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]

Post by 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.
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

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

[WM]

Post by 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

[Zeit Geist]

Post by WM

Post by 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.

That does not follow. You are so sad.

Post by WM
Regards, WM

ZG

[Jew Lover]

Post by Zeit Geist

Post by WM

Post by 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.

That does not follow.

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

Post by Zeit Geist
You are so sad.

If WM is sad, the you are truly tragic.

Post by Zeit Geist

Post by WM
Regards, WM

ZG

[Zelos Malum]

Post by Jew Lover
In other words, there is no infinite set. Period.

A fallacious arguement does nto invalidate infinite sets.

Post by Jew Lover
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.

Post by Jew Lover
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]

Post by Zelos Malum

Post by Jew Lover
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

Post by Jew Lover

Post by Zelos Malum

Post by Jew Lover
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]

Post by j4n bur53
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...

Post by 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/
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.
p q ((p ∧ q) ∧ ((p → ¬q) ∧ (¬q → p)))
F F F
F T F
T F F
T T F

Post by Jew Lover

Post by Zelos Malum

Post by Jew Lover
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]

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.

Post by j4n bur53

Post by j4n bur53
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...

Post by 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/
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.
p q ((p ∧ q) ∧ ((p → ¬q) ∧ (¬q → p)))
F F F
F T F
T F F
T T F

Post by Jew Lover

Post by Zelos Malum

Post by Jew Lover
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.

[Jew Lover]

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

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

[Zelos Malum]

Post by Jew Lover
A fallacious belief does not invalidate logic

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

Post by Jew Lover
Unfortunately, you never even get to the argument part because you have nothing but beliefs

Got nothing of the sort.

Post by Jew Lover
(ZFC crap axioms) and meaningless decrees.

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

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

Neither will you.

Post by Jew Lover
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.

[WM]

Post by 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

[Jew Lover]

Post by WM

Post by 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.

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

Post by WM
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.

Post by 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.
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]

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.

Post by j4n bur53
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
Q(n) <=> N \ n <> {}
forall n e N Q(n)
~Q(N)
Its that simple. Don't apply fallacies if you
don't want to look idiotic.

Post by 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.
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

[WM]

Post by j4n bur53
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?

Post by WM

Post by j4n bur53
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

[Alan Smaill]

Post by WM

Post by j4n bur53
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.

Post by WM
Regards, WM

[WM]

Post by Alan Smaill

Post by WM

Post by j4n bur53
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?

Regards, WM

[Alan Smaill]

Post by WM

Post by Alan Smaill

Post by WM

Post by j4n bur53
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.
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?

Post by WM
Regards, WM

[WM]

Post by Alan Smaill

Post by WM

Post by Alan Smaill

Post by WM

Post by j4n bur53
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.

Post by Alan Smaill
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)])

Post by Alan Smaill
You claim that for an arbitrary (let's say definable in ZF(C)) predicate
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

[j4n bur53]

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

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

Post by WM

Post by Alan Smaill
You claim that for an arbitrary (let's say definable in ZF(C)) predicate
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.

[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?

Post by j4n bur53
In an inconsistent theory, it would hold always you moron.
https://en.wikipedia.org/wiki/Principle_of_explosion

Post by WM

Post by Alan Smaill
You claim that for an arbitrary (let's say definable in ZF(C)) predicate
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.

[Me]

Post by WM
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.

Post by WM
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

[Jew Lover]

Post by Me

Post by WM
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]

Post by Me

Post by WM
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.

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

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.

[Zelos Malum]

Post by WM
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.

Post by WM
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.

Post by WM
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.

Post by WM
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]

Post by Zelos Malum

Post by WM
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.

Post by Zelos Malum
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.

Post by Zelos Malum

Post by WM
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?

Post by Zelos Malum

Post by WM
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?

Post by Zelos Malum
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.

Post by Zelos Malum

Post by WM
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]

Post by WM

Post by Zelos Malum

Post by WM
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.

Post by WM
Regards, WM

[Jew Lover]

Post by Alan Smaill

Post by WM

Post by Zelos Malum

Post by WM
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.

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?

Post by Alan Smaill

Post by WM
Regards, WM

--
Alan Smaill

[jvr]

[...]

Post by WM
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]

Post by jvr
[...]

Post by WM
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)]

Post by jvr
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.

Post by jvr
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

[jvr]

Post by WM

Post by jvr
[...]

Post by WM
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,

"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)]

Post by jvr
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.

Post by jvr
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

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.

[WM]

Post by jvr

Post by WM

Post by jvr
[...]

Post by WM
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,

"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)]

Post by jvr
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.

Post by jvr
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.

Post by jvr
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.

Post by jvr
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.

Post by WM

Post by jvr

Post by WM

Post by jvr
[...]

Post by WM
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,

"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)]

Post by jvr
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.

Post by jvr
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.

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

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.

Post by jvr
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]

Please note that the Cantor-Bernstein-Schroeder

theorem deals with cardinality, and the theorem

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

Post by 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.

Post by WM

Post by jvr

Post by WM

Post by jvr
[...]

Post by WM
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,

"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)]

Post by jvr
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.

Post by jvr
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.

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

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.

Post by jvr
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

[Jew Lover]

Post by 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.

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

Post by j4n bur53

Post by WM

Post by jvr

Post by WM

Post by jvr
[...]

Post by WM
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,

"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)]

Post by jvr
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.

Post by jvr
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.

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

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.

Post by jvr
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

[WM]

Post by 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]

Post by j4n bur53
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]

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)?

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

Post by WM

Post by 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]

Post by j4n bur53
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

[FredJeffries]

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]

Post by FredJeffries
"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

[WM]

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

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

[WM]

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

[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?

Post by WM

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

[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

Post by 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?

Post by WM

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]

Post by 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]

Post by WM

Post by 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.

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

[WM]

Post by Python

Post by WM

Post by 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.

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

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.

Post by WM

Post by Python

Post by WM

Post by 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.

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

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

[WM]

Post by 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]

Post by WM

Post by 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?

[j4n bur53]

Well its "set theory" and not "black pudding and
liversausage theory". What you you expect that

the domain of discourse consists of? Volkswagen
Omlette from Augsburg Crank institute?

Post by WM

Post by 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]

Post by WM

Post by Python

Post by WM

Post by 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.

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

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.

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]

Post by Python

Post by 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?

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.

Post by Python
All natural have more successors (an infinity) than predecessors (a
finite quantity)

And are these successors natural numbers like the predecessors?

Post by Python
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]

Post by WM

Post by Python

Post by 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?

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.

Post by WM

Post by Python
All natural have more successors (an infinity) than predecessors (a
finite quantity)

And are these successors natural numbers like the predecessors?

Post by Python
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]

Post by WM
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.

Post by WM

Post by 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.

[Me]

[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>

[jvr]

[...]

Post by WM
"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"?

Post by WM
"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.

[WM]

Post by jvr
[...]

Post by WM
"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.

Post by jvr
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.

Post by jvr

Post by WM
"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]

Post by WM

Post by jvr
[...]

Post by WM
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,

"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.

Post by WM
"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)]

Post by jvr
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.

Post by jvr
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

[Me]

Post by 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.

Hit: It's accepted because ut can be PROVED in the context of "set threory" (ay ZFC).

Post by WM
The first n FISONs can be omitted
F_(n+1) U F_(n+2) U ... = ℕ .

Right. Since you are dumb like shit, we better write this in symbols:

U_{i e IN\{1,...,n}} F_i = IN

Post by WM
Since this is true for all n, all F_n can be omitted and ...

No, idiot.

Hint: Non sequitur.

[Me]

Post by 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.

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

Post by WM
The first n FISONs can be omitted
F_(n+1) U F_(n+2) U ... = ℕ .

Right.

Post by WM
Since this is true for all n, all F_n can be omitted and ...

No, idiot.

Hint: Non sequitur.

[WM]

Post by Me

Post by 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.

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

Post by Me

Post by WM
The first n FISONs can be omitted
F_(n+1) U F_(n+2) U ... = ℕ .

Right.

Post by WM
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

[FredJeffries]

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

I call your bluff: PROVE IT

[Zeit Geist]

Post by FredJeffries

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

I call your bluff: PROVE IT

Proof? What a proof? How do I start?

ZG

[Jew Lover]

Post by FredJeffries

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

I call your bluff: PROVE IT

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.

Post by Jew Lover

Post by FredJeffries

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

I call your bluff: PROVE IT

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.

[Jew Lover]

Post by j4n bur53
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>

Post by j4n bur53
All these cranks are not able to produce proofs.

Post by Jew Lover

Post by FredJeffries

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

I call your bluff: PROVE IT

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]

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?

Post by Jew Lover

Post by j4n bur53
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>

Post by j4n bur53
All these cranks are not able to produce proofs.

Post by Jew Lover

Post by FredJeffries

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

I call your bluff: PROVE IT

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]

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?

Post by j4n bur53
Its still only an analogy, since the limit in
real numbers, does not behave as the infinite
union in sets.
0.111... = 1.000...
{1,2,3,...} <> {0}
Got it?

Post by Jew Lover

Post by j4n bur53
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>

Post by j4n bur53
All these cranks are not able to produce proofs.

Post by Jew Lover

Post by FredJeffries

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

I call your bluff: PROVE IT

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]

Its only a soft analogy.

Post by j4n bur53
The following function from sets to real numbers,
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.
https://en.wikipedia.org/wiki/Injective_function
What is your analogy moron?

Post by j4n bur53
Its still only an analogy, since the limit in
real numbers, does not behave as the infinite
union in sets.
0.111... = 1.000...
{1,2,3,...} <> {0}
Got it?

Post by Jew Lover

Post by j4n bur53
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>

Post by j4n bur53
All these cranks are not able to produce proofs.

Post by Jew Lover

Post by FredJeffries

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

I call your bluff: PROVE IT

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.

[Jew Lover]

Post by j4n bur53
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.

Post by j4n bur53
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!

Post by j4n bur53
0.111... = 1.000...

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

[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

Post by j4n bur53
0.111... = 1.000...