1. ## Sets

Let A and B be sets. Prove each of the following statements. Remember, using a Venn diagram might help to visualize the problem, but it's not a proof!

a. (A B) ⊆ A
b. A ⊆ A ∪ B
c. A (B - A) = 0

2. a) (A B) ⊆ A

Let x be in (A B), then x is in A and x is in B. Hence x is in A. Thus (A B) ⊆ A

b) A ⊆ A ∪ B

This one seems so easy to prove, I'm not sure what else to say about it:

Let x be in A. If x is in A then x is in A ∪ B. Thus A ⊆ A ∪ B

c) A (B - A) = 0

(B - A) = {x | x is in B and x is not in A}. So intuitively this equality should make sense. You take everything that is in A and intersect with everything that is in B takeaway A, so A and (B-A) should be mutually disjoint sets).

For some reason I can't think of how to prove this one, sorry. Someone else will have to tackle it.

3. Originally Posted by captainjapan
Let A and B be sets. Prove each of the following statements. Remember, using a Venn diagram might help to visualize the problem, but it's not a proof!

a. (A B) ⊆ A
b. A ⊆ A ∪ B
c. A (B - A) = 0

a) xε( Α $\cap$ B) <====> (xεA & xεB) ====> xεA.

b) xεA =====> xεA v xεB <=====> xε( AUB)

c) xε[A $\cap$ (B-A)] <====> xεA & (xεB & ~xεA) <=====>

(xεA & ~xεΑ) & xεB <====> xε ( Α $\cap$ A') & xεB <=====> xε( Φ $\cap$ B) <====> xεΦ.

Because A $\cap$ A' = Φ and , Φ $\cap$ B =Φ

Φ being the empty set