Show that $A^\circ$ is open Assume that $A$ is a subset of a metric space $\textbf X$. The interior $A^\circ$ of $A$ is the set consisting of all interior points of $A$. Show that $A^\circ$ is open.
Proof:
Let $x$ be a point in $A^\circ$, then $x$ is interior point. Is ball $B(x,\varepsilon)$ contained in $A^\circ$? Let $z$ be a point in $B(x,\varepsilon)$. Since $z$ is a point in $A$, then $z$ is either in $A^\circ$ or a boundary point of $A$. If $z$ is boundary point of $A$ then a ball $B(z,\delta)$ will contain both point in $A$ and outside of $A$ for all $\delta >0$. Then for any point $w$ contained in $B(z,\delta)$

I treid to draw the rest of the proof. My question is why do they divide in half. 
 A: More generally, if $A$ is a subset of a topological space $X$, then the interior of $A$ is open. For each $x\in A^\circ$, there is a neighborhood $U_x$ of $x$ with $U_x\subset A$. If $y\in U_x$, then $y\in A^\circ$, so $U_x\subset A^\circ$. It follows that
$$A^\circ = \bigcup_{x\in A}U_x $$
is open as the union of open sets.
A: I will start a mock of proof. And I will let you finish it. Let me know if you see the conclusion or if I should write more details. But I think you almost should have solved it.
You want to show that  $A^{º}$ is open, so you must show that given an arbitrary point of $A^{º}$, let's denote it by p, there exists an open ball $B(p,r)$ for some $r>0$ centered at p and contained in $A^{º}$. (Now think about the definition of Interior point you have written above in the comments and try to follow by yourself).  (I will follow anyway). We know by hypothesis that $p \in A^{º}$ so we know that there exists $B(p,r) \subset A$. We will show that $B(p,r) \subset A^{º}$. Pick a point q in the ball $B(p,r)$, since the ball is open there exist a ball centered in q such that $B(q,r) \subset B(p,r)$ and this shows that q is an interior point. Since $q \in  B(p,r)$ was arbitrary we have finished.
I hope that helps!
