A hidden assumption

doronshadmi said:
jsfisher said:
You didn't address my question. Instead, you expanded your undefined terminology pool.
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Yes I did.

a+b=b+a is not symmetry itself but the result of symmetry\asymmetry complementation.

...<additional gibberish>...
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No, you didn't address my question. I made no mention of "a+b" nor "b+a" nor their equivalence in the referenced question. That is all irrelevant to my question.
 
doronshadmi said:
There is another way to understand the term "the union of all members of X is set X"

You wrote "let X = { {A,B}, {A,C,D} }"

Now you wrote "The union of all members of set X = {A,B} union {A,C,D}."

This is wrong because {A,B} union {A,C,D} is the union of all member of all members od set X.

The right one is:

{{ {A,B}, {A,C,D} } union {{ {A,B}, {A,C,D} } = {{ {A,B}, {A,C,D} }

( http://en.wikipedia.org/wiki/Union_(set_theory) )

Please ahow my mistake .
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For your reference and use, I have highlighted your mistake with bold text.
 
jsfisher said:
For your reference and use, I have highlighted your mistake with bold text.
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ellindsey said:
No.

{A,B} U {A,C,D} is {A,B} U {A,C,D}
{A,B} U {A,C,D} is not { A U B U A U C U D }

{A,B} U {A,C,D} is {A,B,C,D}
{ A U B U A U C U D } is invalid unless A,B,C, and D are all sets themselves.



No.

If X = { {A,B}, {A,C,D} }, then what you have just written is X U X.
This isn't the union of all members of set X. This is the union of X with itself.
The union of all members of set X is still {A,B} U {A,C,D}
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But A,B of {A,B} or A,C,D of {A,C,D} are the members of the members of set X, and my statment is: "the union of all the members of set X is set X.

For example:

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

X={E,F}

The elements (members) of X are E and F, where A,B are the members of E, and A,C,D are the members of F.

Let us quote at http://en.wikipedia.org/wiki/Union_(set_theory):

If A and B 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.

My case is X=AA=BB, so AA U BB = X U X

My statment "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".


In other words you are wrong becuase {A,B}u{A,C,D}= the union of the members of the members of set X={A,B,C,D}≠XUX="the union of all the members of set X=X
 
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doronshadmi said:
But A,B of {A,B} or A,C,D of {A,C,D} are the members of the members of set X, and my statment is: "the union of all the members of set X is set X.
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And your statement is still wrong.

For example:

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

X={E,F}

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".
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Thank you for proof. The members of X are E and F. And so the union of all the members of X would then be the union of E and F.

doronshadmi, if you cannot get this simple stuff, why should anyone take you more advanced gibberish seriously?
 
jsfisher said:
And your statement is still wrong.



Thank you for proof. The members of X are E and F. And so the union of all the members of X would then be the union of E and F.

doronshadmi, if you cannot get this simple stuff, why should anyone take you more advanced gibberish seriously?
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jsfisher you did not get it. This time please read it very carefully.

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

X={E,F}

The elements (members) of X are E and F, where A,B are the members of E, and A,C,D are the members of F.

Let us quote http://en.wikipedia.org/wiki/Union_(set_theory) :

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.

My case is X=AA=BB, so AA U BB = X U X

My statment "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".


In other words you are wrong becuase {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠XUX=the union of the all members of set X=X

In other words jsfisher , you do not understand http://en.wikipedia.org/wiki/Union_(set_theory) .
 
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jsfisher said:
No, you didn't address my question. I made no mention of "a+b" nor "b+a" nor their equivalence in the referenced question. That is all irrelevant to my question.
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Read again your own words:
jsfisher said:
Symmetry

The property of being invariant with respect to some transformation. For example, in normal addition a+b = b+a, so addition is invariant/symmetric under commutation.
Click to expand...
 
doronshadmi said:
Let E={A,B}
Let F={A,C,D}

X={E,F}

The elements (members) of X are E and F, where A,B are the members of E, and A,C,D are the members of F.

Let us quote http://en.wikipedia.org/wiki/Union_(set_theory):

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.
Click to expand...

Yep. Members of X are E and F, and so the union of the members of X is E union F. In other words, the union of the members of X is the union of E and F, and the union of E and F are all the members of E and all the members of F.

My case is X=AA=BB, so AA U BB = X U X
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No, X = { E, F}. X does not equal E and X does not equal F.

My statment "the union of all members of X is set X" = XUX.
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And much like your other statements, you are wrong.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".
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Yep, wrong. No wonder you are so confused.

In other words you are wrong becuase {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠XUX=the union of the all members of set X=X
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Are not the members of X the sets E and F? Didn't you posit that X = {E, F}? So, the members of X are E and F. If the members of X are E and F, then the union of the members of X is precisely the union of E and F.

Stop being dense. It doesn't help your argument.

In other words jsfisher , you do not understand http://en.wikipedia.org/wiki/Union_(set_theory)
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It's that reading comprehension thing again, isn't it?
 
doronshadmi said:
Let us quote http://en.wikipedia.org/wiki/Union_(set_theory):

If A and B 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.
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That URL does not exist as you cited it. Assuming that you mean this page instead, that quote does not exist in the page.

More to the point,

The elements (members) of X are E and F, where A,B are the members of E, and A,C,D are the members of F.
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If E and F are the members of X, then the union of the members of X is (by definition) E ∪ F, which is the set { A, B, C, D }.

By contrast, the union {E} ∪ {F} is indeed the set X. But neither {E} nor {F} are members of X. {E} is a singleton set whose sole member is E, and {E} is of course a (proper) subset of X. But E and {E} are distinct entities (and in fact, much of the modern foundations of mathematics are built on this distinction. This distinction, for example, is formalized as the "axiom of foundation" in ZF set theory.)

This property is also used for the standard definitions of natural numbers, to wit :
  • 0 = {}
  • 1 = {0} = { {} }
  • 2 = {0,1} = { {}. {{}} }
  • et cetera

If, as you suggest, {0} and 0 were the same, then there would only be one number in existence....

Now, of course, this is mere notational convention. But it's the standard notational convention, and if you use non-standard notation, it's your responsibility both to define the notation you use and to justify your variance from the standard. So far, you have done neither.
 
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doronshadmi said:
jsfisher said:
No, you didn't address my question. I made no mention of "a+b" nor "b+a" nor their equivalence in the referenced question. That is all irrelevant to my question.
Click to expand...

Read again your own words:
jsfisher said:
Symmetry

The property of being invariant with respect to some transformation. For example, in normal addition a+b = b+a, so addition is invariant/symmetric under commutation.
Click to expand...
Click to expand...


That's great, except that wasn't the question at issue. The question had to do with how can a single object (in and of itself) be symmetric.


It's that reading comprehension thing, isn't it?
 
drkitten said:
...By contrast, the union {E} ∪ {F} is indeed the set X. But neither {E} nor {F} are members of X...
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You know what makes this so especially ironic? The starting point for much of this nonsense was that the "set concept" was some sort of hidden assumption. Yet, we see that doronshadmi is incapable of distinguishing between A and {A}.
 
drkitten said:
If E and F are the members of X, then the union of the members of X is (by definition) E ∪ F, which is the set { A, B, C, D }.
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Your mistake is very simple E and F are members of X only in this case which is X={E,F}.

If you take them out of X={E,F} then they are no longer considered as the members of X but they are the two independent sets E={A,B} or F={A,C,D}

Now let us look again at http://en.wikipedia.org/wiki/Union_(set_theory) :

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.

My case is X=AA=BB, so AA U BB = X U X

My statement "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".


In other words you are wrong because {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠XUX=the union of the all members of set X=X

jsfisher said:
You know what makes this so especially ironic? The starting point for much of this nonsense was that the "set concept" was some sort of hidden assumption. Yet, we see that doronshadmi is incapable of distinguishing between A and {A}.
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The ironic thing jsfisher is that you think that set E and set F can be unioned out of X={E,F} and still are concidered as X members.
 
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jsfisher said:
You know what makes this so especially ironic? The starting point for much of this nonsense was that the "set concept" was some sort of hidden assumption. Yet, we see that doronshadmi is incapable of distinguishing between A and {A}.
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It's also hardly "hidden," given that the problem with the (naive) concept of a set is exactly what consigned naive set theory to the flames and was the reason that we got that "proper class" nonsense enjoined upon us after something like a twenty-year holy war across the mathematics journals.

That's like this "hidden city" that I just discovered in the middle of Europe. It's called "Berlin" and no one but me has ever seen it before except for the millions of Berliners....
 
doronshadmi said:
Your mistake is very simple E and F are members of X only in this case which is X={E,F}.
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Since that is what was actually posited, then drkitten is correct.

If you take them out of X={E,F} then they are no longer considered as the members of X but they are the two independent sets E={A,B} or F={A,C,D}
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Even if you don't take them out of X = {E, F}, then the are still two independent sets, E = {A, B} and F = {A, C, D}. What's your point?

Now let us look again...My case is X=AA=BB, so AA U BB = X U X
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Let's look again. Your case is X = {E, F}, and not X = E or X = F. Stop being so dense.

My statement "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".
Click to expand...

Your statement continues to be wrong.

In other words you are wrong because {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠XUX=the union of the all members of set X=X
Click to expand...

You really should work on that reading comprehension thing. Do you have other learning disabilities of which we should know?
 
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doronshadmi said:
My statement "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".
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And this is still wrong. The union of X and X is X. But the union of X and X is not the union of all members of X. You seem to be failing to realize that X and {X} are not the same entity.
 
doronshadmi said:
Your mistake is very simple E and F are members of X only in this case which is X={E,F}.

If you take them out of X={E,F} then they are no longer considered as the members of X but they are the two independent sets E={A,B} or F={A,C,D}
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Er, no.

Even if they are independent sets, E is still distinct from {E}, and E U F is still distinct from {E} U {F}.

Furthermore, it is against the rules to "spam" the forum by repetetively posting the same thing. Given that posts 203, 205, and 211 are near-identical, including down to the mis-citation of Wikipedia (the page you cite does not exist), it's obvious that you are simply cutting-and-pasting without actually bothering either to write new material or even read the criticisms. You are at grave risk of drawing moderator attention....

Anyway,...


Now let us look again at http://en.wikipedia.org/wiki/Union_(set_theory) :

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.
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That quote still doesn't exist in the relevant Wikipedia entry, and the citation is still incorrect.


My case is X=AA=BB, so AA U BB = X U X
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And your case is incorrect.

My statement "the union of all members of X is set X" = XUX.
Click to expand...

And your statement is still incorrect.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".
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But {E,F} != "the union of all the members of set X," so your argument still fails.

Repeating an incorrect argument unchanged will not make it correct.



In other words you are wrong because {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠XUX=the union of the all members of set X=X
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Er, no.
 
jsfisher said:
Even if you don't take them out of X = {E, F}, then the are are two independent sets, E = {A, B} and F = {A, C, D}. What's your point?
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No, when they are members they are parts of {E,F}=X, which is not the same as idependent E or independent F, which are not parts of X.
 
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doronshadmi said:
No, when they are members they are parts of {E,F}=X, which is not the same as idependent E or independent F, which are not parts of X.
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So when you write { {A, B}, {A, C, D}}, the {A, B} is no longer {A, B}?

Interesting. What is it, then?
 
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drkitten said:
So when you write { {A, B}, {A, C, D}}, the {A, B} is no longer {A, B}?

Interesting. What is it, then?
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No, set E={A,B} or set F={A,C,D} are concidered as members only if they a included in some set (let us call it X).

This membership is notated as X={E,F}={{A,B},{A,C,D}}

In thise case X={E,F} U X={E,F} = X
 
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doronshadmi said:
No, set E={A,B} or set F={A,C,D} are concidered as members only if they a included in some set (let us call it X).
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Well, good thing the are included in some set. That would be the set X.

This membership is notated as X={E,F}={{A,B},{A,C,D}}
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Yep. That would be the set in question.

In thise case X={E,F} U X={E,F} = X
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Nope. That would be still be the union of X with X, not the union of the members of X. You really need to work on the reading thing.
 
jsfisher said:
Well, good thing the are included in some set. That would be the set X.



Yep. That would be the set in question.



Nope. That would be still be the union of X with X, not the union of the members of X. You really need to work on the reading thing.
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I think that we understand Membership in two different ways.

I understand Membership as a Whole\parts relation where the parts are not considered as members if they are cut out of the whole.

For example: according to your notion of Membership if you cut your ear(=E) and your finger(=F), you still look at them as members of your body(=X) and then you will find the union of the cells of the ear(=E) and the finger(=F) and say that the result of this union is based on two members of your body(=X).

I disagree with this interpretation of Membership and claim that if ear(=E) or finger(=F) are cut out of body(=X) then they a no longer considered as its members.

According to my interpretation of Membership the union of the members of X is X and it is notated as X={E,F} U X={E,F} = X U X = X.

By using my notion of Membership let us look again at http://en.wikipedia.org/wiki/Union_(set_theory) :

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.

My case is X=AA=BB, so AA U BB = X U X

My statement "the union of all members of X is set X" = XUX.

So {E,F}U{E,F}={E,F}="the union of all the members of set X=set X".


In other words, according to my notion of Membership you are wrong because {A,B}U{A,C,D}= the union of the all members of the all members of set X={A,B,C,D}≠the union of the all members of set X = {E,F} U X={E,F} = X U X = X
 
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doronshadmi said:
According to my interpretation of Membership the union of the members of X is X and it is notated as X={E,F} U X={E,F} = X U X = X.
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All well and good, except your interpretation of membership has absolutely nothing to do with the mathematical concept of membership.

Now, if you want to continue to develop your own alternate formulation of Mathematics, care on, but you need to recognize that you are the one abandoning convention. Don't expect us to follow, and don't assume you are right just on the basis of you changing around all the definitions of things.

You are inhabiting a world of your own imagination. I am fairly certain there is a name for that behavior.
 
jsfisher said:
All well and good, except your interpretation of membership has absolutely nothing to do with the mathematical concept of membership.

Now, if you want to continue to develop your own alternate formulation of Mathematics, care on, but you need to recognize that you are the one abandoning convention. Don't expect us to follow, and don't assume you are right just on the basis of you changing around all the definitions of things.

You are inhabiting a world of your own imagination. I am fairly certain there is a name for that behavior.
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There is no such object called the mathematical concept of membership which is independent of the current agreement of the current community of mathematicians.

Your wrong understanding of the real meaning of Membership, prevents the notion of agreement\membership membership which leads you to think that membership exists independently of some agreement.

In other words any agreed term can be changed.
 
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doronshadmi said:
There is no such object called the mathematical concept of membership which is independent of the current agreement of the current community of mathematicians.

In other words any agreed term can be changed.
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Let us know when you get the community of Mathematics to agree to your bizarre re-definition of terms, re-definition with the sole purpose of covering up your basic misunderstanding of concepts.
 
jsfisher said:
Let us know when you get the community of Mathematics to agree to your bizarre re-definition of terms, re-definition with the sole purpose of covering up your basic misunderstanding of concepts.
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Bizarre?


Bizarre is the notion that leads some person to claim that things that are cut out of X are still considered as its members.


This Bizarre notion is the current agreement of the current community of mathematicians.
 
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doronshadmi said:
Bizarre?


Bizarre is the notion that leads some person to claim that things that are cut out of X are still considered as its members.


This Bizarre notion is the current agreement of the current community of mathematicians.
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So, since the set of even whole numbers includes 2 in its membership, the set of primes can not include 2, because it is already in another set? I don't have to "cut out" 2 from either set to consider 2 separately.

You really missed some basic concepts, didn't you?
 
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jsfisher said:
So, since the set of even whole numbers includes 2 in its membership, the set of primes can not include 2, because it is already in another set? I don't have to "cut out" 2 from either set to consider 2 separately.

You really missed some basic concepts, didn't you?
Click to expand...



2 is a member of the set of even whole numbers as long as it is included in {2,4,6,...}

Also 2 is a member the primes of as long as it is included in {2,3,5,7,...}

2 can be an independent element that is not included in any set, and so is any other element.

For example: if 2 represents some set then {{},{{}}} can be independent of { {{}} , {{},{{}}} , ... } which is the set of natural numbers.
 
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doronshadmi said:
2 is a member of the set of even whole numbers as long as it is included in {2,4,6,...}

Also 2 is a member the primes of as long as it is included in {2,3,5,7,...}

2 can be an independent element that is not included in any set, and so is any other element.
Click to expand...

You were going along so well until this last part, then it just fell apart.
 
jsfisher said:
You were going along so well until this last part, then it just fell apart.
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Not at all.

2 is an independent object, exactly as an ear(=E) cut out of body(=X) is no longer considered as its member.

This independency enables the concept of 2 to be a member of as many sets as you wish, as long as it is included as a part (as a member) of each one of them.

2 is an independent object whether it is made only of sets (for example {{},{{}}}) or not
(for example {ur-elementA,{}} or {ur-elementA,ur-elementB}}

2 is a member of the set of whole even numbers = {2,4,6,...}

2 is a member of the set of prime numbers = {2,3,5,7,...}


The global understanding of this independency can be defined as the set {2, {2,4,6,...}, {2,3,5,7,...}} if we wish to do so.

Any given case above is the result of the invariance(notated by "{" and "}")\variance(notated as some content or its absence) complementation, which is the universal principle of the researchable.

I challenge you to avoid the universal principle and still get the researchable.

Please be aware that any relation (also X=X) is the result of the universal principle.
 
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...and you are still wrong about a set being the union of all its members.
 
Dear jsfisher,

There is a one and only one person that can help you to get the universal principle.

His name his jsfisher.

As long as jsfisher tries to understand the universal principle as something out there by ignore himself as a participator of this understanding, he will not get the universal principle.

Each one of us and all of us together are the result of invariance\variance complementation, that without it no reseachable realm (abstract or not) holds.


(By the way do you like Van Gough ?

I personally think that he was a great artist.

http://www.pep-web.org/document.php?id=paq.030.0351a

" It was at Christmas time, 1888, in the Provençal town of Arles, that Vincent van Gogh cut off a piece of his own ear and presented it to a prostitute. From this event has arisen the image of van Gogh as 'the lunatic who cut off his ear'. But there is a contrasting image—'the genius who painted magnificent pictures'. These two divergent concepts have lent some strength to the naïve but popular notion that one must be insane to be a great artist."

I do not think that anyone in our globe has to be educated to see a cut off ear as something that is still considered as a part (a member) of his body , and then to learn how to do some manipulation on it (present it to a prostitute or define the union of its cells) as if it is still a member of his body.

This manipulation is based on insanity, but many people around our globe are educated by this insanity.

One of the main schools of thought that teaches this insanity is the traditional school thought of the mathematical science.

Do you still look at your self as a well educated person of suach a school?)
 
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doronshadmi said:
Dear jsfisher,

There is a one and only one person that can help you to get the universal principle.

His name his jsfisher.
Click to expand...


Thank you for your concern. However, since this universal principle about which you babble is a personal invention of yourself, and since you are incapable of communicating in any effective manner as to essence of this universal principle about which you babble, the fault for my lack of understanding of this universal principle about which you babble falls to you.

Moreover, you have so thoroughly demonstrated severe limitations on your own cognitive and comprehension skills. For example, you are unable to differentiate among a set, its members and its members' members. As a result, you credentials in Mathematics are tarnished. You cannot get the basics right, and so no one here will take you or your gobbledygook seriously on the more advanced topics.
 
jsfisher said:
Thank you for your concern. However, since this universal principle about which you babble is a personal invention of yourself, and since you are incapable of communicating in any effective manner as to essence of this universal principle about which you babble, the fault for my lack of understanding of this universal principle about which you babble falls to you.

Moreover, you have so thoroughly demonstrated severe limitations on your own cognitive and comprehension skills. For example, you are unable to differentiate among a set, its members and its members' members. As a result, you credentials in Mathematics are tarnished. You cannot get the basics right, and so no one here will take you or your gobbledygook seriously on the more advanced topics.
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A paradigm-shift of fandamental concepts is an internal\external complementation.

Anyone who thinks in terms of internal\external dichotomy cannot get it.

Anyone who thinks in terms of internal\external complementation easily gets internal\external dichotomy and understand its built-in irrationality (insanity).

It is all a matter of rational education, and your education probably does not help you to get the Van Gough's case.
 
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doronshadmi said:
A paradigm-shift of fandamental concepts is an internal\external complementation.

Anyone who thinks in terms of internal\external dichotomy cannot get it.

Anyone who thinks in terms of internal\external complementation easily gets internal\external dichotomy and understand its built-in irrationality (insanity).

It is all a matter of rational aducation.
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And still you persist in blaming everyone else for your own inability to define and explain your unique perspective.
 
jsfisher said:
And still you persist in blaming everyone else for your own inability to define and explain your unique perspective.
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I do not blame anyone personally.

I do blame a traditional education that is based on a built-in irrationality (insanity), as explained in the Van Gough's analogy.

You are simply the result of such education, and no one can blame you for this.

Step-by-step thinkers get the distinct side of invariance\variance complementation.

Parallel thinkers get the non-distinct side of invariance\variance complementation.

No one of them is right if he thinks only in terms of Parallel\Step-by-step dichotomy, because a researchable realm is exactly Parallel\Step-by-step complementation.
 
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doronshadmi said:
I do not blame anyone personally.

I do blame a traditional education that is based on built-in irrationality (insanity), as explained in Van Gough's analogy.
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Ah, so you are blaming the traditional education system for your inability to define terms and express yourself. Ok, that makes more sense.

By the way, do you still content that a set is the union of all of its members?
 
jsfisher said:
You didn't address my question. Instead, you expanded your undefined terminology pool.
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I see we have arrived at corndog territory only with math instead of language.

Will we soon see the universal transform that can change one expression into any other?
 
doronshadmi said:
Yes I did.

a+b=b+a is not symmetry itself but the result of symmetry\asymmetry complementation.

Symmetry itself cannot be addressed by using the Multitude, because symmetry itself is total unity where no multitude exists.

If you try to understand Symmetry in terms of one of many things (cardinal=1) or many things (cardinal>1), then you are at symmetry\asymmetry complementation that is not symmetry itself.

If you try to understand Symmetry itself by using cardinal=0 as a complement of cardinal≠0, you are not at Symmetry itself because you are thinking in terms of Nothing\Something complementation.

But Symmetry itself is total unity and it is not researchable.

Asymmetry itself is total isolation and it is not researchable.

The researchable is Symmetry\Asymmetry complementation, but then any researchable thing is the complementation's result.

If you unaware of this notion you are unaware of the fact that any given premise is already a notion that is based on the researchable (which is the result of Symmetry\Asymmetry complementation).

In other words, since mathematics is a researchable thing, it must be aware of researchable\non-researchable complementation.

This awareness is not used as the basis of the mathematical science, and one of the wrong results is to understand the multitude (which is a complementation's result) in terms of the completeness of Symmetry at its self state (which is beyond the researchable and no researchable concept like cardinality has its complete self-state).

You have to be aware of the fact that our dialog is possible because of symmetry\asymmetry complementation, and it cannot be done at Symmetry or Asymmetry self-state.

So the fundamental question of a researchable abstract\non-abstract space is:

How the researchable is possible?

I much as I know, this question was not asked for the past 2500 years of mathematical research, and any axiom that I know is based on non-proved assumptions that are already the result of researchable\non-researchable complementation, without being aware of it.

Furthermore, those axioms are tuned to work only on the asymmetrical side of Symmetry\Asymmetry complementation because Mathematics of the past 2500 years is mostly the result of a step-by-step way of thinking, until this particular way of thinking and Mathematics are synonym.

By understanding this it must not be a surprise to be realized of the fact that Mathematics of the past 2500 years is mostly the result of a step-by-step serial thinkers, and the community known as mathematicians are nothing but a community of persons that are qualified in step-by-step way of thinking, which is nothing but the particular case of an asymmetric point of view of Symmetry\Asymmetry complementation.

In that case we must not be surprised to understand why so many people around the globe do not like Mathematics.

If carefully be examined we discover that most of them are not step-by-step thinkers that are actually forced to do mathematics in term of step-by-step thinkers.

There is no conspiracy of step-by-step of thinkers here but a simple (and devastating) ignorance of Researchable\Non-researchable complementation.

As much as I know my suggested Mathematical framework is the first theory that is aware of the Researchable\Non-researchable complementation.


I'll be more than glade to know that I am wrong about what I wrote above.
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Be glade, you are wrong.

Repeating 'catch phrases" like "step-by-step" makes me wonder about your creativity and therefore your ability to come up with anything original.
 
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doronshadmi said:
I do not blame anyone personally.

I do blame a traditional education that is based on a built-in irrationality (insanity), as explained in the Van Gough's analogy.

You are simply the result of such education, and no one can blame you for this.

Step-by-step thinkers get the distinct side of invariance\variance complementation.

Parallel thinkers get the non-distinct side of invariance\variance complementation.

No one of them is right if he thinks only in terms of Parallel\Step-by-step dichotomy, because a researchable realm is exactly Parallel\Step-by-step complementation.
Click to expand...

Please don't cut off your ear.

You don't hear well enough as it is.
 
jsfisher said:
Ah, so you are blaming the traditional education system for your inability to define terms and express yourself. Ok, that makes more sense.

By the way, do you still content that a set is the union of all of its members?
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By the way do you still content that the a set of union of X any Y that are cut out of set X is still connected to set X?

To claim that X or Y that cut out of set X are still considered as its members, is nothing but a contradiction.

Shell we clap hands for the impressive success of the traditional education method on your "reasonability"?
 
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