# Set theory proof help?

1. Oct 4, 2012

### bonfire09

1. The problem statement, all variables and given/known data
Let A and B be sets. Prove that A U B=(A-B) U (B-A)U (A∩B)

2. Relevant equations

3. The attempt at a solution
If I want to prove the foward direction: A U B⊆(A-B) U (B-A)U (A∩B) then
from my understanding I know that xεA or xεB. And I can assume wolog that xεA. But since I assume that x is an element from A, its already equal to (A-B) U (B-A)U (A∩B)? Since xεA and x is not an element of B or xεb and not an element of A or x is both an element of A and B? is that all i have to show for the foward direction?

Last edited: Oct 4, 2012
2. Oct 4, 2012

### bonfire09

ok i was thinking this for a while. Lets take two sets A={1,2,3,4,5} and {3,4,5,6,7}
so I can divide this proof into 3 cases:
case 1: xεA
case 2: xεB
case 3: xεA and xεB.

am i right?

3. Oct 4, 2012

### happysauce

Im not sure what you've covered so far. Do you know the logical definition of a subset?

4. Oct 4, 2012

### bonfire09

Foward Direction:A U B⊆(A-B) U (B-A)U (A∩B)
Case 1: xεA
Since xεA then xε(A-B). xε(A-B) means that xεA and x is not an element of B.
Case 2: xεB
Since xεB then xε(B-A) which means that xεB and x is not an element of A.
Case 3:xεA and xεB
Since xεA and xεB then xε(A∩B) means that xεA and xεB.
Hence A U B⊆(A-B) U (B-A)U (A∩B).

Reverse direction
(A-B) U (B-A) U (A∩B)⊆A U B
Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

Case 1:xε(A-B)
Since xε(A-B) then this means that xεA and x is not a element of B. Which means that xε(AUB). xε(AUB) means that xεA or xεB. Hence (A-B)⊆A U B.

Case 2: xε(B-A)
Since xε(A-B) then this means that xεB and x is not an element of A. Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

Case 3: xε(A∩B)
Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB. Hence (A∩B)⊆A U B.
Therefore (A-B) U (B-A) U (A∩B)⊆A U B
Im not entirely sure if this proof is right. It seems verbose but then again im taking an introduction class.

Last edited: Oct 4, 2012
5. Oct 5, 2012

### happysauce

Not quite. Let me go through what you did.

Case 1: xεA

"Since xεA then xε(A-B)" (This is not true consider x in both A and B. Then x wont be in A-B but it is in A). xε(A-B) means that xεA and x is not an element of B.
Case 2: xεB
"Since xεB then xε(B-A)" (Again this is not true) which means that xεB and x is not an element of A.
Case 3:xεA and xεB
Since xεA and xεB then xε(A∩B) "means that xεA and xεB (redundant)".
Hence A U B⊆(A-B) U (B-A)U (A∩B).

Reverse direction
(A-B) U (B-A) U (A∩B)⊆A U B
Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

Case 1:xε(A-B)
Since xε(A-B) then this means that xεA "and x is not a element of B (useless info)". Which means that xε(AUB). xε(AUB) "means that xεA or xεB (This step should be before you say x is in AUB). Hence (A-B)⊆A U B.

Case 2: xε(B-A)
Since xε(A-B) then this means that xεB "and x is not an element of A (useless info)". Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

Case 3: xε(A∩B)
Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB (this step should be before concluding xε(AUB)) . Hence (A∩B)⊆A U B.
Therefore (A-B) U (B-A) U (A∩B)⊆A U B
Im not entirely sure if this proof is right. It seems verbose but then again im taking an

The first direction needs more work. The second just needs a little fix. The first direction is harder and it might require showing that A = (A-B)U(A∩B). Also, I'm not sure if your class covered this but, xε(A-B) <=> xε(A∩B$^{c}$) (B$^{c}$) is the complement of B)

6. Oct 5, 2012

### bonfire09

since the definition of AUB is xεA or xεB then I thought that x could be in either set A or set B or both. That was the reason behind my three cases. Oh maybe I should have written it like this for the foward direction

Case 1: xεA and x is not element of B
Case 2: xεA and xεB
Case 3: xεB and x is not an element of A but that is the same as case 1.

7. Oct 6, 2012

### happysauce

I'm not sure what you are trying to justify.
Here is the definition of a subset:
A$\subset$B $\Leftrightarrow$ for any x, x $\in$A $\Rightarrow$ x $\in$B