Isolated Points of Subsets in Euclidean Space

Recall from the Adherent Points of Subsets in Euclidean Space page 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 $s \in B(\mathbf{x}, r)$ for all positive real numbers $r > 0$, i.e., every ball centered at $\mathbf{x}$ contains at least one element from $S$.

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

We will now look at a third type of point called isolated points.

Definition: Let $S \subseteq \mathbb{R}^n$. An Isolated Point is a point $\mathbf{x} \in S$ such that there exists a positive real number $r > 0$ such that $B(\mathbf{x}, r)$ contains no other elements of $S$ other than $\mathbf{x}$.

In other words, an isolated point is a point $\mathbf{x} \in S$ such that there exists a ball centered at $\mathbf{x}$ which contains no other elements from $S$, i.e., there exists an $r > 0$ such that $B(\mathbf{x}, r) \cap S = \{ s \}$.

For example, if we consider the set $S = \{ 0 \} \cup [3, 4)$ then $0 \in S$ is an isolated point of $S$ since the open interval $(-1, 1)$ contains no points of $S$ other than $0$.

Furthermore, if we consider the set $S = \left \{ 1 - \frac{1}{n} : n \in \mathbb{N} \right \}$ then every element in $S$ is an isolated point. To show this, for each $n \in \mathbb{N}$ we have that:

(1)

\begin{align} \quad n - 1 < \frac{2n - 1}{2} < n < \frac{2n + 1}{2} < n + 1 \end{align}

Hence for $n \geq 2$:

(2)

\begin{align} \quad \frac{1}{n+1} < \frac{2}{2n + 1} < \frac{1}{n} < \frac{2}{2n - 1} < \frac{1}{n - 1} \end{align}

And:

(3)

\begin{align} \quad - \frac{1}{n - 1} < - \frac{2}{2n - 1} < - \frac{1}{n} < -\frac{2}{2n + 1} < - \frac{1}{n+1} \end{align}

Therefore:

(4)

\begin{align} \quad 1 - \frac{1}{n - 1} < 1 - \frac{2}{2n - 1} < 1 - \frac{1}{n} < 1 - \frac{2}{2n + 1} < 1 - \frac{1}{n+1} \end{align}

So for each element $1 - \frac{1}{n} \in S$, $n \geq 2$ take the open interval $\left ( 1 - \frac{2}{2n - 1}, 1 - \frac{2}{2n + 1} \right )$. In each of these intervals, the only element from $S$ is $1 - \frac{1}{n}$. If $n = 1$ then take the open interval $(-\frac{1}{4}, \frac{1}{4}$ where $0 \in S$ and no other elements in $S$ are contained in this interval. Therefore every element in $S$ is an isolated point.