If a set $S$ consists of a single point $\{p\}$ in the complex plane $(\mathbb{R^2})$, then is $p$ an interior point, an exterior point, or a boundary point? Is $S$ open, closed, connected, compact, or a region?
Looking closely at the definitions, I believe $p$ is a boundary point as $\forall\epsilon>0, \ B(p,\epsilon)$ contains both points in $S$ and not in $S$.
If this is true, then $S$ must be closed, as $S$ contains all of its boundary points. It is also compact as $S$ is bounded and closed. I don't think it can be connected as no two points can be connected by a polygonal path that lies entirely in $S$ ($S$ consists of only a single point anyway). It also can't be a region as $S$ is not open.
Is this correct?
EDIT:
The set $S=\{p\}$ consists of a single point in the complex plane. Let's consider the set $S^C=\{\mathbb{C}$\ $p$}. Now take an arbitary point $q\in S^c$. The distance from $q$ to $p$ is $|p-q|=\delta$ for some $\delta>0$. Now consider the ball $B\left(q,\frac{\delta}{2}\right)$. Clearly $B\left(q,\frac{\delta}{2}\right)\cap S^C=B\left(q,\frac{\delta}{2}\right)$. Hence $q$ is an interior point. So $S^c$ is open$\Rightarrow S$ is closed.
$S$ is connected is vacuously true. $S$ is also compact as it's closed and bounded. $S$ is not a region (unsure of reason, but it intuitively makes sense).