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.