a problem on the algebra of sets trying to prove that $A \cup ( A \cap B) = A $ for any set $A,B$. 
I am trying to use distributive law for sets, but keep coming to the same form. Is there a way to prove this ? 
 A: Yes, the way to prove this is to prove $A \cup (A \cap B) \subseteq A$ and also that $A \subseteq A \cup (A \cap B)$.  So you have two statements to prove.  I'll help you with the first one and you can do the second one on your own.  
To prove $A \cup (A \cap B) \subseteq A$, we need to show if $x \in A \cup (A \cap B)$, then $x \in A$.
Well, let $x \in A \cup (A \cap B)$, then.  By definition of union, this means $x \in A$ or $x \in A \cap B$.  If $x \in A$, then we are done since that is what we wanted to show.  If on the other hand $x \in A \cap B$, then that means $x \in A$ and $x \in B$, so we get $x \in A$ in this case, too.  Thus, in every case, $x \in A$, so we are done.
Now it is up to you to prove if $x \in A$, then $x \in A \cup (A \cap B)$. 
A: It is clear that $A‎‎\subseteq A \cup ( A \cap B)  $, on the other hand $‎‎A\subseteq A$ and $A \cap B\subseteq A$ thus $A \cup ( A \cap B)\subseteq A $
A: Suppose that $x \in A \cup (A \cap B)$, then $x \in A$ or $x \in A \cap B$  (equivalently $x \in A$ and $x \in B$). In both cases $x \in A$, so $A \cup (A \cap B) \subseteq A$. Conversely you also can see that $A \subseteq A \cup (A \cap B)$, so eventually $A = A \cup (A \cap B)$.
A: It follows because the union is superset of the parts (ie $X \subseteq X \cup Y$)  and join is subset of the parts (ie $X \subseteq X \cap Y$). Also $A \cap A = A$ and join is distributive over unions (ie $X \cap (Y \cup Z) = (X \cup Y) \cap (X \cup Z)$). Consequently:
$A \subseteq A \cup (A \cap B) = (A \cap A) \cup (A \cap B) = A \cap (A \cup B) \subseteq A$
