# Does $A\cap B =\varnothing \Rightarrow B\subseteq \overline{A}$?

How to prove $A\cap B =\varnothing \Rightarrow B\subseteq \overline{A}$? If I going by definitions, there is no $x$ s.t $x\in A$ and $x\in B$. But, what do we can tell about $\overline{A}$? What i'm missing?

• @Battani I think that $\overline{A}$ denotes the complement of $A$. – Crostul Aug 3 '15 at 10:07

Let's assume $A\cap B =\varnothing$ (start hypothesis)

Let $x \in B$

Since $A$ and $B$ are disjoint (start hypothesis), then $x \notin A$

By definition of $\overline A$, since $x \notin A$ then $x \in \overline A ~~~~(= \Omega - A)$

Therefore $B\subseteq \overline A$, because for all $x \in B$, we have $x \in \overline A$

Note that the reciprocal is also true.

Let $x \in B$. We want to prove $x \in \overline{A}$, i.e. that $x \notin A$.

So assume $x \in A$. But then, we have $x \in A$ and $x \in B$, which means $x \in A \cap B$. However, by assumption, $A \cap B$ is a empty. This is a contradiction.

This proves $x \notin A$.

Actually, the converse is also true: $A \cap B$ is empty if and only if $B$ is contained in $\overline{A}$.

• Dear downvoter. Could you please explain what is not clear in my answer? – Martin Brandenburg Aug 3 '15 at 10:19

Suppose $B\nsubseteq \overline{A}$.

Therefore exists an element $x$ so that $x\in B$ and $x \notin \overline{A}$
Then $x\in A$ hence $x\in {A}\cap B$, that's a contradiction.