Proving $ \overset{\circ}{E} \cap \overset{\circ}{F} =\overset{\circ}{( E \cap F )}$? I'm trying to prove $ \overset{\circ}{E} \cap \overset{\circ}{F} =\overset{\circ}{( E \cap F )}$. For clarification, I take $\overset{\circ}{E}$ to mean the interior of E. I believe I've done one direction correctly: > Let $x \in \overset{\circ}{E} \cap \overset{\circ}{F} $. Then $ x \in \overset{\circ}{E}$ and $x \in \overset{\circ}{F}$. Then $\exists R_1, R_2 > 0$ such that $Nr_1(X) \subset E$ and $Nr_2(x) \subset E$. If we let $R = min\\{r_1, r_2\\}$, then it follows that $Nr(P) \subset Nr_1 (p) \cap Nr_2 (p) \subset E \cap F$. Since r is positive, $p \in \overset{\circ}{(E \cap F)} $. However, I'm having difficulty in proving the other direction. Help would be much appreciated!

If $p\in \overset{\circ}{E\cap F}$ then there is some $R>0$ so that $N_R(p)\subset E\cap F$. By definition of intersection, $N_R(p)\subset E$ and $N_R(p)\subset F$. Then you can conclude.