A hidden assumption

[doronshadmi]

It doesn't matter what we choose, 2 continues to be a prime number, an even number, an integer, and so on, all at the same time.
Sigh.

Wrong, 2 is some particular and researchable result of Non-local atom(Symmetry itself)\Local atom (Asymmetry itself) complementation.

No result can be considered as the fundamental state of anything.

2 is not an atom that is unchanged by membership.

It can be understood only from the point of view of an atomic state, which is non-researchable in its self state but can be used as the rigorous point of view of any reseachable thing.

When you get some result right from its atomic state, your resolution of it is increased exactly because you get it from the most general state (the atomic state). As a result there is a difference between 2 as a member and two as simply the result of Non-local atom(Symmetry itself)\Local atom (Asymmetry itself) complementation, which is the common property of the reseachable.

This property is not a member of any set, because it is the cause of Set.

In that case 2 as the result of invariance\variance complementation, is an independent researchable thing that is not necessarily a member of any set.

In the Sieve of Eratosthenes, for example, the numerical value of 2 and its primeness both come up.

As combined properties of 2 that get their meaning if they are related to the Sieve of Eratosthenes.

Indeed. 2 is always cardinal, natural, integer, prime, real, complex, rational and even. It belongs to all of those sets simultaneously; it can't not belong to those sets. You can ignore the sets that are not relevant to a particular theorem, but that says nothing except that you are ignoring those sets because they are not relevant.

The ability to get several properties of a one thing is based on the multitude, and the multitude is the result of invariance\variance complementation.

Again, the concept of Set is the result of invariance\variance complementation, and not vice versa.

What you did (without being aware of it) is to define 2 by using a set of distinct properties as its members.

In that case 2 is not one of the members but the rule that defines the set of the properties of 2.

Now cut out one of the properties of 2 out of the set of the properties of two, and it is not exclusively considered anymore as a member of the set of the properties of 2.

In any case, 2 is a reslt of invariance\variance complementation and not vice versa.

 

[ArmillarySphere]

It's not wrong - that interpretation is the standard. There's no "copying" in the standard notation of mathematics; all ways of arriving at a certain object are equivalent. Writing 4+4 is the same thing as writing 8, writing {A, B} \ {A} is the same as writing {B} and so forth.

And similarly, writing A=2 means A and 2 can be used interchangeably in each and every circumstance where the equality holds.

If you want to redefine mathematics, fine, but there's a lot of work to be done before you've even achieved equal usefulness with the standard definition.

 

[jsfisher]

ahem, I wouldn't want to derail this truly fascinating discussion in any way -- no really I wouldn't. But I feel compelled to point out that 2's primality is important when considering binary approximations of decimal approximations of real numbers (and decimal approximations of binary approximations of real numbers).

Carry on.

Hush, you!

 

[jsfisher]

It doesn't matter what we choose, 2 continues to be a prime number, an even number, an integer, and so on, all at the same time.
Sigh.

Wrong...

Well, that makes it clearer why you misunderstand some very basic stuff, then.

 

[doronshadmi]

Indeed. 2 is always cardinal, natural, integer, prime, real, complex, rational and even. It belongs to all of those sets simultaneously; it can't not belong to those sets. You can ignore the sets that are not relevant to a particular theorem, but that says nothing except that you are ignoring those sets because they are not relevant.

If I made a mistake in my previuos posts by using "explicitly" instead of "exclusively" then I apologize for the confusion that I made because of this mistake.

My argument is wrongly understood by you if you think that I claim that 2 exclusively belongs to any one of the titles that you used above.

2 does not exclusively belong to any of them.

Sine "simultaneously belongs" is also "not exclusively belongs" you actually support my argument.

In that case 2 is independent of any special membership.

So is the case of E={A,B} or F={A,C,D}

X={E,F}

E or F are independent of any special membership like X, exactly as 2 is independent of any special membership.

So the result of E U F is not exclusively related to X, unless we explicitly determine that E and F are X members.

In this case we are not dealing with E U F = {A,B,C,D} but we deal with {E,F}U{E,F}={E,F}=X

 

[drkitten]

My argument is wrongly understood by you if you think that I claim that 2 explicitly belongs to any one of the titles that you used above.

2 does not explicitly belong to any of them.

[Yoda]THAT is why you fail.[/Yoda]

 

[doronshadmi]

all ways of arriving at a certain object are equivalent.

I agree with you, because any reseachable object is the result of the same principle, which is invariance\variane complementation.

 

[doronshadmi]

[Yoda]THAT is why you fail.[/Yoda]

Please refresh your screen and re-read http://www.internationalskeptics.com/forums/showpost.php?p=3577820&postcount=265

 

[ArmillarySphere]

So is the case of E={A,B} or F={A,C,D}

X={E,F}

E or F are independent of any special membership like X, exactly as 2 is independent of any special membership.

So the result of E U F is not exclusively related to X, unless we explicitly determine that E and F are X members.

The bolded part is incorrect. The result of E U F is not exclusively related to X, no matter whether we define them as members or not. The result does indeed have a relation with X as above:

The union of X = U X = E U F = {A, B, C, D}

In this case we are not dealing with E U F = {A,B,C,D} but we deal with {E,F}U{E,F}={E,F}=X

Where did {E, F} enter the picture? E U F is a valid expression no matter what, since both are sets. Let's try it this way:

X U X = {E, F} U {E, F} = {E, F} = X = {{A, B}, {A, C, D}}
U X = E U F = {A, B, C, D}

X U X is not the same thing as the union of X.
X U X is expressed as the union of X with itself. This is not the same thing as the union of X, which is what the axiom of union speaks about.

 

[doronshadmi]

Well, that makes it clearer why you misunderstand some very basic stuff, then.

The very basic stuff is invariant\variant complementation.

2 is the result of this basic stuff.

By using the word "Wrong" about what you said, and I quote:

"It doesn't matter what we choose, 2 continues to be a prime number, an even number, an integer, and so on, all at the same time."

I actually agree with your direction of generalization, but say that your argument about 2 is not its real generalization.

Its real generalization is not less than the argument that 2 is a researchable object, which is the result of invariant\variant complementation, which is also the basis of Membership.

 

[doronshadmi]

The bolded part is incorrect. The result of E U F is not exclusively related to X, no matter whether we define them as members or not. The result does indeed have a relation with X as above:

The union of X = U X = E U F = {A, B, C, D}


Where did {E, F} enter the picture? E U F is a valid expression no matter what, since both are sets. Let's try it this way:

X U X = {E, F} U {E, F} = {E, F} = X = {{A, B}, {A, C, D}}
U X = E U F = {A, B, C, D}

X U X is not the same thing as the union of X.
X U X is expressed as the union of X with itself. This is not the same thing as the union of X, which is what the axiom of union speaks about.

No, the axiom of union is this:

What the axiom is really saying is that, given a set A, we can find a set B whose members are precisely the members of the members of A.
( http://en.wikipedia.org/wiki/Axiom_of_union )

I also did this mistake, when actually I meant to this definition:

If AA and BB are sets, then the union of AA and BB is the set that contains all elements of AA and all elements of BB, but no other elements. The union of AA and BB is usually written "AA ∪ BB".
( http://en.wikipedia.org/wiki/Union_(set_theory) )

E={A,B}
F={A,C,D}

AA=E BB=F , E u F = {A,B,C,D}

In my case X=AA=BB={E,F}u{E,F}=XuX=X

 

[jsfisher]

In that case 2 is independent of any special membership.

No, its set memberships are simply properties of 2. The number has many, many properties, and it is independent of none of them.


By the way, your comments of "exclusive" vs "explicitly" shows a shift in your position. You did in fact say and mean "exclusive" before, culminating in the severed body part analogy. You statements and examples were clear. In effect, you were claiming a sort of Heisenberg Uncertainty Principle for set theory. You are now back-pedaling.

 

[jsfisher]

The union of X = U X = E U F = {A, B, C, D}

No, the axiom of union is this:

Why do you say, "no"? ArmillarySphere is exactly correct applying the Axiom of Union to the example set X.

I also did this mistake, when actually I meant to this definition:

In set theory and other branches of mathematics, the union of a collection of sets is the set that contains everything that belongs to any of the sets, but nothing else.

If AA and BB are sets, then the union of AA and BB is the set that contains all elements of AA and all elements of BB, but no other elements. The union of AA and BB is usually written "AA ∪ BB".
E={A,B}
F={A,C,D}

AA=E BB=E , E u E = {A,B,C,D}

I assume you meant AA=E and BB=F, E U F = {A,B,C,D}

In my case X=AA=BB={E,F}u{E,F}=XuX=X

No, X does not = AA, nor does X = BB. X = {AA, BB} = {E, F}

Is this not obvious to you, that X = {AA, BB}, not AA or BB?

 

[drkitten]

E={A,B}
F={A,C,D}

AA=E BB=F , E u F = {A,B,C,D}

In my case X=AA=BB={E,F}u{E,F}=XuX=X

You're still not being consistent (or correct) in your notation.

If you define E as {A,B} and define AA = E (= {A, B}), then X is not AA.

If you define BB as F (= {A,C,D}), then AA != BB.

If you define AA as {E, F}, then AA is not E.

If you can't even keep your variable assignments straight, then it's small wonder that the rest of your pseudo-theory degenerates into gibberish.

 

[doronshadmi]

You're still not being consistent (or correct) in your notation.

If you define E as {A,B} and define AA = E (= {A, B}), then X is not AA.

If you define BB as F (= {A,C,D}), then AA != BB.

If you define AA as {E, F}, then AA is not E.

If you can't even keep your variable assignments straight, then it's small wonder that the rest of your pseudo-theory degenerates into gibberish.

Why do you say, "no"? ArmillarySphere is exactly correct applying the Axiom of Union to the example set X.



I assume you meant AA=E and BB=F, E U F = {A,B,C,D}



No, X does not = AA, nor does X = BB. X = {AA, BB} = {E, F}

Is this not obvious to you, that X = {AA, BB}, not AA or BB?

You assume right (typo mistake).

No,

AA or BB are not any special set when they are used as a part of the general definiton of Union.

In this case AA or BB can be exchanged by any set name like E,F or X.

 

[drkitten]

In this case AA or BB can be exchanged by any set name like E,F or X.

Yes, but if you can't even keep the meaning of a symbol like AA or E consistent from line to line (or even within a single clause), then you're going to produce wrong conclusions.

Let GNARF be any unmarried male.
By definition, then, all GNARFs are both unmarried and male.
Now let GNARF by any Oscar-winning actress.
Liza Minelli is therefore a GNARF (1972, Cabaret).
Therefore, since Liza is a GNARF and all GNARFS are male, Liza is a male.

 

[ArmillarySphere]

E={A,B}
F={A,C,D}

[with]AA=E BB=F , E u F = {A,B,C,D}

In my case X=AA=BB={E,F}u{E,F}=XuX=X

(bold part added for clarity)
Both statements are correct, but unrelated. They speak of different things.

Indeed, for the second statement the relation of E and F to X is completely unimportant, as XuX = X for any set X.

 

[doronshadmi]

oops...

 

[doronshadmi]

Yes, but if you can't even keep the meaning of a symbol like AA or E consistent from line to line (or even within a single clause), then you're going to produce wrong conclusions.

Let GNARF be any unmarried male.
By definition, then, all GNARFs are both unmarried and male.
Now let GNARF by any Oscar-winning actress.
Liza Minelli is therefore a GNARF (1972, Cabaret).
Therefore, since Liza is a GNARF and all GNARFS are male, Liza is a male.

When a name is used in some general definition, it must be understood as no more than a placeholder ( http://en.wikipedia.org/wiki/Placeholder ).

This is a fundamental notion of general mathematical definitons.

 

[doronshadmi]

(bold part added for clarity)
Both statements are correct, but unrelated. They speak of different things.

Indeed, for the second statement the relation of E and F to X is completely unimportant, as XuX = X for any set X.

Bravo!

Now explain it to jsfisher (he has some problem to understand the concept of placeholder)

If the placeholder itself is generalized, we actually deal with non-researchable\researchable complementation.

And this is exactly what is my theory about.