# Set operation problem

1. Sep 16, 2004

### Yoss

Suppose there are 3 sets: A, B, and C s.t. (U for 'union')

(A - B) U (A - C) U (B - C)

Now, I was wondering if there is the precedence of parentheses over set operations (union in this case).
This is saying "every element in A but not in B" or "every element in A but not in C" or "every element in B but not in C".
I know that nothing in C is contained in this union and that everything in A is minus those that are in both A and B and both A and C. Now, the last difference, B - C, annoys me. Would B - C override A - B and include every element in B but not in C (and not those in both A and B)?

Would this be a case of symmetric difference? For example
(A - B) U (B - A)

would this include everything but the intersection (if A and B are not disjoint or empty and A does not equal B)

(sidenote: Has there been a problem with Latex, it wouldn't show this tag:

[tex]\left(A\setminus B\right)\cup \left(A\setminus C\right)\cup \left(B\setminus C\right)[/tex (last bracket intentionally left out so it would show text))

Last edited: Sep 16, 2004
2. Sep 16, 2004

### arildno

"I know that nothing in C is contained in this union and that everything in A is minus those that are in both A and B and both A and C. Now, the last difference, B - C, annoys me. Would B - C override A - B and include every element in B but not in C (and not those in both A and B)?"
This is plain wrong!
A-B may well have elements also in C. These elements are present in the union
There is no "overriding" present; each member of the union provides its own elements to the union; those elements are not removed from the union even if these elements are lacking from other members of the union.

3. Sep 16, 2004

### Yoss

The definition of A - B (difference) states {(for all x)| x is an element of A and x is not an element of B).
Therefore (A - B) does not contain elements that are in C unless they are in both A and C. But then the second difference, (A - C), removes those elements that are both in A and C. So, thus far (A - B) U (A - C) contains elements exclusive to A.

4. Sep 16, 2004

### Betty-boop

I think you (Yoss) may have the wrong idea of "union". If (A-B) has elements of C, then those elements have become part of the union, regardless of what's in the set (A-C). The parentheses take precedence over the union.

Last edited: Sep 16, 2004
5. Sep 17, 2004

### arildno

"A-B may well have elements also in C. These elements are present in the union
There is no "overriding" present; each member of the union provides its own elements to the union; those elements are not removed from the union even if these elements are lacking from other members of the union."
You have a faulty understanding of the union concept.

6. Sep 17, 2004

### HallsofIvy

Have you tried any examples?
Let A= {every member of the alphabet except a}, B= {every member of the alphabet ecxept b}, C= {every member of alphabet except c}.

Then A-B= {b}, A-C= {c}, B- C= {c}

(A-B)U(A-C)U(B-C)= {b, c}