Two sets are disjoint if and only if one is contained in the complement of the other Prove that $A\cap B = \emptyset$ iff $A\subset B^C$. I figured I could start by letting $x$ be an element of the universe and that $x$ is an element of $A$ and not an element of $B$. 
 A: You must prove both implications, that is: if $A\cap B=\emptyset$ then $A\subset B^c$ and conversely: if $A\subset B^c$ then $A\cap B=\emptyset$.
For the first: A good way to prove that some set is a subset of another one is supposing that $x$ is in the subset and proving that $x$ is in the superset: if $x\in A$, then it must not be in $B$, because $A$ and $B$ have no common elements. Then $x$ is in $B^c$.
For the second: A good way to prove that a set is empty is supposing that $x$ belongs to it and deriving a contradiction: if $x\in A\cap B$ then $x\in B$ and $x\notin B$.
A: $\textbf{Claim: } A\cap B=\emptyset$ iff $A\subset B^c$.
$Proof:\; A\cap B=\emptyset \Leftrightarrow (a\in A\Rightarrow a\notin B)\Leftrightarrow (a\in A\Rightarrow a\in B^c)\Leftrightarrow A\subset B^c$
A: Recall some definitions: Let $\Omega$ denote the universe (At least $A\cup B$).
$$B^C = \{ x \in\Omega | x\notin B\} \\
M\subset C :\Leftrightarrow \forall m\in M: m\in C \\
A\cap B := \{x\in\Omega | x\in A \wedge x\in B\}$$
Now plug this in: $A\subset B^C \Rightarrow A\cap B = \emptyset$ starts with any $x\in \Omega$ we know that $x\in A \Rightarrow x\notin B$ so $\nexists x\in \Omega : x\in A \wedge x\in B \Rightarrow A\cap B = \emptyset$. Now try the other direction.
A: Hint: If $A \cap B = \emptyset $ then what can you say about all elements of $A$? Use this with the definition of $B^C$. $B^C$ is the set of all elements in the universe that are not in $B$, $(U-B)$.
$$ A \cap B = \emptyset \iff \forall x\in A, x\notin B \iff \forall x\in A,x \in B^C \iff A\subset B^C  $$
