Prove by pickin a point that (A n C) is a subset of B if (A-B) n C = null set?

Answer 1

Assume that (A-B) n C = null.
Then (A n B') n C = null (by definition of "-", where B' is the complement of B)
Then (A n C) n B' = null (because intersections are both commutative and associative)
That means that (A n C) and B' are mutually exclusive any element that is in (A n C ) is not in B'.

Let x be in A n C.
Then (from above) x is not in B'.
Then x is in B.
Then (A n C) is a subset of B

Answer 2

I will use _ to indicate the complement.

Write A = A ∩ (B ∪ B_) = (A ∩ B) ∪ (A ∩ B_) = (A ∩ B) ∪ (A - B).

Then (A ∩ C) = ((A ∩ B) ∩ C) ∪ ((A - B) ∩ C)
= (A ∩ B ∩ C) ∪ (null set)
= (A ∩ C) ∩ B

(A ∩ C) = (A ∩ C) ∩ B, so (A ∩ C) ⊆ B.

Alternatively, suppose that (A ∩ C) is not a subset of B and let x ∈ (A ∩ C) ∩ B_. Then x ∈ A, x ∉ B and x ∈ C. So x ∈ A-B and x ∈ C, so x ∈ (A-B) ∩ C. Thus, if (A-B) ∩ C is the null set, no such x can exist, and therefore A ∩ C is a subset of B.