Open and Closed Sets in Metric Spaces

Recall from the Open and Closed Sets in Euclidean Space page that a set $S \subseteq \mathbb{R}^n$ is said to be open if $S =\mathrm{int} (S)$, that is, for every point $\mathbf{a} \in S$ we have that there exists a positive real number $r > 0$ such that the ball centered at $\mathbf{a}$ with radius $r$ is contained in $S$, i.e., $B(\mathbf{a}, r) \subseteq S$.

Furthermore, we said that $S \subseteq \mathbb{R}^n$ is closed if $S^c$ is open.

For any general metric space $(M, d)$, we define open and closeds subsets $S$ of $M$ in a similar manner.

Definition: If $(M, d)$ is a metric space and $S \subseteq M$ then $S$ is said to be Open if $S = \mathrm{int} (S)$ and $S$ is said to be Closed if $S^c$ is open. Moreover, $S$ is said to be Clopen if it is both open and closed.

It is important to note that the definitions above are somewhat of a poor choice of words. A set $S$ may just be open, just closed, open and closed (clopen), or even neither. Unfortunately these definitions are standard and we should note that saying a set is "not open" does not mean it is closed and likewise, saying a set is "not closed" does not mean it is open.

Now consider the whole set $M$. Is $M$ open or closed? Well by definition, for every $a \in M$ there exists a positive real number $r > 0$ such that $B(a, r) \subseteq M$ since the ball centered at $a$ with radius $r$ is defined to be the set of all points IN $M$ that are of a distance less than $r$ of $a$. Therefore $M$ is an open set.

So then the complement of $M$ is $M^c = M \setminus M = \emptyset$ is a closed set. However, it is vacuously true that for all $a \in \emptyset$ there exists a ball centered at $a$ fully contained in $\emptyset$ since $\emptyset$ contains no points to begin with. Therefore $\emptyset$ is also an open set and so $M$ is also a closed set.

This is the case for all metric spaces $(M, d)$. The whole set $M$ and empty set $\emptyset$ are trivially clopen sets!