Prove that for all sets $A$, $B$, and $C$, if $A\cap{B}=\emptyset$ and $A\cap{C}=\emptyset$, then $A\cap({B}\cup {C})=\emptyset$.

Question

Prove that for all sets $A$, $B$, and $C$, if $A\cap{B}=\emptyset$ and $A\cap{C}=\emptyset$, then $A\cap({B}\cup {C})=\emptyset$.

I know that this is obviously true, however I'm not sure how to prove it.

Any help is appreciated, thanks!

Answer 1

$$A\cap({B}\cup {C})=(A\cap B)\cup (A\cap C)=\emptyset\cup\emptyset=\emptyset.$$

Answer 2

Suppose not, suppose if $A \cap B = \emptyset$ and $B \cap C = \emptyset$ then $A \cap (B \cup C) \ne \emptyset$.

Then there must be some $x \in (A \cap B) \cup (B \cap C)$ by the distributive property of intersection over unions.

So then there must be some $x \in (A \cap B)$ or $x \in (B \cap C)$ by the definition of the union.

However, the premise states that both $A \cap B = \emptyset$ and $B \cap C = \emptyset$, so it is impossible for an $x$ to be in either of these sets.

Therefore, we reach a contradiction - it must be the case if $A \cap B = \emptyset$ and $B \cap C = \emptyset$ then $A \cap (B \cup C) = \emptyset$.

EDIT:

For clarification, the trick here is to notice that you can:

1. Distribute $A$ over $B \cup C$ in order to get the two terms $(A \cap B)$ and $(B \cap C$).

2. Use contradiction with the properties of the union to show that an element in the consequent would imply an element in the two sets in the antecedent.

I think in terms of thinking about it, the "enlightening" moment comes from realizing you can distribute the intersection in the way above. The rest follows in a very straight forward, logical fashion.

Answer 3

Proof. A claim as follows may be needed.

Claim (Distributive Law). For any sets $A,B,C$, we have $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$.

Proof of the Claim. This is because:

\begin{align*} x\in A\cap (B\cup C)&\iff x\in A~\text{and}~x\in B\cup C\\ &\iff x\in A~\text{and}~(x\in B~\text{or}~x\in C)\\ &\iff (x\in A~\text{and}~x\in B)~\text{or}~(x\in A~\text{and}~x\in C)\\ &\iff x\in(A\cap B)~\text{or}~x\in(A\cap C)\\ &\iff x\in(A\cap B)\cup(A\cap C). \end{align*}

Now return to the proof as desired:

\begin{array}{rcll} A\cap({B}\cup {C})&=&(A\cap B)\cup (A\cap C)&\text{by Distributive Law}\\ &=&\emptyset\cup\emptyset&\text{by}~A\cap B=\emptyset~\text{and}~A\cap C=\emptyset\\ &=&\emptyset.& \end{array}