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.

Cluster Points

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, $\forall \delta > 0 \: V_{\delta} (c) \: \cap (A \setminus \{ c \} ) \neq \emptyset$.

The definition of a cluster point $c$ says that no matter how small $\delta > 0$ is, we can always find a point other than $c$ in any $\delta$-neighbourhood $V_{\delta} (c)$, or equivalently, every open interval around the point $c$ will contain infinitely many elements of the set $A$. Alternatively, we can prove that a specific point $c$ is a cluster point of the set $A$ if there exists a sequence $(a_n)$ from $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$.

Note: A cluster point $c$ of a subset $A$ of real numbers need not be contained within $A$. Also, a finite subset $A$ will contain no cluster points as it would be impossible for any open interval around $c$ to contain infinitely many points from $A$ regardless of the size of $\delta$ since $A$ would be finite.

Let's look at some examples now.

Example 1

Determine all possible cluster points for the set $A = [2, 3)$.

All values of $x \in [2,3]$ are cluster points of $A$. For example, the point $2$ is a cluster point of $A$ since any delta-neighbourhood around $2$ contains at least one point from $A$ that is different from $2$ as illustrated in the following diagram:

To show that $2$ is a cluster point of $A$, consider $V_{\delta} (2) = \{ x \in \mathbb{R} : \mid x - 2 \mid < \delta \}$, that is the set of $x$ which satisfy the inequality $2 - \delta < x < 2 + \delta$. We want to show that no matter how small $\delta > 0$ is, that there exists a point $a_{\delta} \in [2, 3)$ such that $2 < a_{\delta} < 2 + \delta$ which is equivalent to $0 < a_{\delta} - 2 < \delta$. Since $\delta > 0$, by one of the corollaries to The Archimedean Property it follows that there exists a natural number $n_{\delta} \in \mathbb{N}$ such that $0 < \frac{1}{n_{\delta}} < \delta$. If we choose $a_{\delta} - 2 = \frac{1}{n_{\delta}}$, then $A_{\delta} = 2 + \frac{1}{n_{\delta}}$, and therefore since $0 < a_{\delta} - 2 = \frac{1}{n_{\delta}} < \delta$ then $2 < a_{\delta} = 2 + \frac{1}{n_{\delta}} < 2 + \delta$, and so $2$ is a cluster point of $A$. We note that $2 \neq 2 + \frac{1}{n_{\delta}}$ since $\frac{1}{n_{\delta}} \neq 0$ for all $n \in \mathbb{N}$).

Similarly, we can also show that $3$ is a cluster point of $A$ even though $3 \not \in [2, 3)$ using an analogous argument. We need to show that $V_{\delta} (3)$ contains at least one point from $[2, 3)$ other than $3$ for all $n \in \mathbb{N}$, in other words, we need to show that there exists an $a_{\delta} \in A$ such that $a_{\delta} \in V_{\delta} (3) = \{ x \in \mathbb{R} : \mid x - 3 \mid < \delta \}$ for all $\delta > 0$, that is we need to show that there always exists an $a_{\delta} \in A$ that satisfies the inequality $3 - \delta < x < 3 + \delta$ for all $\delta > 0$. We will show that such an $a_{\delta}$ satisfies $3 - \delta < a_{\delta} < 3$, which is equivalent to $-\delta < a_{\delta} - 3 < 0$ or rather $0 < 3 -a_{\delta} < \delta$. Once again by one of the Archimedean corollaries, since $\delta > 0$, if we let $3 -a_{\delta} = \frac{1}{n_{\delta}}$ then $0 < 3-a_{\delta} = \frac{1}{n_{\delta}} < \delta$ and $-\delta < a_{\delta} - 3 = -\frac{1}{n_{\delta}} < 0$ and $3 - \delta < a_{\delta} = 3 - \frac{1}{n_{\delta}} < 3$, and so $3$ is a cluster point of $A$.

In fact, any point on the interval $[2, 3]$ is a cluster point of $A$.

We can also show that specific points are not cluster points. For example, consider the point $4$ which is not a cluster point of $A$. If we choose $\delta_0 = 1$, then $V_{\delta_0} (4) = \{ x \in \mathbb{R} : \mid x - 4 \mid < 1 \}$ is the set of $x \in \mathbb{R}$ such that $3 < x < 5$. But $A$ is defined as the interval $[2, 3)$, i.e., the set of $x \in \mathbb{R}$ such that $2 ≤ x < 3$, and so there exists no $x \in A$ for which $3 < x < 5$, so $4$ is not a cluster point of $A$ since the $\delta_0 = 1 > 0$ neighbourhood of $4$ contains no points of $A$. (Verify also that if $0 < \delta < 1$ then $V_{\delta} (4)$ contains no points from $A$, that is $V_{\delta}(4) \cap A = \emptyset$).

Example 2

Determine all possible cluster points for the set $A = \{ \frac{1}{n} : n \in \mathbb{N} \}$.

The only cluster point of this set is $0$. Let $\delta > 0$ and look at any delta-neighbourhood of $0$, that is $V_{\delta} (0) = \{ x \in \mathbb{R} : \mid x \mid < \delta \} = (-\delta, \delta)$. To show that $0$ is a cluster point of $A$ we need to find an $n_{\delta} \in N$ such that $x_{n_{\delta}} \in V_{\delta} (0) \cap A \setminus \{0 \}$, that is finding an $n_{\delta} \in \mathbb{N}$ such that $-\delta < \frac{1}{n_{\delta}} < \delta$. Once again by that same Archimedean property, since $\delta > 0$ there exists an $n_{\delta} \in \mathbb{N}$ such that $-\delta < 0 < \frac{1}{n_{\delta}} = x_{n_{\delta}} < \delta$ and so $0$ is a cluster point of $A$.

There are no other cluster points of this set which we will prove in cases.

Case 1: Suppose that $c < 0$. We will show that for some $\delta_0 > 0$ that $V_{\delta_0} (c) \cap A \setminus \{0 \} = \emptyset$. Let $\mid c \mid = \delta_0 > 0$. Then $V_{\delta_0}(c) = \{ x \in \mathbb{R} : \mid x - c \mid < \mid c \mid \ = \delta_0 \} = \emptyset$ since this inequality never holds as $- \mid c \mid < x - c < \mid c \mid$ is equivalent to $-(-c) < x - c < -c$ and $c < x - c < - c$ and so $2c < x < 0$, and there are no $x \in A$ such that $x < 0$ since $\frac{1}{n} > 0$ for all $n \in \mathbb{N}$.

Case 2: Suppose that $c > 1$. This time if we choose $\delta_0 = \mid c - 1 \mid > 0$, then $V_{\delta_0} (c) = \{ x \in \mathbb{R} : \mid x - c \mid < \mid c - 1 \mid \} = \emptyset$ since once again the inequality $- \mid c - 1 \mid < x - c \mid c - 1$ is equivalent to $1 - c < x - c < c - 1$, and $1 < x < 2c - 1$, and there are no $x \in A$ such that $1 < x$ since $\frac{1}{n} ≤ 1$ for all $n \in \mathbb{N}$.

Case 3: Suppose that $0 < c ≤ 1$. By one of the Archimedean corollaries, since $\frac{1}{c} > 0$ then there exists a natural number $n_c \in \mathbb{N}$ such that $n_c - 1 ≤ \frac{1}{c} < n_c$ and so $\frac{1}{n_c} < c ≤ \frac{1}{n_c - 1}$. Choose $\delta_{0} = \mathrm{min} \{ \rvert c - \frac{1}{n_c} \rvert , \rvert \frac{1}{n_c - 1} - c \rvert \}$. Then $V_{\delta} (c) = \{ x \in \mathbb{R} : \mid x - c \mid < \mathrm{min} \{ \rvert c - \frac{1}{n_c} \rvert , \rvert \frac{1}{n_c - 1} - c \rvert \} \} = \emptyset$.