# Accumulation Points of a Set in a Topological Space Examples 1

Recall from the Accumulation Points of a Set in a Topological Space page that if $(X, \tau)$ is a topological space and $S \subseteq X$ then a point $x \in X$ is an accumulation point of $S$ if every open neighbourhood of $X$ contains points of $S$ different from $x$, that is, for all $U \in \tau$ with $x \in U$ we have that $A \cap U \setminus \{ x \} \neq \emptyset$.

We will now look at some examples of accumulation points in topological spaces.

## Example 1

**Let $(X, \tau)$ be a topological space and $S \subseteq X$. Prove that $\{ x \} \in \tau$ if and only if $x$ is not an accumulation point of $S$.**

Suppose that $\{ x \} \in \tau$. Then $\{ x \}$ is an open neighbourhood of $x$. But the open neighbourhood $\{ x \}$ contains no points of $S$ different from $x$. Therefore $x$ is not an accumulation point of any subset $S \subseteq X$.

Now suppose that $x$ is not an accumulation point of $S$. Then there exists an open neighbourhood of $x$ that does not contain any points different from $S$, i.e., $\{ x \} \in \tau$.

## Example 2

**Let $(X, \tau)$ be the discrete topological space, i.e., $\tau = \mathcal P (X)$. What are the accumulation points of $X$?**

Since $(X, \tau)$ is the discrete topological space, we have that for all $x \in X$ that $\{ x \} \in \tau$. So for each $x \in X$, $\{ x \}$ is an open neighbourhood of $x$ and the open neighbourhood $\{ x \}$ contains no points of $X$ different from $x$. Therefore, for all $x \in X$, $x$ is not an accumulation point of any subset $S \subseteq X$. So the discrete topological space has no accumulation points.

## Example 3

**Let $(X, \tau)$ be the indiscrete topological space, i.e., $\tau = \{ \emptyset, X \}$. What are the accumulation points of $X$?**

Let $x \in X$. Then only open neighbourhood of $x$ is $X$. If $X$ contains more than $1$ element, then every $x \in X$ is an accumulation point of $X$. If $X$ contains only $1$ element, then $x$ is not an accumulation point of $X$ and therefore $X$ has no accumulation points.

## Example 4

**Let $X = \{ a, b, c, d, e \}$ and $\tau = \{ \emptyset, \{ a \}, \{ b, c, d, e \}, X \}$. What are the accumulation points of $S = \{ a, b, c \} \subseteq X$? What are the accumulation points of $T = \{ a, b \} \subseteq X$?**

We note that $a \in X$ is not an accumulation point of $S$ since the open set $\{ a \} \in \tau$ does not contain any points in $S$ different from $a$.

However, the points $b, c, d, e \in X$ are accumulation points of $S$ since the only open neighbourhoods of these points are $\{b, c, d, e \}$ and $X$. For $b \in X$, both of these open neighbourhoods contains elements in $S$ (namely $c \in S$) different from $b$. For $c \in X$, both of these open neighbourhoods contain elements in $S$ (namely $b \in S$) different from $c$. For $d \in X$, both of these open neighbourhoods contains elements in $S$ (namely $b, c \in S$), different from $d$. Lastly, for $e \in X$, both of these open neighbourhoods contain elements in $S$ (namely $b, c \in S$), different from $e$.

Now we also note that $a \in X$ is not an accumulation point of $T$ since the open set $\{ a \} \in \tau$ does not (once again) contain any points in $T$ different from $a$. The point $b \in X$ is also not an accumulation point of $T$. The open neighbourhoods of $b \in X$ are $\{b, c, d, e \}$ and $X$. In particular, the open neighbourhood $\{ b, c, d, e \}$ does not contain any elements in $S$ different from $b$.

However, the points $c, d, e \in X$ are accumulation points of $T$ since the only open neighbourhoods of these points are $\{ b, c, d, e \}$ and $X$. For $c \in X$, both of these open neighbourhoods contain elements in $T$ (namely $b \in T$) different from $c$. For $d \in X$, both of these open neighbourhoods contain elements in $S$ (namely $b \in T$) different from $d$. Lastly, for $e \in X$, both of these open neighbourhoods contain elements in $S$ (namely $b \in T$) different from $e$.