Show that $A \cap B = B$ iff $A \cup B = A$, where $A \subseteq B$. 
Show that $A \cap B = B$ iff $A \cup B = A$, where $A \subseteq B$.  

I have tried to do this by element-chasing, but I just end up saying that $x \in A$ and $x \in B$. I am really stuck for hours on this. Can somebody help please?
 A: Let's organize our thoughts. Assume all through this answer that $A \subset B$. Also, assume that $A$ and $B$ are non-empty, so we don't become paranoid. We have four things to do here.


*

*Prove the $\implies$ direction. Assume that $A \cap B = B$. Don't assume that $A \cup B = A$.


1) Check that $A \cup B \subset A$.
2) Check that $A \subset A \cup B$.


*

*Prove the $\impliedby$ direction. Now assume that $A \cup B = A$. Don't assume that $A\cap B = B$.


3) Check that $A \cap B \subset B$.
4) Check that $B \subset A \cap B$.
Notice that items 2) and 3) are always true, independent of our hypothesis that $A \subset B$. That being said, I'll do item 1) and you'll do item 4). 
Take  $x \in A \cup B$. If $x \in A$, there is nothing to do. If $x \in B$, since $A \cap B = B$, we have that $x \in A \cap B$. But $x \in A \cap B$ means, in particular that $x \in A$. So, either way $x \in A$. Since $x$ was arbitrary, we get $A \cup B \subset A$, as wanted.
A: So if $A \cup B = A$, then either $B \subset A $ or $A = B$. In this case, $A = B$. Then $A \cap B = B$.
A: You don't need $\color {red}{A \subseteq B}$. In fact $$A\cap B=B\iff B\subseteq A\iff A\cup B=A.$$ For the first "$\iff$" see here. For the second:
$\Rightarrow :$ Let $x\in A\cup B$ then either $x\in A$ or $x\in B$; in each case $x\in A$. So $A\cup B \subseteq A$, and obviously $A \subseteq A\cup B$.
$\Leftarrow :$ $\forall x\in B, x\in A\cup B=A $
