A hidden assumption

[jsfisher]

You didn't address my question. Instead, you expanded your undefined terminology pool.

Yes I did.

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

...<additional gibberish>...

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.

 

[jsfisher]

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 .

For your reference and use, I have highlighted your mistake with bold text.

 

[doronshadmi]

For your reference and use, I have highlighted your mistake with bold text.

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}

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

 

[jsfisher]

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.

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".

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?

 

[doronshadmi]

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?

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) .

 

[doronshadmi]

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.

Read again your own words:

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.

 

[jsfisher]

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.

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

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.

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".

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

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)

It's that reading comprehension thing again, isn't it?

 

[drkitten]

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.

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.

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.

 

[jsfisher]

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.

Read again your own words:

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.

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?

 

[jsfisher]

...By contrast, the union {E} ∪ {F} is indeed the set X. But neither {E} nor {F} are members of X...

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}.

 

[doronshadmi]

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 }.

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

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}.

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.

 

[drkitten]

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}.

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....

 

[jsfisher]

Your mistake is very simple E and F are members of X only in this case which is X={E,F}.

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}

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

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".

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

You really should work on that reading comprehension thing. Do you have other learning disabilities of which we should know?

 

[ellindsey]

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".

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.

 

[drkitten]

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}

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.

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

And your case is incorrect.

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

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".

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

Er, no.

 

[doronshadmi]

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?

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.

 

[drkitten]

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.

So when you write { {A, B}, {A, C, D}}, the {A, B} is no longer {A, B}?

Interesting. What is it, then?

 

[doronshadmi]

So when you write { {A, B}, {A, C, D}}, the {A, B} is no longer {A, B}?

Interesting. What is it, then?

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

 

[jsfisher]

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).

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}}

Yep. That would be the set in question.

In thise case X={E,F} U X={E,F} = X

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.

 

[doronshadmi]

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.

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