*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# Properties of Cluster Points

Recall the definition of a **cluster point** from the Cluster Points page.

Definition: Let $A$ be a subset of the real numbers. Then the point $c \in \mathbb{R}$ is a cluster point of $A$ if $\forall \delta > 0$, the delta-neighbourhood of $c$, $V_{\delta} (c)$ contains at least one point from $A$ different from $c$. Alternatively a cluster point $c$ can be defined such that $\forall \delta > 0 \: \exists a_{\delta} \in A \setminus \{ c \}$ such that $\mid a_{\delta} - c \mid < \delta$ or rather, $\forall \delta > 0 \: V_{\delta} (c) \: \cap (A \setminus \{ c \} ) \neq \emptyset$. |

Recall that we can also define a cluster point such that for any open interval around $c$ there exists infinitely many points from $A$ in that interval, and further, we can define a cluster point if there exists a sequence $(a_n)$ from $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ such that $\lim_{n \to \infty} a_n = c$.

We will now look at some important properties of cluster points in the following theorems.

On the Cluster Points page, we said that a point $c \in \mathbb{R}$ is a cluster point of $A \subseteq \mathbb{R}$ if there exists a sequence $(a_n)$ from $A$ such that $\lim_{n \to \infty} a_n = c$ and $a_n \neq c$ $\forall n \in \mathbb{N}$. We will now prove this equivalence.

Theorem 1: A point $c \in \mathbb{R}$ is a cluster point of $A \subseteq \mathbb{R}$ if and only if there exists a sequence $(a_n)$ from $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$. |

**Proof:**$\Rightarrow$ Suppose that $c \in \mathbb{R}$ is a cluster point of $A \subseteq \mathbb{R}$. Then for any natural number $n \in \mathbb{N}$, let $\delta = \frac{1}{n}$. Then the $V_{\delta} (c)$ contains at least one point $a_n$ in $A$ that is not $c$, or equivalently, there exists infinitely many points in the open interval about $c$ for any $\frac{1}{n}$ where $n \in \mathbb{N}$. Thus $a_n \in A$, $a_n \neq c$, and $\mid a_n - c \mid < \frac{1}{n}$ implies that $\lim_{n \to \infty} a_n = c$.

- $\Leftarrow$ Now suppose that there exists a sequence $(a_n)$ from $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$. Then it follows by the definition of the limit of a sequence that for any $\delta > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $a_n \in V_{\delta} (c)$. Thus, there are infinitely many points contained in $V_{\delta} (c)$ that are distinct from $c$ and so any delta-neighbourhood about $c$ contains at least one point distinct from $c$, so by definition $c$ is a cluster point of $A$. $\blacksquare$.

We will now look at two somewhat obvious theorems that will become useful later.

Theorem 2: If $A$ is a finite set, then $A$ contains no cluster points. |

**Proof:**By the definition a $c$ to be a cluster point, we must have that for all $\delta > 0$ the delta-neighbourhood $V_{\delta} (c)$ contains at least one point, call it $x \in A$ such that $x \neq c$. Since $A$ is a finite set, then consider the point $x_c \in A$ that is closest to $c$. If we take $\delta_c$ to be less than the distance from $x_c$ to $c$, that is let $0 < \delta_c < \mid x_c - c \mid$ then $V_{\delta_c} (c)$ contains at most only $c$, and so $c$ is not a cluster point of $A$ since there exists a $\delta_c > 0$ such that $V_{\delta_c} (c) \cap ( A \setminus \{ c \} ) = \emptyset$. $\blacksquare$

Theorem 3: The set of natural numbers $\mathbb{N}$ contains no cluster points. |

**Proof:**Suppose we claim that $c$ is a cluster point of $N$.

**Case 1:**Suppose that $c < 0$. Then take $\delta_0 = 1$. Thus $V_{\delta_0} (c) \cap (\mathbb{N} \setminus \{ c \} ) = \emptyset$ as there does not exist any natural number $n$ such that $c - 1 < n < c + 1$ provided that $c < 0$ since then $c + 1 < 1$ and the first natural number is $1$.

**Case 2:**Suppose that $c ≥ 0$. Then take $\delta_0$ to be the distance between $c$ and the next closest natural number $n$. Therefore $V_{\delta_0} (c) \cap (\mathbb{N} \setminus \{ c \}) = \emptyset$.

- We thus conclude that $\mathbb{N}$ contains no cluster points.