The Closure of a Set in a Metric Space in Terms of the Bound. of a Set

# The Closure of a Set in a Metric Space in Terms of the Boundary of a Set

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 $S$, i.e., the set of all points $x \in M$ such that every open ball centered at $x$ contains points of $S$, i.e., $B(x, r) \cap S \neq \emptyset$ for all $r > 0$.

We also noted that for any set that the closure $\bar{S}$ is a closed subset of $M$.

On the The Closure of an Open Ball and Closed Balls in a Metric Space** page we noted that that for any metric space $(M, d)$ and for all $a \in M$ and $r > 0$ that the closure of the open ball centered at $a$ with radius $r$ is contained in the closed ball centered at $a$ with radius $r$, i.e.

(1)
\begin{align} \quad \overline{B(a, r)} \subseteq \bar{B}(a, r) \end{align}

We noted that in general, the reverse inclusion does not always hold.

We will now see that the closure of a set $S$ is equal to the set $S$ union the boundary $\partial S$.

 Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then $\bar{S} = S \cup \partial S$.
• Proof: Let $x \in \bar{S}$. Then $x$ is an adherent point of $S$. So $x$ is either an isolated point of $S$ or an accumulation point of $S$.
• If $x$ is an isolated point then by definition $x \in S$ so $x \in S \cup \partial S$.
• If $x$ is an accumulation point then for all $r > 0$:
(2)
\begin{align} \quad B(x, r) \cap S \setminus \{ x \} \neq \emptyset \end{align}
• Note that $x \in \mathrm{int} S$ or $x \not \in \mathrm{int} S$. If $x \in \mathrm{int} S$ then $x \in S \subseteq S \cup \partial S$ and we're done. If $x \not \in \mathrm{int} S$ then for all $r > 0$, $B(x, r) \not \subseteq S$. Hence, for all $r > 0$, since $B(x, r) \cap S \setminus \{ x \} \neq \emptyset$ there exists an $a \in B(x, r)$ such that $a \in S$, and since $B(x, r) \not \subseteq S$ there exists a $b \in B(x, r)$ such that $a \in S^c$. So $x \in \partial S$ and we're done.
• Thus $\bar{S} \subseteq S \cup \partial S$.
• Now let $x \in S \cup \partial S$. If $x \in S$ then we're done. If $x \in \partial S$ then for all $r > 0$ we have that there exists $a, b \in B(x, r)$ such that $a \in S$ and $b \in S^c$. So $B(x, r) \cap S \neq \emptyset$, so $x$ is an adherent point of $S$. Thus $S \cup \partial S \subseteq \bar{S}$.
• We conclude that $\bar{S} = S \cup \partial S$. $\blacksquare$
 Corollary 1: Let $(M, d)$ be a metric space. Then $S \subseteq M$ is closed if and only if $\partial S \subseteq S$.
• Proof: $\Rightarrow$ Suppose that $S \subseteq M$ is closed. Then $S = \bar{S} = S \cup \partial S$ (by Theorem 1). So $\partial S \subseteq S$.
• $\Leftarrow$ Suppose that $\partial S \subseteq S$. Then by Theorem 1, $S = S \cup \partial S = \bar{S}$, so $S$ is closed. $\blacksquare$