Proving Set Theory Proof: (A-C) \cap (B-C) \cap (A-B) = ∅

[cmajor47]

Homework Statement

Prove that for all sets A, B, and C, (A-C) $$\cap$$ (B-C) $$\cap$$ (A-B) = ∅

Homework Equations

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?

 

[Dick]

Homework Statement

Prove that for all sets A, B, and C, (A-C) $$\cap$$ (B-C) $$\cap$$ (A-B) = ∅

Homework Equations

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

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

 

[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?

 

[Dick]

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