Question

Let A, B and C be sets. Then show that

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)​

Answer 1

Answer:

Since B∩C⊆BB∩C⊆B and B∩C⊆CB∩C⊆C, we have

A∪(B∩C)⊆A∪B

A∪(B∩C)⊆A∪B

and

A∪(B∩C)⊆A∪C

A∪(B∩C)⊆A∪C

This shows that A∪(B∩C)A∪(B∩C) is contained in both A∪BA∪B and A∪CA∪C, so it is contained in their intersection:

A∪(B∩C)⊆(A∪B)∩(A∪C)

A∪(B∩C)⊆(A∪B)∩(A∪C)

This proves containment in one direction.

For the opposite direction, suppose that x∈(A∪B)∩(A∪C)x∈(A∪B)∩(A∪C). There are two possibilities: either x∈Ax∈A or x∉Ax∉A.

If x∈Ax∈A then certainly x∈A∪(B∩C)x∈A∪(B∩C).

On the other hand, if x∉Ax∉A, then xx must be in both BB and CC, since x∈(A∪B)∩(A∪C)x∈(A∪B)∩(A∪C). Consequently, x∈B∩Cx∈B∩C, and therefore x∈A∪(B∩C)x∈A∪(B∩C).

In both cases we have x∈A∪(B∩C)x∈A∪(B∩C). This proves the containment

(A∪B)∩(A∪C)⊆A∪(B∩C)

(A∪B)∩(A∪C)⊆A∪(B∩C)

so we're done.