Accumulation Points of Subsets in Euclidean Space

Recall from the Adherent Points of Subsets in Euclidean Space that if $S \subseteq \mathbb{R}^n$ then a point $\mathbf{x} \in \mathbb{R}^n$ is an adherent point of $S$ if there exists an $s \in S$ such that $s \in B(\mathbb{x}, r)$ for every positive real number $r > 0$, i.e., every ball centered at $\mathbf{x}$ contains at least one element from $S$.

We will now look at a special type of adherent points known as accumulation points.

Definition: Let $S \subseteq \mathbb{R}^n$. An Accumulation Point to $S$ is a point $\mathbf{x} \in \mathbb{R}^n$ such that there exists a point $\mathbf{s} \in S \setminus \{ \mathbf{x} \}$ such that $\mathbf{s} \in B(\mathbf{x}, r)$ for all positive real numbers $r > 0$. The set of all accumulation points to $S$ is called the Derived Set of $S$ denoted $S'$.

In other words, $\mathbf{x} \in \mathbb{R}^n$ is an accumulation point to $S$ if every ball centered at $\mathbf{x}$ contains an element $s \in S$ that is different from $\mathbf{x}$.

A third equivalent definition is that $\mathbf{x} \in \mathbb{R}^n$ is an accumulation point to $S$ if $\mathbf{x}$ is an adherent point to $S \setminus \{ \mathbf{x} \}$.

For example, consider the set $\mathbb{R}$ and let $S = \{ 0 \} \cup [3, 4) \subseteq \mathbb{R}$. For any point $x \in [3, 4]$ every open interval centered at $\mathbf{x}$ will contain an element from $[3, 4) \subset S$ as seen in the following diagram:

Therefore every element $x \in [3, 4)$ is an accumulation point. However, the element $0 \in S$ is NOT an accumulation point. Take the open interval $B(0, 1) = (-1, 1)$. The only element from $S$ in this interval is $0$ itself!

For another example, consider the subset $S \subseteq \mathbb{R}^2$ that is the disk centered at the origin $\mathbf{0} = (0, 0)$ with radius $1$. As we saw on the previous page on adherent points, the points on the unit circle are adherent points and they're in fact also accumulation points.

For a third example, consider the subset $S = \{ 1 - \frac{1}{n} : n \in \mathbb{N} \} \subseteq \mathbb{R}$. The only accumulation point of $S$ is $1$. To prove this, for $r > 0$ consider an open interval centered at $1$, i.e., $(1 - r, 1 + r)$. Then all $x$ in this interval satisfy:

(1)

\begin{align} \quad 1 - r < x < 1 + r \end{align}

For each $r > 0$ there exists $n(r) \in \mathbb{N}$ such that $\frac{1}{r} < n(r)$ by The Archimedean Property. So $\frac{1}{n(r)} < r$ and $0 > -\frac{1}{n(r)} > -r$ so:

(2)

\begin{align} \quad 1 - r < 1 - \frac{1}{n(r)} < 1 + r \end{align}

Since $1 - \frac{1}{n(r)} \neq 1$ for all $n \in \mathbb{N}$ we have that for all $r > 0$ there exists an element $s \in S$ different from $1$ such that $s \in B(1, r)$ so $1$ is an accumulation point of $S$.

Theorem 1: Let $S \subseteq \mathbb{R}^n$. If $\mathbf{x} \in \mathbb{R}^n$ is an accumulation point of $S$ then every ball $B(\mathbf{x}, r)$ contains infinitely many points different from $\mathbf{x}$ in $S$.

• Proof: Let $\mathbf{x} \in \mathbb{R}^n$ be an accumulation point and assume instead that that there exists a positive real number $r > 0$ such that $B(\mathbf{x}, r)$ contains finitely many points from $S$ different from $\mathbf{x}$, say $\mathbf{s_1}, \mathbf{s_2}, ..., \mathbf{s_n} \in S$. Let $\delta$ be defined by:

(3)

\begin{align} \quad \delta = \min \{ \| \mathbf{x} - \mathbf{s_1} \|, \| \mathbf{x} - \mathbf{s_2} \|, ..., \| \mathbf{x} - \mathbf{s_n} \| \} \end{align}

• Notice that $\delta > 0$ since each $\mathbf{s_i}$ is different from $\mathbf{x}$. Now consider the open ball centered at $\mathbf{x}$ with radius $\frac{\delta}{2}$. Then $\mathbf{x}$ is the only element in $B \left (\mathbf{x}, \frac{\delta}{2} \right)$ that is also in $S$, so $\mathbf{x}$ is not an accumulation point - a contradiction.

• Therefore the assumption that there existed a ball centered at $\mathbf{x}$ containing only finitely many points from $S$ different from $\mathbf{x}$ was false. Therefore every ball $B(\mathbf{x}, r)$ contains infinitely many points different from $\mathbf{x}$ in $S$. $\blacksquare$