Limit Point

Definition

Let $A \subseteq R$ and $x \in R$ . Then
$x$ is a limit point of $A$ if every Epsilon-Neighborhood of $x$ intersects $A$ at a point other than $x$ . Namely:

\forall ε > 0 (V_{ε} (x) \cap A \neq {x} \land \emptyset)

For example, if $A = (a, b]$ , then what are the limit points of $A$ ? The limit points of $A$ are:

= a, b

Alternative Definition of Limit Point

Let $A \subset R$ and $x \in R$ . Then $x$ is a limit point of $A$ iff $\exists (a_{n})$ :

a_{n} \in A ∖ {x}

for all $n$ and $a_{n} \to x$ .

Proof

( $\Rightarrow$ ): Suppose $x$ is a limit point of $A$ . Let's apply a sequence of neighborhoods such that we can construct $a_{n}$ . To do this, let $n \in N$ . Then choose:

a_{n} \in (V_{\frac{1}{n}} (x) ∖ {x}) \cap A

we claim that $a_{n} \to x$ . To prove this via the definition let $ϵ > 0$ . Choose $N \in N$ such that $\frac{1}{N} < ϵ$ . Then suppose $n \geq N$ then:

\begin{aligned} | a_{n} - x | & < \frac{1}{n} & (a_{n} \in V_{\frac{1}{n}} (x)) \\ \leq \frac{1}{N} \\ < ϵ \end{aligned}

( $\Leftarrow$ ): Suppose $\exists a_{n}$ where $a_{n} \in A - {x}$ for all $n$ and that $a_{n} \to x$ . Let $ϵ > 0$ so that $V_{ϵ} (x)$ is an epsilon-neighborhood of $x$ . We want to show that $(V_{ϵ} (x) - {x}) \cap A \neq \emptyset$ . We can choose a $N \in N$ such that:

| a_{n} - x | < ϵ

by the definition and that $a_{n} \to x$ . Then definitely $| a_{N} - x | < ϵ$ , so then $N \in V_{ϵ} (x)$ . Furthermore clearly $a_{N} \in A$ while $a_{N} \neq x$ so then $a_{N} \in (V_{ϵ} (x) - {x}) \cap A$ is not empty as desired.

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To illustrate this, consider:

A = {\frac{1}{n} | n \in N}

Then the set of all limit points of $A$ is just ${0}$ . Notice other points like $\frac{1}{2}$ are not in this set since there's no sequence of numbers constructing from the set cannot approach $\frac{1}{2}$ .

Another example is when $A = Q$ . The set of limit points of $A$ are now $R$ .