Set Theory Proof

1. Apr 4, 2009

cmajor47

1. The problem statement, all variables and given/known data
Prove that for all sets A, B, and C, (A-C) $$\cap$$ (B-C) $$\cap$$ (A-B) = ∅

2. Relevant equations

3. The attempt at a solution
Proof: Suppose A, B, and C are sets
Let x $$\in$$ (A-C) $$\cap$$ (B-C) $$\cap$$ (A-B)
Since x $$\in$$ (A-C), by definition of difference, x $$\in$$ A and x $$\notin$$ C
Since x $$\in$$ (B-C), x $$\in$$ B and x $$\notin$$ C
Since x $$\in$$ (A-B), x $$\in$$ A and x $$\notin$$ B
Then by definition of intersection, if x $$\in$$ A then x $$\notin$$ C and x $$\notin$$ B
Also, if x $$\in$$ B then x $$\notin$$ C
Therefore there is no intersection of sets A, B, and C
Therefore, the intersection of (A-C) $$\cap$$ (B-C) $$\cap$$ (A-B) = ∅

Is this proof correct, I feel like I am missing something?

2. Apr 4, 2009

Dick

You are doing fine up to here. Do you see anything contradictory in those conditions you have on x?

3. Apr 4, 2009

cmajor47

It states that x $$\in$$ B and x $$\notin$$ B, which isn't possible.
Do I just say that since this is a contradiction, the intersection is the null set?

4. Apr 4, 2009

Dick

Sure. There is no x that can satisfy those two conditions.