The Closure of a Set in a Metric Space

# The Closure of a Set in a Metric Space

Recall from the Adherent, Accumulation and Isolated Points in Metric Spaces page that if $(M, d)$ is a metric space and $S \subseteq M$ then a point $x \in M$ is said to be an adherent point of $S$ if for all $r > 0$ we have that:

(1)
\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}

In other words, $x \in M$ is an adherent point of $S$ if every ball centered at $x$ contains a point of $S$.

We will now make a very important definition of the set of all adherent points of a set.

 Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. The Closure of $S$ denoted $\bar{S}$ is the set of all adherent points of $S$.

We should note that for any metric space $(M, d)$ and any $S \subseteq M$ then we always have that:

(2)
\begin{align} \quad S \subset \bar{S} \end{align}

This is because for each $s \in S$ and for every $r > 0$, $s \in B(s, r) \cap S$ and so $B(s, r) \cap S \neq \emptyset$.

Let's now look at some examples. Consider the metric space $(\mathbb{R}, d)$ where $d$ is the usual Euclidean metric defined for all $x, y \in \mathbb{R}$ by $d(x, y) = \mid x - y \mid$ and consider the set $S = (0, 1)$. The closure of $S$ is therefore $\bar{S} = [0, 1]$.

For another example, consider the metric space $(M, d)$ where $M$ is any nonempty set and $d$ is the discrete metric defined for all $x, y \in M$ by:

(3)
\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} x = y\\ 1 & \mathrm{if} x \neq y \end{matrix}\right. \end{align}

Consider the singleton set $S = \{ x \}$. Then for any other $y \in M$ we have that $x \not \in B \left ( y, \frac{1}{2} \right )$ and so $B \left ( y, \frac{1}{2} \right ) \cap S = B \left ( y, \frac{1}{2} \right ) \cap \{ x \} = \emptyset$. Therefore the closure of a singleton set with the discrete metric is $\bar{S} = \{ x \}$.

 Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then $\bar{S}$ is closed.
• Proof: Consider the complement $\bar{S}^c$. We will show that this set is open. Let $x \in (\bar{S})^c = M \setminus \bar{S}$. Then $x$ is not an adherent point of $S$, and so for every $x \in (\bar{S})^c$ there exists an $r_x > 0$ such that:
(4)
• We claim that $B \left (x, \frac{r_x}{2} \right ) \cap \bar{S} = \emptyset$ too. • Suppose not. Then there exists a $y \in M$ such that $y \in B \left (x, \frac{r_x}{2} \right ) \cap \bar{S}$ and so $y$ is an adherent point of $S$ with $d(x, y) < \frac{r_x}{2}$.
• Since $y$ is an adherent point of $S$ we have that for all $r > 0$ we have that:
• Take $r_0 = \min \{ d(x, y), \frac{r_x}{2} - d(x, y) \}$. Then $B(y, r_0) \subset B \left (x, \frac{r_x}{2} \right ) \subset B(x, r_x)$. So if $(**)$ holds then $B(y, r_0) \cap S \neq \emptyset$, so, since $B(y, r_0) \subset B (x, r_x )$ we have that $B \left ( x, r_x \right ) \cap S \neq \emptyset$. But this contradicts $(*)$.
• Hence the assumption that $B \left (x, \frac{r_x}{2} \right) \cap \bar{S} \neq \emptyset$ was false. So $B \left (x, \frac{r_x}{2} \right ) \cap \bar{S} = \emptyset$, so, for each $x \in M$ there exists an $\frac{r_x}{2} > 0$ such that:
• Therefore $(\bar{S})^c$ is open and so $\bar{S}$ is closed. $\blacksquare$