## What is an accumulation point?

To answer that question, we first need to define an open neighborhood of a point in $\mathbf{R^{n}}$.

**Definition:** An **open neighborhood** of a point $x \in \mathbf{R^{n}}$ is every open set which contains point *x*.

**Definition:** Let $A \subseteq \mathbf{R^{n}}$. We say that a point $x \in \mathbf{R^{n}}$ is an **accumulation point** of a set *A* if **every** open neighborhood of point *x* contains at least one point from *A* distinct from *x*.

A **derivative set **is a set of all accumulation points of a set *A*. Furthermore, we denote it by $A’$ or $A^{d}$.

An** isolated point **is a point of a set *A* which is not an accumulation point.

**Note:** An accumulation point of a set *A* doesn’t have to be an element of that set.

## Examples

**Example 1: **Consider a set $S = \left<0, 1\right> \subset \mathbf{R}$. Obviously, every point $s \in S$ is an accumulation point of *S*. Furthermore, points $0$ and $1$ are accumulation points of *S* also. Why is that so?

Notice that if $s \in [0, 1]$, then for any $\epsilon > 0$, $\left<s – \epsilon, s + \epsilon \right>$ is an open neighborhood of *s* that intersects $S = \left<0, 1\right>$. Furthermore, that intersection contains an element of *S* which is distinct from *s*.

In conclusion, a set of accumulation points of $S = \left<0, 1\right> \subset \mathbf{R}$ is $[0, 1]$. In other words,

$$\left<0, 1\right>’ = [0, 1].$$

**Example 2:** Singletons, i.e. $\{x\}, x \in \mathbf{R^{n}}$ don’t have accumulation points.

**Example 3: **Consider a set $S = \{x, y, z\}$ and the nested topology $\mathcal{T} = \{\emptyset, S, \{x\}, \{x, y\}\}$. Find the accumulation points of *S*.

**Solution: **Let’s start with the point $x \in S$. The open neighborhood $\{x\} \in \mathcal{T}$ of x doesn’t contain any points distinct from *x*. Therefore, *x *isn’t an accumulation point of *S*. On the other hand, points $y, z \in S$ are accumulation points of *S*.

More precisely, the open neighborhoods of *y* are $\{x, y\}$ and $S = \{x, y, z\}$ and in each of these are points from* S* distinct from *y*. Furthermore, the only open neighborhood of *z* is $X = \{x, y, z\}$ and here are also points from *S* distinct from *z*.

**Example 4: **Prove that the only accumulation point of a set $A = \left \{\frac{1}{n} : n \in \mathbf{N} \right \}$ is $0$.

**Solution:**

Assume that $a \neq 0$ is an accumulation point of a given set. We have three cases.

**(1) $a \in \left<- \infty, 0\right>$**

We can choose $\epsilon = \frac{\mid a\mid}{2}$ such that $\epsilon$ neighborhood only contains negative numbers. In other words, we can find an open neighborhood which doesn’t contain a point from *A* distinct from *a*.

**(2) $a \in \left< 0, 1\right]$**

Similarly, we can choose $n \in \mathbf{N}$ such that $\frac{1}{n + 1} < a \leq \frac{1}{n}$. Then we can find $\epsilon$, for instance $\epsilon = (n + 1)^{-1000}$. Therefore, $\epsilon$ neighborhood will lay between the fractions and again we conclude that *a* is not the accumulation point.

**(3) $a \in \left<1, \infty\right>$**

Here we can also choose $\epsilon = \frac{\mid a – 1\mid}{2}$ such that $\epsilon$ neighborhood only contains number higher than $1$. Therefore, $a \in \left<1, \infty\right>$ is surely not an accumulation point of a given set.

In conclusion, $a \neq 0$ is not an accumulation point of a given set. Therefore, the only accumulation point of that set is number $0$.

**Example 5: **A derivative set of an open ball $K (x, r)$ is closed ball $\overline{K}(x, r)$. In other words,

**$$K (x, r)’ = \overline{K}(x, r).$$**

## Accumulation points of closed sets

**Theorem: ** A set $A \subseteq \mathbf{R^{n}}$ is **closed** if and only if it contains all of its accumulation points.

**Proof:**

We need to prove two directions; **necessity** and **sufficiency**. First, we will prove necessity.

In other words, assume that set *A* is closed. Let $x \in \mathbf{R^{n}}$ be its accumulation point and assume that $x \notin A$. Therefore, $x \in A^{C}$, which is an open set (because *A* is closed) containing *x* that does not intersect *A*.

Hence, this contradicts the fact that *x* is an accumulation point of a set *A*. In conclusion, *x* must be an element of *A*.

Next, we will prove sufficiency.

Suppose* A* contains all of its accumulation points. We will show that $A^{C}$ is an open set. If $x \in A^{C}$ and hence is not an accumulation point of *A*, then there exists an open set *U* containing *x* such that $A \cap U = \emptyset$.

But in that case, $x \in U \subset A^{C}$ which means that $A^{C}$ is an open set.