Proof of Partition Property for Subset A in Universal Set U

[TheMathNoob]

Homework Statement

Assume {B, C, D} is a partition of the universal set U, A is a subset of U and A is not a subset of B complement, A is not a subset of C complement, A is not a subset of D complement. Prove that {A ∩ B, A ∩ C, A ∩ D} is a partition of A.

Homework Equations

The Attempt at a Solution

I know that this is right intuitively. I know how explain it with words, but I don't know how mathematically.

A is not a subset of B complement, A is not a subset of C complement, A is not a subset of D complement implies that A has to be distributed among the three subsets. There is nothing left of A because B,C and D is a partition of the universal set. Therefore, the union of the pieces in which A overlaps with B,C,D is A. This pieces are going to be disjoint. Mathematically, I can prove that

(AnB)n(AnC)n(AnD)=An(BnCnD)= An empty set =empty set

 

[Ssnow]

You must prove that the intersection of every two pairs of your new family $\{A\cap B, A\cap C, A\cap D\}$ is empty (using the fact that $\{B,C,D\}$ is a partition of $U$ and that in fact $A$ is not contained in the complement of every set of the partition (that is the same that $A$ has intersection noempty with every set in the partition...)) and that the union of all is $A$. You will use the distributive law for intersection and union ...