Introduction to Topology 3: Limit Points, Boundary Points, and Sequential Limits

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 $x\leq -1$.

4 thoughts on “Introduction to Topology 3: Limit Points, Boundary Points, and Sequential Limits

  1. Kyle

    metric 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

    Reply
    1. John Lee Post author

      Rigorously 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.

      Reply
  2. lee Post author

    Baba,

    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.

    Reply

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