The Openness of Open Balls / Closedness of Closed Balls in Met. Sps.

# The Openness of Open Balls and Closedness of Closed Balls in Metric Spaces

Recall from the Open and Closed Balls in Metric Spaces page that if $(M, d)$ is a metric space and $a \in M$, then the open ball centered at $a$ with radius $r > 0$ is the set of all points $x \in M$ whose distance is less than $r$ from $a$ with respect to the metric $d$, that is:

(1)
\begin{align} \quad B(a, r) = \{ x \in M : d(x, a) < r \} \end{align}

Furthermore, the closed ball centered at $a$ with radius $r > 0$ is the set of all points $x \in M$ whose distance is less than or equal to $r$ from $a$ with respect to the metric $d$, that is:

(2)
\begin{align} \quad \bar{B}(a, r) = \{ x \in M : d(x, a) \leq r \} \end{align}

Also recall from the Open and Closed Sets in Metric Spaces page that a general subset $S \subseteq M$ is said to be open if $S = \mathrm{int}(S)$ and closed if $S^c$ is open.

We will now see that all open balls are open sets and all closed balls are closed sets in a metric space - a concept that seems obvious from the terminology we used but must of course be proven.

 Theorem 1: Let $(M, d)$ be a metric space. Then for all $a \in M$ and for all $r > 0$, $B(a, r)$ is an open set.
• Proof: Let $a \in M$ and $r > 0$. To show that $B(a, r)$ is an open set we must show that $B(a, r) = \mathrm{int} (B(a, r))$. Let $x \in B(a, r)$. There are two cases to consider.
• Suppose that $x = a$. Then for any positive real number $r_0 > 0$ such that $r > r_0$ we have that $B(a, r_0) \subseteq B(a, r) = S$, so $x = a \in \mathrm{int} (B(a, r))$.
• Now suppose that $x \in B(a, r)$ and $x \neq a$. Since $B(a, r)$ is the set of all points $x \in M$ that are of a distance of less than $r$ from $a$, then if $x$ is of a distance of less than $\frac{r}{2}$ from $a$ then $r_0 = d(x, a) > 0$ is such that $B(a, r_0) \subseteq B(a, r)$.
• If $x$ is of a distance greater than $\frac{r}{2}$ (and less than $r$) from $a$ then $r_0 = r - d(x, a) > 0$ is such that $B(a, r_0) \subseteq B(a, r)$.
• Therefore, let $r_0 > 0$ be given as:
(3)
\begin{align} \quad r_0 = \mathrm{min} \{ d(x, a), r - d(x, a) \} > 0 \end{align}
• Then for every $x \in B(a, r)$ we have that there exists an $r_0 > 0$ such that $B(a, r_0) \subseteq B(a, r)$, so $B(a, r)$ is an open set. $\blacksquare$
 Theorem 2: Let $(M, d)$ be a metric space. Then for all $a \in M$ and for all $r > 0$, $\bar{B}(a, r)$ is a closed set.
• Proof: Let $a \in M$ and $r > 0$ and consider the closed ball $\bar{B}(a, r)$. By definition the closed ball centered at $a$ with radius $r > 0$ in set notation is:
(4)
\begin{align} \quad \bar{B}(a, r) = \{ x \in M : d(x, a) \leq r \} \end{align}
• Now consider the complement of this set:
(5)
\begin{align} \quad \bar{B}(a, r)^c = \{ x \in M : d(x, a) > r \} \end{align}
• We wish to show that this set is open. Let $y \in \bar{B}(a, r)^c$. Then $d(y, a) = q > r$ for some positive real number $q > 0$. Let $r^* = \frac{q - r}{2}$. Then we have that $B(y, r^*) \cap \bar{B}(a, r) = \emptyset$, so $B(y, r^*) \subseteq \bar{B}(a, r)^c$. Hence $y \in \mathrm{int}(\bar{B}(a, r)^c)$ and so:
(6)
\begin{align} \quad \bar{B}(a, r)^c = \mathrm{int}(\bar{B}(a, r)^c) \end{align}
• Hence $\bar{B}(a, r)^c$ is an open set, so $\bar{B}(a, r)$ is a closed set. $\blacksquare$