In this lecture, we study the topological nature of some familiar notions from analysis such as limit points and limits of sequences.

Throughout this lecture, we assume that the nonempty set $S$ is a topological space.

Definition. A point $x\in S$ is called a limit point of $A\subset S$ if for any open set $U$ containing $x$, $(U-\{x\})\cap A\ne\emptyset$. The set of limiting points of $A$ is called the derived set of $A$ and is denoted by $A’$.

Exercise. If $A\subset B\subset S$, then show that $A’\subset B’$.

Definition. Let $S$ be a space. $x\in S$ is called a boundary point of $A\subset S$ if for any open set $U$ containing $x$, $U\cap A\ne\emptyset$ and $U\cap(S\setminus A)\ne\emptyset$. The set of boundary points of $A$ is called the boundary of $A$ and is denoted by $B(A)$.

In Lecture 1, we studied the notion of the closure of a set. The following theorem relates the closure, limit points and boundary points of a set.

Theorem. Let $A\subset S$. Then
$$\bar A=A\cup A’=A\cup B(A).$$

Proof. First we show that $\bar A=A\cup B(A)$. Clearly, $A\cup B(A)\subset\bar A$. If $x\not\in\bar A$. Then there exists an open set $U$ containing $x$ such that $U\cap A=\emptyset$. This implies that $x\not\in A$ and $x\not\in B(A)$.

Now we show that $\bar A=A\cup A’$. Clearly, $A\cup A’\subset\bar A$. If $x\not\in A\cup A’$, then there exists an open set $U$ containing $x$, $(U-\{x\})\cap A=\emptyset$. Since $x\not\in A$, $U\cap A=\emptyset$. Thus, $x\not\in\bar A$.

Note that $A\cup A’=A\cup B(A)$ does not necessarily mean that $A’\subset B(A)$ or $B(A)\subset A’$ as shown in the following example.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$ and let $A=(0,1)\cup\{2\}$. Then $\bar A=[0,1]\cup\{2\}$, $A’=[0,1]$, and $B(A)=\{0,1,2\}$.

Definition. Let $A,B\subset S$. $A$ is dense in $B$ if $B\subset\bar A$. We say $A$ is dense in $S$ if $\bar A=S$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Let $A=(a,b)$ and $B=[a,b]$. Then $A$ is dense in $B$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Q}}=\mathbb{R}$, $\mathbb{Q}$ is dense in $\mathbb{R}$.

Definition. $A\subset S$ is said to be nowhere dense in $S$ if $\bar A$ contains no member of $\tau\setminus\{\emptyset\}$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Z}} =\mathbb{Z}$ contains no open interval, $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$.

Exercise. If $p\geq 2$ is an integer, a $p$-adic rational is a real number $r=\frac{k}{p^n}$ for some nonnegative integer $k$ and positive integer $n$. Show that the set of $p$-adic rationals in $I=[0,1]$ is dense in $I$.

Definition. $A\subset S$ is said to be perfect if $A$ is closed and $A\subset A’$.

Exercise. The Cantor ternary set $K$ is the set of all $x\in [0,1]$ having a ternary expansion $x=\frac{t_1}{3}+\frac{t_2}{3^2}+\cdots+\frac{t_n}{3^n}+\cdots$ with $t_n\ne 1$, for all $n\in\mathbb{N}$. Intuitively, $K$ can be thought of as the set obtained from $[0,1]$ following the successive removal of all open middle thirds. Show that $K$ is uncountable, perfect, and nowhere dense in $[0,1]$.

Definition. Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $S$ and $x\in S$. We say that $\{x_n\}_{n\in\mathbb{N}}$ converges to $x$ and write $x_n\rightarrow x$ if for any open set $U$ containing $x$, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. As a sequential limit, $x$ is also denoted by $\displaystyle\lim_{n\to\infty}x_n$.

Note that a sequence may have no limits, a unique limit, or several limits, depending upon the topology on $S$.

Example. Let $\tau$ be the cofinite topology on $\mathbb{R}$ i.e.
$$\tau=\{\emptyset\}\cup\{U\subset \mathbb{R}: \mathbb{R}\setminus U\ \mbox{is finite}\}.$$
Let $x_n=n$, $n\in\mathbb{N}$. Let $x\in\mathbb{R}$ and $x\in U\in\tau$. Then since $\mathbb{R}\setminus U$ is finite, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. Hence, $x_n\rightarrow x$ for all $x\in\mathbb{R}$.

Theorem. Let $A\subset S$ and $x\in S$.

1. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ such that $x_n\rightarrow x$, then $x\in\bar A$.
2. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence of distinct points in $A$ such that $x_n\rightarrow x$, then $x\in A’$.

Proof.

1. Assume the hypothesis. Let $U$ be an open set containing $x$. Then there exists a nutural number $N$ such that $x_n\in U$ for all $n\geq N$. Since $\{x_n\}_{n\in\mathbb{N}}\subset A$, $G\cap A\ne\emptyset$. Hence, $x\in\bar A$.
2. Left as an exercise.

A limit point is not necessarily a sequential limit, and a sequential limit is not necessarily a limit point as seen in the following example.

Example. Let $S=\{a,b,c\}$ and $\tau=\{\emptyset,\{a,b\},\{c\},S\}$. Let $x_1=a$, $x_2=b$, and $x_n=c$ for all $n\geq 3$. Clearly, $x_n\rightarrow c$. However, $c$ cannot be a limit point since $c\in\{c\}\in\tau$ and $(\{c\}-\{c\})\cap\{a,b,c\}=\emptyset$. $a$ and $b$ are limit points but they cannot be sequential limits since $a,b\in\{a,b\}\in\tau$ and $x_n\notin\{a,b\}$ for all $n\geq 3$.

Exercise. Let $\tau=\{\emptyset\}\cup\{(a,\infty):a\in\mathbb{R}\}\cup\{\mathbb{R}\}$. Verify that $\tau$ is a topology on $\mathbb{R}$ and establish in $(\mathbb{R},\tau)$ the following sequential convergence and divergence:

1. If $x_n=n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\in\mathbb{R}$.
2. If $x_n=-n$, for each $n\in\mathbb{N}$, then $\{x_n\}_{n\in\mathbb{N}}$ does not converge in $\mathbb{R}$.
3. If $x_n=(-1)^n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\leq -1$.