On Monday, September 6, 2021 at 9:37:55 AM UTC-3, WM wrote:
> 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.

Nope. "address" is just another of the many many many ways you have of saying "write down" . Just because you cannot write down an element of |N_F does not mean it is "nothing"

--
William Hughes

William schrieb am Montag, 6. September 2021 um 15:48:07 UTC+2:
> On Monday, September 6, 2021 at 9:37:55 AM UTC-3, WM wrote:
> > 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.
> Nope. "address" is just another of the many many many ways you have of saying "write down" . Just because you cannot write down an element of |N_F does not mean it is "nothing"
>
How would you handle an element and show its existence if you cannot write down or address it?

Regards, WM

On 9/6/2021 11:59 AM, WM wrote:
> William schrieb am Montag, 6. September 2021 um 15:48:07 UTC+2:
>> On Monday, September 6, 2021 at 9:37:55 AM UTC-3, WM wrote:
>>> 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.
>> Nope. "address" is just another of the many many many ways you have of saying "write down" . Just because you cannot write down an element of |N_F does not mean it is "nothing"
>>
> How would you handle an element and show its existence if you cannot write down or address it?
>
> Regards, WM
>

call it a variable

On Monday, September 6, 2021 at 1:59:34 PM UTC-3, WM wrote:
> William schrieb am Montag, 6. September 2021 um 15:48:07 UTC+2:
> > On Monday, September 6, 2021 at 9:37:55 AM UTC-3, WM wrote:
> > > 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.
> > Nope. "address" is just another of the many many many ways you have of saying "write down" . Just because you cannot write down an element of |N_F does not mean it is "nothing"
> >
> How would you handle an element and show its existence if you cannot write down or address it?

Add the term "handle" to the list of the many many many ways you have or saying "write down". Your question is the silly, how can you write and element of |N_F down if the element cannot be written down. The idea that "can be written down" is some magical property without which an element of |N_F cannot have any other property is nonsense.

--
William Hughes

On Monday, September 6, 2021 at 9:59:34 AM UTC-7, WM wrote:
> >
> How would you handle an element and show its existence if you cannot write down or address it?
>
> Regards, WM

Prove it doesn't not exist.

That's called an existence proof.

It usually follows deduction....

Really these are simple enough concepts that
eventually a general audience becomes well-versed
in such matters of critical thinking that require fundamentally
a handle on reason, which is generally useful in all matters.

Again, "Cantor proves the line is drawn", and, it's particularly
a central fixture in mathematics _that looks and acts exactly like time_,
then for the uncountable complete ordered field, then, for the rationals
being large, the difference between the space of functions N <-> [0,1],
R <-> R, and Q[a,b] <-> R[a,b], with line continuity, field continuity,
and signal continuity. Or, "time draws the line and crests the signal".

Which is much better than its closed counterpart "don't know better".

William schrieb am Montag, 6. September 2021 um 19:09:24 UTC+2:
> On Monday, September 6, 2021 at 1:59:34 PM UTC-3, WM wrote:

> > How would you handle an element and show its existence if you cannot write down or address it?
> Add the term "handle" to the list of the many many many ways you have or saying "write down". Your question is the silly, how can you write and element of |N_F down if the element cannot be written down.

How do you show that it is and individual and differs from other individuals, i.e., it is not dark?

Regards, WM

On Monday, September 6, 2021 at 2:28:41 PM UTC-3, WM wrote:
> William schrieb am Montag, 6. September 2021 um 19:09:24 UTC+2:
> > On Monday, September 6, 2021 at 1:59:34 PM UTC-3, WM wrote:
>
> > > How would you handle an element and show its existence if you cannot write down or address it?
> > Add the term "handle" to the list of the many many many ways you have or saying "write down". Your question is the silly, how can you write and element of |N_F down if the element cannot be written down.
> How do you show that it is and individual and differs from other individuals, i.e., it is not dark?

All elements of |N_F have the property that they differ from other elements. Calling an element of |N_F "dark" is silly.

On 9/5/2021 4:28 PM, WM wrote:
> 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.

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

....which implies...

If INT(Ends) is NOT empty,
then Ends is NOT the collection of all FISONable end segments.

....which implies...

If Ends is the collection of all FISONable end segments,
then INT(Ends) is empty.

....where...

INT(Ends) is the intersection of collection Ends
and so
conatains only such elements as are in each set in Ends.

Ends is an arbitrary collection of FISONable-natural end segments
and so
contains only FISONable-natural end segments.

E(m+1) is a FISONable-natural end segment beginning at m+1.
and so
is non-empty, closed upward, and contains only FISONable naturals.

m is an arbitrary FISONable natural
and so
there exists a steppable {0,...,m} in which,
for adjacent i,j, j = i+1.

> This is independent of how many endsegments there are.

True. Instead, it is dependent on the fact that
each FISONable natural is not-in at least one
FISONable-natural end segment.

Because Ak e N: E(k+1) = E(k)\{k}
the intersection is empty.

WM wrote on 9/6/2021 :
> William schrieb am Montag, 6. September 2021 um 15:48:07 UTC+2:
>> On Monday, September 6, 2021 at 9:37:55 AM UTC-3, WM wrote:
>>> 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.
>> Nope. "address" is just another of the many many many ways you have of
>> saying "write down" . Just because you cannot write down an element of |N_F
>> does not mean it is "nothing"
>>
> How would you handle an element and show its existence if you cannot write
> down or address it?
>
> Regards, WM

Pi times the exponential e is a real number even if you don't think so.

On 9/6/2021 12:28 PM, WM wrote:
> William schrieb am Montag, 6. September 2021 um 19:09:24 UTC+2:
>> On Monday, September 6, 2021 at 1:59:34 PM UTC-3, WM wrote:
>
>>> How would you handle an element and show its existence if you cannot write down or address it?
>> Add the term "handle" to the list of the many many many ways you have or saying "write down". Your question is the silly, how can you write and element of |N_F down if the element cannot be written down.
>
> How do you show that it is and individual and differs from other individuals, i.e., it is not dark?
>
> Regards, WM
>

do you believe there is a number between 100 and 200 without writing it down ?

William schrieb am Montag, 6. September 2021 um 19:56:52 UTC+2:

> All elements of |N_F have the property that they differ from other elements.

Can you show an example of two individual elements which are not nameable?

Regards, WM

On 9/6/2021 1:55 PM, WM wrote:
> William schrieb am Montag, 6. September 2021 um 19:56:52 UTC+2:
>
>> All elements of |N_F have the property that they differ from other elements.
>
> Can you show an example of two individual elements which are not nameable?
>
> Regards, WM
>

back to counting rocks again...

what if two rocks are identical in every way ? are they still two separate rocks ?

how would you distinguish them, or name them ?

I have 2 marbles that fell out.
They are exactly the same.
How do I distinguish them?
Should I give them back to you ?
Would you miss them ? which one would you miss the most ? How do you know ?

Jim Burns schrieb am Montag, 6. September 2021 um 20:21:13 UTC+2:

> Because Ak e N: E(k+1) = E(k)\{k}
> the intersection is empty.

The intersection is the last endsegment. If there is no last endsegment, then every endegment has infinitely many elements and there is only a current intersection which is the same as the current endegment.

Regards, WM

On Monday, September 6, 2021 at 8:52:09 PM UTC+2, Serg io wrote:

> do you believe there is a number between 100 and 200 without writing it down ?

Rather: Do you believe there is a number between 10^80^100 and 10^80^100 + 2 without writing it down? :-)

How many elements (natural numbers) are in the set {n e IN: n < 10^80^100}?

On 9/6/2021 2:02 PM, WM wrote:
> Jim Burns schrieb am Montag, 6. September 2021 um 20:21:13 UTC+2:
>
>> Because Ak e N: E(k+1) = E(k)\{k}
>> the intersection is empty.
>
> The intersection is the last endsegment. If there is no last endsegment, then every endegment has infinitely many elements and there is only a current intersection which is the same as the current endegment.
>
> Regards, WM
>

The last endsegment (the movie)

https://tubitv.com/movies/463838/the-last-unicorn

Greg Cunt schrieb am Montag, 6. September 2021 um 21:10:54 UTC+2:

> How many elements (natural numbers) are in the set {n e IN: n < 10^80^100}?

According to ideal mathematics there are 10^80^100 - 1 elements.
According to MatheRealism there are as many as you can individually define. The few first and the few last are simple:
1, 2, 3, ..., 10^80^100 -3, 10^80^100 - 2, 10^80^100 - 1.
But many are missing between these ends.

Regards, WM

On Monday, September 6, 2021 at 9:02:07 PM UTC+2, WM wrote:
> Jim Burns schrieb am Montag, 6. September 2021 um 20:21:13 UTC+2:
> >
> > Because Ak e N: E(k+1) = E(k)\{k} the intersection [of the set of all endelements ] is empty.

INTERSECTION {E(k) : k e IN} = {}

Hints:
(1) Ak e IN: E(k) is infinite.
(2) There is no "last" ensegment in {E(k) : k e IN}

> The intersection is the last endsegment.

There is no last endsegment in {E(k) : k e IN}.

> If there is no last endsegment, then every endsegment has infinitely many elements

Well, since each and every endsegment has infinitely many elements by definition, your claim is certainly true.

"If New York is a big town, then 1 + 1 = 2." => true.

> and there is only a current intersection which is the same as the current endegment.

Please define the notion of "a current intersection" and the notion of "the current endegment".

Is it floating? A fluent?
See: https://en.wikipedia.org/wiki/Fluent_(mathematics)

On Monday, September 6, 2021 at 9:15:03 PM UTC+2, Serg io wrote:
> On 9/6/2021 2:02 PM, WM wrote:
> > Jim Burns schrieb am Montag, 6. September 2021 um 20:21:13 UTC+2:
> >
> >> Because Ak e N: E(k+1) = E(k)\{k}
> >> the intersection is empty.
> >
> > The intersection is the last endsegment. If there is no last endsegment, then every endegment has infinitely many elements and there is only a current intersection which is the same as the current endegment.
> >
> > Regards, WM
> >
> The last endsegment (the movie)
> >
> https://tubitv.com/movies/463838/the-last-unicorn

The last winter (the movie)

https://en.wikipedia.org/wiki/The_Last_Winter_(2006_film)

Great film!

Greg Cunt schrieb am Montag, 6. September 2021 um 21:18:54 UTC+2:
> On Monday, September 6, 2021 at 9:02:07 PM UTC+2, WM wrote:

> > If there is no last endsegment, then every endsegment has infinitely many elements
>
> Well, since each and every endsegment has infinitely many elements by definition, your claim is certainly true.
>
> > and there is only a current intersection which is the same as the current endegment.
> Please define the notion of "a current intersection" and the notion of "the current endegment".

It is the last one considered.
>
> Is it floating? A fluent?

Yes, since there is no last one to be considered.

Regards, WM

On Monday, September 6, 2021 at 3:55:25 PM UTC-3, WM wrote:
> William schrieb am Montag, 6. September 2021 um 19:56:52 UTC+2:
>
> > All elements of |N_F have the property that they differ from other elements.
> Can you show an example of two individual elements which are not nameable?

Add "show" and "nameable" to the many many many ways you have of saying "written down"' This is a silly question asking to write down two elements that cannot be written down. The fact that you cannon write down an element that you cannot write down does not mean the element cannot have other properties. In particular, any element of |N_F will have the properties that every element of |N_F has.

--
William Hughes

On Monday, 6 September 2021 at 16:39:36 UTC-3, WM wrote:
> Greg Cunt schrieb am Montag, 6. September 2021 um 21:18:54 UTC+2:
> > On Monday, September 6, 2021 at 9:02:07 PM UTC+2, WM wrote:
>
> > > If there is no last endsegment, then every endsegment has infinitely many elements
> >
> > Well, since each and every endsegment has infinitely many elements by definition, your claim is certainly true.
> >
> > > and there is only a current intersection which is the same as the current endegment.
> > Please define the notion of "a current intersection" and the notion of "the current endegment".
> It is the last one considered.
> >
> > Is it floating? A fluent?
> Yes, since there is no last one to be considered.

.... proving again to the world (well, readers of sci.math) that you have no clue about intersections of infinite collections, or stepwise processes, and why your stepwise approach is doomed to failure.

On Monday, September 6, 2021 at 4:02:07 PM UTC-3, WM wrote:

> The intersection is the last endsegment. If there is no last endsegment, then every endegment has infinitely many elements and there is only a current intersection which is the same as the current endegment.

Piffle. Add "current intersection" to the list of nonsense changing things like "Potentially infitite set"

--
William Hughes

On 9/6/2021 2:39 PM, WM wrote:
> Greg Cunt schrieb am Montag, 6. September 2021 um 21:18:54 UTC+2:
>> On Monday, September 6, 2021 at 9:02:07 PM UTC+2, WM wrote:
>
>>> If there is no last endsegment, then every endsegment has infinitely many elements
>>
>> Well, since each and every endsegment has infinitely many elements by definition, your claim is certainly true.
>>
>>> and there is only a current intersection which is the same as the current endegment.
>> Please define the notion of "a current intersection" and the notion of "the current endegment".
>
> It is the last one considered.

oh, so some poor dude, has to consider all the endsegments? even to the very last one ?

>>
>> Is it floating? A fluent?
>
> Yes, since there is no last one to be considered.

so the dude never finishes, he has lifetime employment.

>
> Regards, WM
>

Serg io wrote:

>>> Is it floating? A fluent?
>>
>> Yes, since there is no last one to be considered.
>
> so the dude never finishes, he has lifetime employment.

you math guys gotta learn to sing *The Internationale*.

la laaaa, lala lala laaaa la, la la lala lala lala.

On 9/6/2021 4:56 PM, Serg io wrote:
> On 9/6/2021 2:39 PM, WM wrote:

>>
>> It is the last one considered.
>
> oh, so some poor dude, has to consider all the endsegments?
> even to the very last one ?
>
>>> Is it floating? A fluent?
>>
>> Yes, since there is no last one to be considered.
>
> so the dude never finishes, he has lifetime employment.

Or longer.

http://www.lasfsinc.info/index.php?option=com_content&task=view&id=161&Itemid=519

Death Will Not Release You