Interior, exterior, and boundary points of $\{(x, y) : x^{2} + y^{2} = 1\}$ What is the boundary of $S = \{(x, y) \mid x^2 + y^2 = 1\}$ in $\mathbb{R}^2$? I believe the answer is $\emptyset$, but it could also just be $S$ itself. 
I know that the union of interior, exterior, and boundary points should equal $\mathbb{R}^{2}$. I thought that the exterior would be $\{(x, y) \mid x^2 + y^2 \neq 1\}$ which means that the interior union exterior equals $\mathbb{R}^{2}$. But since each of these sets are also disjoint, that leaves the boundary points to equal the empty set.
Is this correct? 
 A: Your definition as in the comments: $\partial S$ is the set of points $x$ in $\mathbb R^2$ such that any open ball around $x$ intersects $S$ and $S^c$. 
A sketch with some small details left out for you to fill in:
First, for any $s\in S$, any open ball $B$ around $s$ intersects $S$ trivially. Furthermore, the point $(1+\epsilon)s \notin S$ is an element of $B$, for sufficiently small $\epsilon>0$. Thus, we conclude $S\subseteq \partial S$.
Conversely, suppose $s\notin S$. Since $S$ is closed, there exists an open ball around $s$ that does not intersect $S$. Thus, $s\notin \partial S$. We conclude that $ S ^c  \subseteq \partial S^c$. 
Hence, $S = \partial S$.

The OP in comments has said he requires proof that $S$ is closed without using preimages. Let $s$ be any point not in $S$. If $|s|<1$, a small enough ball around $s$ won't have points of size $\ge 1$. If $|s|>1$, a small enough ball around $s$ won't have points of size $\le 1$. I leave the details(triangle inequality) to you.
A: The closure of $S$ is $S$ itself. And its interior is the emptyset. Hence the boundary of $S$ is $S$ itself.
Why is $S$ its own closure? Because $S$ is a closed subset of $\mathbb R^2$. $S$ is closed as it is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y) \mapsto x^2+y^2$. Another way to see that $S$ is closed is to prove that its complementary set is open. For this, take a point $M = (x,y) \in \mathbb R^2 \setminus S$ and prove that the open disk $D$ centered on $M$ with radius $r = \vert 1- \sqrt{x^2+y^2}\vert$ is included in $\mathbb R^2 \setminus S$.
And the interior is empty as no open ball is included in $S$.
