William schrieb am Freitag, 10. September 2021 um 16:38:04 UTC+2:
> 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.
> A Peano set exists, it does not change, there is no element missing, no element that can be added.

Then it is impossible that all fractions 1/1, 2/1, 3/1, ... are enumerated and all fractions
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... are enumerated too.

> The word "complete" is one of your nonsense words.

No it is Cantor's.

Regards, WM

William schrieb am Freitag, 10. September 2021 um 16:47:24 UTC+2:
> 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.
>
The latter clause is irrelevant. As long as the endsegments are infinite, the intersection is infinite too. Inclusion monotony. As long as endesegmnets are infinite they share their contents with all predecessors.

Regards, WM

On Friday, September 10, 2021 at 6:17:54 PM UTC+2, WM wrote:
> 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.
> >
> 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}.

Please prove this argument.

On Friday, September 10, 2021 at 6:20:14 PM UTC+2, WM wrote:
> 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 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.

Let n be an element in IN_F and let m be an element in IN_F\{n}. Then m =/= n.

No need to "know" n and/or "m".

On Friday, September 10, 2021 at 6:22:51 PM UTC+2, WM wrote:

> > The word "complete" is one of your nonsense words.
> >
> No it is Cantor's.

Oh, Cantor is active in this NG? Cool!

On Friday, September 10, 2021 at 6:26:12 PM UTC+2, WM wrote:

> As long as the endsegments are infinite, the intersection is infinite too.

Actually, the intersection of (the set of) all endsegments is empty.

Greg Cunt schrieb am Freitag, 10. September 2021 um 18:49:34 UTC+2:
> On Friday, September 10, 2021 at 6:26:12 PM UTC+2, WM wrote:
>
> > As long as the endsegments are infinite, the intersection is infinite too.
> Actually, the intersection of (the set of) all endsegments is empty.

Actually it is the minimum endsegment. If all endsegments are infinite, the minimum endsegment is infinite too.

Regards, WM

On Friday, September 10, 2021 at 6:52:16 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Freitag, 10. September 2021 um 18:49:34 UTC+2:
> > On Friday, September 10, 2021 at 6:26:12 PM UTC+2, WM wrote:
> > >
> > > As long as the endsegments are infinite, the intersection is infinite too.
> > >
> > Actually, the intersection of (the set of) all endsegments is empty.
> >
> Actually it is the minimum endsegment.

The MINIMAL ENDSEGMENT??? Huh???!!!

There IS NO MINIMAL ENDSEGMENT, you silly crank!

Hint: For all n e IN: E(n+1) c E(n) & E(n+1) =/= E(n).

!!! STRICT inclusion monotony !!!

> If all endsegments are infinite, the minimum endsegment is infinite too.

There IS NO MINIMAL ENDSEGMENT, you silly crank!

HOLY SHIT!!!

On Friday, September 10, 2021 at 1:26:12 PM UTC-3, WM wrote:
> William schrieb am Freitag, 10. September 2021 um 16:47:24 UTC+2:
> > 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.
> >
> The latter clause is irrelevant. As long

Indeed. "As long" , so for every finite initial subsequence of the sequence, the intersection is not empty. However, we are considering a property, not of every finite initial subsequence, but a property of the sequence.

--
William Hughes

On 9/10/2021 8:19 AM, WM wrote:
> 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.

One must align all identified sheeps to be counted (including the last one) into rows and vertices (not columns) pointed to magnetic north.
When that is done, then you count each sheep once, once thou shall countith said sheep, not twice, but only once.
Remember to increase the count by 1 each time;
not 1,1,1,1,1,1...
but 1,2,3,4,5,6...

>
>> Where can we find them?
>
> In school books for instance.

there are lots of books on "how to count rocks", and "how to count sheeps using rocks".

>
> Regards, WM
>

On 9/10/2021 8:16 AM, 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.
>
> Regards, WM
>

then there are no "anothers"

On Friday, September 10, 2021 at 1:22:51 PM UTC-3, WM wrote:
> William schrieb am Freitag, 10. September 2021 um 16:38:04 UTC+2:
> > 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.
> > A Peano set exists, it does not change, there is no element missing, no element that can be added.
> Then it is impossible that all fractions 1/1, 2/1, 3/1, ... are enumerated and all fractions
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... are enumerated too.

Piffle. (More of your "used up" nonsense). A Peano set exists and does not change.

--
William Hughes

On 9/10/2021 11:20 AM, WM wrote:
> 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
>

dude, m and n are spelt differently! or didn't you notice ?

Just pour your M+N's into your mask, and eat them like a horse.

On Friday, September 10, 2021 at 1:20:14 PM UTC-3, WM wrote:
> 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.

Piffle. We know that n exists and we know that m is the successor of n, so we know that m exists and is not n.

--
William Hughes

On 9/10/2021 11:17 AM, WM wrote:
> 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
>

that is not a proof at all.

anyone can prove there are no elements in the intersection.

On 9/10/2021 11:26 AM, WM wrote:
> William schrieb am Freitag, 10. September 2021 um 16:47:24 UTC+2:
>> 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.
>>
> The latter clause is irrelevant. As long as the endsegments are infinite, the intersection is infinite

wrong.

On 9/10/2021 12:00 PM, Greg Cunt wrote:
> On Friday, September 10, 2021 at 6:52:16 PM UTC+2, WM wrote:
>> Greg Cunt schrieb am Freitag, 10. September 2021 um 18:49:34 UTC+2:
>>> On Friday, September 10, 2021 at 6:26:12 PM UTC+2, WM wrote:
>>>>
>>>> As long as the endsegments are infinite, the intersection is infinite too.
>>>>
>>> Actually, the intersection of (the set of) all endsegments is empty.
>>>
>> Actually it is the minimum endsegment.
>
> The MINIMAL ENDSEGMENT??? Huh???!!!
>
> There IS NO MINIMAL ENDSEGMENT, you silly crank!
>
> Hint: For all n e IN: E(n+1) c E(n) & E(n+1) =/= E(n).
>
> !!! STRICT inclusion monotony !!!
>
>> If all endsegments are infinite, the minimum endsegment is infinite too.
>
> There IS NO MINIMAL ENDSEGMENT, you silly crank!
>
> HOLY SHIT!!!
>

Minimal Ants

Last Endsegment Ants of the Holy Infinity

Silly Crank Ants

On 9/10/2021 8:23 AM, 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.
>
> Regards, WM
>

Quackers

On 9/10/2021 9:45 AM, Ross A. Finlayson wrote:
> 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.

what ?

>
> 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.

What ?

>
> 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.

Buzz word BOT spew.

>
> What theorists would have not exist....
>

On 9/8/2021 2:48 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 8. September 2021 um 07:07:01 UTC+2:
>> tisdag 7 september 2021 kl. 20:47:43 UTC+2 skrev WM:
>
>>> Mathematics needs physical abilities to be practiced
>> We do not need to be able to write everything down to argue about them!
>
> And you refrain also from thinking, as it appears.
>
> Regards, WM
>

WM's folly is not staying awake in Math classes.

On 9/10/2021 9:23 AM, 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.

If there is no last end segment, then
the intersection of all end segments
is not an end segment.

Proof.
Assume the intersection is an end segment.
There is a last end segment.
Contradiction.

----
If the intersection is not an _end segment_
all of the _end segments_ being infinite implies
what? about _things that aren't end segments_

> Note: Inclusion monotony.
> The intersection cnnot shrink below all endsegments.

The intersection of all (closed upwards) end segments
is closed upwards.

That leaves two possibilities:
(i)
Their intersection is an end segment,
that is, closed upwards and non-empty.
(ii)
Their intersection is the empty set,
that is, closed upwards (vacuously) and empty.

If there is no last end segment,
only non-end-segment closed upwards sets
can be their intersection.
And what are these sets? The empty set.

If there is no last end segment,
then the intersection of all end segments is empty.

Serg io explained on 9/10/2021 :
> On 9/10/2021 8:19 AM, WM wrote:
>> 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.
>
> One must align all identified sheeps to be counted (including the last one)
> into rows and vertices (not columns) pointed to magnetic north. When that is
> done, then you count each sheep once, once thou shall countith said sheep,
> not twice, but only once. Remember to increase the count by 1 each time; not
> 1,1,1,1,1,1... but 1,2,3,4,5,6...
>
>>
>>> Where can we find them?
>>
>> In school books for instance.
>
> there are lots of books on "how to count rocks", and "how to count sheeps
> using rocks".

Sadly, none for teaching rocks how to do math.

Greg Cunt schrieb am Freitag, 10. September 2021 um 18:44:29 UTC+2:
> On Friday, September 10, 2021 at 6:17:54 PM UTC+2, WM wrote:
> > 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.
> > >
> > 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}.
>
> Please prove this argument.

No proof requird because it is the definition of the endsegments.

Regards, WM

Greg Cunt schrieb am Freitag, 10. September 2021 um 18:46:54 UTC+2:
> On Friday, September 10, 2021 at 6:20:14 PM UTC+2, WM wrote:
> > 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 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.
> Let n be an element in IN_F and let m be an element in IN_F\{n}. Then m =/= n.
>
Wrong.

Regards, WM

William schrieb am Freitag, 10. September 2021 um 20:19:16 UTC+2:
> On Friday, September 10, 2021 at 1:26:12 PM UTC-3, WM wrote:
> > William schrieb am Freitag, 10. September 2021 um 16:47:24 UTC+2:
> > > 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.
> > >
> > The latter clause is irrelevant. As long as the endsegments are infinite, the intersection is infinite too.
> Indeed. "As long" , so for every finite initial subsequence of the sequence, the intersection is not empty. However, we are considering a property, not of every finite initial subsequence, but a property of the sequence.

It consist of only finite initial subsequences.

Regards, WM