# Proof with disjoint and sets.

Prove that $A$ and $B$ are disjoint if and only if $A\subseteq B'$

What I know:

If $S\cap T =\emptyset$ then $S$ and $T$ are said to be disjoint.

The intersection of two sets $S$ and $T$ is the set $S\cap T$, consisting of all elements that are both in $S$ and $T$, hence $S\cap T =$ {$x | x\in S$ AND $x\in T$}.

Maybe something along the lines of

if $A\subset B'$: $$A\cap B' = A$$ therefore $$A\cap B = \left(A \cap B'\right)\cap B = A \cap \left(B'\cap B\right) = A \cap \emptyset = \emptyset$$

if $B' \subset A$ then there $$A\cap S = A = A\cap(B'\cup B) = (A \cap B') \cup (A \cap B) = B' \cup (A \cap B)$$ if $A \cap B = \emptyset$ then $A = B'$ which contradicts the first statement of $B' \subset A$

for $A = B'$ it is easy to sure that this disjoint.

Here is a simple proof. Let's start in the forward direction.

Suppose $A$ and $B$ are disjoint. Then $\forall a \in A$, $a \notin B \Longrightarrow a \in B'$. Thus we have that $A \subseteq B'$, because every element of $a$ must also be in $B'$.

Now to prove the reverse statement, let $A \subseteq B'$. Then $\forall a \in A$, we also have $a \in B' \Longrightarrow a \notin B$. Similarly, if $b \in B$ then $b \notin B'$, so especially we cannot have $b \in A$. Therefore $A$ and $B$ are disjoint.