On 9/8/2021 2:46 PM, WM wrote:
> Serg io schrieb am Mittwoch, 8. September 2021 um 06:26:47 UTC+2:
>> On 9/7/2021 2:01 PM, WM wrote:
>>> Serg io schrieb am Dienstag, 7. September 2021 um 20:56:00 UTC+2:
>
>>>>> But how do you know that n is a natural number if you can't define it.
>>>> you simply say "n is a natural number".
>>>
>>> But you can check that its second row is empty and its upper left corner is no vertex.
>>>
>> n doesn't need a "row" nor a "vertex" to be a natural number,
>
> On the contrary. Objects which have rows or vertices are with certainty no natural numbers. So your simply saying "n is a natural number" is nonsense.
>
> Regards, WM
>

no, you are nonsense, and what you call your math is nonsense.

what are the row and vertex of 17 ?

On 9/8/2021 3:38 PM, WM wrote:
> 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".

An end segment is a non-empty, closed upwards set.
A non-empty, closed upwards set is an end segment.
All end segments are end segments.

> "All" can only be proved by showing the last.

That you (WM) do not know how to do something is not
proof that it can't be done.

Describe an end segment.
Derive further claims by truth-preserving inferences.
Those further claims are true of all end segments.

If there is no last end segment of all end segments,
then the intersection of all end segments is
a closed upwards set which is empty or non-closed-upwards.

On Wednesday, September 8, 2021 at 4:43:40 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 8. September 2021 um 02:22:26 UTC+2:
> > On Tuesday, September 7, 2021 at 3:56:41 PM UTC-3, WM wrote:
> >
> > > If you have an unwritable object, how do you find out that it is an element of |N_F?

We know that an unwritable object exists. You continue with the silly claim that you have to check what you know.

> We know only that if actual infinity is a meaningful notion dark or unwritable or undefinable elements exist.

Piffle.

--
William Hughes

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.

Piffle. 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. No natural number has the property that it has a row. The idea that you can only be sure that n does not have a row by writting it down is stilly.

--
William Hughes

On Thursday, September 9, 2021 at 1:00:13 AM UTC+2, William wrote:

> No natural number has the property that it has a row.

Not so fast. There could very well be such natural numbers. Of course, we don't know, but we can't exclude it from a logical point of view. (Please don't confuse "the natural numbers" with their usual set-theoretic surrogates.)

Hint: "Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true." (Bertrand Russell)

On Wednesday, September 8, 2021 at 11:11:34 PM UTC+2, Serg io wrote:
> On 9/8/2021 2:43 PM, WM wrote:
> >
> > my question was: If you have two objects, how can you find out that they differ and in what way?
> >
> call one a, the other b.

Indeed. Assume that a set M contains exactly 2 elements: card(M) = 2.

Then we may start the following way: Let /a/ be an element in M.

Now we may proceed: Let /b/ be an element in M \ {a}.

Not that difficult. From this we can PROVE: a =/= b.

| "how can you find out that they differ and in what way?"

It's called "proof". By (giving) a proof.

On Wednesday, September 8, 2021 at 11:12:39 PM UTC+2, Serg io wrote:
> On 9/8/2021 2:46 PM, WM wrote:
> >
> > Objects which have rows or vertices are with certainty no natural numbers.

Really?! How can we KNOW that? Just because we know certain "features" of "the" natural numbers, does that (logically) imply that "they" CAN'T HAVE other features too? I don't think so.

Hint: "If I subsume under my points arbitrary systems of things, e.g. the system: love, law, chimney sweep ..., and then just assume all my axioms as relationships among these things, then my theorems, e.g. also the Pythagorean theorem, are true of these things, too."

(D. Hilbert, transl. from Gottlob Freges Briefwechsel mit D. Hilbert , E. Husserl , B. Russell sowie ausgewählte Einzelbriefe Freges, Felix Meiner Verlag, 1980, p. 13)

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.
>
> Regards, WM

Where does it say that "all" must mean a last?

onsdag 8 september 2021 kl. 21:48:13 UTC+2 skrev WM:
> 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

You are talking about yourself, you make up rules for no reason!

zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:22 UTC+2:
> onsdag 8 september 2021 kl. 21:48:13 UTC+2 skrev WM:
> > 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.
> >
> You are talking about yourself, you make up rules for no reason!

No I only mentioned that using your brain is using a physical tool.

Regards, WM

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.

Regards, WM

Greg Cunt schrieb am Donnerstag, 9. September 2021 um 03:02:08 UTC+2:
> On Wednesday, September 8, 2021 at 11:12:39 PM UTC+2, Serg io wrote:
> > On 9/8/2021 2:46 PM, WM wrote:
> > >
> > > Objects which have rows or vertices are with certainty no natural numbers.
> Really?! How can we KNOW that?

Because we or our ancestors have constructed them without rows and vertices.

Regards, WM

Jim Burns schrieb am Mittwoch, 8. September 2021 um 23:22:06 UTC+2:

> Describe an end segment.
> Derive further claims by truth-preserving inferences.

As long as infinite endsegments exist, they have an infinite amount of elements with all predecessors and with all infinite successors. If the sequence of infinite endsegments is infinite, then we have a contradiction: To enumerate the endsegments E(n), we need aleph_0 indices n. It is impossible to index their contents.

> Those further claims are true of all end segments.
> If there is no last end segment of all end segments,
> then the intersection of all end segments is

infinite. No infinite endsegment can cause a finite intersection together with its predecessors and its infinte successors. Inclusion monotony.

Regards, WM

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 and m.

> No natural number has the property that it has a row. The idea that you can only be sure that n does not have a row by writing it down is stilly.

You can only be sure by having n defined. What cannot be written and cannot be defined in any other way is not part of mathematics or is dark.

Regardes, WM

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. Saying that n is "dark" is silly.

--
William Hughes

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.

--
William Hughes

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?

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? Where can we find them?

On Thursday, September 9, 2021 at 3:09:49 PM UTC+2, WM wrote:

> > [Since] there is no last end segment of all end segments,
> > [...] the intersection of all end segments is
> >
> infinite.

Nope. Empty.

> No infinite endsegment can cause <bla bla bla>

Yeah, whatever, Mückenheim.

On 9/9/2021 8:02 AM, WM wrote:
> Greg Cunt schrieb am Donnerstag, 9. September 2021 um 03:02:08 UTC+2:
>> On Wednesday, September 8, 2021 at 11:12:39 PM UTC+2, Serg io wrote:
>>> On 9/8/2021 2:46 PM, WM wrote:
>>>>
>>>> Objects which have rows or vertices are with certainty no natural numbers.
>> Really?! How can we KNOW that?
>
> Because we or our ancestors have constructed them without rows and vertices.
>
> Regards, WM
>

?? totally illogical !!

"Objects which have rows or vertices are with certainty no natural numbers, because we or our ancestors have constructed them without rows and vertices."

I plant potatos in a square pattern in the ground, but *they are not potatoes* because your ancesters planted potatoes in a row.

On 9/8/2021 7:12 PM, Greg Cunt wrote:
> On Thursday, September 9, 2021 at 1:00:13 AM UTC+2, William wrote:
>
>> No natural number has the property that it has a row.
>
> Not so fast. There could very well be such natural numbers. Of course, we don't know, but we can't exclude it from a logical point of view. (Please don't confuse "the natural numbers" with their usual set-theoretic surrogates.)
>
> Hint: "Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true." (Bertrand Russell)
>

Idea!

New Natural Numbers! can only be in ROWS !

ah... which row ?

what if the row is circular ?

what if the rows for a square ?

what if it is a degenerate row like all in one place

why not have New New Natural Numbers !!! which have a position in 3D and a time which they exist at that point ?

On 9/9/2021 9:09 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 8. September 2021 um 23:22:06 UTC+2:

>> Describe an end segment.
>> Derive further claims by truth-preserving inferences.
>
> As long as infinite endsegments exist,

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

( To a sequence of end segments
( E(n) = { n,n+1,n+2,n+3,... }
( E(n+1) = { n+1,n+2,n+3,n+4,... }
( E(n+2) = { n+2,n+3,n+4,n+5,... }
( E(n+3) = { n+3,n+4,n+5,n+6... }
( ...
( ( there corresponds a sequence of difference sets
( D(n) = E(n)\E(n+1) = {n}
( D(n+1) = E(n+1)\E(n+2) = {n+1}
( D(n+2) = E(n+2)\E(n+3) = {n+2}
( D(n+3) = E(n+3)\E(n+4) = {n+3}
( ...
( ( Since there are is no last E(n), there is no last D(n).
( The D(n) are pairwise disjoint and non-empty.
( ( Each E(n) is the union of the infinitely-many non-empty
( pairwise disjoint D(n) which follow.
( E(n) = Union{ D(n),D(n+1),D(n+2),... }
( E(n+1) = Union{ D(n+1),D(n+2),D(n+3),... }
( E(n+2) = Union{ D(n+2),D(n+3),D(n+4),... }
( E(n+3) = Union{ D(n+3),D(n+4),D(n+5),... }
( ...
( ( Each E(n) is infinite.

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

> As long as infinite endsegments exist,
> they have an infinite amount of elements with all predecessors
> and with all infinite successors.
> If the sequence of infinite endsegments is infinite,
> then we have a contradiction:
> To enumerate the endsegments E(n), we need aleph_0 indices n.
> It is impossible to index their contents.

Let's pretend that, instead of indexes, you're making that
argument about letters. Your argument becomes

| It is impossible to write
| | If the sequence of infinite endsegments is infinite,
| | then we have a contradiction:
| because we have already used various letters.
| Instead, we can at best write
| | If the s qu nc of i t d gm ,
| | w av r c :

It's a crazy argument. It's true enough that one can't
write very much without using letter smore than once,
but it assumes that letters can't be used more than once.
That is flat out crazy.

Your argument is based on the assumption that,
if we index end segments, we can't also index contents.
That is flat out crazy.

>> Those further claims are true of all end segments.
>> If there is no last end segment of all end segments,
>> then the intersection of all end segments is
>
> infinite.
> No infinite endsegment can cause a finite intersection
> together with its predecessors and its infinte successors.
> Inclusion monotony.

An end segment is a non-empty, closed upwards set.

If the intersection of all end segments is a
non-empty, closed upwards set,
then there is a last end segment.

The intersection of closed upwards sets is closed upwards.

With or without a last end segment,
the intersection of all end segments (closed upwards)
is closed upwards.

If the intersection of all end segments is non-empty,
then the intersection is closed upwards, making it
an end segment, and there is a last end segment.

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

On 9/8/2021 2:38 PM, WM wrote:
> 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.

nope. There is no Last endsegment, Miss Leader.

>
> Regards, WM
>

On 9/9/2021 8:09 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 8. September 2021 um 23:22:06 UTC+2:
>
>> Describe an end segment.
>> Derive further claims by truth-preserving inferences.
>
> As long as infinite endsegments exist, they have an infinite amount of elements with all predecessors and with all infinite successors. If the sequence of infinite endsegments is infinite, then we have a contradiction: To enumerate the endsegments E(n), we need aleph_0 indices n. It is impossible to index their contents.

Wrong, Fail, go back to school.

oo = k * oo

>
>> Those further claims are true of all end segments.
>> If there is no last end segment of all end segments,
>> then the intersection of all end segments is
>
> infinite. No infinite endsegment can cause a finite intersection together with its predecessors and its infinte successors. Inclusion monotony.

your repeated always wrong posts are loser monotony

>
> Regards, WM
>

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.

>
> Regards, WM
>