On 9/9/2021 8:51 AM, Greg Cunt wrote:
> On Thursday, September 9, 2021 at 3:01:10 PM UTC+2, WM wrote:
>
>> In linear order nobody should believe the claim of all without being shown the last element.
>
> FASCINATING! You mean the last natural number?
>

yes they made a movie about it;

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

On Thursday, September 9, 2021 at 9:34:35 PM UTC+2, Serg io wrote:
> On 9/9/2021 8:51 AM, Greg Cunt wrote:
> > On Thursday, September 9, 2021 at 3:01:10 PM UTC+2, WM wrote:
> > >
> > > In linear order nobody should believe the claim of all without being shown the last element.
> > >
> > FASCINATING! You mean the last natural number?
> >
> yes they made a movie about it;
>
> https://tubitv.com/movies/463838/the-last-unicorn

I don't think so. Imho it's

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

Greg Cunt presented the following explanation :
> On Thursday, September 9, 2021 at 9:34:35 PM UTC+2, Serg io wrote:
>> On 9/9/2021 8:51 AM, Greg Cunt wrote:
>>> On Thursday, September 9, 2021 at 3:01:10 PM UTC+2, WM wrote:
>>>>
>>>> In linear order nobody should believe the claim of all without being shown
>>>> the last element.
>>>>
>>> FASCINATING! You mean the last natural number?
>>>
>> yes they made a movie about it;
>>
>> https://tubitv.com/movies/463838/the-last-unicorn
>
> I don't think so. Imho it's
>
> https://en.wikipedia.org/wiki/The_Last_Winter_(2006_film)

I have a DVD set of "The Last Samurai" -- it is a finite set.

Serg io was thinking very hard :
> On 9/9/2021 8:01 AM, WM wrote:
>> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
>> UTC+2:
>>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
>>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
>>>>
>>>>> If there is no last end segment of all end segments,
>>>>> then the intersection of all end segments is
>>>> Then there is no "all endsegments". "All" can only be proved by showing
>>>> the last.
>>>>
>>> Where does it say that "all" must mean a last?
>>
>> In linear order nobody should believe the claim of all without being shown
>> the last element.
>
>
> all natural numbers => 1,2,3,...
>
> so according to WM do not believe in Natural Numbers unless WM can show you
> the Last one.

Sure, there's the last one, this current one, and the next one.

On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
> Serg io was thinking very hard :
> > On 9/9/2021 8:01 AM, WM wrote:
> >> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
> >> UTC+2:
> >>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
> >>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
> >>>>
> >>>>> If there is no last end segment of all end segments,
> >>>>> then the intersection of all end segments is
> >>>> Then there is no "all endsegments". "All" can only be proved by showing
> >>>> the last.
> >>>>
> >>> Where does it say that "all" must mean a last?
> >>
> >> In linear order nobody should believe the claim of all without being shown
> >> the last element.
> >
> >
> > all natural numbers => 1,2,3,...
> >
> > so according to WM do not believe in Natural Numbers unless WM can show you
> > the Last one.
> Sure, there's the last one, this current one, and the next one.

Cool. So, in an inductive set it should indeed be sufficient to be shown the last one, because you then have this current one, as well as the next one! Is that what the great perfessor is taking about?

On Friday, September 10, 2021 at 1:19:15 AM UTC+2, Gus Gassmann wrote:
> On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
> > Serg io was thinking very hard :
> > >
> > > so according to WM do not believe in Natural Numbers unless WM can show you the Last one.
> > >
> > Sure, there's the last one, this current one, and the next one.
> >
> Cool. So, in an inductive set it should indeed be sufficient to be shown the last one, because you then have this current one, as well as the next one! Is that what the great perfessor is taking about?

Exactly!

Here's a quote from a recent post:
===============================================

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

WM> It is the last one considered.

GC> Is it floating? A fluent?

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

===============================================

I'd say this helps a lot. lol

This might help too: https://www.youtube.com/watch?v=0a5WyAjL1MM

Once one of my colleagues uttered: "It's tuesday, I guess. Who knows?!"

Gus Gassmann pretended :
> On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
>> Serg io was thinking very hard :
>>> On 9/9/2021 8:01 AM, WM wrote:
>>>> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
>>>> UTC+2:
>>>>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
>>>>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
>>>>>>
>>>>>>> If there is no last end segment of all end segments,
>>>>>>> then the intersection of all end segments is
>>>>>> Then there is no "all endsegments". "All" can only be proved by showing
>>>>>> the last.
>>>>>>
>>>>> Where does it say that "all" must mean a last?
>>>>
>>>> In linear order nobody should believe the claim of all without being shown
>>>> the last element.
>>>
>>>
>>> all natural numbers => 1,2,3,...
>>>
>>> so according to WM do not believe in Natural Numbers unless WM can show you
>>> the Last one.
>> Sure, there's the last one, this current one, and the next one.
>
> Cool. So, in an inductive set it should indeed be sufficient to be shown the
> last one, because you then have this current one, as well as the next one! Is
> that what the great perfessor is taking about?

Who knows what he's talking about, he hasn't made any sense for
decades.

On Thursday, September 9, 2021 at 7:25:37 PM UTC-7, FromTheRafters wrote:
> Gus Gassmann pretended :
> > On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
> >> Serg io was thinking very hard :
> >>> On 9/9/2021 8:01 AM, WM wrote:
> >>>> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
> >>>> UTC+2:
> >>>>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
> >>>>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
> >>>>>>
> >>>>>>> If there is no last end segment of all end segments,
> >>>>>>> then the intersection of all end segments is
> >>>>>> Then there is no "all endsegments". "All" can only be proved by showing
> >>>>>> the last.
> >>>>>>
> >>>>> Where does it say that "all" must mean a last?
> >>>>
> >>>> In linear order nobody should believe the claim of all without being shown
> >>>> the last element.
> >>>
> >>>
> >>> all natural numbers => 1,2,3,...
> >>>
> >>> so according to WM do not believe in Natural Numbers unless WM can show you
> >>> the Last one.
> >> Sure, there's the last one, this current one, and the next one.
> >
> > Cool. So, in an inductive set it should indeed be sufficient to be shown the
> > last one, because you then have this current one, as well as the next one! Is
> > that what the great perfessor is taking about?
> Who knows what he's talking about, he hasn't made any sense for
> decades.

It's always "infinite induction exhausts" never satisfying related rates problems
or the countable and uncountable in functional analysis and infinite domains.

(WM is _plainly_ mostly Hodges' hopeless and sometimes "infinite integers
as discredited as possible".)

Is it a paradox that the differential via countable additivity both integrates
under any continuous functions but there are uncountably many points on it?

Is it not because there aren't uncountably partitions of the unit interval because
each would contain a rational?

Doesn't well-ordering the reals have a subset that is well-ordering uncountably many,
in their order?

Counterexamples....

I know what's he's talking about, just doing it wrong, not enough, and the wrong way.

Ross A. Finlayson was thinking very hard :
> On Thursday, September 9, 2021 at 7:25:37 PM UTC-7, FromTheRafters wrote:
>> Gus Gassmann pretended :
>>> On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
>>>> Serg io was thinking very hard :
>>>>> On 9/9/2021 8:01 AM, WM wrote:
>>>>>> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
>>>>>> UTC+2:
>>>>>>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
>>>>>>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
>>>>>>>>
>>>>>>>>> If there is no last end segment of all end segments,
>>>>>>>>> then the intersection of all end segments is
>>>>>>>> Then there is no "all endsegments". "All" can only be proved by
>>>>>>>> showing the last.
>>>>>>>>
>>>>>>> Where does it say that "all" must mean a last?
>>>>>>
>>>>>> In linear order nobody should believe the claim of all without being
>>>>>> shown the last element.
>>>>>
>>>>>
>>>>> all natural numbers => 1,2,3,...
>>>>>
>>>>> so according to WM do not believe in Natural Numbers unless WM can show
>>>>> you the Last one.
>>>> Sure, there's the last one, this current one, and the next one.
>>>
>>> Cool. So, in an inductive set it should indeed be sufficient to be shown
>>> the last one, because you then have this current one, as well as the next
>>> one! Is that what the great perfessor is taking about?
>> Who knows what he's talking about, he hasn't made any sense for
>> decades.
>
>
> It's always "infinite induction exhausts" never satisfying related rates
> problems or the countable and uncountable in functional analysis and
> infinite domains.
>
> (WM is _plainly_ mostly Hodges' hopeless and sometimes "infinite integers
> as discredited as possible".)
>
> Is it a paradox that the differential via countable additivity both
> integrates under any continuous functions but there are uncountably many
> points on it?
>
> Is it not because there aren't uncountably partitions of the unit interval
> because each would contain a rational?
>
> Doesn't well-ordering the reals have a subset that is well-ordering
> uncountably many, in their order?
>
>
> Counterexamples....
>
>
> I know what's he's talking about, just doing it wrong, not enough, and the
> wrong way.

Yeah, I guess it is easy to know what he is talking about when you
consider his goal is to use his misunderstandings about "series" to
show Cantor's diagonal method of 'listing; the rationals is flawed
because Cantor ignores WM's 'Dark Numbers' which he says cannot be
mapped because they can't be exactly expressed as a finite sequence.

He tries to shoehorn in some 'Dark Numbers' into ZFC and when it
doesn't work he exclaims "ZFC is inconsistent" because it ignores his
"Dark Numbers".

FromTheRafters schrieb am Freitag, 10. September 2021 um 12:15:23 UTC+2:

> He tries to shoehorn in some 'Dark Numbers' into ZFC and when it
> doesn't work he exclaims "ZFC is inconsistent" because it ignores his
> "Dark Numbers".

I try to save as much as possible of ZFC. Without dark numbers this is impossible.

Regards, WM

William schrieb am Donnerstag, 9. September 2021 um 15:41:02 UTC+2:
> On Thursday, September 9, 2021 at 10:29:13 AM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 9. September 2021 um 01:00:13 UTC+2:
> > > On Wednesday, September 8, 2021 at 4:46:14 PM UTC-3, WM wrote:
> > > . Objects which have rows or vertices are with certainty no natural numbers. So your simply saying "n is a natural number" is nonsense.
> > > If you say that n is a natural number that cannot be written down (we know such a number exists) the n has all the properties of a natural number.
> > But you cannot know anything further about n
> We do not need to. We know existence and we know that n has all the properties of an element of |N_F.

You caqnnot distinguish it from another such element.

Regards, WM

William schrieb am Donnerstag, 9. September 2021 um 15:46:00 UTC+2:
> On Thursday, September 9, 2021 at 10:09:49 AM UTC-3, WM wrote:
> > No [single element of the sequence] can cause a finite intersection
>
> True. A finite intersection is a property of the sequence, not a property of an element of the sequence.

The sequence is inclusion monotonic. Therefore its intersection cannot be empty if no endsegment is empty.

Regards, WM

Greg Cunt schrieb am Donnerstag, 9. September 2021 um 15:51:46 UTC+2:
> On Thursday, September 9, 2021 at 3:03:01 PM UTC+2, WM wrote:
>
> > Because we or our ancestors have constructed them without rows and vertices.
> Where did they construct them?

On many occasions. Counting sheep for intance.

> Where can we find them?

In school books for instance.

Regards, WM

William schrieb am Donnerstag, 9. September 2021 um 15:58:45 UTC+2:
> On Thursday, September 9, 2021 at 10:01:10 AM UTC-3, WM wrote:
>
> >In linear order nobody should believe the claim of all without being shown the last element.
> A Peano set is linearly ordered, and a Peano set does not have a last element.

It is not complete. There is no "all".

Regards, WM

Jim Burns schrieb am Donnerstag, 9. September 2021 um 20:57:23 UTC+2:

> If there is no last end segment,
> then no end segment is finite.

Then the intersection is infinite. Note: Inclusion monotony. The intersection cnnot shrink below all endsegments.

Regards, WM

On Friday, September 10, 2021 at 3:18:07 PM UTC+2, WM wrote:

> The sequence is inclusion monotonic. Therefore its intersection cannot be empty if no endsegment is empty.

Please prove this argument.

On Friday, September 10, 2021 at 3:20:47 PM UTC+2, WM wrote:
> William schrieb am Donnerstag, 9. September 2021 um 15:58:45 UTC+2:
> > On Thursday, September 9, 2021 at 10:01:10 AM UTC-3, WM wrote:
> > >
> > > In linear order nobody should believe the claim of all without being shown the last element.
> > >
> > A Peano set is linearly ordered, and a Peano set does not have a last element.
> >
> It is not complete. There is no "all".

So we can't state anything concerning all elements in IN in Muckenmath?

Not even: For all n,m e IN: n + n = m + n

or: For all n e IN: 0 <= n .

etc. FASCINATING!

Well, at least in the context of MATHEMATICS we can.

On Friday, September 10, 2021 at 3:23:14 PM UTC+2, WM wrote:
> Jim Burns schrieb am Donnerstag, 9. September 2021 um 20:57:23 UTC+2:
> >
> > If there is no last end segment, then no end segment is finite.
> >
> Then the intersection is infinite.

Which intersection?

Hint: The intersection of (the set of) all end segments is empty.

On Friday, 10 September 2021 at 10:23:14 UTC-3, WM wrote:
> Jim Burns schrieb am Donnerstag, 9. September 2021 um 20:57:23 UTC+2:
>
> > If there is no last end segment,
> > then no end segment is finite.
> Then the intersection is infinite. Note: Inclusion monotony. The intersection cnnot shrink below all endsegments.

Nice, Muckenheim! Circular logic AND quantifier dyslexia in one single line. Add to that a complete lack of understanding about inclusion monotony, and you got today's Trifecta. Keep up the good work. (NOT!)

On Friday, September 10, 2021 at 10:16:28 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 9. September 2021 um 15:41:02 UTC+2:
> > On Thursday, September 9, 2021 at 10:29:13 AM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 9. September 2021 um 01:00:13 UTC+2:
> > > > On Wednesday, September 8, 2021 at 4:46:14 PM UTC-3, WM wrote:
> > > > . Objects which have rows or vertices are with certainty no natural numbers. So your simply saying "n is a natural number" is nonsense.
> > > > If you say that n is a natural number that cannot be written down (we know such a number exists) the n has all the properties of a natural number.
> > > But you cannot know anything further about n
> > We do not need to. We know existence and we know that n has all the properties of an element of |N_F.
> You caqnnot distinguish it from another such element.

Piffle. We know n has a successor and no element of |N_F is equal to its successor.

--
William Hughes

On Friday, September 10, 2021 at 10:20:47 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 9. September 2021 um 15:58:45 UTC+2:
> > On Thursday, September 9, 2021 at 10:01:10 AM UTC-3, WM wrote:
> >
> > >In linear order nobody should believe the claim of all without being shown the last element.
> > A Peano set is linearly ordered, and a Peano set does not have a last element.
> It is not complete.

Piffle. A Peano set exists, it does not change, there is no element missing, no element that can be added.. The word "complete" is one of your nonsense words.

--
William Hughes

On Friday, September 10, 2021 at 3:15:23 AM UTC-7, FromTheRafters wrote:
> Ross A. Finlayson was thinking very hard :
> > On Thursday, September 9, 2021 at 7:25:37 PM UTC-7, FromTheRafters wrote:
> >> Gus Gassmann pretended :
> >>> On Thursday, 9 September 2021 at 18:07:28 UTC-3, FromTheRafters wrote:
> >>>> Serg io was thinking very hard :
> >>>>> On 9/9/2021 8:01 AM, WM wrote:
> >>>>>> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00
> >>>>>> UTC+2:
> >>>>>>> onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
> >>>>>>>> Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
> >>>>>>>>
> >>>>>>>>> If there is no last end segment of all end segments,
> >>>>>>>>> then the intersection of all end segments is
> >>>>>>>> Then there is no "all endsegments". "All" can only be proved by
> >>>>>>>> showing the last.
> >>>>>>>>
> >>>>>>> Where does it say that "all" must mean a last?
> >>>>>>
> >>>>>> In linear order nobody should believe the claim of all without being
> >>>>>> shown the last element.
> >>>>>
> >>>>>
> >>>>> all natural numbers => 1,2,3,...
> >>>>>
> >>>>> so according to WM do not believe in Natural Numbers unless WM can show
> >>>>> you the Last one.
> >>>> Sure, there's the last one, this current one, and the next one.
> >>>
> >>> Cool. So, in an inductive set it should indeed be sufficient to be shown
> >>> the last one, because you then have this current one, as well as the next
> >>> one! Is that what the great perfessor is taking about?
> >> Who knows what he's talking about, he hasn't made any sense for
> >> decades.
> >
> >
> > It's always "infinite induction exhausts" never satisfying related rates
> > problems or the countable and uncountable in functional analysis and
> > infinite domains.
> >
> > (WM is _plainly_ mostly Hodges' hopeless and sometimes "infinite integers
> > as discredited as possible".)
> >
> > Is it a paradox that the differential via countable additivity both
> > integrates under any continuous functions but there are uncountably many
> > points on it?
> >
> > Is it not because there aren't uncountably partitions of the unit interval
> > because each would contain a rational?
> >
> > Doesn't well-ordering the reals have a subset that is well-ordering
> > uncountably many, in their order?
> >
> >
> > Counterexamples....
> >
> >
> > I know what's he's talking about, just doing it wrong, not enough, and the
> > wrong way.
> Yeah, I guess it is easy to know what he is talking about when you
> consider his goal is to use his misunderstandings about "series" to
> show Cantor's diagonal method of 'listing; the rationals is flawed
> because Cantor ignores WM's 'Dark Numbers' which he says cannot be
> mapped because they can't be exactly expressed as a finite sequence.
>
> He tries to shoehorn in some 'Dark Numbers' into ZFC and when it
> doesn't work he exclaims "ZFC is inconsistent" because it ignores his
> "Dark Numbers".

Yeah that's wrong, given usual formalisms and facts like "irrational numbers
exist and are dense in the reals", the antidiagonal argument, nested intervals
and other results in topology of numbers about uncountability. That said, the
rationals are HUGE in a sense, i.e. in a usual sense much larger a "set" than
the integers, a set in th etopology of the numbers, in all the supersets of the
rationals and the integers as a subset, of the rationals, by their values.

Many people have a number model of values with attributes of being integer,
rational, real, ..., for each value, and that R > Q > N means subset and also size.
(In those labels of those sets under that extensionality.) For many people R-values
is enough numbers for the purposes of scalars in vector spaces.

Then, for something like "ZF is inconsistent for example according to Russell
a model of it as a set contains itself so it's not consistent its axioms those
that restrict comprehension i.e. well-foundedness and 'ordinary' omega".
is a way to re-frame usual paradoxes as what arrived at those axioms already.

What theorists would have not exist....

On Friday, September 10, 2021 at 10:18:07 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 9. September 2021 um 15:46:00 UTC+2:
> > On Thursday, September 9, 2021 at 10:09:49 AM UTC-3, WM wrote:
> > > No [single element of the sequence] can cause a finite intersection
> >
> > True. A finite intersection is a property of the sequence, not a property of an element of the sequence.
> The sequence is inclusion monotonic. Therefore its intersection cannot be empty if no endsegment is empty.

No, the intersection, a property of the sequence, cannot be empty if no endsegment is empty and there is a last endsegment.

--
William Hughes

Greg Cunt schrieb am Freitag, 10. September 2021 um 15:32:37 UTC+2:
> On Friday, September 10, 2021 at 3:18:07 PM UTC+2, WM wrote:
>
> > The sequence is inclusion monotonic. Therefore its intersection cannot be empty if no endsegment is empty.
> Please prove this argument.

I did many times. If the intersection is not infinite, then there must be a first infinite endsegment with finite intersection because of stepwise loss:
∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Regards, WM

William schrieb am Freitag, 10. September 2021 um 16:30:39 UTC+2:
> On Friday, September 10, 2021 at 10:16:28 AM UTC-3, WM wrote:

> > > We do not need to. We know existence and we know that n has all the properties of an element of |N_F.
> > You cannot distinguish it from another such element.
> We know n has a successor and no element of |N_F is equal to its successor.

But you don't know n. Therefore you cannot know whether m is n.

Regards, WM