Eram semper recta schrieb am Mittwoch, 30. März 2022 um 07:11:35 UTC+2:
> On Wednesday, 30 March 2022 at 07:51:39 UTC+3, Eram semper recta wrote:
> > On Tuesday, 29 March 2022 at 16:44:31 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Dienstag, 29. März 2022 um 08:09:23 UTC+2:
> > >
> > > > p/q means number where p and q are symbols or place-holders for parts of the number (name).
> > > >
> > > > p and q are NOT variables any more than say the x in x + 3 = 5 is a "variable".
> > > Are you serious?
> > Of course.
> > > x = 2, no other value possible.
> > Exactly! Therefore x is a CONSTANT.
> > > p and q: every value possible.
> > No, not every value.

> The fact that you misread so much of what I wrote is typical of the mainstream who never pay any attention to detail.

I did not misread but read exactly what you say below:
>
> p/q is a symbolic fraction just like dy/dx. Both are treated exactly like numbers. It is 100% correct to write dy/dx = dy/dt x dt/dx.

Yes p/q is a symbolic fraction, but not an individual like 2.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 30. März 2022 um 07:04:58 UTC+2:
> tisdag 29 mars 2022 kl. 15:51:08 UTC+2 skrev WM:
> > Jim Burns schrieb am Montag, 28. März 2022 um 22:07:25 UTC+2:
> >
> > > Note that a finite number of words can describe infinitely-
> > > -many things.
> > But only collectively, not individually. If you describe the set you don't describe its elements, unless there is only one or none.
> >
> your "collectively" and "individually" is meaningless.

n ∈ ℕ describes natural numbers collectively. 2 ∈ ℕ describes an individual.

Regards, WM

david...@gmail.com schrieb am Mittwoch, 30. März 2022 um 06:08:44 UTC+2:
> On Sunday, March 27, 2022 at 2:31:42 PM UTC-7, WM wrote:
> > david...@gmail.com schrieb am Sonntag, 27. März 2022 um 18:48:34 UTC+2:
> > > On Saturday, March 19, 2022 at 9:35:13 AM UTC-7, WM wrote:
> > > > There is no permutation of the Xs in
> > > >
> > > > XOOO...
> > > > XOOO...
> > > > XOOO...
> > > > XOOO...
> > > > ...
> > > >
> > > > such that all positions are covered by Xs.
> > >
> > > Would you agree that there is permutation of the X's and 0's such that all the 0's end up in the first column and all the other columns are "covered" by X's?
> > No, that is obviously wrong. Here are the fractions m/n written out:
>
> The problem is that it's not so obvious you're right.

You deleted the detailled explanation. In the first column there are as many places as natural numbers. So all fractions must be accomodated in the first column, but without forgetting those initially residing there. Hence, only exchanges of two fractions at any one step are possible. In this way never all fractions will enter the first column by finite steps - and others are not available.
>
> You say your argument is simple, and yet hardly anyone here understands it! In what way is Poincare's argument not sufficiently simple?

My argument gives a concrete example. Of course it is clear that his sentence is absolutely true:

If we take Cantor's ideas about infinite sets seriously, we would conclude that there cannot be more than "countably" many ideas, since there are not more than "countably" many finite strings of symbols.

But then he goes on to conlude that countability should be thrown away because uncountability is nonsense:

That last conclusion is so absurd that the whole idea of "countable" should be thrown away.

I prove this by a concrete example: The fractions will never be collected in the first column.
>
> I think you have some intuition that the rest of us don't share, and that you haven't been able to clearly articulate. Could I suggest that maybe you should work on clearly articulating that intuition?

Here is the simplest example I can think of:

The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain the union of all their predecessors:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
....

That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments too. This sequence however does not contain an actually infinite finite initial segment. If ℕ is actually infinite, it must contain more in addition than all finite initial segments. What is this additional contents?
> Anyway, you've made up a very interesting listing of quotes from mathematicians who speak about the lack of necessity for Cantor's theory of infinite sets (is it still available on your website?). Do you still have plans to publish that? I think it would make an interesting reference work.

After having grown over five years it is available now in its final version and will remain so:
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

It is not necessary to waste paper. Acces is public and details can easier be found by the find function, for instance the quotes of David Petry.

Regards, WM

On Wednesday, 30 March 2022 at 08:42:44 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 30. März 2022 um 07:05:15 UTC+2:
> > tisdag 29 mars 2022 kl. 16:02:25 UTC+2 skrev WM:
>
> > > But it is still true that there are undefinable elements in actual infinity (or not existing elements in potential infinity).
> > >
> > You have yet to show any "dark" stuff exists
> Here is a simple proof:
>
> The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain the union of all their predecessors:
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> ...
>
> That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments. This sequence however does not contain an infinite finite initial segment.

The usual bullshit again. All it means is that an infinite (yes, infinite) number of finite initial segments can cover all of ℕ. The fact that the finite initial segments of natural numbers are called "finite" is, well, not an accident. Every natural number is finite, and yet ℕ contains infinitely many of them. That you can't understand that is really your fault alone.

Gus Gassmann schrieb:

> The usual bullshit again.

Sure, and you knew about that: for about 30 years now...

On Wednesday, 30 March 2022 at 08:47:46 UTC-3, WM wrote:
[...]
> n ∈ ℕ describes natural numbers collectively.

Not so. If you want to describe them collectively, you *must* put a quantifier: ALL n ∈ ℕ. n is just as individual as 2 or 3. But quantifiers have not been good to you lately (if ever).

On Wednesday, 30 March 2022 at 09:24:13 UTC-3, Tom Bola wrote:
> Gus Gassmann schrieb:
>
> > The usual bullshit again.
>
> Sure, and you knew about that: for about 30 years now...

Wow! Are you stalking me now?

Alan Mackenzie schrieb am Dienstag, 29. März 2022 um 21:04:50 UTC+2:
> WM <askas...@gmail.com> wrote:

> As something to think about, consider a random digit generator generating
> a decimal representation. I don't think you would deny the sequence it
> produces exists. However there is no formula for it - it can't be
> "given".

Here you touch a borderlin case.
The sequence will never become actually infinite. So we cannot claim that a certain digit sequence exists. Of course every digit will have a successor, but they are not yet fixed.
In the case SUM 1/n! we cannot calculate more than any digit at a finite position either, but we can be sure that the digit discovered at position n will be independent of who calculates it.

> >> Just because there may not be a formula known for some decimal
> >> representation, how does that imply non-existence?
>
> > What do you understand by existence?
> In mathematics, something exists if it can be mathematically described
> and doesn't give rise to contradictions.

Therefore every digit sequence that can be described by a finite formula like 1/3 or SUM 1/n! does exist. But sequences without formula are necessarily finite because infinity cannot be finished or completed.

> > Does something that cannot be known exist?
> Some somethings, yes. The random decimal expansion exists.

Yes, this notion exists. But the digit sequence is not yet fixed and will never be fixed completely.
>
> > As a dark item?
>
> No. I don't know what you mena by "dark item".

Here is an example: The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain the union of all their predecessors:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
....

That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments too. This sequence however does not contain an actually infinite finite initial segment. If ℕ is actually infinite, it must contain more in addition than all finite initial segments. What is this additional contents? It is undefinable or dark, although you can define a potentially infinite collection of segments.

Regards, WM

Alan Mackenzie schrieb am Dienstag, 29. März 2022 um 21:39:24 UTC+2:
> WM <askas...@gmail.com> wrote:

> > Of course. But all numbers described individually belong to a
> > potentially infinite set. Most numbers will never be identified =
> > described individually.
> This is of minor importance in mathematics. We can discuss the
> properties of prime numbers, for example, without specifying each prime
> number individually.

Of course that is correct. Every natural number, specified or not, has a unique prime decomposition. But if not specified, it cannot be mapped. It appears so in a simple mapping like ℕ to ℕ, but it is wrong as we see when ℕ to ℕ U {a} is erroneously accepted.

Regards, WM

Gus Gassmann schrieb:

> On Wednesday, 30 March 2022 at 09:24:13 UTC-3, Tom Bola wrote:
>> Gus Gassmann schrieb:
>>
>>> The usual bullshit again.
>>
>> Sure, and you knew about that: for about 30 years now...
>
> Wow! Are you stalking me now?

You are pissing dirt into the newsnet.

horand....@gmail.com schrieb am Mittwoch, 30. März 2022 um 14:22:28 UTC+2:
> On Wednesday, 30 March 2022 at 08:42:44 UTC-3, WM wrote:

> > The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain the union of all their predecessors:
> >
> > {1} = {1}
> > {1} U {1, 2} = {1, 2}
> > {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> > {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> > ...
> >
> > That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments. This sequence however does not contain an infinite finite initial segment.
> All it means is that an infinite (yes, infinite) number of finite initial segments can cover all of ℕ.

The "infinite number" is not an actually infinite number. Therefore they cannot cover an actually infinite ℕ.

There are as many finite initial segments as they have contents:

o
oo
ooo
oooo
ooooo
....

Never an actually infinite contents ==> never an actually infinite finite initial segment.

> The fact that the finite initial segments of natural numbers are called "finite" is, well, not an accident.

This fact proves that their union cannot grow to actual infinity because it is always contained in a finite initial segment. You know, the sequence contains all unions and it does not contain an infinite finite initial segment.

> Every natural number is finite, and yet ℕ contains infinitely many of them.

Again the ambivalence of infinities, desired and even dictated by matheologians because their belief requires this confusion. As we see here all natural numbers definable by finite initial segments do not make an actually infinite set ℕ.

> That you can't understand that is really

the proof that I do not thoughtlessly believe in confusing infinities.

Regards, WM

horand....@gmail.com schrieb am Mittwoch, 30. März 2022 um 14:25:13 UTC+2:
> On Wednesday, 30 March 2022 at 08:47:46 UTC-3, WM wrote:
> [...]
> > n ∈ ℕ describes natural numbers collectively.
> Not so. If you want to describe them collectively, you *must* put a quantifier:

n is a not fixed natural number. Therefore every natural number can feel addressed.

> ALL n ∈ ℕ. n is just as individual as 2 or 3.

No. 2 is less than 3. For n this cannot be determined.

> But quantifiers have not been good

They have obviously confused your thinking and that of most matheologians.

Regards, WM

The imbecile clown WM drivels his abhoring dirt again and again:

> They have obviously confused your thinking and that of most matheologians.

Piss already off you imbecile clown.

On Wednesday, 30 March 2022 at 14:45:44 UTC+3, WM wrote:
> Eram semper recta schrieb am Mittwoch, 30. März 2022 um 07:11:35 UTC+2:
> > On Wednesday, 30 March 2022 at 07:51:39 UTC+3, Eram semper recta wrote:
> > > On Tuesday, 29 March 2022 at 16:44:31 UTC+3, WM wrote:
> > > > Eram semper recta schrieb am Dienstag, 29. März 2022 um 08:09:23 UTC+2:
> > > >
> > > > > p/q means number where p and q are symbols or place-holders for parts of the number (name).
> > > > >
> > > > > p and q are NOT variables any more than say the x in x + 3 = 5 is a "variable".
> > > > Are you serious?
> > > Of course.
> > > > x = 2, no other value possible.
> > > Exactly! Therefore x is a CONSTANT.
> > > > p and q: every value possible.
> > > No, not every value.
> > The fact that you misread so much of what I wrote is typical of the mainstream who never pay any attention to detail.

> I did not misread but read exactly what you say below:

You did misread what I wrote because the claim "x = 2, no other value possible. " is very much in line with what I was saying.

> >
> > p/q is a symbolic fraction just like dy/dx. Both are treated exactly like numbers. It is 100% correct to write dy/dx = dy/dt x dt/dx.
> Yes p/q is a symbolic fraction, but not an individual like 2.

No. p/q is very much an individual fraction. Stop thinking about what I write to you in terms of the set theory bullshit that is thrown at you from all sides from the numerous retards on this newsgroup.

p/q is NOT a "formula". Try to understand the difference between number and formula.

>
> Regards, WM

Eram semper recta schrieb am Mittwoch, 30. März 2022 um 15:18:28 UTC+2:
> > Yes p/q is a symbolic fraction, but not an individual like 2.
> No. p/q is very much an individual fraction.

For 2 you can find out whether it is larger than 10, for p/q you cannot find out whether it is larger than 10.

Regards, WM

On 3/30/2022 6:42 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 30. März 2022 um 07:05:15 UTC+2:
>> tisdag 29 mars 2022 kl. 16:02:25 UTC+2 skrev WM:
>
>>> But it is still true that there are undefinable elements in actual infinity (or not existing elements in potential infinity).
>>>
>> You have yet to show any "dark" stuff exists
>
> Here is a simple proof:
>
> The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain the union of all their predecessors:
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> ...
>
> That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments.

what you said was : "the union of all finite initial segments is composed of all finite initial segments"

> This sequence however does not contain an infinite finite initial segment.

how do you define "infinite finite" ?

> If ℕ is actually infinite, it must contain more than all finite initial segments in addition.

wrong. Just assume a natural number, k, which is in ℕ and show it must be in a finite initial segment.

> What is this additional contents?

Grade F smoke-able Troll Food.

>
> Regards, WM

Jim Burns schrieb am Mittwoch, 30. März 2022 um 03:30:19 UTC+2:
> On 3/29/2022 10:02 AM, WM wrote:

> > You intend to study all fractions and their indices.
> > But you cannot find the dark fractions.
> > So you claim that they have left the matrix.
> I claim that
> only r/c
> for which collections exist with counting-orders
> which begin at 1 and end at r and at c
> are r/c
> for which collections exist with counting-orders
> which begin at 1 and end at r and at c

That is true. But most places of the matrix don't share this privilege.
>
> If you change what's in the matrix as we proceed through
> the discussion, you are equivocating.

Every positive fraction is and remains in the matrix. But after a while you can't find Bob. That proves his change to dark places.
> >
> > You want to forget them.
> Do your dark fractions have r/c
> for which collections exist with counting-orders
> which begin at 1 and end at r and at c ?

No, dark fractions don't have numbers with FISON-ends.
>
> If they do, r/c refers to them, too.
>
> If they don't, r/c has never referred to them.

Only a tine part of the matrix has definable co-ordinates:

11, 12, 13, ..., Dark....................................................
21, 22, 23, ..., Dark....................................................
31, 32, 33, ..., Dark....................................................
....... ...... ....... Dark......................................................
Dark...........................................................................
....................................................................................

> >> Your argument is that
> >> Cantor is wrong because
> >> his system implies undefinable points,
> >> so, if his system is used,
> >> there will be points undefinable.
> >
> > My main point is that he claims that
> > all fractions are definable and can be enumerated.
> k = (r+c-1)*(r+2-2)/2 + r

please check the formula

> >> Perhaps I missed it, but
> >> you've never answered the question
> >> "and then what?"
> >
> > Then the notion of countable sets is nonsense,
> A finite non-empty collection
> cannot match element-for-element
> any of its proper subsets.

It can if the finity is not fixed.
>
> We can say that more compactly as
> "The set of FISONs is countably infinite"

They contain the union of all their predecessors:

{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
....

That implies that the sequence of *all* finite initial segments contains the union of all finite initial segments too. This sequence however does not contain an actually infinite finite initial segment. If ℕ is actually infinite, it must contain more in addition than all finite initial segments. What is this additional contents? It is undefinable or dark, although you can define a potentially infinite collection of segments.

Regards, WM

WM drivels:

> Only a tine part of the matrix has definable co-ordinates:
>
> 11, 12, 13, ..., Dark....................................................
> 21, 22, 23, ..., Dark....................................................
> 31, 32, 33, ..., Dark....................................................
> ...... ...... ....... Dark.....................................................
> Dark..........................................................................
> ..................................................................................

ROFL
On weekdays 12 o'clock is fixing in London what's the biggest non-dark natural...

On 3/30/2022 7:45 AM, WM wrote:
> Alan Mackenzie schrieb am Dienstag, 29. März 2022 um 21:39:24 UTC+2:
>> WM <askas...@gmail.com> wrote:
>
>>> Of course. But all numbers described individually belong to a
>>> potentially infinite set. Most numbers will never be identified =
>>> described individually.
>> This is of minor importance in mathematics. We can discuss the
>> properties of prime numbers, for example, without specifying each prime
>> number individually.
>
> Of course that is correct. Every natural number, specified

"specified" is red herring, Distractor.

>or not, has a unique prime decomposition. But if not specified,

"specified" is red herring, WM the Distractor.

> It appears so in a simple mapping like ℕ to ℕ, but it is wrong as we see when ℕ to ℕ U {a} is erroneously accepted.

Wrong again, as your {a} is already in ℕ, without error.

>
> Regards, WM

On 3/30/2022 6:47 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 30. März 2022 um 07:04:58 UTC+2:
>> tisdag 29 mars 2022 kl. 15:51:08 UTC+2 skrev WM:
>>> Jim Burns schrieb am Montag, 28. März 2022 um 22:07:25 UTC+2:
>>>
>>>> Note that a finite number of words can describe infinitely-
>>>> -many things.
>>> But only collectively, not individually. If you describe the set you don't describe its elements, unless there is only one or none.
>>>
>> your "collectively" and "individually" is meaningless.
>
> n ∈ ℕ describes natural numbers collectively. 2 ∈ ℕ describes an individual.
>
> Regards, WM

Wrong.

n ∈ ℕ means n is an element of the set of natural numbers.

2 ∈ ℕ means 2 is an element of the set of natural numbers.

You cannot change math definitions to suit your imaginatory Darkies, without incurring failure.

It happens that WM formulated :
> zelos...@gmail.com schrieb am Mittwoch, 30. März 2022 um 07:05:15 UTC+2:
>> tisdag 29 mars 2022 kl. 16:02:25 UTC+2 skrev WM:
>
>>> But it is still true that there are undefinable elements in actual infinity
>>> (or not existing elements in potential infinity).
>>>
>> You have yet to show any "dark" stuff exists
>
> Here is a simple proof:
>
> The finite initial segments {1, 2, 3, ..., n} of natural numbers ℕ contain
> the union of all their predecessors:
>
> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
> ...
>
> That implies that the sequence of *all* finite initial segments contains the
> union of all finite initial segments. This sequence however does not contain
> an infinite finite initial segment. If ℕ is actually infinite, it must
> contain more than all finite initial segments in addition. What is this
> additional contents?

Induction. If you have an element, you also have a next element.
Nothing more is added to this notion to terminate the step-by-step
process you are sure to envision. It keeps going -- no element is left
in darkness.

On 3/30/2022 8:06 AM, WM wrote:
> horand....@gmail.com schrieb am Mittwoch, 30. März 2022 um 14:25:13 UTC+2:
>> On Wednesday, 30 March 2022 at 08:47:46 UTC-3, WM wrote:
>> [...]
>>> n ∈ ℕ describes natural numbers collectively.
>> Not so. If you want to describe them collectively, you *must* put a quantifier:
>
> n is a not fixed natural number. Therefore every natural number can feel addressed.
>
>> ALL n ∈ ℕ. n is just as individual as 2 or 3.
>
> No. 2 is less than 3. For n this cannot be determined.

that is not what the symbol ∈ is about.

WM the Distractor.

>
>> But quantifiers have not been good
>
> They have obviously confused your thinking and that of most matheologians.
>
> Regards, WM

On 3/30/2022 9:02 AM, WM wrote:
> Eram semper recta schrieb am Mittwoch, 30. März 2022 um 15:18:28 UTC+2:
>
>>> Yes p/q is a symbolic fraction, but not an individual like 2.
>> No. p/q is very much an individual fraction.
>
> For 2 you can find out whether it is larger than 10, for p/q you cannot find out whether it is larger than 10.
>
> Regards, WM

distraction. this is 4rd grade stuff

WM explained on 3/30/2022 :

> There are as many finite initial segments as they have contents:

Wrong, the contents will always be finite since they are defined as
such.

Clown WM drivels:

> {1} = {1}
> {1} U {1, 2} = {1, 2}
> {1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4}
> {1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
*)
> ...
>
> That implies that the sequence of *all* finite initial segments contains
> the union of all finite initial segments too. This sequence however does
> not contain an actually infinite finite initial segment.

But it happened infinite many times there *).

Although each natural number n is finite does IN contain infinite many ones.

You abhoring idiot, of course, are way too imbecile to get this simple definition.