# The Closure of an Open Ball and Closed Balls in a Metric Space

Recall from The Closure of a Set in a Metric Space page that if $(M, d)$ is a metric space and $S \subseteq M$ then the closure of $S$ denoted $\bar{S}$ is defined to be the set of all adherent points of $S$.

Awhile back, we saw on the Open and Closed Balls in Metric Spaces page we defined the closed ball centered at a point $a$ with radius $r > 0$ to be the set of points $x \in M$ if a distance of less than or equal to $r$ from $a$, that is:

(1)One might be tempted to ask whether the closure of an open ball $B(a, r)$ is equal to the corresponding closed ball $\bar{B}(a, r)$. Unfortunately the answer is no in general. The following example will show that $\overline{B(a, r)} \neq \bar{B}(a, r)$ in general.

Consider the metric space $(M, d)$ where $M$ contains more than $1$ element and $d$ is the discrete metric defined for all $x, y \in M$ by:

(2)Let $a \in M$ and consider the open ball centered at $a$ with radius $1$:

(3)Notice that the only point that is a distance of less than $1$ from $a$ is $a$ itself, and so:

(4)Note that the closure of $B(a, 1)$ is the set of all adherent points of $B(a, 1)$ which is simply:

(5)Now consider the closed ball centered at $a$ with radius $1$:

(6)Then every point in $M$ is of a distance of $0$ or a distance of $1$ from $a$ and therefore:

(7)Since $M$ is an infinite set we clearly see that $\overline{B(a, 1)} \neq \bar{B}(a, 1)$, and so in general, the closure of an open ball need not equal the corresponding closed ball.

Nevertheless, we do see that $\overline{B(a, 1)} \subseteq \bar{B}(a, 1)$ which we prove in generality in the following proposition.

Proposition 1: Let $(M, d)$ be a metric space. Then for every $a \in M$ and for all $r > 0$, $\overline{B(a, r)} \subseteq \bar{B}(a, r)$. |

**Proof:**Let $a \in M$ and $r > 0$.

- Take $x \in \overline{B(a, r)}$. Then $x \in B(a, r)$. So $d(a, x) < r$. But this implies that $d(x, a) \leq r$, so $x \in \bar{B} (a, r)$. Thus $\overline{B(a, r)} \subseteq \bar{B}(a, r)$. $\blacksquare$