Question

Proof: Suppose A, B, and C are any sets. [To show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), we must show that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) and that (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C).] g

Answer 1

I suppose you're supposed to prove that set intersection is distributive across a union,

`A\cap(B\cup C)=(A\cap B)\cup(A\cap C)`

Two sets are equal if they are subsets of one another. To prove a set
`X` is a subset of another set
`Y`, you have to show that any element
`x\in X` also belongs to
`Y`.

Let
`x\in A\cap(B\cup C)`. By definition of intersection, both
`x\in A` and
`x\in B\cup C`. By definition of union, either
`x\in B` or
`x\in C`. If
`x\in B`, then clearly
`x\in A\cap B`; if
`x\in C`, then
`x\in A\cap C`. Either way,
`x\in(A\cap B)\cup(A\cap C)`. Hence
`A\cap(B\cup C)\subseteq(A\cap B)\cup(B\cap C)`.

The proof in the other direction uses the same sort of reasoning. Let
`x\in(A\cap B)\cup(A\cap C)`. Then either
`x\in A\cap B` or
`x\in A\cap C`. If
`x\in A\cap B`, then both
`x\in A` and
`x\in B`; if
`x\in A\cap C`, then both
`x\in A` and
`x\in C`. So certainly
`x\in A`, and either
`x\in B` or
`x\in C` so that
`x\in B\cup C`. Hence
`(A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C)`.

Both sets are subsets of one another, so they are equal.