# Thread: Interesting Proof for Sets

1. ## Interesting Proof for Sets

If C is a subset of A and D is a subset of B, prove D-A is a subset of B-C.

I'm not sure how to do this. I know that I need to prove that every element in D-A is also in B-C, but how?

2. Originally Posted by veronicak5678
If C is a subset of A and D is a subset of B, prove D-A is a subset of B-C.

I'm not sure how to do this. I know that I need to prove that every element in D-A is also in B-C, but how?
So, you know what you want to get to, so just start, and follow your nose. where did you get stuck?

We need to show that $\displaystyle \displaystyle x \in (D - A) \implies x \in (B - C)$, so...

Assume $\displaystyle \displaystyle x \in (D - A)$. Then, $\displaystyle \displaystyle x \in D$ but $\displaystyle \displaystyle x \not \in A$. But $\displaystyle \displaystyle C \subseteq A$ and so $\displaystyle \displaystyle x \in C \implies x \in A$, or equivalently, by the contrapositive, $\displaystyle \displaystyle x \not \in A \implies x \not \in C$. Thus we have that $\displaystyle \displaystyle x \not \in C$....

How do you think you should proceed now?

3. I can prove with the same logic that any x in D-A is in B because D is a subset of B. We now have x is in B, not in C, or B-C.

Thanks for your help! Sometimes I just have trouble getting started.

4. Originally Posted by veronicak5678
I can prove with the same logic that any x in D-A is in B because D is a subset of B. We now have x is in B, not in C, or B-C.

Thanks for your help! Sometimes I just have trouble getting started.
yup

so, you knew what you needed to prove. It was an implication, $\displaystyle \displaystyle P \implies Q$. Those you can prove one of three ways.

Directly: Assume $\displaystyle \displaystyle P$ and show $\displaystyle \displaystyle Q$ follows.

By the contrapositive (prove $\displaystyle \displaystyle \sim Q \implies \sim P$): Assume $\displaystyle \displaystyle \sim Q$ and show $\displaystyle \displaystyle \sim P$ follows.

By contradiction: Assume $\displaystyle \displaystyle P \wedge \sim Q$ is true, and show that a contradiction arises.

With set proofs like this, i find that the direct proof is often the best way to go. So just assume the antecedent is true (the first statement in the implication, the P), and try to get to the consequent (the Q). So that's how you'd start if you don't know what to do. just jump in and hope for the best.