Accumulation Points of a Set in a Topological Space

Accumulation Points of a Set in a Topological Space

Recall from the The Open Neighbourhoods of Points in a Topological Space page that if $(X, \tau)$ is a topological space and $x \in X$ then a open set $U \in \tau$ is called an open neighbourhood of $x$ if $x \in U$.

We will now define a very important type of point of a set in a topological spaces known as accumulation points.

 Definition: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is called an Accumulation Point of $A$ if every open neighbourhood of $x$ contains a point in $A$ different from $x$. The Set of all Accumulation Points of $A$ (sometimes called the Derived Set of $A$) is denoted by $A'$.

The terms "Limit Point" and "Cluster Point" are sometimes used to mean "accumulation point".

 Proposition 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A \cup A'$ is closed.
• Proof: Let $x \in X \setminus (A \cup A')$. Then $x \not \in A$ and $x \not \in A'$. So $x$ is not in $A$ and $x$ is not an accumulation point of $A$. Since $x$ is not an accumulation point of $A$ there exists an open neighbourhood $U_x$ of $x$ such that $U_x$ does not contain a point in $A$ different from $x$. But also, $x \not \in A$, and so $U_x \cap A = \emptyset$. Since such a neighbourhood $U_x$ exists for every $x \in X \setminus (A \cup A')$ we see that $X \setminus (A \cup A')$ is open, and thus $A \cup A'$ is closed. $\blacksquare$

Let's now look at some examples.

Example 1

Consider the finite set $X = \{ a, b, c \}$ and the nested topology $\tau = \{ \emptyset, \{ a \}, \{ a , b \}, \{ a, b, c \} \}$. Let $A = X$ itself. The point $a \in X$ is NOT an accumulation point because the open neighbourhood $\{ a \} \in \tau$ of $a$ does not contain any points different from $a$.

However, the points $b, c \in X$ are accumulation points of $X$. The open neighbourhoods of $b$ are $\{ a, b \}$ and $\{a, b, c \}$, and in each of these open neighbourhoods there exists points different from $b$. Similarly, the only open neighbourhood of $c$ is $\{ a, b , c \}$ and there exists points different from $c$ in this open neighbourhood.

Example 2

Consider the set of real numbers $\mathbb{R}$ with the usual Euclidean topology. Let $A = (0, 1)$. Then every $x \in A$ is an accumulation point of $A$. Also, $0$ and $1$ (which are not in $A$) are accumulation points of $A$. In fact, $A' = [0, 1]$.

To partially verify this, observe that if $x \in [0, 1]$ then for any $\epsilon > 0$, $(x - \epsilon, x + \epsilon)$ is an open neighbourhood of $x$ that intersects $A = (0, 1)$ for which the intersection contains a member of $A$ different from $x$. This sufficiently shows that $[0, 1] \subseteq A'$ for if $U$ is any open neighbourhood of $x$ then $U$ must contain a set of the form $(x - \epsilon, x + \epsilon)$ for some $\epsilon > 0$.

Example 3

Consider the set of real numbers $\mathbb{R}$ where the topology $\tau$ is given by:

(1)
\begin{align} \quad \tau = \{ \emptyset, \mathbb{R} \} \cup \{ (-n, n) : n \in \mathbb{N} \} \end{align}

Let $A = \mathbb{R}$. Consider the point $0 \in \mathbb{R}$. Then $(-1, 1)$, $(-2, 2)$, …, $(-n, n)$, …, $\mathbb{R}$ are all of the open neighbourhoods of $0$. Each of these open intervals contains points different from $0$, so $0 \in \mathbb{R}$ is an accumulation point in $\mathbb{R}$ with the topology $\tau$.

In fact, every element $x \in \mathbb{R}$ is an accumulation point. To prove this, fix $x \in \mathbb{R}$. Then $\mid x \mid > 0$ and by The Archimedean Property there exists a smallest natural number $n(x) \in \mathbb{N}$ (dependent on $x$) such that $\mid x \mid < n(x)$, so:

(2)
\begin{align} \quad -n(x) < x < n(x) \end{align}

By the density of the real numbers, there exists a $\xi \neq x$ such that $-n(x) < x < \xi < n(x)$. So $\xi \in (-n(x), n(x))$. All other larger open neighbourhoods of $x$ are of the form:

(3)
\begin{align} \quad (-n(x) - k, n(x) + k), \: k \in \mathbb{N} \end{align}

Furthermore, the open neighbourhoods of $x$ are nested:

(4)
\begin{align} \quad x, \xi \in (-n(x), n(x)) \subset (-n(x) - 1, n(x) + 1) \subset (-n(x) - 2, n(x) + 2) \subset ... \subset (-n(x) - k, n(x) + k) \subset ... \subset \mathbb{R} \end{align}

Therefore, for each open neighbourhood of $x$ there exists a $\xi \in \mathbb{R}$, $\xi \neq x$ such that $\xi$ is contained in each of these open neighbourhoods. Therefore every $x \in \mathbb{R}$ is an accumulation point of $\mathbb{R}$ with this topology.