(3.2.4) A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_{\epsilon}(x)$ of $x$ intersects the set $A$ at some point other than $x$.

So for any $\epsilon > 0$, whenever we take the intersection of this neighborhood of $x$ with the set $A$, we ‘ll find points close to $x$ but not $x$ itself. $x$ is also called a cluster point or an accumulation point. The next theorem makes it a little easier to understand. In fact it’s equivalent to the definition above and we’ll prove this.

(3.2.5) A point $x$ is a limit point of a set $A$ if and only if $x = \lim a_n$ for some sequence $a_n$ contained in $A$ satisfying $a_n \neq x$ for all $n \in \mathbf{N}$.

So “$x$ is a limit point of $A$” means that $x$ is the limit of a sequence in $A$. The only thing we don’t want is to have any term of that sequence be $x$ itself. The reason for this is then every member of the set could potentially be a limit point since we can create the sequence $\{x,x,x,x,x,...\}$ and we don’t want that.

Proof

$\Rightarrow$: Assume that $x$ is a limit point of $A$ and so for every $\epsilon > 0$, the intersection of $V_{\epsilon}(x)$ with the set $A$ is nonempty and has an element that is not $x$. We then want to prove that there is a sequence of points $a_1, a_2, a_3, ...$ from $A/\{x\}$ such that $(a_n) \rightarrow x$. Now for each $n \in \mathbf{N}$, pick a point

$$ \begin{align*} a_n \in V_{1/n}(x) \cap A, \end{align*} $$

This means that the first element in the sequence is in $V_1(x)$, the second element is in $V_{1/2}(x)$. The third element is in $V_{1/3}(x)$ and so on (We are basically getting closer and closer to $x$). This sequence $(a_n)$ coverages to $x$. To see why, let $\epsilon > 0$. We can choose $N$ such that $N > \frac{1}{\epsilon}$. So now for any $n > N$,

$$ \begin{align*} |a_n - x| &< \frac{1}{n} \quad \text{ All the elements in $V_{1/n}$ must be within $1/n$ at most from $x$}\\ &< \frac{1}{N} \\ &< \frac{1}{1/\epsilon} = \epsilon. \end{align*} $$

$\Leftarrow$: Assume that $(a_n) \rightarrow x$ where $a_n \in A$ and $a_n \neq x$. Let $V_{\epsilon}$ be arbitrary. We want to prove that there some element from $(a_n)$ in that neighborhood $V_{\epsilon}$. But since $(a_n) \rightarrow x$, then we know that there exists an $N \in \mathbf{N}$ such that if $n \geq N$, we must have $|a_n - x| \leq \epsilon$. Therefore, for any $n \geq N$, we know that we have $a_n \in V_{\epsilon}$.

$\blacksquare$.

Other Definitions and Properties

Fo the absolute value function definition and other properties, see this.
For the definitions of sequences, subsequences and what it means to for a sequence to converge, see this.
For the definitions of series, and what it means to for a series to converge, see this.
For the "show the limit" template and an example, see this.

References: