A strange belief

[WM]

It is generally accepted that the union of the set of FISONs is

{1} U {1, 2} U {1, 2, 3} U ... = ℕ
or briefly
F_1 U F_2 U F_3 U ... = ℕ.

The first n FISONs can be omitted

F_n U F_(n+1) U F_(n+2) U ... = ℕ .

Since this is true for all n, all F_n can be omitted and we find that no FISONs are necessary but

U{ } = ℕ.

This result makes the mathematical reality of the transfinite set ℕ quite doubtful. Therefore it is not acceptable for adherents of transfinite set theory.

But the following accepted definition of being required

F_n is required <==> U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= ℕ

shows that, according to this definition, the set of required FISONs is empty too.

This is a beautiful result because it forces adherents of transfinite set theory to believe that re-inserting the FISONs F_1, F_2, ..., F_(n-1) into the union U{F(n+1), F_(n+2), ...} is of any significance, such that for some n:

U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= U{F(n+1), F_(n+2), ...}

We do not want to spoil their belief. But students who are not yet closely connected with set theory will perhaps give it some thought whether to get in closer touch with this persuasion.

Regards, WM

[Jew Lover]

In other words, there is no infinite set. Period.

[WM]

Am Mittwoch, 27. Februar 2019 21:21:39 UTC+1 schrieb Jew Lover:


> In other words, there is no infinite set. Period.

Otherwise we would have to accept that

F_1 =/= ℕ
F_1 U F_2 =/= ℕ
F_1 U F_2 U F_3 =/= ℕ
...
but
F_1 U F_2 U F_3 U ... = ℕ when constructing ℕ bei the union of all FISONs.

On the other hand

F_1 =/= ℕ
F_1 U F_2 =/= ℕ
F_1 U F_2 U F_3 =/= ℕ
...
and
F_1 U F_2 U F_3 U ... =/= ℕ when destructing ℕ by removing all FISONs.

Regards, WM

[j4n bur53]

Are you again using:

forall n e N P(n) => P(N) ?

There is no such inference rule on FOL=+ZFC.

You are crazy as usual. Look see, take as predication
the predication of being non-empty, using von Neuman
ordinals, i.e. n={0,..,n-1}:

Q(n) <=> N \ n <> {}

Of course we have:

forall n e N Q(n)

But we don't have, i.e. it is:

~Q(N)

Its that simple. Don't apply fallacies if you
don't want to look idiotic.

[j4n bur53]

Or to sum up:

"set theory doesn't claim that if something
holds for all elements below omega, that
it automatically also holds for omega"

Alan Smaill already told you that your subject
to some circular reasoning. i.e. you don't
read Cantor correctly.
https://groups.google.com/d/msg/sci.math/X6_CKMTq00Y/QtIFmLCSBgAJ

In particular you are jumping to conclusions
what the logical content of an induction axiom
means, especially transfinite induction.

[Zeit Geist]

That does not follow. You are so sad.

> Regards, WM

ZG

[Jew Lover]

Only a deluded fool would try to refute any of this. There are many of those in the Church of Academia.

>
> Regards, WM

[Jew Lover]

It follows precisely. This is the kind of logic a two year old will understand, but not an orangutan of set theory.

> You are so sad.

If WM is sad, the you are truly tragic.

>
> > Regards, WM
>
> ZG

[Me]

On Wednesday, February 27, 2019 at 7:44:10 PM UTC+1, WM wrote:

> It is generally accepted that the union of the set of FISONs is
>
> {1} U {1, 2} U {1, 2, 3} U ... = ℕ
> or briefly
> F_1 U F_2 U F_3 U ... = ℕ.

Hint: It's accepted because it can be PROVED in the context of "set theory" (say ZFC).

> The first n FISONs can be omitted
>

> F_(n+1) U F_(n+2) U ... = ℕ .

Right.

> Since this is true for all n, all F_n can be omitted and ...

No, idiot.

Hint: Non sequitur.

[Zelos Malum]

>Since this is true for all n, all F_n can be omitted and we find that no FISONs are necessary but

This is where you go wrong, Ax(P(x)) is not the same as P(|N), you have only gotten the former, not the latter.

>This result makes the mathematical reality of the transfinite set ℕ quite doubtful. Therefore it is not acceptable for adherents of transfinite set theory.

The only thing it makes doubtful is your understanding of mathematics which we already know you have none.

>We do not want to spoil their belief. But students who are not yet closely connected with set theory will perhaps give it some thought whether to get in closer touch with this persuasion.

I think a better thing would you taking some courses and talk to professors that are knowledgable.

[Zelos Malum]

>In other words, there is no infinite set. Period.

A fallacious arguement does nto invalidate infinite sets.

>It follows precisely. This is the kind of logic a two year old will understand, but not an orangutan of set theory.

It does not, there is no theorem in FOL that Ax(P(x)) => P(|N)

It does not exist so it is fallacious to use it.

>Only a deluded fool would try to refute any of this. There are many of those in the Church of Academia.

Why? it is extremely trivial to refute it.

[Jew Lover]

On Thursday, 28 February 2019 02:05:13 UTC-5, Zelos Malum wrote:
> >In other words, there is no infinite set. Period.
>
> A fallacious arguement does nto invalidate infinite sets.

A fallacious belief does not invalidate logic. Chuckle. Unfortunately, you never even get to the argument part because you have nothing but beliefs (ZFC crap axioms) and meaningless decrees.

YOU will not dictate what is correct knowledge. Not now and not in a thousand years.

[j4n bur53]

You mean invalidate, like in invalid 3=<4. Ha Ha?

Do you have any clue what you are saying?
A fallacy is an invalid logical inference.
As long as you cannot show why a fallacy should

nevertheless hold, you didn't show anything
mathematically. You only conjecture something.
Unfortunately not all fallacies can be repaired,

some are broken beyond repair. Anyway there
is a simple medicin for morons like WM and JG,
why don't you play around with a truth table generator?

http://web.stanford.edu/class/cs103/tools/truth-table-tool/

For example this fallacy can be repaired:

p q ((p → q) → (¬p → ¬q))
F F T
F T F
T F T
T T T

If manage to show that for your p,q the situation
p=F and q=T cannot happen on some other grounds.

But repairing this one will be like impossible:

p q ((p ∧ q) ∧ ((p → ¬q) ∧ (¬q → p)))
F F F
F T F
T F F
T T F

[j4n bur53]

Well you can also repair the below:

> p q ((p ∧ q) ∧ ((p → ¬q) ∧ (¬q → p)))
> F F F
> F T F
> T F F
> T T F

In the empty set of possible worlds.
Oh empty set doesn't exist in WM Muckamatik.

How convenient...

[j4n bur53]

Everything where the outcome of the truth
table is not always T, is a fallacy.

Otherwise when the outcome of the truth
table is always T, its a tautology,

and you are allowed to use the schema.

j4n bur53 schrieb:

[Jew Lover]

On Thursday, 28 February 2019 07:24:33 UTC-5, moron Jan Burse driveled:

> You mean invalidate, like in invalid 3=<4...

Shut up moron. 3=<4 is used by intellectually challenged fools like you.

[WM]

Am Mittwoch, 27. Februar 2019 22:37:47 UTC+1 schrieb j4n bur53:
> Are you again using:
>
> forall n e N P(n) => P(N) ?
>
> There is no such inference rule on FOL=+ZFC.

Nevertheless set theory proves it. If you remove all elements from |N then you have removed |N. Or does some empty shell remain?
>
Regards, WM

[j4n bur53]

An example is not a prove. For a prove you
would need to have P arbitrary.

Where did you learn logic? In some dump?

[WM]

Am Donnerstag, 28. Februar 2019 00:44:12 UTC+1 schrieb Me:
> On Wednesday, February 27, 2019 at 7:44:10 PM UTC+1, WM wrote:
>
> > It is generally accepted that the union of the set of FISONs is
> >
> > {1} U {1, 2} U {1, 2, 3} U ... = ℕ
> > or briefly
> > F_1 U F_2 U F_3 U ... = ℕ.
>
> Hint: It's accepted because it can be PROVED in the context of "set theory" (say ZFC).

It can also be proved that ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø

>
> > The first n FISONs can be omitted
> >
> > F_(n+1) U F_(n+2) U ... = ℕ .
>
> Right.
>
> > Since this is true for all n, all F_n can be omitted and ...
>
> No,

But all FISONs can be omitted, when the useless smaller ones are re-inserted?

Regards, WM

[WM]

Am Donnerstag, 28. Februar 2019 08:01:43 UTC+1 schrieb Zelos Malum:
> >Since this is true for all n, all F_n can be omitted and we find that no FISONs are necessary but
>
> This is where you go wrong, Ax(P(x)) is not the same as P(|N), you have only gotten the former, not the latter.

If you remove all elements from |N then you have removed |N.
>

Regards, WM

[WM]

Am Donnerstag, 28. Februar 2019 08:05:13 UTC+1 schrieb Zelos Malum:


> It does not, there is no theorem in FOL that Ax(P(x)) => P(|N)

In fact it is rather nonsensical. But set theory proves it. If you remove all elements from |N then you have removed |N.
>
Regards, WM

[j4n bur53]

Also your nonsense is even a counter example,
to your fallacy. This means its a row in the
truth table with "F":

/* WMs Law, a Fallacy */

forall n e N P(n) => P(N)

Just use for P, the following predication, where
x can be either a finite Neuman ordinal x={0,..,n-1}=n,
or the least limit Neuman ordinal omega={0,1,2,...}=N.

Q(x) = N \ x <> {}

The we have:

forall n e N Q(n)

And also:

~Q(N)

So you produced your own counter example, showing
that your WM law is a fallacy. Thank you!

[j4n bur53]

To avoid confusion, I guess I have
to wrote the WMs Law counter example with
parenthesis:

Q(x) = (N \ x <> {})

Or in words:

Q(x) is true iff
x subtracted from N is non-empty

[WM]

Am Donnerstag, 28. Februar 2019 14:49:08 UTC+1 schrieb j4n bur53:
> Also your nonsense is even a counter example,
> to your fallacy. This means its a row in the
> truth table with "F":
>
> /* WMs Law, a Fallacy */
>
> forall n e N P(n) => P(N)

Simply learn what set theory is based upon.

If you remove all elements from |N then you have removed the set |N.

If you biject all elements of Q with elements of |N, then you biject the sets Q and |N.

That's set theory.

It seems that you have a good taste for what is nonsense but a bad knowledge of set theory.

Regards, WM

[j4n bur53]

Well I take your word. "If you remove all

elements from |N then you have removed the set |N."

Now take:

Q(x) is true iff
x subtracted from N is non-empty

Therefore:

Q(N)=F, well easy, N\N = {},
and it is not {}<>{}

But also:

forall n e N Q(n)=T, well easy, N\n = {n+1,...},
and it is {n+1,..}<>{}

This violates your WMs Law. Because your WMs
Law would say, before applied to Q, if it were a law,
only the rows with "T" would be possible:

forall n e N P(n) P(N) forall n e N P(n) => P(N)
F F T
F T T
T T T

But we have just produced a counter example,
namely we found forall n e N Q(n)=T and Q(N)=F,
violating the law:

forall n e N Q(n) Q(N) forall n e N Q(n) => Q(N)
T F F

So you produced your own counter example. Thank you!

[j4n bur53]

WM, are you L^3 type? Lausig Lange Leitung?

[FredJeffries]

On Thursday, February 28, 2019 at 5:45:20 AM UTC-8, WM wrote:

> It can also be proved that ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø

I call your bluff: PROVE IT

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Mittwoch, 27. Februar 2019 22:37:47 UTC+1 schrieb j4n bur53:
>> Are you again using:
>>
>> forall n e N P(n) => P(N) ?
>>
>> There is no such inference rule on FOL=+ZFC.
>
> Nevertheless set theory proves it.

Then show us a proof in ZF(C).
That would settle the matter.

You may use any result from standard texts of set theory.


>>
> Regards, WM

--
Alan Smaill

[Zeit Geist]

Proof? What a proof? How do I start?

ZG

[Jew Lover]

The proof is simple and is in accordance with your mainstream ideas:

There are infinitely many FISONs F_i. All we need to believe is that an infinite process is possible. This is not a problem in your misguided theory because you claim that 3.14159... is the infinite decimal representation of the measure of that size known as pi.

Now pi - pi = 0. That is,

pi -(3+1/10+4/100+1/1000+...) = 0

which is only possible if 3+1/10+4/100+1/1000+... is pi.

ℕ - (F_1 U F_2 U F_3 U ...) = {}

which is only possible if F_1 U F_2 U F_3 U ...= ℕ.

Therefore, ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø.

[j4n bur53]

Thats an analogy, but not a proof.

All these cranks are not able to produce proofs.

[Jew Lover]

On Thursday, 28 February 2019 11:44:16 UTC-5, j4n bur53 wrote:
> Thats an analogy, but not a proof.

Just because you cannot understand simple proofs, does not mean they are invalid. One look at you and your last 10000 posts reveals you are a crank.

Now pi - pi = 0. That is,

pi -(3+1/10+4/100+1/1000+...) = 0

If you like, you can turn (3+1/10+4/100+1/1000+...) into a set of all the significant digits: { 3; 1; 4; 1; 5; 9; ...}.

You can't deny such a set exists which is not in accordance with your superstitious ZFC crap axioms.

Let FISOPDIGs (Finite sequences of pi digits).

The rest will be obvious to anyone who isn't blind.

which is only possible if { 3; 1; 4; 1; 5; 9; ...} is pi.

pi - { 3; 1; 4; 1; 5; 9; ...} = {}


which is only possible if {3} U {3; 1} U {3; 1; 4} U ...= pi.

Therefore, pi \ {3} U {3; 1} U {3; 1; 4} U ... = Ø.

If you claim this is false, then you cannot claim that

pi = { 3; 1; 4; 1; 5; 9; ...}

Jan Burse: troll, incredibly stupid moron, unemployed Swiss programmer and <whatever derisive term you can think of here>

[WM]

Am Donnerstag, 28. Februar 2019 15:09:10 UTC+1 schrieb j4n bur53:
> Well I take your word. "If you remove all
> elements from |N then you have removed the set |N."
> Now take:

>...

>
> This violates your WMs Law.

Then set theory is violated because then also this would be wrong: "If you combine all natural numbers (elements of |N) then you have built the set |N."

> So you produced your own counter example. Thank you!

It's set theory that produces its own counter examples.

[WM]

I will do it, but first let me know whether you will aceept it.

What is the union of all FISONs, F_1 U F_2 U F_3 U ..., in your opinion?

REgards, WM

[WM]

This one is enough: If you biject all elements of Q with elements of |N, then you biject the sets Q and |N.

Do you know that result?

Regards, WM

[j4n bur53]

Its still only an analogy, since the limit in

real numbers, does not behave as the infinite
union in sets.

Here is an example (in real numbers, base 2):

0.111... = 1.000...

But you don't have (in sets):

{1,2,3,...} <> {0}

Got it?

[j4n bur53]

The following function from sets to real numbers,

is not an injection:

f : P(N) -> R

f(s) := Lim n->oo Sum_i=0^n s(i)*2^(-i)

You don't have s<>t => f(s)<>f(t).

So when there is no injection, then there is
hardly an isomorphism, then its hardly an analogy.

See also:

https://en.wikipedia.org/wiki/Injective_function

What is your analogy moron?

[j4n bur53]

You still don't get what it means "if a theory
doesn't claim a law". It doesn't mean that there
cannot be some cases where the rule can

nevertheless be observed. It is not that when a
theory doesn't claim a law, it automatically claims
the negation of a law.

What is wrong with you? Too much booze again?

Are all people in Augsburg that debilated?
Do you even understand a single word english
or are you dumb like a stone?

[j4n bur53]

Its only a soft analogy.

[FredJeffries]

On Thursday, February 28, 2019 at 10:12:22 AM UTC-8, WM wrote:
> Am Donnerstag, 28. Februar 2019 16:11:28 UTC+1 schrieb FredJeffries:
> > On Thursday, February 28, 2019 at 5:45:20 AM UTC-8, WM wrote:
> >
> > > It can also be proved that ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø
> >
> > I call your bluff: PROVE IT
>
> I will do it, but first let me know whether you will aceept it.

Why should that make any difference? According to you I am "brain-damaged".

But I am not fool enough to say that I will "aceept[sic]" any alleged "proof" before I see it.

> What is the union of all FISONs, F_1 U F_2 U F_3 U ..., in your opinion?

Until I see a (convincing) proof or refutation, I have no "opinion" about ANY (reputed) mathematical statement.

> REgards, WM

[Jew Lover]

On Thursday, 28 February 2019 13:54:22 UTC-5, j4n bur53 wrote:
> Its still only an analogy,

It's a proof, you stupid. And yes, it is a proof for an analogy which infers that WM's proof is correct.

> since the limit in real numbers, does not behave as the infinite
> union in sets.

Really?!!! What about Dedekind Cuts and Cauchy sequences? Gosh, you are a moron, aren't you!

>
> Here is an example (in real numbers, base 2):
>
> 0.111... = 1.000...

This is nonsense. Not even worthy to be called anything else.

[Jew Lover]

On Thursday, 28 February 2019 15:15:42 UTC-5, FredJeffries wrote:
> On Thursday, February 28, 2019 at 10:12:22 AM UTC-8, WM wrote:
> > Am Donnerstag, 28. Februar 2019 16:11:28 UTC+1 schrieb FredJeffries:
> > > On Thursday, February 28, 2019 at 5:45:20 AM UTC-8, WM wrote:
> > >
> > > > It can also be proved that ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø
> > >
> > > I call your bluff: PROVE IT
> >
> > I will do it, but first let me know whether you will aceept it.
>
> Why should that make any difference?

Because deceptive reptiles like you will never accept a proof and even if a proof is provided, you will then retort that you did not accept the initial assumption.

Get it stupid?

> According to you I am "brain-damaged".

Also according to me. Chuckle.

[j4n bur53]

This is how real numnbers are constructed. If they
were the same as sets, you wouldn't need the construction.

https://en.wikipedia.org/wiki/Construction_of_the_real_numbers

[Me]

On Thursday, February 28, 2019 at 2:41:28 PM UTC+1, WM wrote:

> If you remove all elements from |N

Actually, you can't (literally) REMOVE elements FROM a set.

But we can consider the set difference between, say, IN and IN.

> then <bla>. Or does some empty shell remain?

Exactly, it's called the empty set:

IN \ IN = {} .

See: http://mathworld.wolfram.com/SetDifference.html
and: https://proofwiki.org/wiki/Definition:Set_Difference

[Me]

On Thursday, February 28, 2019 at 3:01:37 PM UTC+1, WM wrote:

> Simply learn what set theory is based upon.

Look, idiot, you know SHIT about set theory.

Hence you better shut up!

[No one]

On Wednesday, 27 February 2019 14:44:10 UTC-4, WM wrote:
> It is generally accepted that the union of the set of FISONs is
>
> {1} U {1, 2} U {1, 2, 3} U ... = ℕ
> or briefly
> F_1 U F_2 U F_3 U ... = ℕ.
>
> The first n FISONs can be omitted
>
> F_n U F_(n+1) U F_(n+2) U ... = ℕ .

Congratulations. You have managed to prove that
for all n [F_n U F_(n+1) U F_(n+2) U ...] = [F_1 U F_2 U F_3 U ... ].

As Alan Smaill pointed out using your own writing, this does not imply that
{for ℕ [ {} = (F_1 U F_2 U F_3 U ... ) ] }.

It's very sad that you are too stupid to understand your own writings and/or too dishonest to admit that you just plagiarized the passage on p.52.

> Since this is true for all n, all F_n can be omitted and we find that no FISONs are necessary but
>

> U{ } = ℕ.
>
> This result makes the mathematical reality of the transfinite set ℕ quite doubtful. Therefore it is not acceptable for adherents of transfinite set theory.
>
> But the following accepted definition of being required
>
> F_n is required <==> U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= ℕ
>
> shows that, according to this definition, the set of required FISONs is empty too.
>
> This is a beautiful result because it forces adherents of transfinite set theory to believe that re-inserting the FISONs F_1, F_2, ..., F_(n-1) into the union U{F(n+1), F_(n+2), ...} is of any significance, such that for some n:
>
> U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= U{F(n+1), F_(n+2), ...}
>
> We do not want to spoil their belief. But students who are not yet closely connected with set theory will perhaps give it some thought whether to get in closer touch with this persuasion.
>
> Regards, WM

[Jew Lover]

On Thursday, 28 February 2019 18:06:49 UTC-5, Me wrote:
> On Thursday, February 28, 2019 at 2:41:28 PM UTC+1, WM wrote:
>
> > If you remove all elements from |N
>
> Actually, you can't (literally) REMOVE elements FROM a set.

Chuckle. What a moron. You do it all the time in your bogus calculus!

Watch how you remove the point (c, f(c)) from the function f and hence from the domain and range:

f(x) * (x-c)/(x-c)

Obviously, if you can generate all the elements of a set, then you can remove all the elements too. Doh!


Thus, to remove all the elements of f on the interval [a,b], you simply
multiply f(x) by (x - d)/(x - d) where d = a + {[b-a]/n} as n increases indefinitely. Since you believe in infinity, you should have no problem completing this task in your syphilitic brain. Chuckle.

To remove all, you simply take the interval (-oo, oo) and you are done. LMAO.

[Jew Lover]

On Thursday, 28 February 2019 18:06:49 UTC-5, Me wrote:

> On Thursday, February 28, 2019 at 2:41:28 PM UTC+1, WM wrote:
>
> > If you remove all elements from |N
>
> Actually, you can't (literally) REMOVE elements FROM a set.
>
> But we can consider the set difference between, say, IN and IN.
>
> > then <bla>. Or does some empty shell remain?
>
> Exactly, it's called the empty set:
>
> IN \ IN = {} .

Er, no idiot. That's not what the argument is saying. It's you who knows shit about set theory or anything else. Yeah, follow your advice and shut up.

[K_h]

"WM" wrote in message
news:71103447-d101-4bff...@googlegroups.com...

Am Donnerstag, 28. Februar 2019 15:09:10 UTC+1 schrieb j4n bur53:
> > This violates your WMs Law.
>
> Then set theory is violated because then also this would be
> wrong: "If you combine all natural numbers (elements of |N)
> then you have built the set |N."

Hello WM. It has been a while. The statement for all n e N, P(n) --> P(N)
cannot be true generally because simple counter examples can be offered.
For instance all natural numbers greater than 1 have a predecessor but
w=omega=N does not have a predecessor (although in this example P(0) would
be false too since it has no predecessor). But you get the idea.

x

[Zelos Malum]

>A fallacious belief does not invalidate logic

Of course not, that is why your beliefs do not invalidate logic.

>Unfortunately, you never even get to the argument part because you have nothing but beliefs

Got nothing of the sort.

>(ZFC crap axioms) and meaningless decrees.

In deductive logic one must always begin with axioms, from nothing you can only conclude nothing.

>YOU will not dictate what is correct knowledge. Not now and not in a thousand years.

Neither will you.

>Shut up moron. 3=<4 is used by intellectually challenged fools like you.

3 <= 4 is perfectly valid, there is nothign in the definitions that says it is wrong.

[Zelos Malum]

>Nevertheless set theory proves it. If you remove all elements from |N then you have removed |N. Or does some empty shell remain?

At no point have you removed all elements in your arguement. All your arguement says is that for any given n, you can remove all that is less than it and the union remains the same. That is not the same as removing all elements.

>If you remove all elements from |N then you have removed |N.

Except you have not removed all elements, you have only shown that any arbitrary finite set of elements can be removed. A whole different proposition.

>In fact it is rather nonsensical. But set theory proves it. If you remove all elements from |N then you have removed |N.

It does not, you are proving Ax(P(x)), not once P(|N).

This is your inability to do logic that you cannot understand the differens between those propositions.

>Simply learn what set theory is based upon.

We know it already, learn how logic works so you can make an arguement that isn't fundamentally fallacious.

[WM]

Am Freitag, 1. März 2019 05:57:42 UTC+1 schrieb K_h:
> "WM" wrote in message
> news:71103447-d101-4bff...@googlegroups.com...
>
> Am Donnerstag, 28. Februar 2019 15:09:10 UTC+1 schrieb j4n bur53:
> > > This violates your WMs Law.
> >
> > Then set theory is violated because then also this would be
> > wrong: "If you combine all natural numbers (elements of |N)
> > then you have built the set |N."
>
> Hello WM. It has been a while. The statement for all n e N, P(n) --> P(N)
> cannot be true generally because simple counter examples can be offered.

Of course! the union of all natural numbers does not yield a complete fixed set larger than all FISONs. That is impossible by the fact that for every n there is an n+1. See my animation on slide 57 of my lesson https://www.hs-augsburg.de/~mueckenh/HI/HI12.PPT: It is impossible to collect all natural numbers within the blue cube, if number n may leave the intermediate reservoir (the yellow hatbox) only after n+1 has arrived there. A completely natural requirement.

> For instance all natural numbers greater than 1 have a predecessor but
> w=omega=N does not have a predecessor (although in this example P(0) would
> be false too since it has no predecessor). But you get the idea.

I got it long ago and try to extinguish it. It is impossible to enumerate all rational numbers. See Not enumerating all positive rational numbers (formal proof) on p. 254 of https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

But set theory uses this falsity.

Regards, WM

[WM]

Am Freitag, 1. März 2019 07:35:44 UTC+1 schrieb Zelos Malum:
> >Nevertheless set theory proves it. If you remove all elements from |N then you have removed |N. Or does some empty shell remain?
>
> At no point have you removed all elements in your arguement.

At no point have you collected all elements in any argument. But we have to assume that this is possible in set theory.

> All your arguement says is that for any given n, you can remove all that is less than it and the union remains the same. That is not the same as removing all elements.

Same with enumerating "all" rationals or with Cantor's list.

>
> >If you remove all elements from |N then you have removed |N.
>
> Except you have not removed all elements, you have only shown that any arbitrary finite set of elements can be removed. A whole different proposition.

Why does collecting all elements show more?

>
> >In fact it is rather nonsensical. But set theory proves it. If you remove all elements from |N then you have removed |N.
>
> It does not, you are proving Ax(P(x)), not once P(|N).

How should that be possible?

>
> This is your inability to do logic that you cannot understand the differens between those propositions.

Tell me where the difference appears and in particular why it does.

>
> >Simply learn what set theory is based upon.
>
> We know it already, learn how logic works so you can make an arguement that isn't fundamentally fallacious.

Teach me where the difference appears and in particular why and how something is "simultaneous" and something is not.

Regards, WM

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Freitag, 1. März 2019 07:35:44 UTC+1 schrieb Zelos Malum:
>> >Nevertheless set theory proves it. If you remove all elements from
>> > |N then you have removed |N. Or does some empty shell remain?
>>
>> At no point have you removed all elements in your arguement.
>
> At no point have you collected all elements in any argument. But we
> have to assume that this is possible in set theory.

In particular, at no point have you collected all elements when using
proof by induction over the natural numbers. But in Mueckenheimia we
have to assume that the conclusion follows, by some miracle.

>
> Regards, WM

--
Alan Smaill

[Alan Smaill]

Indeed, by definition of biject.
No need for your bogus inference rule in this case.

But strangely you failed to address the question asked.

You claim that for an arbitrary (let's say definable in ZF(C)) predicate
P, the following is provable in set theory:

all n: |N P(n) => P(|N)

You are being asked for a proof in ZF(C) that this *always* holds.

That is after all your claim, isn't it?

[jvr]

[...]

> Teach me where the difference appears and in particular why and how something is "simultaneous" and something is not.

It is more likely that I can teach a donkey to recite the Psalms of David,
but let us demonstrate once more where the real difficulty lies:

Let S_n, n = 1,2, ..., be a sequence of sets s.t. S_n <subset> S_{n+1} for
every n.
Now clearly S = S_1 v S_2 v .... = S_n v S_{n+1} v .... for every n.
Notice that nothing is ever "removed from" S.

Your claim is that from this relationship you can deduce, using nothing
but the axioms of set theory, that S is empty.

All that can be said about such an obviously false conclusion is that, if
it is provable, then the axioms are inconsistent. But you haven't given
a proof. So where is your problem?

[WM]

Am Freitag, 1. März 2019 12:20:08 UTC+1 schrieb Alan Smaill:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Donnerstag, 28. Februar 2019 16:20:06 UTC+1 schrieb Alan Smaill:
> >> WM <wolfgang.m...@hs-augsburg.de> writes:
> >>
> >> > Am Mittwoch, 27. Februar 2019 22:37:47 UTC+1 schrieb j4n bur53:
> >> >> Are you again using:
> >> >>
> >> >> forall n e N P(n) => P(N) ?
> >> >>
> >> >> There is no such inference rule on FOL=+ZFC.
> >> >
> >> > Nevertheless set theory proves it.
> >>
> >> Then show us a proof in ZF(C).
> >> That would settle the matter.
> >>
> >> You may use any result from standard texts of set theory.
> >
> > This one is enough: If you biject all elements of Q with elements of
> > |N, then you biject the sets Q and |N.
> >
> > Do you know that result?
>
> Indeed, by definition of biject.

By conclusion from every FISON to the whole set.

> No need for your bogus inference rule in this case.

It has been used to obtain that definition.

if there is never an obstacle or halt in this process of assignment, then both infinite sets are in bijection. ("und es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor, p. 239])

All well-ordered sets can be compared. They have the same number if they, by preserving their order, can be uniquely mapped or counted onto each other. ("Dabei nenne ich zwei wohlgeordnete Mengen von demselben Typus und schreibe ihnen gleiche Anzahl zu, wenn sie sich unter Wahrung der festgesetzten Rangordnung ihrer Elemente gegenseitig eindeutig aufeinander abbilden, oder wie man sich gewöhnlich ausdrückt, aufeinander abzählen lassen." [G. Cantor, letter to W. Wundt (5 Oct 1883)])

>
> You claim that for an arbitrary (let's say definable in ZF(C)) predicate
> P, the following is provable in set theory:
>
> all n: |N P(n) => P(|N)
>
> You are being asked for a proof in ZF(C) that this *always* holds.

It is the basis of set theory. See above. But since set theory is inconsistent, it does not hold in general.

For instance when removing FISONs from the union of all FISONs (before the union has been executed), then it does not hold. But fortunately we find then that

∃n ∈ ℕ: U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} =/= U{A(n+1), A_(n+2), ...}.

That shows that in fact it does never hold.
>
Regards, WM

[Jew Lover]

Don't you assume the same thing when you claim that pi = 3.14159... ?

A decimal representation is a *measure*. So are you not assuming the completed measure in this case? And if so, how can the measure be anything but a rational number in spite of the fact that we can prove there is no rational number which represents pi?

[WM]

Am Freitag, 1. März 2019 12:20:40 UTC+1 schrieb jvr:
> [...]
> > Teach me where the difference appears and in particular why and how something is "simultaneous" and something is not.
>
> It is more likely that I can teach a donkey to recite the Psalms of David,

than you can any intelligent person your theorems like the following:

"You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;" [Jürgen Rennenkampff in "How many different paths can exist in the complete infinite Binary Tree?", sci.logic (15 Jul 2018)]

"There are points in the complement C that are neither endpoints of intervals nor interior points of C." [Jürgen Rennenkampff in "Clusters and Cantor dust", sci.logic (5 Jan 2019)]

> Your claim is that from this relationship you can deduce, using nothing
> but the axioms of set theory, that S is empty.

I use in fact that the union of FISONs is |N.

>
> All that can be said about such an obviously false conclusion is that, if
> it is provable, then the axioms are inconsistent. But you haven't given
> a proof.

This proof has been established in collaboration with Franz Fritsche. If not all FISONs A_n can be removed from the set of FISONs to be unioned without changing the result of the union, then

∃n ∈ ℕ: U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} =/= U{A(n+1), A_(n+2), ...}.

But such a number n does not exist.

Q.E.D.

REgards, WM

[j4n bur53]

In an inconsistent theory, it would hold always you moron.

https://en.wikipedia.org/wiki/Principle_of_explosion

[j4n bur53]

The principle of explosion is even valid in
intuitionistic logic, not only in classical

logic. This here is valid in
intuitionistic logic:

A & ~A -> B

Whats the motto of you cranks? Not a single day
without some new nonsense?

[jvr]

As pointed out before, this is a simple logical error.
Take S = A v B v C. We cannot "remove all three summands without changing the result".
Nevertheless it is possible that A v B = B v C = C v A. I.e. none of the
summands is "necessary".
I.e. your conclusion "there exists n in N etc" is false.

[jvr]

[...]

>
> than you can any intelligent person your theorems like the following:
>
> "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;" [Jürgen Rennenkampff in "How many different paths can exist in the complete infinite Binary Tree?", sci.logic (15 Jul 2018)]

Indeed, Doktor Mückenheim, retired Professor of General Studies, that is
a fact. Would you like to know why I stipulated "in the context"?

>
> "There are points in the complement C that are neither endpoints of intervals nor interior points of C." [Jürgen Rennenkampff in "Clusters and Cantor dust", sci.logic (5 Jan 2019)]

This is an elementary topological fact known since the days of Cantor.
Any math undergraduate who has learned a little topology can prove it.

[Jew Lover]

On Friday, 1 March 2019 07:14:27 UTC-5, WM wrote:
> Am Freitag, 1. März 2019 12:20:40 UTC+1 schrieb jvr:
> > [...]
> > > Teach me where the difference appears and in particular why and how something is "simultaneous" and something is not.
> >
> > It is more likely that I can teach a donkey to recite the Psalms of David,
>
> than you can any intelligent person your theorems like the following:
>
> "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;" [Jürgen Rennenkampff in "How many different paths can exist in the complete infinite Binary Tree?", sci.logic (15 Jul 2018)]

For all n terms the sequence 3; 3.1; 3.14; ... < pi, but somehow at infinity, it magically morphs into an imaginary number called "pi". Chuckle.

Euler's Blunder S = Lim S is the HIV of mathematics.

[WM]

Am Freitag, 1. März 2019 14:01:59 UTC+1 schrieb jvr:
> On Friday, March 1, 2019 at 1:14:27 PM UTC+1, WM wrote:
> > Am Freitag, 1. März 2019 12:20:40 UTC+1 schrieb jvr:
> > > [...]
> > > > Teach me where the difference appears and in particular why and how something is "simultaneous" and something is not.
> > >
> > > It is more likely that I can teach a donkey to recite the Psalms of David,
> >
> > than you can any intelligent person your theorems like the following:
> >
> > "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;" [Jürgen Rennenkampff in "How many different paths can exist in the complete infinite Binary Tree?", sci.logic (15 Jul 2018)]
> >
> > "There are points in the complement C that are neither endpoints of intervals nor interior points of C." [Jürgen Rennenkampff in "Clusters and Cantor dust", sci.logic (5 Jan 2019)]
> >
> > > Your claim is that from this relationship you can deduce, using nothing
> > > but the axioms of set theory, that S is empty.
> >
> > I use in fact that the union of FISONs is |N.
> > >
> > > All that can be said about such an obviously false conclusion is that, if
> > > it is provable, then the axioms are inconsistent. But you haven't given
> > > a proof.
> >
> > This proof has been established in collaboration with Franz Fritsche. If not all FISONs A_n can be removed from the set of FISONs to be unioned without changing the result of the union, then
> >
> > ∃n ∈ ℕ: U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} =/= U{A(n+1), A_(n+2), ...}.
> >
> > But such a number n does not exist.
> >
> > Q.E.D.

> As pointed out before, this is a simple logical error.
> Take S = A v B v C. We cannot "remove all three summands without changing the result".

The error is this: A union of FISONs can never yield an actually infinite set larger than every FISON.

> Nevertheless it is possible that A v B = B v C = C v A. I.e. none of the
> summands is "necessary".

If the summands are well-ordered, then this is not possible. You had argued with this fallacy two years ago already:

When choosing the order

{3, 1} U {1, 2} U {2, 3} = {1, 2, 3}

we could drop {3, 1} but then {1, 2} is the first set necessary, because otherwise 1 would be missing in the union.

I am indebted to Mr. J. Rennenkampff for the instructive, although not inclusion monotonic, examples (*) and (**) [J. Rennenkampff in "Von seinen Jüngern verleugnet", de.sci.mathematik (20 Apr 2016)]

https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 209.

> I.e. your conclusion "there exists n in N etc" is false.

This conclusion is false. Note that Cantor's theorem B concerns ordinal numbers and not the enumerated objects. We have to choose an order. In case of FISONs we can use the natural order or understand FISONs as ordinals according to von Neumann.

Regards, WM

[j4n bur53]

What is Cantors Theorem B. Could you please enlighten
us what you mean by Cantors Theorem B?

You use it all the time, but you never defined it so
far if I am not totally wrong.

[j4n bur53]

Please note that the Cantor-Bernstein-Schroeder

theorem deals with cardinality, and the theorem

doesn't involve any ordinals. So what is Theorem B?

[WM]

Am Freitag, 1. März 2019 14:16:44 UTC+1 schrieb jvr:
> [...]
> >
> > than you can any intelligent person your theorems like the following:
> >
> > "You are apparently trying to appeal to a theorem that says that if a_n < A for all n then lim a_n =< A. This statement is false in the context;" [Jürgen Rennenkampff in "How many different paths can exist in the complete infinite Binary Tree?", sci.logic (15 Jul 2018)]
>
> Indeed, Doktor Mückenheim, retired Professor of General Studies

O, I am just alive and kicking and giving lectures.

> that is
> a fact. Would you like to know why I stipulated "in the context"?

There is no context where the theorem could be false. Every context where it appears to be false is false.

> > "There are points in the complement C that are neither endpoints of intervals nor interior points of C." [Jürgen Rennenkampff in "Clusters and Cantor dust", sci.logic (5 Jan 2019)]
>
> This is an elementary topological fact known since the days of Cantor.

It is an elementary mistake comitted by matheologians who cannot think enough to understand that an infinite digit sequence does never define a real number or a point but who are too lazy or too much deceiving themselves to prove this impossibility to themselves.

Regards, WM

[Jew Lover]

On Friday, 1 March 2019 10:08:46 UTC-5, moron driveled:

> What is Cantors Theorem B. Could you please enlighten
> us what you mean by Cantors Theorem B?
>
> You use it all the time, but you never defined it so
> far if I am not totally wrong.

You can either be wrong or not. "Totally wrong" only happens in FOoL. In sound logic, one cannot be wrong proportionally. Chuckle.

[WM]

Am Freitag, 1. März 2019 16:08:46 UTC+1 schrieb j4n bur53:
> What is Cantors Theorem B. Could you please enlighten
> us what you mean by Cantors Theorem B?

"Unter den Zahlen der Menge (') gibt es immer eine kleinste." [Cantor, p. 200]
Daß es in jeder Menge (') transfiniter Zahlen immer eine kleinste gibt, läßt sich folgendermaßen einsehen. [Cantor, p. 208f]
Satz B. "Jeder Inbegriff von verschiedenen Zahlen der ersten und zweiten Zahlenklasse hat eine kleinste Zahl, ein Minimum." [Cantor, p. 332]
"Würde nun der Index ' nicht alle Zahlen der zweiten Zahlenklasse durchlaufen, so müßte es eine kleinste Zahl  geben, die er nicht erreicht." [Cantor, p. 349]
"Aber aus den in § 13 über wohlgeordnete Mengen bewiesenen Sätzen folgt auch leicht, daß jede Vielheit von Zahlen, d. h. jeder Teil von  eine kleinste Zahl enthält." [Cantor, p. 444]

>
> You use it all the time, but you never defined it so
> far if I am not totally wrong.

I have defined it frequently, but obviously not yet often enough. Mr. Rennenkampff has not yet understood that is concerns ordinal numbers and not objects without order, although I told him years ago about this secret.

Regards, WM

[j4n bur53]

Thats not a definition. Thats a quote from Cantor.
Could you please define what Theorem B formally
means inside ZFC?

Quotes from some verbal bla bla are not mathematical
definitions. Also you are not citing a theorem B itself,
but also some proof attempst.

Could you only define the statement of the Theorem B,
not some proof?

Could it be that you mean:

"In mathematics, the well-ordering principle
states that every non-empty set of positive
integers contains a least element."
https://en.wikipedia.org/wiki/Well-ordering_principle

Which would be:

forall M (M subset |N /\ M <> {} =>
exists x (x e M /\ forall y (y e M => x =< y)))

Or could it be that you mean:

"Every well-ordered set is uniquely order
isomorphic to a unique ordinal number, called the
order type of the well-ordered set. The well-ordering
theorem, which is equivalent to the axiom of choice,
states that every set can be well ordered."
https://en.wikipedia.org/wiki/Well-order#Ordinal_numbers

Which would be:

(X,=<) wellordered <=> forall M (M subset X /\ M <> {} =>
exists x (x e M /\ forall y (y e M => x =< y)))

forall X exists =< (X,=<) wellordered

[j4n bur53]

In:

Which would be:

(X,=<) wellordered <=> forall M (M subset X /\ M <> {} =>
exists x (x e M /\ forall y (y e M => x =< y)))

forall X exists =< (X,=<) wellordered

I have left out the ordinal thingy. Its not
necessary to formulate the well-ordering theorem.

So you have already two choices, and the 3rd
choice would include also some theorems about
ordinals.

What do you wish is Theorem B, choice 1), 2)
or choice 3)?

[FredJeffries]

On Friday, March 1, 2019 at 10:39:26 AM UTC-8, j4n bur53 wrote:
> Thats not a definition. Thats a quote from Cantor.
> Could you please define what Theorem B formally
> means inside ZFC?

He SEEMS to be referring to the Theorem B of section 16 (our Professor seems oblivious to the fact that Cantor repeats the indexing A, B, C, ... in each section) which appears on page 171 of "Contributions to the Founding of the Theory of Transfinite Numbers" which simply states "Every totality of different numbers of the first and second number-classes has a least number".

https://books.google.com/books?id=W1gNAAAAYAAJ

[Me]

On Friday, March 1, 2019 at 8:18:19 PM UTC+1, FredJeffries wrote:

> "Every totality of different numbers of the first and second number-classes
> has a least number".

Yeah! We (except Mückenheim, of course) have a word for this property:

https://en.wikipedia.org/wiki/Well-order

[K_h]

"WM" wrote in message
news:22ffe9b0-bb08-4f39...@googlegroups.com...

>
> Of course! the union of all natural numbers does not yield a complete
> fixed set larger than all FISONs.

There are an infinite number of natural numbers. Proof: the number of
natural numbers is not finite. By definition, anything which is not finite
is infinite. So N is an infinite set. QED. Trivially, the union of all
initial segments of the natural numbers yields the set N:

{0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N

The union of all natural numbers also yields the set N:

{0} U {1} U {2} U {3} U {4} U {5} U {6} U {7} U {8} U {9} U ... = N

In both cases there are an infinite number of sets in the union. Nobody
ever claimed that one of these unions would result in a bigger set than the
other.

k

[WM]

Am Freitag, 1. März 2019 19:39:26 UTC+1 schrieb j4n bur53:
> Thats not a definition. Thats a quote from Cantor.

It is Cantor's theorem B.

Satz B. "Jeder Inbegriff von verschiedenen Zahlen der ersten und zweiten Zahlenklasse hat eine kleinste Zahl, ein Minimum." [Cantor, p. 332]

> Could you please define what Theorem B formally
> means inside ZFC?

It means precisely that what is written above - and need not be translated for you.

>
> Also you are not citing a theorem B itself,
> but also some proof attempst.

Try to sober up.

Regards, WM

[WM]

Am Freitag, 1. März 2019 21:24:17 UTC+1 schrieb K_h:
> "WM" wrote in message
> news:22ffe9b0-bb08-4f39...@googlegroups.com...
> >
> > Of course! the union of all natural numbers does not yield a complete
> > fixed set larger than all FISONs.
>
> There are an infinite number of natural numbers.

Of course. But there is no fixed quantity, namely a set that is the union of all FISONs and is larger than all FISONs.

>
> {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N

No that is blatantly wrong. A union of inclusion-monotonic sets cannot be larger than all sets. |N may be understood as the limit but not as the union. It is like 0.999... that is an infinite sequence convergi ng towards the limit 1.

>
> The union of all natural numbers also yields the set N:
>
> {0} U {1} U {2} U {3} U {4} U {5} U {6} U {7} U {8} U {9} U ... = N

That is the original mistake but not so easy to contradictc as the first one:

In {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N all FISONs can be omitted without changing the result, yielding U{ } = |N.

Regards, WM

[WM]

Am Freitag, 1. März 2019 20:33:51 UTC+1 schrieb Me:
> On Friday, March 1, 2019 at 8:18:19 PM UTC+1, FredJeffries wrote:
>
> > "Every totality of different numbers of the first and second number-classes
> > has a least number".
>
> Yeah! We (except Mückenheim, of course) have a word for this property:
>

But you do not understand what it means. Or have you learnt meanwhile that

∀n ∈ ℕ: U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} = U{A(n+1), A_(n+2), ...}.

Regards, WM

[FredJeffries]

On Thursday, February 28, 2019 at 12:15:42 PM UTC-8, FredJeffries wrote:
> On Thursday, February 28, 2019 at 10:12:22 AM UTC-8, WM wrote:
> > Am Donnerstag, 28. Februar 2019 16:11:28 UTC+1 schrieb FredJeffries:
> > > On Thursday, February 28, 2019 at 5:45:20 AM UTC-8, WM wrote:
> > >
> > > > It can also be proved that ℕ \ (F_1 U F_2 U F_3 U ... ) = Ø
> > >
> > > I call your bluff: PROVE IT
> >
> > I will do it, but first let me know whether you will aceept it.
>
> Why should that make any difference? According to you I am "brain-damaged".
>
> But I am not fool enough to say that I will "aceept[sic]" any alleged "proof" before I see it.
>
> > What is the union of all FISONs, F_1 U F_2 U F_3 U ..., in your opinion?
>
> Until I see a (convincing) proof or refutation, I have no "opinion" about ANY (reputed) mathematical statement.
>
> > REgards, WM

So, once again, we'll just consider your refusal to supply a "proof" to be an admission that you CANNOT do so and that your claims are complete nonsense. Thank you for clearing up THAT issue.

[K_h]

"WM" wrote in message
news:1b25cb2c-a0d0-4412...@googlegroups.com...

>
> Of course. But there is no fixed quantity, namely a set that is the
> union of all FISONs and is larger than all FISONs.

N is the set that is the union of all of its initial segments and N is
larger than any of its initial segments. To see this, just note that
CARD(N)=ALEPH_0 and card(n)=n for any natural number n and n<ALEPH_0 for any
natural number n.

> >
> > {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ...
> > = N
>
> No that is blatantly wrong. A union of inclusion-monotonic sets cannot
> be larger than all sets. |N may be understood as the limit but not as
> the union. It is like 0.999... that is an infinite sequence converging
> towards the limit 1.

{0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N

No, what I wrote is blatantly correct. It is a theorem of ZF that if L is a
limit ordinal then UL=L (the union of the sets in L equals L itself). A
proof of this can be found on page 210 of the textbook "Classic Set Theory"
by Derek Goldrei. In this example, think of it this way: pick any natural
number you want out of the set N (the right-hand side of the above equation)
and you can find the first set which includes it on the left-hand side of
the same above equation. In fact, once you have found the first set that
includes it then all subsequent sets in this ordered union also includes it.
It is true that no set on the left-hand side of the equation includes all
the members in N but there are an infinite number of such sets which is why
the equation is true. Note that each member of N can be found on the
left-hand side by simply looking at the last member in each set. Also note
that the below equation is true:

{0} U {1} U {2} U {3} U {4} U {5} U {6} U {7} U {8} U {9} U ... = N

because

{0} U {1} U {2} U {3} U {4} U {5} U {6} U {7} U {8} U {9} U ... =

{0,1,2,3,4,5,6,7,8,9,...}

and {0,1,2,3,4,5,6,7,8,9,...} = N.

> In {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... =
> N
> all FISONs can be omitted without changing the result, yielding U{ } = |N.

No. If each set in the union is omitted then no union is taking place at
all. U{}={} since there are no sets in the empty set and taking the union
of nothing yields nothing.

k

[j4n bur53]

But how would you translate it into ZFC. There
is not "Zahlen der ersten und zweiten Zahlenklasse"
directly in ZFC.

https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory#Axioms

Where do you see "Zahlen der ersten und zweiten
Zahlenklasse" in the axioms of ZFC. Could
you enlighten us?

[j4n bur53]

Maybe should open a GitHub repository for WMs PDF.
And then start some issue tracking.

Issue #1: Botched Infinity Axiom.

Ha Ha

Could do the same for bird brain John Garbageiel.
In his case we would have:

Issue #1: 3=<4 Invalid Doesn't Make Sense

[WM]

Am Sonntag, 3. März 2019 01:19:39 UTC+1 schrieb K_h:
> "WM" wrote in message

> > Of course. But there is no fixed quantity, namely a set that is the
> > union of all FISONs and is larger than all FISONs.
>
> N is the set that is the union of all of its initial segments and N is
> larger than any of its initial segments.

Therefore it cannot be the union.

> > No that is blatantly wrong. A union of inclusion-monotonic sets cannot
> > be larger than all sets. |N may be understood as the limit but not as
> > the union. It is like 0.999... that is an infinite sequence converging
> > towards the limit 1.
>
> {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N
>
> No, what I wrote is blatantly correct.

It yields the contradiction that also when subtracting FISONs from the set {{0}, {0,1}, {0,1,2}, {0,1,2,3} ...} under the premise that the union remains unchanged, no FISON remains. Therefore either |N is more than the union or |N is nothing, i.e., the empty set.

> It is a theorem of ZF that if L is a
> limit ordinal then UL=L (the union of the sets in L equals L itself).

That theorem is wrong.

Look: Uneducated mathematicians believe that the limit of a sequence is always a term of the sequence.

Educated mathematicians understand that the limit of a series is not one of the partial sums (although they often are not aware of that fact without being told).

Even very well educated mathematicians do not understand that the infinite union of inclusion monotonic sets cannot be a fixed set larger than each of the unioned sets. But it is fact. The infinite union of FISONs has the limit |N, but it is not |N.

Analogy: The sequence 1, 1/2, 1/3, ..., if finitely defined by (1/n), has the limit 0. This limit exists by the finite definition and does not change when you delete terms of the sequence, even when you delete all.

> A
> proof of this can be found on page 210 of the textbook "Classic Set Theory"
> by Derek Goldrei.

It cannot be proved, since it is wrong.

> In this example, think of it this way: pick any natural
> number you want out of the set N

That is already the first mistake. You cannot pick "any" number. Each one that you can pick belongs to a tiny finite initial segment which is followed by infinitely many natural numbers. Try to refute me.

> (the right-hand side of the above equation)
> and you can find the first set which includes it on the left-hand side of
> the same above equation.

Ex falso quodlibet. Before going on show me that you can pick a natural number that has more predecessors than successors.

> > In {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... =
> > N
> > all FISONs can be omitted without changing the result, yielding U{ } = |N.
>
> No. If each set in the union is omitted then no union is taking place at
> all.

Obviously that is not required if |N is not that union.

> U{}={} since there are no sets in the empty set and taking the union
> of nothing yields nothing.

Every set of ordinals has a first element. The set of all FISONs is not required. If there are FISONs required, then this required subset has a first element. That, by the way, is basic for set theory.

REgards, WM

[WM]

Am Sonntag, 3. März 2019 03:17:45 UTC+1 schrieb j4n bur53:
> But how would you translate it into ZFC. There
> is not "Zahlen der ersten und zweiten Zahlenklasse"
> directly in ZFC.

It is impossible to translate reasoning about contracictions into ZFC since ZFC is the language of Zero Findable Contradictions and is excluding anything of mathematical value.

Regards, WM

[Python]

Herr Mueckenheim you've just admitted that your sophistry need
equivocation and ZFC is precise enough to prevent your equivocation.

[Me]

On Sunday, March 3, 2019 at 2:00:42 PM UTC+1, WM wrote:
> Am Sonntag, 3. März 2019 01:19:39 UTC+1 schrieb K_h:

> > N is [...] the union of all of its initial segments and N is

> > larger than any of its initial segments.
> >
> Therefore it cannot be the union.

"Therefore" for you obviously means: "be aware of the following nonsense".

Hint:

> > {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N
> >

> It yields the contradiction that <bla>

No, it doesn't "yield a contradiction". At least none that is (presently) known.

> > It is a theorem of ZF that if L is a
> > limit ordinal then UL=L (the union of the sets in L equals L itself).
> >
> That theorem is wrong.

Huh?!

What do you mean by "wrong"?

"I don't like it!" Or: "I can't prove it in Mückenmatic!"? Or what?

Hint: It holds in the context of SET THEORY, you know.

> The sequence 1, 1/2, 1/3, ... [...] has the limit 0.
> This limit [...] does not change when you delete terms of the sequence,

> even when you delete all.

Fascinating! Do you teach this type of stuff?

> > A proof of this can be found on page 210 of the textbook "Classic Set
> > Theory" by Derek Goldrei.
>

> It cannot be proved, since <bla>

Can't you read, or what's the matter with you, imbecile?

*** A proof of this can be found on page 210 of the textbook "Classic Set Theory" by Derek Goldrei ***

> show me that you can pick a natural number that has more predecessors than
> successors.

There is no such number, your requirement cannot be satisfied by any natural number. Hence he can't perfom that feat. :-)

> > If each set in the union is omitted then no union is taking place at
> > all.

Right:. U{} = {}.

> Obviously that is <bla>

Shut up, idiot!

> Every set of ordinals has a first element.

No, idiot: Every NON EMPTY set of ordinals has a first element.

[WM]

Am Sonntag, 3. März 2019 14:38:35 UTC+1 schrieb Me:
> On Sunday, March 3, 2019 at 2:00:42 PM UTC+1, WM wrote:


> > > {0} U {0,1} U {0,1,2} U {0,1,2,3} U {0,1,2,3,4} U {0,1,2,3,4,5} U ... = N
> > >
>

> No, it doesn't "yield a contradiction". At least none that is (presently) known.

What about the different results of
U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} =/= |N
and
U{A(n+1), A_(n+2), ...} =/= |N
?

>
> > > It is a theorem of ZF that if L is a
> > > limit ordinal then UL=L (the union of the sets in L equals L itself).
> > >
> > That theorem is wrong.
>

> What do you mean by "wrong".

Its proof is based on a fallacy.

>

> > The sequence 1, 1/2, 1/3, ... [...] has the limit 0.
> > This limit [...] does not change when you delete terms of the sequence,
> > even when you delete all.
>
> Fascinating! Do you teach this type of stuff?

Is that necessary? I never met a person who doubted that.

>
> > > A proof of this can be found on page 210 of the textbook "Classic Set
> > > Theory" by Derek Goldrei.
> >

> > It cannot be proved.

>
> *** A proof of this can be found on page 210 of the textbook "Classic Set Theory" by Derek Goldrei ***

It may be sufficient to convince stupids who do not understand that it is impossible to choose every natural number.

>
> > show me that you can pick a natural number that has more predecessors than
> > successors.
>
> There is no such number, your requirement cannot be satisfied by any natural number. Hence he can't perfom that feat.

Why then do matheologians boastfully claim they could choose every natural number?

>
> > Every set of ordinals has a first element.
>

> No: Every NON EMPTY set of ordinals has a first element.

That shows that the set of FISONs U{A(n+1), A_(n+2), ...} that is required to yield the union |N is empty.

Regards, WM

[WM]

Is ZFC expressing and acknowledging the fact that the universal quantifier concerns only natural numbers that have more successors than predecessors, i.e., almost none?

Regards, WM

[j4n bur53]

Nope, in ZFC, the forall ranges over sets from
the domain. Read Zermelos 1908 paper to see
what a set theory domain means.

[j4n bur53]

Can you tell us in a video what you mean?

When will there be the first video, live
from Augsburg Crank institute? Explaining

us Volkswagen Omlette? Maybe a video in
the quality of this, much better than

bird brain John Gabriels videos?

Was ist ein Punkt? (Teil 1)
https://www.youtube.com/watch?v=lGtL8Vdw5tM

On Wednesday, February 27, 2019 at 7:44:10 PM UTC+1, WM wrote:
> It is generally accepted that the union of the set of FISONs is
>
> {1} U {1, 2} U {1, 2, 3} U ... = ℕ
> or briefly
> F_1 U F_2 U F_3 U ... = ℕ.
>
> The first n FISONs can be omitted
>
> F_n U F_(n+1) U F_(n+2) U ... = ℕ .
>
> Since this is true for all n, all F_n can be omitted and we find that no FISONs are necessary but
>
> U{ } = ℕ.
>
> This result makes the mathematical reality of the transfinite set ℕ quite doubtful. Therefore it is not acceptable for adherents of transfinite set theory.
>
> But the following accepted definition of being required
>
> F_n is required <==> U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= ℕ
>
> shows that, according to this definition, the set of required FISONs is empty too.
>
> This is a beautiful result because it forces adherents of transfinite set theory to believe that re-inserting the FISONs F_1, F_2, ..., F_(n-1) into the union U{F(n+1), F_(n+2), ...} is of any significance, such that for some n:
>
> U{F_1, F_2, ..., F_(n-1), F(n+1), F_(n+2), ...} =/= U{F(n+1), F_(n+2), ...}
>
> We do not want to spoil their belief. But students who are not yet closely connected with set theory will perhaps give it some thought whether to get in closer touch with this persuasion.
>
> Regards, WM

[j4n bur53]

Well it would be better, if Prof. Bedürftig
would talk a little faster...

[Python]

Stop your idiotic babbling Muckenheim. Universal quantifier on all
naturals concerns all naturals -period.

All natural have more successors (an infinity) than predecessors (a
finite quantity), so it is not "almost none" but the contrary. Moreover,
as ALL of your sophistry, is is IRRELEVANT.

[WM]

Am Sonntag, 3. März 2019 15:39:18 UTC+1 schrieb j4n bur53:
> Nope, in ZFC, the forall ranges over sets from
> the domain.

Look, this shows that Matheologians are even less intelligent than the least intelligent.

Note: Everybody, even if not very intelligent realizes, that you can only uwe natural numbers wnich have less predecessors than successors.

And even the less intelligent will understand that after 10 or 100 or 1000 trials, according to their level of intelligence.

But set theorists will not even understand that after an uncountable sequence of flops and claim they could chose all natural numbers.

It must be a miserable feeling to belong to this sect of fools. You have my sympathy.

Regards, WM

[Python]

Idiotic crank Mueckenheim wrote:
> Am Sonntag, 3. März 2019 15:39:18 UTC+1 schrieb j4n bur53:
>> Nope, in ZFC, the forall ranges over sets from
>> the domain.
>
> Look, this shows that Matheologians are even less intelligent than the least intelligent.
>
> Note: Everybody, even if not very intelligent realizes, that you can only uwe natural numbers wnich have less predecessors than successors.

Does it depends of the time of the day, crank Muckenheim? At 15:15
Central European Time you stated otherwise:

"... natural numbers that have more successors than predecessors,
i.e., almost none?"

Maybe it is dependant of the level of alcohol in your blood, Herr Crank?

[WM]

Am Sonntag, 3. März 2019 19:44:21 UTC+1 schrieb Python:

> > Is ZFC expressing and acknowledging the fact that the universal quantifier concerns only natural numbers that have more successors than predecessors, i.e., almost none?
>
> Stop your idiotic babbling Muckenheim. Universal quantifier on all
> naturals concerns all naturals -period.

Chuckle, probably you will never understand. But every person of medium intelligence can tell you that it is impossible to choose a natural number that has more predecessors than successors.

>
> All natural have more successors (an infinity) than predecessors (a
> finite quantity)

And are these successors natural numbers like the predecessors?

> Moreover,
> as ALL of your sophistry, is is IRRELEVANT.

Every person of medium intelligence knows that there are inaccessible numbers if in every case the chosen number has more successors than predecessors. It must be a hard piece of work and willpower to condition the own brain to the contrary unless it is stupid by nature.

Regards, WM

[Python]

Über-Crank Mueckenheim wrote:
> Am Sonntag, 3. März 2019 19:44:21 UTC+1 schrieb Python:
>
>>> Is ZFC expressing and acknowledging the fact that the universal quantifier concerns only natural numbers that have more successors than predecessors, i.e., almost none?
>>
>> Stop your idiotic babbling Muckenheim. Universal quantifier on all
>> naturals concerns all naturals -period.
>
> Chuckle, probably you will never understand. But every person of medium intelligence can tell you that it is impossible to choose a natural number that has more predecessors than successors.

You are the one who pretented that there is "almost none" "natural
numbers that have more successors that predecessors", just read your
own post above... Now after a few more drinks you've changed your mind.
Nice. But it is still irrelevant by the way.

>> All natural have more successors (an infinity) than predecessors (a
>> finite quantity)
>
> And are these successors natural numbers like the predecessors?
>
>> Moreover,
>> as ALL of your sophistry, is is IRRELEVANT.
>
> Every person of medium intelligence knows that there are inaccessible numbers if in every case the chosen number has more successors than predecessors. It must be a hard piece of work and willpower to condition the own brain to the contrary unless it is stupid by nature.

You are high on drugs, Herr Crank Mueckenheim. Sad.

[Me]

On Sunday, March 3, 2019 at 3:15:58 PM UTC+1, WM wrote:

> [Each and every] natural number [has] more successors than predecessors

Indeed. Hint: Each and every natural number has INFINITELY many successors, but only FINITELY MANY predecessors.

In ZFC we can prove:

An e IN: card({0, ..., n-1}) = n
and
An e IN: card({n+1, n+2, ...}) = aleph_0 .

Moreover we can prove

An e IN: n < aleph_0 .

> i.e. <nonsense deleted>

[Me]

On Sunday, March 3, 2019 at 3:09:08 PM UTC+1, WM wrote:

> What's about the different results of

>
> U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} =/= |N
>
> and
>
> U{A(n+1), A_(n+2), ...} =/= |N

Huh? Two false statments. So what?

Hint: An e IN: U{A_1, A_2, ..., A_(n-1), A(n+1), A_(n+2), ...} = |N
and: An e IN: U{A(n+1), A_(n+2), ...} = |N .

> > > The sequence 1, 1/2, 1/3, ... [...] has the limit 0.
> > > This limit [...] does not change when you delete terms of the sequence,

> > > even when you delete all. (Mückenheim)

> > >
> > Fascinating! Do you teach this type of stuff?
> >

> [...] I never met a person who doubted that.

Oh really? Fascinating!

> > > Every set of ordinals has a first element. (Mückenheim)

> > >
> > No: Every NON EMPTY set of ordinals has a first element.

Got it?

[Me]

On Sunday, March 3, 2019 at 8:45:04 PM UTC+1, WM wrote:

> it is impossible to choose a natural number that has more predecessors than
> successors.

Indeed. Since each and every number has more successors than predecessors.

Python:

> > All natural have more successors (an infinity) than predecessors (a
> > finite quantity)
> >
> And are these successors natural numbers like the predecessors?

Of course, idiot. In IN this is indeed the case.