2021 um 23:48:51 UTC+2:
>> On 9/3/2021 4:47 PM, WM wrote:
>>> Jim Burns schrieb

>>>> | In short,
>>>> | if INT(Ends) is NOT empty,
>>>> | then Ends is NOT the collection of all end segments of
>>>> | FISONable naturals.
>>>
>>> The intersection cannot be empty without an empty endsegment
>>> because of inclusion monotony.
>>
>> Why?
>
> Because of inclusion monotony every endsegment is a subset of
> its predecessors.

Why this next bit?

> 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.

It is an excess requirement for for each element to be
not-in *the same* set in the collection.

This is not about potential infinity or actual infinity.
It is about intersection.

>> ----
>> What is the first claim you object to?
>>
>> (i)
>> If INT(Ends) is NOT empty,
>> then a FISONable natural m is in INT(Ends).
>
> No. "Not empty" means containing a natnural number but
> not necessarily a FISONable one.

Replacing...

| Let _Ends_ be an arbitrary collection of end segments of
| FISONable naturals.
| Let INT(Ends) be its intersection.
| In short,
| if INT(Ends) is NOT empty,
| then Ends is NOT the collection of all end segments of
| FISONable naturals.

INT(Ends) cannot contain anything which is not-in each set
in Ends. Each set (end segment) in Ends cannot contain
anything which is not FISONable.

Therefore,
if m is in INT(Ends), then m is FISONable.

( If m is in INT(Ends), then E(m+1) is not-in Ends. )

----
Think about what it implies that, in order to find something
to object to, you need to change what my argument is.

What if you changed your mind?
People do that, you know. It's not a tragedy.

Serg io wrote:

>> Wrong. You clearly have no understanding of intersections, end
>> segments,
>> or infinity. I truly pity your students for having to sit through your
>> drivel year after year.
>>
> agree, WM is doing so badly on very simple stuff. Teacher ? NFW. I
> would walk out of class in the first 15 min. and never go back.

TRUCK CARRYING MODERNA JABS CRASHES - DEPT OF DEFENSE IN HAZMAT SHOW UP..

the toxic spike_protein is the poison from the stinging scorpion like
creatures, from the Bible.

Here comes your covid_19 permit code

█▀▀▀▀▀█ █ ██████ ██▄▄ █▀▀▀▀▀█
█ ███ █ ▀▀▀▄▄▀█▄ ▄▀▄█ █ ███ █
█ ▀▀▀ █ ▀▄█▄ █▄▀ ▀▄▀▄ █ ▀▀▀ █
▀▀▀▀▀▀▀ ▀▄▀ █ █ ▀ █▄█ ▀▀▀▀▀▀▀
█▄▄▀▀█▀▀█▄▀▄█▀█▀▀▀▄▄ ▀ ▄█ ▀▀▀
▄ █▄▄▀▀▄█ █▀▄ █▄ █ ▀ ▀ ▄▀▄
▄█ ▀█ ▀ ▀▄▀▀▀▀▀ ▀ ▄▀█ ██▄█▄▀█
██▀▀ ▀▀█▀ ▄▄█▀▀ █ ▄█ ▄▄▄▄█
▄▀██▀▀▀██▄███ █▀▄▄██▄ ▄▄ ▀▀
██▀██ ▀▀▄ █ █▀█▄▀▄█▄█ ▄ █▄▀
▀▀▀▀ ▀ █▄ ▄█ ▄▀▀█▀█▀▀▀██▀ ▄
█▀▀▀▀▀█ █ ███ ▀█████ ▀ █ ▀▄▄
█ ███ █ █▀█ ▄█▄▄█▄▄██▀▀▀▀█▄▀█
█ ▀▀▀ █ ▀ █ ▄▀▀▀ ▀▀ █ ███▄▀█
▀▀▀▀▀▀▀ ▀▀ ▀ ▀ ▀▀ ▀

Serg io wrote:

> gad, so there is no knowledge of infinity, nor limits, nor equations
> with WM, no wonder he is shepherded around by his imaginary Darkies, WM
> has no mathematical structures to rely on.

Wow! Vaccine-Caused Deaths Reported as "Unvaxxed" COVID Deaths!
https://www.bitchute.com/video/klxSY2Jlb6f4/

these mass murderers are so evil.

On 9/4/2021 3:26 PM, H2O wrote:
> Serg io wrote:
>
>>> Wrong. You clearly have no understanding of intersections, end
>>> segments,
>>> or infinity. I truly pity your students for having to sit through your
>>> drivel year after year.
>>>
>> agree, WM is doing so badly on very simple stuff. Teacher ? NFW. I
>> would walk out of class in the first 15 min. and never go back.
>
> TRUCK CARRYING MODERNA JABS CRASHES - DEPT OF DEFENSE IN HAZMAT SHOW UP..
>
> the toxic spike_protein is the poison from the stinging scorpion like
> creatures, from the Bible.
>
> Here comes your covid_19 permit code
>
>
>
> █▀▀▀▀▀█ █ ██████ ██▄▄ █▀▀▀▀▀█
> █ ███ █ ▀▀▀▄▄▀█▄ ▄▀▄█ █ ███ █
> █ ▀▀▀ █ ▀▄█▄ █▄▀ ▀▄▀▄ █ ▀▀▀ █
> ▀▀▀▀▀▀▀ ▀▄▀ █ █ ▀ █▄█ ▀▀▀▀▀▀▀
> █▄▄▀▀█▀▀█▄▀▄█▀█▀▀▀▄▄ ▀ ▄█ ▀▀▀
> ▄ █▄▄▀▀▄█ █▀▄ █▄ █ ▀ ▀ ▄▀▄
> ▄█ ▀█ ▀ ▀▄▀▀▀▀▀ ▀ ▄▀█ ██▄█▄▀█
> ██▀▀ ▀▀█▀ ▄▄█▀▀ █ ▄█ ▄▄▄▄█
> ▄▀██▀▀▀██▄███ █▀▄▄██▄ ▄▄ ▀▀
> ██▀██ ▀▀▄ █ █▀█▄▀▄█▄█ ▄ █▄▀
> ▀▀▀▀ ▀ █▄ ▄█ ▄▀▀█▀█▀▀▀██▀ ▄
> █▀▀▀▀▀█ █ ███ ▀█████ ▀ █ ▀▄▄
> █ ███ █ █▀█ ▄█▄▄█▄▄██▀▀▀▀█▄▀█
> █ ▀▀▀ █ ▀ █ ▄▀▀▀ ▀▀ █ ███▄▀█
> ▀▀▀▀▀▀▀ ▀▀ ▀ ▀ ▀▀ ▀
>
>
>

ha! yo mama tattooed that on yo face!
you Bug Face !

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, it still has the property that it can be distinguished. That is your claim?

Regards, WM

William schrieb am Samstag, 4. September 2021 um 16:13:42 UTC+2:
> On Saturday, September 4, 2021 at 9:48:28 AM UTC-3, Gus Gassmann wrote:
> > On Saturday, 4 September 2021 at 09:43:50 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. No empty intersection.

Regards WM

William schrieb am Samstag, 4. September 2021 um 16:07:51 UTC+2:
> On Saturday, September 4, 2021 at 9:45:37 AM UTC-3, WM wrote:
> > William schrieb am Samstag, 4. September 2021 um 02:19:47 UTC+2:
> > > On Friday, September 3, 2021 at 5:41:24 PM UTC-3, WM wrote:
> > > > Then you could remove all real numbers individually such that nothing remains. Then necessarily a last one would be removed.
> > > Niope, You can remove all without removing a last
> > This is only possible collectively. It is not possible one by one or individually.
> Nope. It is only not possible if there must be a last step. "one by one or individually" does not mean there has to be a last step.

The step to zero is the last step. This step can be done because it is possible to reach zero.

Regards, WM

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, 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.

Regards, WM

On Saturday, September 4, 2021 at 11:00:34 PM UTC+2, WM wrote:

> If there is no empty endsegment or no last endsegment, then there will remain natural numbers in the intersection.

Nope. Your claim is false.

Hint: There is no empty endsegment and there is no "last" endsegment, but the intersection of (the set of) all endsegments is empty.

What's the matter with you, Mückenheim? Lost in delusion?

“When we are lost in delusion, it's hard to see even the most obvious truths.” ― Jack Kornfield,

On 9/4/2021 3:52 PM, 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, it still has the property that it can be distinguished. That is your claim?
>
> Regards, WM
>

guys, that is not making any sense.

n is n is n

if n is a variable, it is a variable. n element of |N_F

If you want to say a "number cannot be written down" just say the number is unknown or a variable.

if you say "distinguished", you go back to using WM's daffynition, in imagination land, which is unworkable on its face (there is no requirement to
recording or preserve the number, raps, beeps, chicken pecks, or taps or writing down... of the number)

and you leave out the context, which shows you are not serious.

Ants that cannot be written down
Ants without context
Not Serious Ants that do not make sense

On 9/4/2021 3:55 PM, WM wrote:
> William schrieb am Samstag, 4. September 2021 um 16:07:51 UTC+2:
>> On Saturday, September 4, 2021 at 9:45:37 AM UTC-3, WM wrote:
>>> William schrieb am Samstag, 4. September 2021 um 02:19:47 UTC+2:
>>>> On Friday, September 3, 2021 at 5:41:24 PM UTC-3, WM wrote:
>>>>> Then you could remove all real numbers individually such that nothing remains. Then necessarily a last one would be removed.
>>>> Niope, You can remove all without removing a last
>>> This is only possible collectively. It is not possible one by one or individually.
>> Nope. It is only not possible if there must be a last step. "one by one or individually" does not mean there has to be a last step.
>
> The step to zero is the last step.

There is no last step.

> This step can be done because it is possible to reach zero.

no

>
> Regards, WM
>

On 9/4/2021 3:54 PM, WM wrote:
> William schrieb am Samstag, 4. September 2021 um 16:13:42 UTC+2:
>> On Saturday, September 4, 2021 at 9:48:28 AM UTC-3, Gus Gassmann wrote:
>>> On Saturday, 4 September 2021 at 09:43:50 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. No empty intersection.
>
> Regards WM
>

dream on

On 9/4/2021 4:00 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, 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.
>
> Regards, WM
>

no, your position on this has been proved false.