# The Derived Set 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 accumulation point of $S$ if for all $r > 0$ we have that:

(1)In other words, every ball centered at $x$ contains a point of $S$ different from $\{x \}$.

We will now look at an important definition of the set of all accumulation points of a set in a metric space.

Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then Derived Set of $S$ denoted $S'$ is the set of all accumulation points of $S$. |

For example, consider the metric space $(\mathbb{R}, d)$ where $d$ is the usual Euclidean metric on $\mathbb{R}$ defined for all $x, y \in \mathbb{R}$ by $d(x, y) = \mid x - y \mid$, and consider the set $S = (0, 1) \cup \{ 2 \}$.

Clearly every $x \in (0, 1)$ is an accumulation point of $S$ since for all $r > 0$, $B(x, r) \cap (0, 1) \setminus \{ x \} = \emptyset$. Furthermore, the points $0$ and $1$ are also accumulation points of $S$. However, $2$ is not an accumulation point of $S$ since $B \left (2, \frac{1}{2}\right ) \cap S \setminus \{ 2 \} = \emptyset$. Therefore:

(2)The following theorem gives us a connection between the closure and derived set of a set $S$ in a metric space.

Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then $\bar{S} = S \cup S'$. |

**Proof:**Let $x \in \bar{S}$. Then $x$ is an adherent point of $S$, so $x$ is either an accumulation point or an isolated point. If $x$ is an accumulation point then $x \in S'$, and if $x$ is an isolated point then $x \in S$, so in either case, $x \in S \cup S'$ so $\bar{S} \subseteq S \cup S'$.

- Now let $x \in S \cup S'$. Then $x$ in $S$ or $x \in S'$. If $x \in S'$ then $x$ is an accumulation point of $S$ and so $x$ is an adherent point of $S$ too, so $x \in \bar{S}$. If $x \in S \setminus S'$ then $x$ must be an isolated point, and so $x$ is also an adherent point of $S$, so $x \in \bar{S}$. In both cases, $x \in \bar{S}$ so $\bar{S} \supseteq S \cup S'$.

- Hence we conclude that $\bar{S} = S \cup S'$. $\blacksquare$

*On The Closure of a Set in a Metric Space in Terms of the Boundary of a Set page we also saw that $\bar{S} = S \cup \partial S$.*