Help with set proof: $A \cap B = A $ if and only if $A \subseteq B $. 
$A \cap B = A $ if and only if $A \subseteq B $

It's been a while since I've done this sort of proof. I can't think of how I would prove this statement. I'm too used to numerical proofs.
What kind of proof would I use and how?
 A: Suppose $A\not\subset B$, then there exists an $a\in A$ such that $a\notin B$. Hence $a\notin A\cap B$. Thus $A\cap B\neq A$.
Conversely, if $A\cap B=A$ then $A\subset A\cap B$, so for each $a\in A$ we have that $a\in A\cap B$, which implies that $a\in B$. Thus $A\subset B$.
A: Hints:


*

*You have to show both ways (as mentioned in Hawks' comment).

*Use proof by contradiction.

*Consider one statement is true, then take an element from that set and consider that does not belong to the other. Finally, show that it is not possible.

A: To prove $A\cap B=A \Rightarrow A\subseteq B$ you can use the identity $B=(A\cap B)\cup(B\setminus A)$. If $A\cap B=A$, then $B=A\cup(B\setminus A)$. Hence $A\subseteq B$.
A: $(\Rightarrow ):$
  $$A= A \cap B\subseteq B$$
$(\Leftarrow ):$
By definition, $A \cap B\subseteq A$. To see converse note that: For all $x\in A$, by assumption $x\in B$. So $x\in A\cap B$
A: Let $A\cap B=A$ Show that $A\subseteq B$.
Let $A\subseteq B$, show that $A\cap B=A$.
Where is your problem?
