On 9/25/2021 8:29 PM, Serg io wrote:
> On 9/25/2021 5:03 PM, FromTheRafters wrote:

>> You don't understand what you just wrote.
>> Each FISON is finite by definition and has a least and
>> a greatest element. Subtracting finite proper subsets
>> from the naturals does not change the superset's cardinality.
>
> WM does not understand, this is re-beating a dead stale horse,
> "partitioning the real numbers",
> what I cant figure out is how he stays stuck on his
> fisons and endsegments, this is very simple stuff,
> covered in depth in 15 min in a real class.

| It is difficult to get a man to understand something
| when his salary depends on his not understanding it.
| -- Upton Sinclair 1878–1968. American novelist and social reformer.

I doubt that WM's salary is at risk here.
But something similar is, I'd bet. His ego?

On Friday, September 24, 2021 at 10:36:54 PM UTC+2, Jim Burns wrote:
> On 9/24/2021 3:17 PM, Transfinity wrote:

> > n is neither a number nor a circle or a cube.

Depends on the definition of /n/.

> It's possible that n is either, neither, or both.

I beg to differ: n can't be "both" (in one and the same context). I mean, if n is a (natural) number (due to von Neumann), it certainly won't be a circle or a cube, and if it is a circle or a cube, it certainly won't be a number (due to von Neumann).

> Your placeholders are a pointless complication.
> Unless the point is to complicate, in order to hide
> your ignorance. I think that is the point.

He's mixing up (1) n with "n" and (2) "n" used/adopted as a /constant/ with "n" used/adopted as a variable.

Lit. concerning (1): https://en.wikipedia.org/wiki/Use%E2%80%93mention_distinction

(2):
(a) Let n be a natural number. << Here "n" is a /constant/ (used in a certain context ("namespace"), usually a proof).
(b) For all n e IN: n >= 0. << Here "n" is a (bound) variable (WM: "placeholder")

On Saturday, September 25, 2021 at 11:39:13 PM UTC+2, Serg io wrote:

> Did you know that "n ∈ ℕ" says "n is a element of the set of natural numbers" ?

He's just too dumb to get that.

On Saturday, September 25, 2021 at 10:40:03 PM UTC+2, WM wrote:

> [It's a] fact that n with no further specifications than ["Let] n ∈ ℕ["] is not a number.

Oh really?! On the other hand "n ∈ ℕ" is just a formal expression for "n is an element in IN". And since IN is the set of all natural numbers, that just means "n is a natural number". (You may look that up in Halmos' "Naive Set Theory".)

You see, if I state (in a mathematical text): "Let n be a natural number." THEN n IS a natural number (in that context). In other words, then "n" refers to a natural number.

Hint: You are mixing up (1) n with "n" and (2) "n" used/adopted as a /constant/ with "n" used/adopted as a variable.

Lit. concerning (1): https://en.wikipedia.org/wiki/Use%E2%80%93mention_distinction

(2):
(a) Let n be a natural number. << Here "n" is a /constant/ (used in a certain context ("namespace"), usually a proof).
(b) For all n e IN: n >= 0. << Here "n" is a (bound) variable (your term: "placeholder")

On Saturday, September 25, 2021 at 4:58:58 AM UTC+2, Serg io wrote:

> in math we say "let n be a natural number" that means that n is a natural number.

Right.

> n is called a "variable",

Nope. At least not IN THIS context.

Actually, "n" in math is used for both purposes, i.e. as a constant or a variable. Not really "clean" (from a logcal point of view), but that's math (in practice).

Again,

(a) "Let n be a natural number." << Here "n" is a /constant/ (used in a certain context ("namespace"), usually a proof).
(b) "For all n e IN: n >= 0." << Here "n" is a (bound) variable (WM: "placeholder")
> and since we said "let n be a natural number", we know throughout the problem that n is a natural number.

Exactly. From a technical point of view "Let n be a natural number" is a "definition" which introduces the constant "n". (In a formal ND proof "n" might be called an "arbitrary name", due to Lemmon.)

On Friday, 24 September 2021 at 16:57:04 UTC-3, Transfinity wrote:
> Gus Gassmann schrieb am Freitag, 24. September 2021 um 21:05:15 UTC+2:
>
> > The intersection over infinitely many end segments is empty.
> Infinitely many endsegments are more than every finite set of endsegments? What is in the infinite set that is not in at least one of all finite sets making the intersection infinite?

I tell you what is not in it: Elements. The intersection over an infinite number of end segments is *EMPTY*. In particular no natural number n is an element of its successor's end segment E(n+1). It is clear that your attempts at deceit, lying, misdirection and confusion are getting more desparate, and your dementia is showing itself more and more clearly.

On Friday, 24 September 2021 at 17:00:26 UTC-3, Serg io wrote:
> On 9/24/2021 11:46 AM, WM wrote:
> > FromTheRafters schrieb am Freitag, 24. September 2021 um 00:08:40 UTC+2:
> >> WM has brought this to us :
> >>> Gus Gassmann schrieb am Donnerstag, 23. September 2021 um 15:36:27 UTC+2:
> >>>> "Infinite" means that there is *no end* to the successor operation.
> >>>
> >>> Nevertheless there is an end to the natural numbers, namely omega. Like the
> >>> zero ends the sequence of unit fractions.
> >> Stop lying.
> >
> > Start thinking.
> start Learning Math!
>
> WHY CANT YOU USE EQUATIONS ?

Two reasons: One, he never bothered to learn and his dementia prevents it; and two, equations expose his ambiguity, lies, deceits and sleights of hand.

On Sunday, September 26, 2021 at 12:12:40 AM UTC+2, FromTheRafters wrote:
> WM formulated on Saturday :
> >
> > [It's a] fact that n with no further specifications than n ∈ ℕ is not a number.
> >
> Elements of the set of natural numbers are not numbers? [...]

Note that "n+1 is not a natural number too" [WM].

On 9/25/2021 8:51 PM, Greg Cunt wrote:
> On Friday, September 24, 2021 at 10:36:54 PM UTC+2,
> Jim Burns wrote:
>> On 9/24/2021 3:17 PM, Transfinity wrote:

>>> n is neither a number nor a circle or a cube.
>
> Depends on the definition of /n/.
>
>> It's possible that n is either, neither, or both.
>
> I beg to differ: n can't be "both" (in one and the same context).
> I mean, if n is a (natural) number (due to von Neumann),
> it certainly won't be a circle or a cube, and if it is a circle
> or a cube, it certainly won't be a number (due to von Neumann).

Hooo-boy. My bad.
I wasn't paying enough attention. I thought it said
"n is either a number or a square or a cube."

( I thought that I was actually being a bit _clever_ by
( mentioning numbers that are both squares and cubes.
( Such as sixth powers.
( Oops.

One point, though.

In the context of geometric figures, yes,
if it is circle or a cube,then it's not a number.
However, in the context of arithmetic operations,
it could be both a cube and a number.

A very small point, almost entirely useless except to
underline the importance of context.

On 9/24/2021 12:42 PM, WM wrote:
> zelos...@gmail.com schrieb
> am Freitag, 24. September 2021 um 07:08:11 UTC+2:

>> if I say n is a natural number, it is a natural number.
>
> If you say you have proved the Riemann hypothesis,
> then you have proved it?
> If you say that set theory is mathematics,
> then it is mathematics?
> You need not say that you are a fool.

| n is a natural number
is merely a better way of saying
| That is a natural number.

Suppose I say
| That is a natural number.
Are you going to object and say
| No! "That" is a pronoun!

Well, maybe _you_ will.

The sort of work "n" does for us can be done without variables.
Before they were developed , it _was_ done without variables.
However, it's done better (more clearly, more easily) _with_
variables.

For example.

When Pierre de Fermat wrote his Last Theorem in the margin of
Diophantus's _Arithmetica_ variables were in development
in Europe at that time. What he wrote was

| Cubum autem in duos cubos, aut quadratoquadratum in duos
| quadratoquadratos & generaliter nullam in infinitum ultra
| quadratum potestatem in duos eiusdem nominis fas est dividere

In English,

| It is impossible to separate a cube into two cubes, or a
| fourth power into two fourth powers, or in general, any power
| higher than the second, into two like powers.

This is a much better way to express that:

| An > 2, ~Ex,y,z: x^n + y^n = z^n

"Cubum autem in duos cubos" refers to natural numbers.
n,x,y,z are used to say the same as the Latin.
They also refer to natural numbers.

"That" is a pronoun.
That is a natural number, when I refer to a natural number.

"n" is variable name, or a placeholder, or call it what you like.
n is a natural number, if I say it is.

On Sunday, September 26, 2021 at 7:14:45 PM UTC+2, Jim Burns wrote:

> The sort of work "n" does for us can be done without variables.

Indeed, espercially, when "n" is used as a constant.

Example: "Let n be a natural number."

[This amounts to a introduction of a constant, namely, "n".]

Hint: In Hilbert's epsilon calculus there's actually an operator which could be used in a FORMAL definition introducing "n":

n = epsilon x(x e IN) .

See: https://en.wikipedia.org/wiki/Epsilon_calculus

On the other hand, HERE

An e IN: n + n = 2*n

"n" is indeed a variable.

> "n" is variable [...], or a placeholder, or call it what you like.
> n is a natural number, if I say it is.

Indeed.

On 9/24/2021 12:37 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 23. September 2021 um 23:04:42 UTC+2:

>> *What we do not want to do* is to _re-define_ our topic
>
> My definition is the valid one.

<WM>
> My definition:
> Any natnumbers must have its UPF known in principle
> or knowable.

Bullshit.
A natural number has a unique prime factorization,
whether it's known, whether it's not known,
whether it can be known, whether it cannot be known.

>> It is a _theorem_ not a _definition_ that
>> a natural number >= 2 has a unique prime factorization.
>
> That is what we use for the definition of natural numbers
> if some fools claim that the expression "natural number"
> is a natural number.

Try to keep up.

I claim that I can refer to a natural number.
I can make claims about it (which refer to it)
which we know are true, even if we no one knows
which natural number I refer to.

( Other claims, we know are false. Still others could be
( either true of false without more information.

Starting from a claim for each natural number, we can derive
more claims true for each natural number.

On Sunday, September 26, 2021 at 7:44:50 PM UTC+2, Jim Burns wrote:
> On 9/24/2021 12:37 PM, WM wrote:
> >
> > some fools claim that the expression "natural number" is a natural number.

Really? Who are these fools?

Clearly, WM is a fool himself.

Though HE claims that the natural number n is a "placeholder". :-)

Assuming that IN only contains natural numbers as elements, the question arises how an n with n e IN can be a placeholder. Are some natural numbers placeholders?

fredag 24 september 2021 kl. 15:48:09 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
> > torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:
>
> > > My definition: Any natnumbers must have its UPF known in principle or knowable.
> > >
> > if I say n is a natural number, it is a natural number.
> It is not difficult to prove you wrong. You claim that every natural number has a decimal representation. Name the decimal representation of n. Fail.
>
> Regards, WM

n has a decimal representation, doesn't mean I have to give it.

fredag 24 september 2021 kl. 18:49:35 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:12:49 UTC+2:
> >of course the set N is not followed by anything, it is a meaningless thing to ask because the successor function cannot take N in it as it is not defined for it!
> It cannot take undefinable numbers. It can take every set of definable numbers.
> > The natural numbers IN the set N however ALWAYS have successors,
> Consider all natural numbers which can be last of a FISON F(n). They are by definition the definable natnumbers.
>
> ∀n ∈ ℕ: ℕ \ F(n) =/= { }
>
> When subtracting them from ℕ infinitely many numbers will remain.
>
> Obviously collectively more can be subtracted, namely all:
>
> ℕ \ ℕ = { }
> Regards, WM

a function can take any member of a set it is defined for and nothing else. S is defined on N but not for N.

fredag 24 september 2021 kl. 18:58:07 UTC+2 skrev Transfinity:
> Serg io schrieb am Freitag, 24. September 2021 um 01:48:30 UTC+2:
> >
> > There is no element common to all endsegments.
> But all have infinitely many elements in common with E(1). So they must be disjoint. Do you claim that?
>
> Regards, WM

Why would they need to be disjoint?

zelos...@gmail.com schrieb am Montag, 27. September 2021 um 06:48:17 UTC+2:
> fredag 24 september 2021 kl. 15:48:09 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
> > > torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:
> >
> > > > My definition: Any natnumbers must have its UPF known in principle or knowable.
> > > >
> > > if I say n is a natural number, it is a natural number.
> > It is not difficult to prove you wrong. You claim that every natural number has a decimal representation. Name the decimal representation of n. Fail.
> >
> n has a decimal representation, doesn't mean I have to give it.

If you claim that n has a decimal representation and this is doubted, then you have to prove it by giving it.

Regards, WM

Greg Cunt schrieb am Sonntag, 26. September 2021 um 20:20:26 UTC+2:
> On Sunday, September 26, 2021 at 7:44:50 PM UTC+2, Jim Burns wrote:
> > On 9/24/2021 12:37 PM, WM wrote:
> > >
> > > some fools claim that the expression "natural number" is a natural number.
> Really? Who are these fools?

Those who claim that the abbreviation of natural number, namely n, is a natural number. One of them is you.
>
> Assuming that IN only contains natural numbers as elements, the question arises how an n with n e IN can be a placeholder. Are some natural numbers placeholders?

No, there is a clear distinction.

Regards, WM

Jim Burns schrieb am Sonntag, 26. September 2021 um 19:44:50 UTC+2:

> I claim that I can refer to a natural number.
> I can make claims about it (which refer to it)
> which we know are true, even if we no one knows
> which natural number I refer to.

In that case use n.
>
> ( Other claims, we know are false. Still others could be
> ( either true of false without more information.
>
> Starting from a claim for each natural number, we can derive
> more claims true for each natural number.

If n stands for each natural number it has no unique prime decomposition. But every natural number has.

Regards, WM

On Monday, September 27, 2021 at 9:37:01 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Sonntag, 26. September 2021 um 20:20:26 UTC+2:
> > On Sunday, September 26, 2021 at 7:44:50 PM UTC+2, Jim Burns wrote:
> > > On 9/24/2021 12:37 PM, WM wrote:
> > > >
> > > > some fools claim that the expression "natural number" is a natural number.
> > > >
> > Really? Who are these fools?
> >
> Those who claim that the abbreviation of natural number, namely n, is a natural number.

Wait a second, that's a DIFFERENT claim! So what exactly is the claim that "some fools claim"?

> One of them is you.

Can you prove your idiotic claim by a quote?

> > So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?

Now YOUR claim was:

> They are placeholders.

But

> > assuming that IN only contains natural numbers as elements, the question arises how an n with n e IN can be a placeholder. Are some natural numbers placeholders?
>
> No, there is a clear distinction.

Pleas answer my question. If n is a placeholder and n is an element in IN, doesn't that mean that there are placeholders in IN?

FromTheRafters schrieb am Sonntag, 26. September 2021 um 00:03:30 UTC+2:
> WM brought next idea :
> > FromTheRafters schrieb am Freitag, 24. September 2021 um 21:12:12 UTC+2:
> >> WM used his keyboard to write :
> >
> >>> Consider all natural numbers which can be last of a FISON F(n). They are by
> >>> definition the definable natnumbers.
> >> Since each and every natural number has that property,
> >
> > Wrong. Provably wrong:
> > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> > But it is possible that you can't understand what follows.
> Each FISON is finite by
> definition and has a least and a greatest element. Subtracting finite
> proper subsets from the naturals does not change the superset's
> cardinality.
But subtracting all finite proper subsets does change the superset's cardinality. That is just the point, the consequences of which you can't understand.

Regards, WM

On Monday, September 27, 2021 at 9:41:25 PM UTC+2, WM wrote:

> If n stands for each natural number

But it does not "stands for each natural number". After introducing "n" with

Let n be a natural number greater that 1

n is just a natural number greater than 1. With other word's "n" denotes a certain (but unknown) natural number greater than 1.

Hence n has a unique prime factorization.

Since

> [...] every natural number [> 1] has [one].

Right.

Greg Cunt schrieb am Montag, 27. September 2021 um 22:02:57 UTC+2:
> On Monday, September 27, 2021 at 9:37:01 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Sonntag, 26. September 2021 um 20:20:26 UTC+2:
> > > On Sunday, September 26, 2021 at 7:44:50 PM UTC+2, Jim Burns wrote:
> > > > On 9/24/2021 12:37 PM, WM wrote:
> > > > >
> > > > > some fools claim that the expression "natural number" is a natural number.
> > > > >
> > > Really? Who are these fools?
> > >
> > Those who claim that the abbreviation of natural number, namely n, is a natural number.
> Wait a second, that's a DIFFERENT claim! So what exactly is the claim that "some fools claim"?
> > One of them is you.
> Can you prove your idiotic claim by a quote?

Of course:
>
> > > So n is in IN

> Now YOUR claim was:
>
> > They are placeholders.
>
> But
>
> > > assuming that IN only contains natural numbers as elements, the question arises how an n with n e IN can be a placeholder. Are some natural numbers placeholders?
> >
> > No, there is a clear distinction.
> Pleas answer my question. If n is a placeholder and n is an element in IN,

n is not an element of |N.

Regards, WM

Gus Gassmann schrieb am Sonntag, 26. September 2021 um 03:16:34 UTC+2:
> On Friday, 24 September 2021 at 16:57:04 UTC-3, Transfinity wrote:
> > Gus Gassmann schrieb am Freitag, 24. September 2021 um 21:05:15 UTC+2:
> >
> > > The intersection over infinitely many end segments is empty.
> > Infinitely many endsegments are more than every finite set of endsegments? What is in the infinite set that is not in at least one of all finite sets making the intersection infinite?
> I tell you what is not in it: Elements. The intersection over an infinite number of end segments is *EMPTY*.

The intersection over infinite endsegments is infinite.

> In particular no natural number n is an element of its successor's end segment E(n+1).

What are the elements of the infinite endsegments? Are they also in |N?

Regards, WM

On Monday, September 27, 2021 at 9:41:25 PM UTC+2, WM wrote:

> If n stands for each natural number

But it does not "stands for each natural number". After introducing "n" with

Let n be a natural number greater that 1

n is just a natural number greater than 1. With other words, "n" denotes a certain (but unknown) natural number greater than 1.

Hence n has a unique prime factorization.

Since

> [...] every natural number [> 1] has [one].

Right.