A hidden assumption

[doronshadmi]

You know, the bolded part, above, has no meaning. "Comparison" is definitely the wrong word.

A membership rule is essentially a function, as in C(A) with value of either "A is a member" or "A isn't a member" of some set. There is no binary operator (comparison or otherwise) that interacts with C and A.


ETA: Ben_M beat me to this point. Doh!!

One of my problems here is to express myself in English.

In order to avoid confusion let us use "Association" instead of "Comparison".

All I wish to say is this:

Set B is the result of rule C\element A association.

Without this association nothing is researchable, in this case.

I have found that there are at least two cases that are non-researchable at their self state (where nothing is associated), which are:

Total unity, total isolation.

Furthermore, I can conclude that exactly because I exist in a realm which is not total unity and not total isolation, but it is a synthesis of these extremes, where this synthesis enables to look at each one of these extremes form both internal and external (associated) points of view (I understand Unity because of Unity\Isolation simultaneous (and associated) point of view, and I understand Isolation because of Isolation\Unity simultaneous (and associated) point of view).

As I understand it, Cardinality can be formalized only if there is associater\associated association.

I logically addressed the associater by NXOR connective and the associated by XOR connective.

In that case rule C is logically understood in terms of NXOR connective and element A is logically understood in terms of XOR connective.

As a result the researchable object is not less than an outcome of a NXOR\XOR association.


Let no association be called Selfness. In that case Cardinality (which is a fundamental concept of any Set theory, standard or not) cannot be formalized (as a general concept).


For example:

Z is an axiomatic framework that is based on one and only one axiom:

"One is a natural number"

I claim that this formal framework is not researchable (nothing is deduced except the axiom).


If you disagree with me than please provide a formal (and researchable) example where no association is involved.

 

[ben m]

One of my problems here is to express myself in English.

This is a good reason to learn mathematical notation.

I have found that there are at least two cases that are non-researchable at their self state (where nothing is associated), which are:

Gibberish. You need to start over.

"Set B" is defined as "The set of A for which C(A) is true." You can also define another set B' as "The set of A for which C(A) is false." That's it. I'm finished defining Set B and Set B'. There's nothing else I can do with my C(X) function; I've just evaluated it for every possible object.

You keep mentioning XOR and XNOR. These are binary operations that act on Boolean values. (T xor F = T, T xor T = F, but 3 xor 10 is meaningless. Since C(X) is a Boolean function, C(A) xor C(A') is a reasonable thing to evaluate for elements A and A', but simply A xor A' is meaningless.) What pair of Boolean values do you want to apply XOR to? The result of applying XOR is another Boolean value, X xor Y = Z. What does this value mean?

 

[doronshadmi]

"Set B" is defined as "The set of A for which C(A) is true." You can also define another set B' as "The set of A for which C(A) is false." That's it.

Please look at it from a general point of view. In both cases there is associater(rule C)\associated(element A) association. Without this association set B cannot formally be determined.

I'm finished defining Set B and Set B'. There's nothing else I can do with my C(X) function; I've just evaluated it for every possible object.

You used the term "evaluated". In that case there is an evaluator\evaluated association.

You keep mentioning XOR and XNOR. These are binary operations that act on Boolean values. (T xor F = T, T xor T = F, but 3 xor 10 is meaningless.
What pair of Boolean values do you want to apply XOR to? The result of applying XOR is another Boolean value, X xor Y = Z. What does this value mean?

I am looking for the association's general principle.

Here is the relevant part from my first post:

Let a 2-valued framework be represented by A B.

A B system is:

A B
A A
A B
B A
B B

This system can be reduced (without a loss of generality) to
symmetric (AA or BB) \ asymmetric (AB or BA) association, represented as:

AB
XX is AB symmetry --> A=B (AB is the same)
XY is AB asymmetry --> A≠B (AB is not the same)

A=B can be reduced (without a loss of generality) to a single value X (non-local atom).

A≠B cannot be reduced to a single value without a loss of detail X(local atom) or Y(local atom).

Furthermore, in order to conclude that A≠B, they must share the same framework.

So 2-valued framework is at least X (symmetry or sameness) \ XY (asymmetry or difference) association.

I have found that Symmetry\Asymmetry association holds (is indeed a general principle) for any |Z|-valued framework, where |Z| any possible cardinal (finite or not).

Symmetry is expressed as the non-local state (invariant under exchange) of |Z|-valued framework and Asymmetry is expressed as the local state (variant under exchange) of |Z|-valued framework.

Since the multiple is Symmetry\Asymmetry association's outcome, then no infinitely many elements (distinct or not) get the magnitude of the non-local atom (total symmetry). In that case any magnitude, which is based on the multiple, is incomplete.

If you know how to formally address it by avoiding confusion with already formalized mathematical concepts, then please share it with us.

I hope that my notions become clearer during our dialog.

 

[ben m]

Please look at it from a general point of view. In both cases there is associater (rule C)\associated(element A) association. Without this association set B cannot formally be determined.

Tell me if this translation is right: "If I have a function C() and a collection of elements A, and I want to determine whether A is a member of B, I have to be able to evaluate C(A). We are interested in cases where C(A) can be evaluated, in which case I say "C and A are associated", as opposed to cases where C(A) cannot be evaluated, in which I say "C and A are not associated"". Is that right?

If C(X) is "(X > 10)" and A is 4, are C and A associated or not?
If C(X) is "(X > 10)" and A is "green", are C and A associated or not?

Please do not cut-and-paste from older posts; your original posts failed to communicate anything. Rethink your ideas in the language I'm using and state them in a way we can understand.

 

[doronshadmi]

Tell me if this translation is right: "If I have a function C() and a collection of elements A, and I want to determine whether A is a member of B, I have to be able to evaluate C(A). We are interested in cases where C(A) can be evaluated, in which case I say "C and A are associated", as opposed to cases where C(A) cannot be evaluated, in which I say "C and A are not associated"". Is that right?

If C(X) is "(X > 10)" and A is 4, are C and A associated or not?
If C(X) is "(X > 10)" and A is "green", are C and A associated or not?

Please do not cut-and-paste from older posts; your original posts failed to communicate anything. Rethink your ideas in the language I'm using and state them in a way we can understand.

It was not a clear cut-n-paste, I have fixed it because of the dialog.

As for your fine questions:

In order to determine if C(X) is valuable (or not), C() and A must be associated.

Association does not mean C(A).

Association means Association(C(),A) which returns C(X) is valuable (or not).

Actually this is exactly how you are doing in order to understand my framework.

You associate between my ideas and your well-defined framework in order to conclude if it is valuable (or not, and if not it is called gibberish or nonsense and I am a crank).

 

[ben m]

In order to determine if C(X) is valuable (or not), C() and A must be associated.

Association does not mean C(A).

Association means Association(C(),A) which returns C(X) is valuable (or not).

OK, fine. You could write such a function in C---"bool CheckTypeAgreement(void* C, void* A)" and have it tell you whether C is compatible with A's type. Great. If I have bool C(int) and float A, and int A', then CheckTypeAgreement(C,A) = false and CheckTypeAgreement(C,A') = true.

Now, using exactly the notation of Association(f,X), f(X), X, what were those AA AB BA BB tables supposed to mean?

 

[kallsop]

'0 0' or '1 1' means that in,out the same.

'0 1 or ' 1 0' means that in,out not the same.

Now try to compare between in,out without any common thing.

In this case '0' is totally isolated of '1' and nothing can be compared and concluded.

I compare "0" to "1" and conclude they are not common and are different.

Now, how do I claim the prize?

 

[doronshadmi]

OK, fine. You could write such a function in C---"bool CheckTypeAgreement(void* C, void* A)" and have it tell you whether C is compatible with A's type. Great. If I have bool C(int) and float A, and int A', then CheckTypeAgreement(C,A) = false and CheckTypeAgreement(C,A') = true.

In that case you are using Association(CheckTypeAgreement(C,A) = false , CheckTypeAgreement(C,A') = true) which enables you to determine, which result is valuable or holds (you call it true) and which result is non-valuable or does not hold (you call it false).

You can continue as much as you like, but you cannot avoid association if you wish to distinguish between the valuable(what holds) and the non-valuable(what does not hold).

Now, using exactly the notation of Association(f,X), f(X), X, what were those AA AB BA BB tables supposed to mean?

If association is inevitable (as I claim) in order to get a researchable framework, then any researchable framework
is at least Association( associater , associated ).

Let AA or BB represent sameness.

Let AB or BA represent difference.

In that case a 2-valued framework notated as A B is based on Association( sameness , difference ).

If you find my notion valueable and you have a better way to formally define it, then please feel free to do it.

 

[ben m]

In that case you are using Association(CheckTypeAgreement(C,A) = false , CheckTypeAgreement(C,A') = true) which enables you to determine, which result is valuable or holds (you call it true) and which result is non-valuable or does not hold (you call it false).

First of all, you've reinvented "association" again. A moment ago it was being used to see if an operator was happy with an input value. Now you're using it on two Boolean values. Please keep it straight. Either you allowed me to misdefine "Association" before, or you're misusing it now, or you've got two or three different types of "association" for which you're using a single word. This is not nitpicking, this is a very serious problem---it sounds like you yourself do not know what "associations" are actually doing, just that they're supposed to "compare things" and "make regular math wrong" in some vague way.

In any case, we avoiding any of these problems by definition. I have chosen, or invented, a CheckTypeAgreement "association" function which returns "true" when "the result is valuable or holds" and returns "false" when "the result is not valuable or does not hold". If you don't know what true and false mean here, that only means that you did not construct the CheckTypeAgreement function to correspond to your "association".

If you want to dispute the notion that "true" and "false" are, in fact, "associated", or that they can be chosen/defined in such a way that they're "associated", then your notion of "association" sounds pretty useless.

 

[ben m]

Let AA or BB represent sameness.

Let AB or BA represent difference.

In that case a 2-valued framework notated as A B is based on Association( sameness , difference ).

Still gibberish. I don't care what AB represents yet---first I need to know what A and B are! Two elements? Two sets? An operator (like C()) and an element? Two operators? The results of two Association(X,Y) tests? If they're two elements, are you putting them into a multiset {A,A}? Are you comparing them with an Boolean equality test (A==A)?

 

[doronshadmi]

Still gibberish. I don't care what AB represents yet---first I need to know what A and B are! Two elements? Two sets? An operator (like C()) and an element? Two operators? The results of two Association(X,Y) tests? If they're two elements, are you putting them into a multiset {A,A}? Are you comparing them with an Boolean equality test (A==A)?

Actually it does not matter what A B are and how they are represented.

When I re-examine my notion according to the dialog with you, I understand that a researchable framework is not less than Association(A,B) were A or B are placeholders ( http://en.wikipedia.org/wiki/Placeholder ) for any possible association that leads to some conclusion.

 

[ben m]

Actually it does not matter what A B are and how they are represented.

In this ganaral level any researchable framework is not less than Association(A,B) were A or B are placeholders ( http://en.wikipedia.org/wiki/Placeholder ) for any possible association that leads to some conclusion.

Are you being obscure on purpose? Do you mean "let Association(A,A) and Association(B,B) represent sameness"?

You've still got the problem that "Association(A,A)" doesn't mean anything yet. You've only explained "Association(C(),A)" between a function and a member.

 

[doronshadmi]

Are you being obscure on purpose? Do you mean "let Association(A,A) and Association(B,B) represent sameness"?

You've still got the problem that "Association(A,A)" doesn't mean anything yet. You've only explained "Association(C(),A)" between a function and a member.

I am doing my best in order to be understood.

Let us say that A,B placeholders' values are understood in terms of Symmetry (A is B, and vice versa) or in terms of Asymmetry (A is not B, and vice versa).

In that case AA or BB means that placeholder_A value is placeholder_B value,
and AB or BA means that placeholder_A value is not placeholder_B value.

In that case that the framework is Association(Symmetry,Asymmetry), where Association(Symmetry) is not reseachable
and so is Association(Asymmetry).

If only one placehoder is used (for example Association(A), where A is a placeholder) we get A's self state, but self state is not researchable exactly like a framework that has only one axiom.

(By the way, Association() means no association. For example: no axioms at all)

 

[ben m]

Let us say that A,B placeholders' values are understood in terms of Symmetry (A is B, and vice versa) or in terms of Asymmetry A is not B, and vice versa).

In that case AA or BB means that placeholder_A value is placeholder_B value,
and AB or BA means that placeholder_A value is not placeholder_B value.

In that case the 2-valued framework is Association(Symmetry,Asymmetry).

Um. 1) So you're taking the abstract set {AA, BB, AB, BA} and dividing it into two sets, {AA, BB} and {AB, BA} which you name "symmetry" and "asymmetry". That's fine, you can define anything you like in a mathematical proof.

2) Then you announce that you want to plug these two sets into a function, "Association", even though you haven't explained what "Association(set,set)" actually means or does in general.

3) You want to call "Association(set1, set2)" a "two valued framework"---what's "association(set,set)" do? How is it a framework? What is a framework?

You might as well have said, "Let us say that cats and 5 are 'uffish' and greenness and Liberty Mutual are 'polymorphic'. In that case the two-valued framework is AddMoreSalt(uffish,polymorphic)"

 

[doronshadmi]

Um. 1) So you're taking the abstract set {AA, BB, AB, BA} and dividing it into two sets, {AA, BB} and {AB, BA} which you name "symmetry" and "asymmetry". That's fine, you can define anything you like in a mathematical proof.

2) Then you announce that you want to plug these two sets into a function, "Association", even though you haven't explained what "Association(set,set)" actually means or does in general.

3) You want to call "Association(set1, set2)" a "two valued framework"---what's "association(set,set)" do? How is it a framework? What is a framework?

You might as well have said, "Let us say that cats and 5 are 'uffish' and greenness and Liberty Mutual are 'polymorphic'. In that case the two-valued framework is AddMoreSalt(uffish,polymorphic)"

You have missed the point.

A set is the result of an Association and not vice versa.

The association (notated as the outer "{" and "}") is used but not explicitly defined by any Set theory that I know.

My framework explicitly defines it as the fundamental property of any researchable framework, where a set is one of its results.

Furthermore, you have missed the notion of a placeholder, and looked only on the results that can be expressed as collections of distinct objects (known as sets).

If you think only in terms of collections of distinct objects, then Association(A,B) will produce for you only collections of distinct objects, but then you have missed the generalization.

By the way, I think that this is the main reason why my generalization is not understood, because it produces what you wished to be produced, but than you see only the wished outcome and miss the generalization.


This is exactly why I claim that it is indeed the universal principle of any researchable framework, because any researchable framework is produced by it.

 

[ben m]

You have missed the point.

If you think only in terms of collections of distinct objects, then Association(A,B) will produce for you only collections of distinct objects, but then you have missed the generalization.

Doronshadmi, here's a general principle: if you can't produce any particular cases from your supposed "generalization", then the generalization is probably wrong. "The sum of two odd numbers always is even ... for example, 3+5 = 8" is good. "The sum of two potzgrebies is always grokkable, and that's all I can say" is bad.

So, please, let's think in terms of collections of distinct objects. Maybe your theory can do a whole Universe of things, but ordinary sets are *in* that Universe and your theory has to be able to describe them as a special case. Remember, you've written 20 posts on your "general" theory and nobody understands it. If you keep leaping ahead, we're going to continue not getting it.

• Association(A,B) still doesn't mean anything, you haven't defined it at all.
• Association(C(),A) you have defined, but C() is the thing that "produces collections of distinct objects like B = {A1, A2, A3}"

 

[jsfisher]

A set is the result of an Association and not vice versa.

The association (notated as the outer "{" and "}") is used but not explicitly defined by any Set theory that I know.

You circled back to this many times. My sense is that you are somewhat focussed on the braces used in our notation rather than the underlying abstract concepts.

You also seem to be saying that the elements of a set must have some underlying association between them, and that is what makes them elements of the same set. I suppose that may be weakly true, but it is not particularly useful a concept. Consider the set of the number 6, my favorite pen, and the color yellow. The only association the three elements have with each other is the membership in the same set. That makes the association concept somewhat circular.

The conventional notion of set is that a set is a well-defined collection of distinct things.

If you want to think of a set as some sort of container, well, ok, it is not necessary, but ok. The idea is supported by the notion of a set. Notice, too, that the notion does no imply any association among the members.

(By the way, the very formal ZFC formulation doesn't define 'set' at all. A set is just an implied consequence of the ZFC axioms.)

 

[doronshadmi]

Doronshadmi, here's a general principle: if you can't produce any particular cases from your supposed "generalization", then the generalization is probably wrong.

On the contrary, any particular case that you wish can be produced by Association(placeholder_A,placeholder_B)

For example: let Association be +, let placeholder_A be 3, let placeholder_B be 5.

So +(3,5)=8

You circled back to this many times. My sense is that you are somewhat focussed on the braces used in our notation rather than the underlying abstract concepts.

You also seem to be saying that the elements of a set must have some underlying association between them, and that is what makes them elements of the same set. I suppose that may be weakly true, but it is not particularly useful a concept. Consider the set of the number 6, my favorite pen, and the color yellow. The only association the three elements have with each other is the membership in the same set. That makes the association concept somewhat circular.

The conventional notion of set is that a set is a well-defined collection of distinct things.

If you want to think of a set as some sort of container, well, ok, it is not necessary, but ok. The idea is supported by the notion of a set. Notice, too, that the notion does no imply any association among the members.

(By the way, the very formal ZFC formulation doesn't define 'set' at all. A set is just an implied consequence of the ZFC axioms.)

I do not think that a collection of distinct members exists without a collector.

The "Collector" (which is a better concept, in this case) enables the association between distinct elements.

Furthermore, ZFC axioms cannot fully work if they are not collected. Please try to deduce anything from them by totally avoiding any association between them (after all each axiom is independent of any other axiom, so it can be done. The name of the collector in the case of axioms is known as "Consistency".)

If you do that you will find that ZFC is a very weak theory.

As for a set, please define a set that its cardinality > 1 without any association between its members.

Let me address it this way: an association is like the memory of a set that enables to simultaneously collect distinct members. This memory is not one of the members that are collected but it is an underlying property that without it any element is totally separated from any other element (I have changed "member" to "element" because without this memory no two things can be considered as members of a one object, known as set).

Maybe some analogy can help here:

If a distinct member is a bead-like element, then the memory (the collector) is a string-like element.

In that case a non-empty set is not less than a necklace that the cardinality of its members (the beads-like element) > 0

By using this analogy, the empty set is only the string-like element (a collector with no members), where a set with cardinal=1 is the string-like element with a single bead.

By using Logic, I have found that the invariant property of Symmetry is the string-like element, where the variant property of Asymmetry is a bead-like element.

Please look at 2-valued logic, again:

T F
F F (Symmtery's invariance (non-locality))
F T (Asymmetry's variance (locality))
T F (Asymmetry's variance (locality))
T T (Symmetry's invariance (non-locality))

It can be reduced to:

T F
X X (Symmtery's invariance (non-locality))
X Y (Asymmetry's variance (locality))

No collection of infinitely many local elements have the magnitude of the non-local (string-like) element.

 

[ben m]

I do not think that a collection of distinct members exists without a collector.

What property does your new concept ("collector") have that our old concept ("set") does not have?

By using Logic, I have found that the invariant property of Symmetry is the string-like element, where the variant property of Asymmetry is a bead-like element.

Please look at 2-valued logic, again:

T F
F F (Symmtery's invariance (non-locality))
F T (Asymmetry's variance (locality))
T F (Asymmetry's variance (locality))
T T (Symmetry's invariance (non-locality))

OK, I think I finally understand.

I think you've thinking your way vaguely towards discovering the D2 dihedral group. Your two-element truth table has the same properties as this group: imagine drawing TT as the point (1,1) on an X-Y plot, TF as (1,-1), etc.. So, you've noticed that the four members of this group are all related by rotations and reflections. You've noticed that if you ignore reflections, D2 has two subgroups (each isomorphic to D1) related by rotations---a rotation gets you from TT to FF (or from 1,1 to -1,-1) or from TF to FT (or from 1,-1 to -1,1), but you need a reflection to get from (1,1) to (1,-1) and such. So your "symmetry's invariance" is one copy of D1, with two components related by rotation; your "asymmetry's variance" is another copy of D1 with two components related by rotation; reflections take one of these D1s into the other; that reflection itself has the D1 symmetry.

But you can just as well do it the other way, calling "TF and TT" a D1 group related by a reflection, and "FT and FF" another D1 group related by a reflection, and there's a D1-like rotation that gets you from one group to the other. By focusing on the words "Symmetry" and "non-local atom" and whatnot, you've overlooked the fact that this set of subgroups has exactly the same properties as your preferred "XX vs XY" version.

So, my advice to you is to forget (for the moment, anyway) about your set/connector/etc. ideas, buy a textbook on group theory, and teach yourself about the easy-to-understand finite groups: dihedral groups, permutation groups. Keep in mind that group theory is fairly old, and any time you think you're going to do something new you are quite likely to be rediscovering something old. Good luck.

 

[ben m]

On the contrary, any particular case that you wish can be produced by Association(placeholder_A,placeholder_B)

For example: let Association be +, let placeholder_A be 3, let placeholder_B be 5.

So +(3,5)=8

Except that this is not what you did with "associations" earlier---what the heck happened to "association" telling you whether or not a result "was valuable" or not?