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. $\blacksquare$