About interior of the frontier (proof-checking)

Question

Let $M$ be a metric space, and $A \subset M$ an open set. Show that $\stackrel{o}{\widehat{\partial A}} = \emptyset$.

($\stackrel{o}{\widehat{\partial A}}$ is the interior of the frontier)

I thought the following: Given $x \in \partial A$, I have to show that: $x$ cannot be in $\stackrel{o}{\widehat{\partial A}}$ $\iff$ $B(x,r)$ can't be contained in $\partial A$ $\iff$ exists a point in $B(x,r)$ that doesn't belong in $\partial A$ $\iff$ there's an open ball centered in that last point that is contained in $A$ or $A^c$.

My solution: Let $x \in \partial A$. For all $r > 0$, the open ball $B(x,r)$ has points from $A$ and $A^c$. Then, take $y \in B(x,r) \cap A$. Since $A$ is open, exists $s > 0$ for which $B(y,s) \subset \left(B(x,r) \cap A \right) \subset A$. Then $B(x,r) \not\subset \partial A$, for any $x$, and so $\stackrel{o}{\widehat{\partial A}} = \emptyset$ $\hspace{2cm} \blacksquare$

Is that okay? Can I improve something?

Answer 1

The proof is correct. For fun, here's a version without symbols:

If the boundary of a set contains an open ball, that ball intersects the set itself. But an open set is disjoint from its boundary.