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 $x\in S$.

- If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ such that $x_n\rightarrow x$, then $x\in\bar A$.
- 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*.

- 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, $x\in\bar A$.
- 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:

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

Kylemetric spaces are more specific than topological ones; in fact every metric space is Hausdorff and the definition of the limit extends naturally from the topological definition to the real analysis definition

John LeePost authorRigorously speaking, a metric space $(X,d_x)$ itself is not a topological space but the metric $d_X$ induces a topology on $X$. The open $n$-balls $B(x,\epsilon)=\{y\in X: d_X(x,y)< \epsilon\}$ for $\epsilon>0$ are basic open sets for the topology. The resulting topological space is indeed Hausdorff.

leePost authorBaba,

In general, you cannot define a limit point without open sets. An exception is a metric space. In a metric space $(X,d)$, $x\in X$ is a limit point of $A\subset X$ if given $\epsilon>0$ there exists $y\in A$ such that $d(y,x)<\epsilon$. An equivalent definition is that $x\in X$ is a limit point of $A\subset X$ if there exists a sequence $\{x_n\}\subset A$ such that $x_n$ converges to $x$. Using limit points one can define a closet set. First we define the closure $\bar A$ of $A\subset X$ as $\bar A=A\cup A'$ and then we say $A\subset X$ is closed if $A=\bar A$. Once we know what closed sets are, we know open sets as $U\subset X$ being open means the compliment $X\setminus U$ is closed. However, this is not the usual way to define open sets in a metric space. In a metric space, we can first define basic open sets (open $d$-balls) using the given metric and then obtain open sets generated by these basic open sets.

BabaHow can you use open sets to define a limit point if limit points are what define open sets?