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Accumulation point of a set

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, 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.