Criteria for a Set to be Closed in a Metric Space

Criteria for a Set to be Closed 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 every ball centered at $x$ contains points of $S$, that is, for all $r > 0$ we have that $B(x, r) \cap S \neq \emptyset$.
• A point $x \in M$ is said to be an accumulation point of $S$ if every ball centered at $x$ contains points of $S$ different from $x$, that is, for all $r > 0$ we have that $B(x, r) \cap S \setminus \{ x \} \neq \emptyset$.
• A point $x \in S$ is said to be an isolated point of $S$ if there exists a ball centered at $x$ that contains no other points of $S$, that is, there exists an $r_0 > 0$ such that $B(x, r_0) \cap S = \emptyset$.

We will now look at a nice theorem which gives us criteria for the set $S$ to be closed with regards to adherent and accumulation points.

 Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then the following statements are equivalent: a) $S$ is a closed set in $M$. b) $S$ contains all of its adherent points. c) $S$ contains all of its accumulation points.
• Proof of $a) \implies b)$: Let $S$ be a closed set. Then $S^c = M \setminus S$ is an open set. Suppose that $S$ does not contain all of its adherent points. Then there exists an $x \in S^c = M \setminus S$ such that all ball centered at $x$ contains points of $S$, i.e., for all $r > 0$ we have that:
(1)
\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}
• Since $S^c$ is an open set we have that there exists a ball centered at $x$ with radius $r_0 > 0$ such that:
(2)
\begin{align} \quad x \in B(x, r_0) \subseteq S^c = M \setminus S \end{align}
• But $x \not \in S$ and $B(x, r_0) \subseteq M \setminus S$ so $(B,x, r_0) \cap S = \emptyset$ which is a contradiction to the assumption that $S$ does not contain all of its adherent points. Hence if $S$ is closed then $S$ contains all of its adherent points.
• Proof of $b) \implies c)$ Suppose that $S$ contains all of its adherent points. Every accumulation point is an adherent point, so $S$ also contains all of its accumulation points.
• Proof of $c) \implies a)$: Suppose that $S$ contains all of its accumulation points and assume that $S$ is not closed. Then $S^c$ is not open. So there exists a point $x \in S^c$ such that $x \not \in \mathrm{int} (S^c)$, i.e., for every ball centered at $x$ with radius $r$ we have that:
(3)
\begin{align} \quad B(x, r) \not \subseteq S^c \end{align}
• So for all $r > 0$ we must have that $B(x, r) \cap S \neq \emptyset$. Since $x \in S^c$, we have that $x \not \in S$ and so:
(4)
\begin{align} \quad B(x, r) \cap S = B(x, r) \cap S \setminus \{ x \} \neq \emptyset \end{align}
• But then $x \in S^c$ is an accumulation point of $S$ that is not contained in $S$ which is a contradiction. Hence the assumption that $S$ was not closed is false and so $S$ is closed. $\blacksquare$