Accumulation Points

Definition (Accumulation Points)

Given a set $A$ in $X$, the derived set is

$$A^{\prime }=\{x\in X\mid \forall N_x,N_x\cap (A-\{x\})\ne \varnothing \}$$

These points are called accumulation points. Note that $N_x$ is the neighborhood of $x$, and that $x$ may or may not be in $A$.

Also know as limit points. Related: Definition (Neighborhood in a Topology)

Example 1

$$A=\{0\}\cup \left\{\frac{1}{n}\mid n\ge 1\right\}\subseteq \mathbb{R}$$

We see that for $x=0$, every neighborhood of $x$ must contain some other point in $A$. We prove by Archimedean Property. So, $\overline{A}=A$.

Here, the only accumulation point is $0$.

Example 2

$$B=\left\{\frac{1}{n}\mid n\ge 1\right\}$$

Again, for $x=0$ every neighborhood must contain some other point in $B$. But as $0\notin B$, we see $\overline{B}=A$.

Likewise, the only accumulation point is $0$.

Theorem (Relationship with Accumulation Points and Closure)

$$\overline{A}=A\cup A^{\prime }$$

where $\overline{A}$ is the closure of $A$.

Proof: $(\supseteq )$: Certainly $A\subseteq \overline{A}$. If $x\in A^{\prime }$ then every neighborhood of it meets $A\backslash \{x\}$ which means $x\in \overline{A}$.

$(\subseteq )$: Let $x\in \overline{A}$. If $x\in A$, then $x\in A\cup A^{\prime }$. If $x\notin A$ then every neighborhood of $x$ meets $A$, since it is a point of the closure. But then every neighborhood of $x$ meets $A=A\backslash \{x\}$. Hence $x\in A^{\prime }$.

Theorem (Closure with Accumulation Points)

$A$ is closed $⟺$ it contains all of its accumulation points.