Prove that $(A \cap B)^\circ = A^\circ \cap B^\circ$ Let $(X,d)$ be a metric space and let $A, B \subset X$  
How can I show that  $(A \cap B)^\circ = A^\circ \cap B^\circ$ ?
Please just tell me a Hint. ($A^\circ$ and $B^\circ$ are sets of interior points of A and B)

Use $x \in A^\circ \iff \exists r >0 : B(x,r) := \{ y\in X \mid d(x,y) < r \} \subset A$.
