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

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- Oct 29th 2008, 05:24 PMcaptainjapanSets
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 - Oct 29th 2008, 05:55 PMsuperevilcube
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. - Oct 29th 2008, 06:17 PMpoutsos.B

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

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

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

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

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

Φ being the empty set