Are "dark numbers" simply in the shadow ?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

these are "undiscovered numbers" unseen by man.

(what about aliens? bet they havent "seen" most of them either, so an
infinity of "undiscovered" and/or "unseen numbers" will always remain.)

Ganzhinterseher

2020-02-07 08:08:27 UTC

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

these are "undiscovered numbers" unseen by man.
(what about aliens? bet they havent "seen" most of them either, so an
infinity of "undiscovered" and/or "unseen numbers" will always remain.)

Some presently undefined or unseen numbers can become defined and seen but infinitely many, precisely: aleph_0, oo them will forever remain undefined, i.e., dark because of this fact:

Every endsegment in

∩{E(1), E(2), E(3), ...} = { } (*)

can eject only one natnumber because of

∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

But for all definable (i.e., not dark) endsegments we have

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo .

If (*) is true, then also finite sets must have been passed between the infinite endsegments and the empty set.

Compare a long staircase. If the bottom is resached from above, then also the first stairs have been passed. This is a finite argument for a finite set and cannot be circumvented by infinity-blather.

But finite sets are not passed by definable endsegments.

Either there are dark natural numbers or there is no finished infinity and the set in (*) is never completely empty but only empty "up to every finite k".

Regards, WM

Chris M. Thomasson

2020-02-07 08:26:26 UTC

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

these are "undiscovered numbers" unseen by man.
(what about aliens? bet they havent "seen" most of them either, so an
infinity of "undiscovered" and/or "unseen numbers" will always remain.)

Every endsegment in
∩{E(1), E(2), E(3), ...} = { } (*)
can eject only one natnumber because of
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
But for all definable (i.e., not dark) endsegments we have
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo .
If (*) is true, then also finite sets must have been passed between the infinite endsegments and the empty set.
Compare a long staircase. If the bottom is resached from above, then also the first stairs have been passed. This is a finite argument for a finite set and cannot be circumvented by infinity-blather.
But finite sets are not passed by definable endsegments.
Either there are dark natural numbers or there is no finished infinity and the set in (*) is never completely empty but only empty "up to every finite k".

There is no finished infinity, that's an oxymoron. Just like there is no
largest integer.

Chris M. Thomasson

2020-02-07 08:29:54 UTC

[...]

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished infinity
and the set in (*) is never completely empty but only empty "up to
every finite k".

There is no finished infinity, that's an oxymoron. Just like there is no
largest integer.

Think of looking at a fractal from afar. Some of them have defined limit
shapes. However, this is not finished. When we zoom in we can see more
and more detail that was totally invisible to us before the zoom.

Dark numbers can be akin to unseen features in a fractal zoom.

Ganzhinterseher

2020-02-07 09:17:53 UTC

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished infinity and the set in (*) is never completely empty but only empty "up to every finite k".

There is no finished infinity, that's an oxymoron. Just like there is no
largest integer.

Of course not. But an estimated 100000 of mathematicians and logicians believe there is "all finite numbers" because the cloudy "infinite" is not easy to refute. But they have not understood yet that the precise complement of "all" is the empty set, which is not cloudy and not dependent on infinite mathematics but clearly reached by finite mathematics in a stepwise process

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

and not by cloudy delusions.

Regards, WM

Ross A. Finlayson

2020-02-15 19:38:25 UTC

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

these are "undiscovered numbers" unseen by man.
(what about aliens? bet they havent "seen" most of them either, so an
infinity of "undiscovered" and/or "unseen numbers" will always remain.)

There is no finished infinity, that's an oxymoron. Just like there is no
largest integer.

Must disagree, that "there's no finished infinity",
though of course "the unbounded" is not finished by
bounding.

Stephan Koerner wrote a book "The Philosophy of Mathematics,
An Introduction". It really picks up after LEJ Brouwer,
who coined intuitionism after constructivism (and it's a
constructivism, and for logicism).

"In speaking of a real number the classical analyst is
committed to the assumption that it is 'possible' to
pick out a subclass from the _actual_ totality of all
natural numbers. In speaking of all real numbers, he
is not only committed to assuming the actual totality
of all natural numbers but also the _greater actual_
infinite totality of all subclasses of this class.
The assumption of such totalities implied in speaking
of a real number, or even of all real numbers,
transcends the finite point of view and the
employment of finite methods."

Then, for integers and their properties,
and their resulting space and all the
properties of the space, has that integers
are as neatly point-like and infinitesimal
parts of the entire space, as they are wholes
to the infinitesimal parts of a number between
zero and one (all of them).

Ross Finlayson

2024-07-28 15:55:07 UTC

Post by Ross A. Finlayson

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

these are "undiscovered numbers" unseen by man.
(what about aliens? bet they havent "seen" most of them either, so an
infinity of "undiscovered" and/or "unseen numbers" will always remain.)

There is no finished infinity, that's an oxymoron. Just like there is no
largest integer.

Must disagree, that "there's no finished infinity",
though of course "the unbounded" is not finished by
bounding.
Stephan Koerner wrote a book "The Philosophy of Mathematics,
An Introduction". It really picks up after LEJ Brouwer,
who coined intuitionism after constructivism (and it's a
constructivism, and for logicism).
"In speaking of a real number the classical analyst is
committed to the assumption that it is 'possible' to
pick out a subclass from the _actual_ totality of all
natural numbers. In speaking of all real numbers, he
is not only committed to assuming the actual totality
of all natural numbers but also the _greater actual_
infinite totality of all subclasses of this class.
The assumption of such totalities implied in speaking
of a real number, or even of all real numbers,
transcends the finite point of view and the
employment of finite methods."
Then, for integers and their properties,
and their resulting space and all the
properties of the space, has that integers
are as neatly point-like and infinitesimal
parts of the entire space, as they are wholes
to the infinitesimal parts of a number between
zero and one (all of them).

It's not so much that dark numbers are "in" the shadow,
as, dark numbers "are" the shadow, the shadow light of
what must be a "real infinity".

Zelos Malum

2020-02-10 07:29:52 UTC

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished infinity and the set in (*) is never completely empty but only empty "up to every finite k".

Or here is a much more reasonable suggestion, you do not fucking know how to argue with logic.

Ganzhinterseher

2020-02-10 09:26:46 UTC

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished infinity and the set in

∩{E(1), E(2), E(3), ...} = { } (*)

is never completely empty but only empty "up to every finite k".

Post by Zelos Malum
Or here is a much more reasonable suggestion, you do not fucking know how to argue with logic.

We need no sufggestions but mathematics. The simplest poof of your claim would be to find a counter example for

∀k ∈ ℕ_def: |∩{E(1), E(2), ..., E(k)}| = |∩{E(k) | k ∈ ℕ_def}|

There is not much logic required. A counter example shows that a theorem is false.

Why can't you find any?

Regards, WM

me2020

2020-02-10 09:50:11 UTC

INTER{E(1), E(2), E(3), ...} = {}
is never completely empty but

Huh?! {} is "never completely empty" in your psychotic version of set theory?

Ak e IN_def: |INTER{E(1), E(2), ..., E(k)}| = |INTER{E(k) | k e IN_def}|

Look, dumbo. If E(k) := {n e IN : n >= k} for all k e IN,and IN_def c IN, then (for any k_0 e IN)

|INTER{E(1), E(2), ..., E(k_0)}| = |INTER{E(k) | k e IN_def}|

implies that IN_def is finite. On the other hand, we have for all k e IN:

|INTER{E(1), E(2), ..., E(k)}| =/= |INTER{E(k) | k e IN}| ,
since
Ak e IN: |INTER{E(1), E(2), ..., E(k)}| = aleph_0 ,

|INTER{E(k) | k e IN}| = 0 ,
and
aleph_0 =/= 0 .

What's the matter with you, you psychotic asshole full of shit?

Sergio

2020-02-11 04:49:40 UTC

Post by me2020

INTER{E(1), E(2), E(3), ...} = {}
is never completely empty but

Huh?! {} is "never completely empty" in your psychotic version of set theory?

Ak e IN_def: |INTER{E(1), E(2), ..., E(k)}| = |INTER{E(k) | k e IN_def}|

Look, dumbo. If E(k) := {n e IN : n >= k} for all k e IN,and IN_def c IN, then (for any k_0 e IN)
|INTER{E(1), E(2), ..., E(k_0)}| = |INTER{E(k) | k e IN_def}|
|INTER{E(1), E(2), ..., E(k)}| =/= |INTER{E(k) | k e IN}| ,
since
Ak e IN: |INTER{E(1), E(2), ..., E(k)}| = aleph_0 ,
|INTER{E(k) | k e IN}| = 0 ,
and
aleph_0 =/= 0 .
What's the matter with you, you psychotic asshole full of shit?

his numbers are {VantaBlack}

Ganzhinterseher

2020-02-11 13:23:59 UTC

Post by me2020

INTER{E(1), E(2), E(3), ...} = {}
is never completely empty but

Huh?! {} is "never completely empty"

No. INTER{E(1), E(2), E(3), ...} is never empty.

Look, there are two statements: Every natnumber k is removed by endsegment E(k+1). Every natnumber is succeeded by infinitely many.

Every element is removed but the set does not get empty. This is not a contradiction. Don't trust your intuition.

Regards, WM

Zelos Malum

2020-02-12 06:41:27 UTC

Post by Ganzhinterseher
No. INTER{E(1), E(2), E(3), ...} is never empty.

Are you fucking stupid? That is emtpy because no element is in all the sets!

Post by Ganzhinterseher
You are wrong. I do not use E(|N).

I didn't say that, you are however using trying to remove ALL of them, which is |N

Post by Ganzhinterseher
You are wrong. I do not use E(|N).

False

Post by Ganzhinterseher
The set of definable endsegments is not actually infinite. Find any definable natnumber that make the equation wrong

You have already admitted |N_def = |N

Ganzhinterseher

2020-02-12 16:57:42 UTC

Post by Ganzhinterseher
You are wrong. I do not use E(|N).

I didn't say that, you are however using trying to remove ALL of them, which is |N

No, there is a vast difference in set theory between all elements and the set.
All elements are finite, all FISONs are finite, but the set is infinite.

Post by Ganzhinterseher
You are wrong. I do not use E(|N).

False

If you can't distinguish the set from all its elements, then go to school again.

Post by Ganzhinterseher
The set of definable endsegments is not actually infinite. Find any definable natnumber that make the equation wrong

You have already admitted |N_def = |N

Never.
By definition we have ∩{E(k) | k ∈ ℕ_def} =/= 0
In set theory we have ∩{E(k) | k ∈ ℕ} = 0
Impossible for same sets.

Regards, WM

FromTheRafters

2020-02-12 17:39:40 UTC

Post by Ganzhinterseher
You are wrong. I do not use E(|N).

I didn't say that, you are however using trying to remove ALL of them, which is |N

No, there is a vast difference in set theory between all elements and the set.

If that is so (it isn't btw) then how would you use the set B as a
subset of set A and the set A also as a subset of set B to prove set A
is "equal to" set B?

It doesn't work with "Proper Subsets" which always leave an element
behind when building such a subset.

Ganzhinterseher

2020-02-13 11:17:09 UTC

Post by Ganzhinterseher
No, there is a vast difference in set theory between all elements and the set.

If that is so (it isn't btw) then how would you use the set B as a
subset of set A and the set A also as a subset of set B to prove set A
is "equal to" set B?

For infinite sets this cannot be proved by lists of elements (extensionally) but only by formulas (intensionally).

Post by FromTheRafters
It doesn't work with "Proper Subsets" which always leave an element
behind when building such a subset.

Consider all elements of ℕ. They differ from ℕ in being finite wheras ℕ is infinite. This is also true for all FISONs:

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

None is ℕ and they all are not ℕ. Proof by inclusion monotony.

Regards, WM

FromTheRafters

2020-02-13 11:59:58 UTC

Post by Ganzhinterseher
No, there is a vast difference in set theory between all elements and the set.

If that is so (it isn't btw) then how would you use the set B as a
subset of set A and the set A also as a subset of set B to prove set A
is "equal to" set B?

For infinite sets this cannot be proved by lists of elements (extensionally)
but only by formulas (intensionally).

So what? The only difference between the set and all of its elements is
that the set also contains the null set as a subset, hardly a vast
difference.

Post by FromTheRafters
It doesn't work with "Proper Subsets" which always leave an element
behind when building such a subset.

Consider all elements of ℕ. They differ from ℕ in being finite wheras ℕ is
{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
...
None is ℕ and they all are not ℕ. Proof by inclusion monotony.

So, an infinite collection (set) of finite objects (elements) no
problem. Can you also have finite collections of infinite objects in
your world?

Ganzhinterseher

2020-02-13 14:46:44 UTC

Post by Ganzhinterseher
No, there is a vast difference in set theory between all elements and the set.

If that is so (it isn't btw) then how would you use the set B as a
subset of set A and the set A also as a subset of set B to prove set A
is "equal to" set B?

For infinite sets this cannot be proved by lists of elements (extensionally)
but only by formulas (intensionally).

So what? The only difference between the set and all of its elements is
that the set also contains the null set as a subset, hardly a vast
difference.

No, the difference is that the elements are infinitely many but the set is a unit.

Post by FromTheRafters
It doesn't work with "Proper Subsets" which always leave an element
behind when building such a subset.

So, an infinite collection (set) of finite objects (elements) no
problem.

The set is a unit but is lager than all FISONs. We can subtract every FISON from ℕ - but an infinite set remains:

∀n ∈ ℕ: ℕ \ {1, 2, 3, ..., n} =/= { }.

Regards, WM

FromTheRafters

2020-02-13 15:50:38 UTC

Post by Ganzhinterseher
No, there is a vast difference in set theory between all elements and the set.

If that is so (it isn't btw) then how would you use the set B as a
subset of set A and the set A also as a subset of set B to prove set A
is "equal to" set B?

For infinite sets this cannot be proved by lists of elements
(extensionally) but only by formulas (intensionally).

So what? The only difference between the set and all of its elements is
that the set also contains the null set as a subset, hardly a vast
difference.

No, the difference is that the elements are infinitely many but the set is a unit.

A unit with a cardinality equal to the cardinality of the set I can
build from all of the elements in the aforementioned set.

Again, so what?

[...]

Ganzhinterseher

2020-02-15 11:04:03 UTC

Post by FromTheRafters
A unit with a cardinality equal to the cardinality of the set I can
build from all of the elements in the aforementioned set.

No, it cannot. All definable natnumbers only build a potentailly infinite set. This can be seen in the clearest way by the chain of endsegments. Every definable endsegment is a last E(k) in

∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

and fails to get an empty intersection. (You cannot define an endsegment that fils this test, can you?) But the larger, infinite set of endsegments in

∩{E(k) | k ∈ ℕ } = { }

cannot be constructed because then we would get a contradiction from

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

because then the last steps before the empty set would be observable.

Regards, WM

Zelos Malum

2020-02-13 06:44:40 UTC

Post by Ganzhinterseher
If you can't distinguish the set from all its elements, then go to school again.

Coimes from the one that fails basic fucking induction!

Post by Ganzhinterseher
By definition we have ∩{E(k) | k ∈ ℕ_def} =/= 0

Ganzhinterseher

2020-02-13 11:10:24 UTC

Post by Ganzhinterseher
By definition we have ∩{E(k) | k ∈ ℕ_def} =/= 0

No, that is something you've assumed, you've never proven it.

It is the definition.

Induction is valid for potentially infinite sets. Induction cannot span actual infinite sets. Therwise induction would prove that all FISONs can be removed from |N without emptying |N.

Induction proves that all FISONs are finite

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

and never complete an infinite set ℕ.

But if ℕ were the same as ℕ_def, then we had the contradiction

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} =/= { }
and
∩{E(k) | k ∈ ℕ} = { }
violating mathematics
∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Regards, WM

Zelos Malum

2020-02-13 11:30:20 UTC

Post by Ganzhinterseher
It is the definition.

Then you got a contradiction because by induction, it is equal to |N

Post by Ganzhinterseher
Induction is valid for potentially infinite sets. Induction cannot span actual infinite sets. Therwise induction would prove that all FISONs can be removed from |N without emptying |N.

It applies to ANY subset of |N!

So your |N_def is |N because it is subject to induction!

Post by Ganzhinterseher
and never complete an infinite set ℕ.

Oh wow, sets never meant to be |N is not |N, HOW REMARKIBLE!

Your |N_def is subject to induction, it satisfies all the induction criteria (by your own fucking admission!) then it is equal to |N, be honest for once in your life.

Ganzhinterseher

2020-02-13 14:46:52 UTC

Post by Zelos Malum
Your |N_def is subject to induction, it satisfies all the induction criteria

Therefore it is a potentially infinite set.

Post by Zelos Malum
(by your own fucking admission!) then it is equal to |N

The set |N is not created by Peano but by Zermelo. But |N_def is created by Peano. If both were the same, then we had the contradiction that

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

would result in

∩{E(k) | k ∈ ℕ } = { }

violating mathematics:

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

Regards, WM

Zelos Malum

2020-02-14 07:45:09 UTC

Post by Ganzhinterseher
Therefore it is a potentially infinite set.

Therefore it is equal to |N

Post by Ganzhinterseher
The set |N is not created by Peano but by Zermelo. But |N_def is created by Peano.

Peano axiomized Natural numbers nicely. |N and your |N_def are the same FUCKING SIT!

Post by Ganzhinterseher
If both were the same, then we had the contradiction that

Those are not contradictions beyond you not understanding basic logic!

Ganzhinterseher

2020-02-15 11:06:42 UTC

Post by Ganzhinterseher
Therefore it is a potentially infinite set.

Therefore it is equal to |N

Post by Ganzhinterseher
The set |N is not created by Peano but by Zermelo. But |N_def is created by Peano.

Peano axiomized Natural numbers nicely. |N and your |N_def are the same FUCKING SIT!

No. All definable natnumbers only build a potentailly infinite set. This can be seen in the clearest way by the chain of endsegments. Every definable endsegment is a last E(k) in

∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

and fails to get an empty intersection. You cannot define an endsegment that fails this test, can you? Either do it or withdraw you claim.

Regards, WM

Zelos Malum

2020-02-17 07:38:48 UTC

Post by Ganzhinterseher
No. All definable natnumbers only build a potentailly infinite set.

It is equal to |N by induction, get the fuck over it alraedy you lying sack of shit.

Ganzhinterseher

2020-02-17 10:38:23 UTC

No. All definable natnumbers only build a potentially infinite set.

It is equal to |N by induction,

By induction we find: If (1, 2, 3, ..., n) =/= |N
then If (1, 2, 3, ..., n+1) =/= |N.

That is induction. And logic.

Regards, WM

Mostowski Collapse

2020-02-17 13:00:35 UTC

Challenge, take rational points on
a sphere, like here:

https://de.wikipedia.org/wiki/Pythagoreisches_Quadrupel#Erzeugung_von_pythagoreischen_Quadrupeln

Can we construct a dumbinable function
(definable in MathRealism of Prof
Muckefunk), that does the

Banach-Tarski paradox?

No. All definable natnumbers only build a potentially infinite set.

It is equal to |N by induction,

By induction we find: If (1, 2, 3, ..., n) =/= |N
then If (1, 2, 3, ..., n+1) =/= |N.
That is induction. And logic.
Regards, WM

Zelos Malum

2020-02-18 06:59:51 UTC

Ganzhinterseher

2020-02-18 11:03:55 UTC

Nevertheless a sober mind will recognize that by induction we can prove that every definable natural number does not spread an infinite set but is in a finite set:

(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...

whereas all natural numbers

(1, 2, 3, ...)

make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.

Regards, WM

h***@gmail.com

2020-02-18 14:25:57 UTC

(1)
(2, 1)
(3, 2, 1) (*)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...
whereas all natural numbers
(1, 2, 3, ...)
make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.

...proving once again that there are at least three things mathematical of which the good professor has zero understanding:

1. The definition of N.
N is the smallest set that contains 1 and with every k also contains its successor S(k). Nothing more, nothing less. (This, in and of itself, makes N an infinite set.)

2. The principle of induction.
What it says is that if a particular statement P(n) is true for n = 1 and P(n) implies P(n+1), then P(k) holds for every k in N. (NOT for N itself, Herr Professor!) In particular, if P(n) is the statement "{1,...,k} is a finite set" then (assuming he can prove the inductive step) then {1,...,k} is a finite set for every k in N. Again, nothing more, nothing less. In particular, it says nothing about P(N) and does not prove that {1,2,...} is a finite set.

3. Quantifiers.
The good Professor routinely and consistently treats "For every ... there exists ..." as synonymous with "There exists... for every...". This quantifier dyslexia leads him to the ridiculously erroneous assertion that the finiteness of every line {1,2,...k} in his sequence (*) implies that {1,2,...} must be finite also.

Python

2020-02-18 15:56:37 UTC

Post by h***@gmail.com

1. The definition of N.
N is the smallest set that contains 1 and with every k also contains its successor S(k). Nothing more, nothing less. (This, in and of itself, makes N an infinite set.)
2. The principle of induction.
What it says is that if a particular statement P(n) is true for n = 1 and P(n) implies P(n+1), then P(k) holds for every k in N. (NOT for N itself, Herr Professor!) In particular, if P(n) is the statement "{1,...,k} is a finite set" then (assuming he can prove the inductive step) then {1,...,k} is a finite set for every k in N. Again, nothing more, nothing less. In particular, it says nothing about P(N) and does not prove that {1,2,...} is a finite set.
3. Quantifiers.
The good Professor routinely and consistently treats "For every ... there exists ..." as synonymous with "There exists... for every...". This quantifier dyslexia leads him to the ridiculously erroneous assertion that the finiteness of every line {1,2,...k} in his sequence (*) implies that {1,2,...} must be finite also.

Adjunkt Lekturer Crank Wolfgang Mueckenheim, from Hochschule Ausburg,
is absolutely incapable of any kind of abstraction. It is not only a
shame and a disgrace that he'd been teaching in an academic institution
but even very surprising he ever found his way, as a student, ahead of
high school.

A career as a stand-up comedian would have been more suitable to him:

"Es gibt kaum jemanden hier außer mir, der mathematisch denken kann."

(WM, de.sci.mathematik, 27.08.2019)

"There is hardly anyone here other than me who can think
mathematically."

Zelos Malum

2020-02-19 06:46:53 UTC

Post by h***@gmail.com

Pretty much, he fail so many basic things

Ganzhinterseher

2020-02-19 10:50:28 UTC

Post by h***@gmail.com

Post by Ganzhinterseher
(1)
(2, 1)
(3, 2, 1) (*)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...
whereas all natural numbers
(1, 2, 3, ...)
make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.

N is the smallest set that contains 1 and with every k also contains its successor S(k). Nothing more, nothing less. (This, in and of itself, makes N an infinite set.)

The same validity can be claimed for the axiom that all triples of points lie on one straight line. Nonsense in nonsense out.

All definable natnumbers provably fail to span an actually infinite set.

Post by h***@gmail.com
if P(n) is the statement "{1,...,k} is a finite set" then (assuming he can prove the inductive step) then {1,...,k} is a finite set for every k in N. Again, nothing more, nothing less. In particular, it says nothing about P(N) and does not prove that {1,2,...} is a finite set.

It says that |N is not a set that contains only definable natural numbers because all fail to span an actually infinite set.

Post by h***@gmail.com
The finiteness of every line {1,2,...k} in his sequence (*) implies that {1,2,...} must be finite also.

The universal quantifier says something about all elements. If all definable elements fail to span actual infinite |N, then |N contains more or does not exist.

Matheologians believe that all flops together will succeed. But for all definable natnumbers we can exclude an actually infinite set.

Regards, WM

Sergio

2020-02-18 15:30:36 UTC

(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...
whereas all natural numbers
(1, 2, 3, ...)
make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.
Regards, WM

two different approaches,

both different ways of constructing different sets...

Sergio

2020-02-18 16:51:45 UTC

(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...

that is infinite collection of finite sets, you could just have one per
set too, more direct example...

(1)
(2)
(3)
.....

Post by Ganzhinterseher
whereas all natural numbers
(1, 2, 3, ...)

that is one infinite set.

Post by Ganzhinterseher
make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.
Regards, WM

Ganzhinterseher

2020-02-19 10:46:14 UTC

Post by Ganzhinterseher
(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...

that is infinite collection of finite sets, you could just have one per
set too, more direct example...
(1)
(2)
(3)
.....

Post by Ganzhinterseher
whereas all natural numbers
(1, 2, 3, ...)

that is one infinite set.

No, it is wrong in both cases to believe that all definable natnumbers form an actually infinite set. But it is easier to see in my first example, that they fail.

You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore an actually infinite set requires more: dark numbers.

Regards, WM

Zelos Malum

2020-02-19 06:46:07 UTC

Any sane mind will also know thats wholy irrelevant and you're pulling a non-sequitor you dishonest scumbag.

Ganzhinterseher

2020-02-19 10:47:19 UTC

Post by Ganzhinterseher
(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...
whereas all natural numbers
(1, 2, 3, ...)
make up a larger set than all finite lines above. You cannot define two or more definable numbers that fail to be together in one of the finite lines. Therefore the larger set requires more natnumbers.

Any sane mind will also know thats wholy irrelevant

Why? Isn't it fact that you cannot define two or more definable numbers that fail to be together in one of the finite lines? If so, then the larger set requires more natnumbers.

Regards, WM

Zelos Malum

2020-02-20 10:38:54 UTC

Post by Ganzhinterseher
Why? Isn't it fact that you cannot define two or more definable numbers that fail to be together in one of the finite lines? If so, then the larger set requires more natnumbers.

No you fucking moron, because you conjure up an arbitrary irrelevant criteria that would "disprove it" but in fact, that is not logically connected at fucking all.

There need not be any specific natural number to fail.

The set of all natural numbers is larger than any FISON, but that does not in anyway require that there are any more natural numbers than you can find in some FISON.

Ganzhinterseher

2020-02-21 07:43:07 UTC

There need not be any specific natural number to fail.

Then all fail, since all natural numbers are specific.

Post by Zelos Malum
The set of all natural numbers is larger than any FISON,

Here we talk about all FISONs, not about any single one. For all FISONs we have

E(1) =/= { }
E(2) ∩ E(1) =/= { }
E(3) ∩ E(2) ∩ E(1) =/= { }
E(4) ∩ E(3) ∩ E(2) ∩ E(1) =/= { }
E(5) ∩ E(4) ∩ E(3) ∩ E(2) ∩ E(1) =/= { }
...

Post by Zelos Malum
but that does not in anyway require that there are any more natural numbers than you can find in some FISON.

Logic says that all definable natnumbers fail to complete the set |N and to give E(1) ∩ E(2) ∩ E(3) ∩ ... = { }.

If you refuse logic, then your claims are irelevant.

The empty set differs drastically from all non-empty sets.

Regards, WM

Python

2020-02-21 11:26:27 UTC

Post by Ganzhinterseher
The empty set differs drastically from all non-empty sets.

oh, really, Crank Adjunkt Lekturer Wolfgang Mueckenheim, from
Hochschule Augsburg? How come? Does it differ less drastically
from all other empty sets?

Does it differ only slightly drastically from singletons? From
finite sets? From infinite sets?

You build most of you sophistries by deliberately omit some
kind of (somewhat drastic) difference between kind of sets
(finite or infinite sets typically), and now you insist in
a drastic difference you are even able to state properly...

You are a disgrace, a shame, a scandal in German Academy. You
are abusing students, abusing your authority, hurting human
basic decency. You'll be removed once of your position and
you should be publicly ashamed for what you've done during all
these years of LIES, FALLACIES and ABUSE.

Zelos Malum

2020-02-24 06:48:22 UTC

Post by Ganzhinterseher
Then all fail, since all natural numbers are specific.

Everyone fail, but all together they do not.

1 cell is not sentient, but all our brain cells together are.

Post by Ganzhinterseher
If you refuse logic, then your claims are irelevant.

The one refusing logic is you, you do not understand even the most basic of logic here and definition.

Just because

E(1) ∩ E(2) ∩ E(3) ∩ ... ∩ E(n) = E(n)

it does NOT in ANYWAY mean that

E(1) ∩ E(2) ∩ E(3) ∩ ... = { }

Must be false, how many times do I have to say, that finite cases can be different from infinite?

Ganzhinterseher

2020-02-24 17:43:07 UTC

Post by Ganzhinterseher
Then all fail, since all natural numbers are specific.

Everyone fail, but all together they do not.

All definable natnumbers together are in a FISON.
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

Since ∀k ∈ ℕ_def: ℕ \ {1, 2, 3, ..., k} =/= Ø the set ℕ contains undefinable natnumbers.

Post by Zelos Malum
Just because
E(1) ∩ E(2) ∩ E(3) ∩ ... ∩ E(n) = E(n)
it does NOT in ANYWAY mean that
E(1) ∩ E(2) ∩ E(3) ∩ ... = { }
Must be false,

You forgot to find a set of individually definable Endsegments which does not fit into a FISON. Therefore the set {E(1), E(2), E(3), ... } contains undefinable endegments.

REgards, WM

Python

2020-02-24 17:58:28 UTC

Post by Ganzhinterseher
Then all fail, since all natural numbers are specific.

Everyone fail, but all together they do not.

All definable natnumbers together are in a FISON.
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

But it doesn't follow that there is a FISON containing
all natnumbers, so your usual arguments ends in sophistry,
as usuel, Crank Adjunkt Lekturer Wolfgang Mueckenheim, from
Hochschule Augsburg.

Ganzhinterseher

2020-02-24 19:16:10 UTC

Post by Python

Post by Ganzhinterseher
All definable natnumbers together are in a FISON.
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

But it doesn't follow that there is a FISON containing
all natnumbers,

What ever may follow from your insane theories: You cannot find any set of individually definable natnumbers that is larger than every FISON. That proves that |N_def is not larger than every finite set and certainly smaller than the set |N which is larger than every FISON.

Regards, WN

Python

2020-02-24 19:26:35 UTC

Post by Python

Post by Ganzhinterseher
All definable natnumbers together are in a FISON.
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

But it doesn't follow that there is a FISON containing
all natnumbers,

What ever may follow from your insane theories: You cannot find any set of individually definable natnumbers that is larger than every FISON

than every (each) fison, sure. Than all unioned together none, as it is
this set himself.

Even your sophistries are getting dumber and dumber Crank Adjunkt
Lekturer Wolgang Mueckenheim from Hochschule Augburg, and piece of
dirt abusing students.

Ganzhinterseher

2020-02-24 19:43:24 UTC

Post by Ganzhinterseher
You cannot find any set of individually definable natnumbers that is larger than every FISON
than every (each) fison, sure. Than all unioned together

No union required. Try to find a collection of definable natnumbers that goes beyond a FISON that is not followed by almost all FISONs. Fail. Understand that the union of all natnumbers, if existing, consists almost purely of undefinable natnumbers.

Regards, WM

Python

2020-02-24 20:24:34 UTC

Post by Ganzhinterseher
You cannot find any set of individually definable natnumbers that is larger than every FISON
than every (each) fison, sure. Than all unioned together

This is your garbage, Crank Wolfgang Mueckenheim, from Hochschule
Augbusrg, it doesn't even make sense.

You are piece of dirt that should be expelled from German Academy and
put in jail for the crime of abusing students.

Ganzhinterseher

2020-02-24 22:16:27 UTC

Post by Python

Post by Ganzhinterseher
No union required. Try to find a collection of definable natnumbers that goes beyond a FISON that is not followed by almost all FISONs. Fail. Understand that the union of all natnumbers, if existing, consists almost purely of undefinable natnumbers.

it doesn't even make sense.

You are not bright enough to understand? Here is a simplified version:

There are infinitely many finite initial segments of natural numbers (FISONs):

(1)
(2, 1)
(3, 2, 1)
(4, 3, 2, 1)
(5, 4, 3, 2, 1)
...

All natural numbers which you or other people will ever be able to define individually will fit into one of these FISONs, say F(n). Infinitely many FISON will follow upon this very FISON. All of them will be proper subsets of |N:

F(1) c F(2) c ... c F(n) c F(n+1) c ... c |N.

There is no necessity to union anything, let alone all FISONs to accomodate all individually definable natural numbers together.

Regards, WM

Zelos Malum

2020-02-25 06:34:36 UTC

Post by Ganzhinterseher
All definable natnumbers together are in a FISON.

Nope, there is no FISON that contains all definable natural numbers because, as we already proved, its the same as natural numbers.

Post by Ganzhinterseher
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

fitting into A FISON is not the same as there being a FISON that has ALL of them. THe statement is about the latter you dishonest scumbag.

Post by Ganzhinterseher
You forgot to find a set of individually definable Endsegments which does not fit into a FISON.

Because I don't need to.

Post by Ganzhinterseher
Therefore the set {E(1), E(2), E(3), ... } contains undefinable endegments.

Non-sequitor again, it really is your favourite logical fallacy!

Ganzhinterseher

2020-02-25 12:03:44 UTC

Post by Ganzhinterseher
All definable natnumbers together are in a FISON.

Nope, there is no FISON that contains all definable natural numbers

Then define numbers individually which do not fit into a FISON.

Post by Ganzhinterseher
Proof: You cannot find a set of individually definable natnubers which do not fit into a FISON.

fitting into A FISON is not the same as there being a FISON that has ALL of them.

Not all of them are required. One of the first FISONs is enough, which is followed by infinitely many FISONs which are not required.

Post by Zelos Malum
THe statement is about the latter

and it is correct in this form. Only few of the first FISONs are required. One of them accomodates all you can define. Infinitely many FISONs are following and are not required.

Post by Ganzhinterseher
You forgot to find a set of individually definable Endsegments which does not fit into a FISON.

Because I don't need to.

You would need to find more definable natnumbers if you would refute my theorem that all definable natnunbers fit into one of the first FISONs.

Post by Ganzhinterseher
Therefore the set {E(1), E(2), E(3), ... } contains undefinable endegments.

Non-sequitor again

Stop your silly claims. Find a set or sequence or collection of individually definable natnumbers which does not fit into one of the first FISONs, or confess that you cannot find them.

Regards, WM

Zelos Malum

2020-02-26 06:35:27 UTC

Post by Ganzhinterseher
Then define numbers individually which do not fit into a FISON.

No need to because it proves nothing.

All natural numbers fit into a fison, but there is no fison with all natural numbers.

Post by Ganzhinterseher
Not all of them are required. One of the first FISONs is enough, which is followed by infinitely many FISONs which are not required.

Correct, but that does not mean anything you want it to.

Post by Ganzhinterseher
and it is correct in this form. Only few of the first FISONs are required. One of them accomodates all you can define. Infinitely many FISONs are following and are not required.

No it isn't, nothing you do is correct, your go to logical fallacy is always non-sequitor.

Post by Ganzhinterseher
You would need to find more definable natnumbers if you would refute my theorem that all definable natnunbers fit into one of the first FISONs.

No I do not, because all I need to do is show where you've done a logical error and I have pointed it out already.

Post by Ganzhinterseher
Stop your silly claims. Find a set or sequence or collection of individually definable natnumbers which does not fit into one of the first FISONs, or confess that you cannot find them.

It is not silly, you are engaging in non-sequitors constantly.

I don't need to find any of that to show you're wrong. You don't get to decide what shows you wrong, logic does and we can logicalyl show you're wrong at every turn

3 properties of yoru |N_def shows that |N_def=|N, get over it already and be honest for once in your life.

Ganzhinterseher

2020-02-26 11:06:37 UTC

Post by Ganzhinterseher
Then define numbers individually which do not fit into a FISON.

No need to because it proves nothing.

It proves that all definable natnunbers fit into one of the first FISONs.

Post by Zelos Malum
All natural numbers fit into a fison, but there is no fison with all natural numbers.

Then find individually definable natnumbers which do not fit into one of the first FISONs. Fail. Remain stupid.

Post by Ganzhinterseher
Not all of them are required. One of the first FISONs is enough, which is followed by infinitely many FISONs which are not required.

Correct, but that does not mean anything you want it to.

It means what I said and what you cannot disprove by an example.

Post by Zelos Malum
I don't need to find any of that to show you're wrong.

I said that you cannot find any of that. That is the only point. You cannot find any and you know it but you claim that I am wrong. Typical in matheology.

Regards, WM

Mostowski Collapse

2020-09-16 14:23:37 UTC

For long time, I thought WMs psychosis
can be explained by modal logic. Now
I have the feeling free logic is enough.

How does WMs psychosis come into play?
Well the U(_) does not only allow an
empty universe of existent objects. But

it also splits the FOL universe V into
existent objects U and non-existent objects
V\U. What can happen is now, that a

function maps an existent object to
a non-existent object. Eh voila we get
a little nice "Dark Number" Paradox:

/* ∀xN(x) => ∀xN(s(x)) is FOL provable */
?- prove((all(X,n(X))-:all(Y,n(s(Y)))),3,N).
Y = '$0',
N = 1

/* ∀xN(x) => ∀xN(s(x)) is not free logic provable */
?- freeify((all(X,n(X))-:all(Y,n(s(Y)))),F), prove(F,3,N).
No

Free Logic to FOL
With Beckert and Posegga operators
https://gist.github.com/jburse/5ec1b5d77b7dc7525e3337d7d61ec335#file-free-pl

https://groups.google.com/d/msg/sci.logic/7FR2ImeMhIY/PWOH6ZPAAgAJ

I said that you cannot find any of that. That is the only point. You cannot find any and you know it but you claim that I am wrong. Typical in matheology.

Zelos Malum

2020-02-11 08:24:46 UTC

Post by Ganzhinterseher
We need no sufggestions but mathematics

We do and we have it, you're the one fucking failing in mathematics.

Anyone with a rudimentary knowledge of logic adn mathematics knows you cannot argue from An e |N(P(n) to P(|N), it is what you are doing and it is a pure example of a non-sequitor.

Post by Ganzhinterseher
There is not much logic required. A counter example shows that a theorem is false.

1: your N_def is just |N so just cut that shit out.

2: your equation is wrong, the finite intersection is not equal to the infinite intersection.

Ganzhinterseher

2020-02-11 13:24:27 UTC

Post by Ganzhinterseher
We need no sufggestions but mathematics

We do and we have it,
Anyone with a rudimentary knowledge of logic adn mathematics knows you cannot argue from An e |N(P(n) to P(|N), it is what you are doing and it is a pure example of a non-sequitor.

You are wrong. I do not use E(|N).

Post by Ganzhinterseher
There is not much logic required. A counter example shows that a theorem is false.

1: your N_def is just |N

Hardly. |∩{E(k) | k ∈ ℕ_def}| = aleph_0
|∩{E(k) | k ∈ ℕ}| = 0

Post by Zelos Malum
2: your equation is wrong, the finite intersection is not equal to the infinite intersection.

The set of definable endsegments is not actually infinite. Find any definable natnumber that make the equation wrong.

Regards, WM

Python

2020-02-10 08:32:54 UTC

Crank Wolfgang Mueckenheim, Ganzhinterseher wrote:
...

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished
infinity and the set in (*) is never completely empty but only
empty "up to every finite k".

"Empty up to every k"?! What a strange concept of yours. I really
wonder how you could make sense of such a idiotic thing, Crank
Adjunkt Lekture Wolfgang Mueckenheim, from Hochschule Augsburg.

Ganzhinterseher

2020-02-10 09:26:54 UTC

Post by Python
...

Post by Ganzhinterseher
Either there are dark natural numbers or there is no finished
infinity and the set in (*) is never completely empty but only
empty "up to every finite k".

"Empty up to every k"?!

That is potential infinity of sober mathematics, contrary to tranfinity drunkenness.

Post by Python
What a strange concept of yours.

It is the same as Cauchy used.

Post by Python
I really
wonder how you could make sense of such a idiotic thing

That is strange.

Regards

Mostowski Collapse

2021-11-16 09:40:00 UTC

"I'll stop wearing black when they make a darker color."
- Wednesday Addams

Are there darker numbers than dark numbers?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

Mostowski Collapse

2021-11-16 09:43:18 UTC

LoL, this guy made my day:

I Painted My Entire Room With Musou Black

Post by Mostowski Collapse
"I'll stop wearing black when they make a darker color."
- Wednesday Addams
Are there darker numbers than dark numbers?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

Serg io

2021-11-16 14:29:54 UTC

so,

someone had to paint all those numbers black. or they were born black ?

How do they know they are black if they cannot see themselves ?

Perhaps they are not black at all, but just "the unborn" numbers ?

The parents of 10 are 2 and 5, as they multiply...

Post by Mostowski Collapse
I Painted My Entire Room With Musou Black
http://youtu.be/p6q54q2iam8

Post by Mostowski Collapse
"I'll stop wearing black when they make a darker color."
- Wednesday Addams
Are there darker numbers than dark numbers?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

Mostowski Collapse

2021-11-16 15:04:20 UTC

2 divides 10, 5 divides 10, symbolically we can write 2 | 10, 5 | 10, etc..
The divides relation over natural numbers is a well founded relation.
The dark numbers after n are not n+1, n+2, n+3, .... The next

dark number is 2*n, then comes 3*n, 4*n, etc.. in general
n and m have different dark numbers following them, the
common dark numbers are only at multiples of n*m/GCD(n,m),

its a structure with more than fifthy shades of gray.

Post by Serg io
so,
someone had to paint all those numbers black. or they were born black ?
How do they know they are black if they cannot see themselves ?
Perhaps they are not black at all, but just "the unborn" numbers ?
The parents of 10 are 2 and 5, as they multiply...

Post by Mostowski Collapse
I Painted My Entire Room With Musou Black
http://youtu.be/p6q54q2iam8

Post by Mostowski Collapse
"I'll stop wearing black when they make a darker color."
- Wednesday Addams
Are there darker numbers than dark numbers?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

Mostowski Collapse

2021-11-16 15:18:40 UTC

Corr.: Sorry, my bad
The divides relation (|)/2 is not well founded,
since the divides relation is not a total order.
For two natural numbers a, b we do not have:

a | b v b | a

Nevertheless it has no infinite decending
sequences of natural numbers. So it has
something in common with sets. Sets are

also not in total order by the subset relation,
but for example ordinals have this property.
A lot of exercises lurking for Danny Boy.

We can say the divides relation is regular?
Thats possibly not a good question, because
the divides relation is a little more powerful

than the membership relation, it is transitive.

Post by Mostowski Collapse
2 divides 10, 5 divides 10, symbolically we can write 2 | 10, 5 | 10, etc..
The divides relation over natural numbers is a well founded relation.
The dark numbers after n are not n+1, n+2, n+3, .... The next
dark number is 2*n, then comes 3*n, 4*n, etc.. in general
n and m have different dark numbers following them, the
common dark numbers are only at multiples of n*m/GCD(n,m),
its a structure with more than fifthy shades of gray.

Post by Mostowski Collapse
I Painted My Entire Room With Musou Black
http://youtu.be/p6q54q2iam8

Post by Mostowski Collapse
"I'll stop wearing black when they make a darker color."
- Wednesday Addams
Are there darker numbers than dark numbers?

Post by Sergio
also, how dark are dark numbers ?
like "black" dark, or "brown" dark ? or darkish ?
better to call them "unseen numbers" ?

Wade Earl

2021-11-16 16:57:27 UTC