After open sets and interior, we study how a set is approached by nearby points. Limit points and closure formalize the idea that a point may be forced into a set by approximation. These concepts are central in the study of closed sets, dense sets, boundary, subspaces, and completeness. In metric spaces, closure can also be described by sequences, making it especially useful for analysis. :::definition[Limit Point] Let $(X,d)$ be a metric space and let $A\subset X$. A point $x\in X$ is called a $limit point$ of $A$ if every open ball with centre $x$ contains a point of $A$ different from $x$. Equivalently, $$\begin{equation*} (S(x,r)\setminus\{x\})\cap A\neq\varnothing \end{equation*}$$ for every $r>0$. ::: :::definition[Derived Set] Let $A$ be a subset of a metric space $(X,d)$. The set of all limit points of $A$ is called the $derived set$ of $A$ and is denoted by $A'$. ::: :::example[Limit Point of a Sequence Set] Let $$\begin{equation*} A=\left\{1,\frac12,\frac13,\ldots,\frac1n,\ldots\right\}\subset\mathbb{R}. \end{equation*}$$ Then $0$ is a limit point of $A$, because every open interval around $0$ contains $1/n$ for all sufficiently large $n$. ::: :::example[Derived Set of an Open Interval] For $A=(0,1)\subset\mathbb{R}$, $$\begin{equation*} A'=[0,1]. \end{equation*}$$ Every point of $[0,1]$ is approached by points of $(0,1)$, and no point outside $[0,1]$ is a limit point of $(0,1)$. ::: :::example[Limit Points in a Discrete Metric Space] Let $(X,d)$ be a discrete metric space. Since $S(x,1/2)=\{x\}$ for every $x\in X$, no subset of a discrete metric space has a limit point. ::: :::theorem[Limit Point Balls Contain Infinitely Many Points] Let $(X,d)$ be a metric space and let $A\subset X$. If $x$ is a limit point of $A$, then every open ball centred at $x$ contains infinitely many points of $A$. ::: :::proof Given that $x$ is a limit point of $A$. To prove that every open ball centred at $x$ contains infinitely many points of $A$. Suppose, if possible, that some ball $S(x,r)$ contains only finitely many points of $A$ different from $x$, say $a_1,\ldots,a_n$. Since $a_i\neq x$, $d(x,a_i)>0$ for every $i$. Put $$\begin{equation*} \rho=\frac12\min\{d(x,a_1),\ldots,d(x,a_n),r\}. \end{equation*}$$ Then $\rho>0$, and $S(x,\rho)$ contains no point of $A$ different from $x$. This contradicts the definition of limit point. Hence every ball centred at $x$ contains infinitely many points of $A$. $\square$ ::: :::theorem[Sequential Characterization of Limit Points] Let $(X,d)$ be a metric space and let $A\subset X$. A point $x\in X$ is a limit point of $A$ if and only if there exists a sequence $\{x_n\}$ of distinct points of $A$ such that $x_n\to x$. ::: :::proof Given that $(X,d)$ is a metric space and $A\subset X$. To prove the equivalence. Suppose $x$ is a limit point of $A$. For each $n\in\mathbb{N}$, choose $$\begin{equation*} x_n\in A\cap S\left(x,\frac1n\right),\qquad x_n\neq x. \end{equation*}$$ The points may be chosen distinct because every ball around a limit point contains infinitely many points of $A$. Then $d(x_n,x)<1/n$, so $x_n\to x$. Conversely, suppose there exists a sequence $\{x_n\}$ of distinct points of $A$ with $x_n\to x$. Let $r>0$. For sufficiently large $n$, $d(x_n,x)<r$. Thus $S(x,r)$ contains a point of $A$ different from $x$. Hence $x$ is a limit point of $A$. $\square$ ::: :::definition[Closure] Let $(X,d)$ be a metric space and let $A\subset X$. The $closure$ of $A$ is $$\begin{equation*} \overline{A}=A\cup A'. \end{equation*}$$ Thus the closure consists of all points of $A$ together with all limit points of $A$. ::: :::theorem[Sequential Characterization of Closure] Let $(X,d)$ be a metric space and let $A\subset X$. For $x\in X$, the following are equivalent: $x\in\overline{A}$; every open ball centred at $x$ meets $A$; there exists a sequence $\{x_n\}$ in $A$ such that $x_n\to x$. ::: :::proof Given that $(X,d)$ is a metric space and $A\subset X$. To prove the equivalence. If $x\in\overline{A}$, then either $x\in A$ or $x$ is a limit point of $A$. In both cases every ball centred at $x$ meets $A$. Suppose every ball centred at $x$ meets $A$. For each $n$, choose $x_n\in A\cap S(x,1/n)$. Then $d(x_n,x)<1/n$, so $x_n\to x$. Suppose there exists a sequence $\{x_n\}$ in $A$ such that $x_n\to x$. If infinitely many terms equal $x$, then $x\in A$. Otherwise every ball around $x$ contains a point of $A$ different from $x$, so $x$ is a limit point. Therefore $x\in\overline{A}$. $\square$ ::: :::theorem[Algebra of Closures] Let $(X,d)$ be a metric space and let $A,B\subset X$. Then $$\begin{equation*} \overline{A\cup B}=\overline{A}\cup\overline{B} \end{equation*}$$ and $$\begin{equation*} \overline{A\cap B}\subset\overline{A}\cap\overline{B}. \end{equation*}$$ ::: :::proof Given that $A,B\subset X$. To prove the closure relations. Since $A\subset A\cup B$ and $B\subset A\cup B$, monotonicity gives $\overline{A}\subset\overline{A\cup B}$ and $\overline{B}\subset\overline{A\cup B}$. Hence $\overline{A}\cup\overline{B}\subset\overline{A\cup B}$. The set $\overline{A}\cup\overline{B}$ is closed and contains $A\cup B$, so it contains $\overline{A\cup B}$. Hence equality holds. Since $A\cap B\subset A$ and $A\cap B\subset B$, we get $\overline{A\cap B}\subset\overline{A}\cap\overline{B}$. $\square$ ::: :::solved-problem Let $A=\{(x,y):y=\sin(1/x),\ 0<x\leq1\}\subset\mathbb{R}^2$. Show that $$\begin{equation*} \overline{A}=A\cup\{(0,y):-1\leq y\leq1\}. \end{equation*}$$ ::: :::solution Since $-1\leq\sin(1/x)\leq1$, every limit point on the vertical axis has the form $(0,y)$ with $-1\leq y\leq1$. Conversely, for any $y\in[-1,1]$, choose $t_n\to\infty$ such that $\sin t_n\to y$ and put $x_n=1/t_n$. Then $x_n\to0$ and $$\begin{equation*} (x_n,\sin(1/x_n))=(x_n,\sin t_n)\to(0,y). \end{equation*}$$ Thus every point $(0,y)$ with $-1\leq y\leq1$ belongs to the closure. Hence the stated formula holds. $\square$ ::: :::exercise [1] Find the derived set of $\{1/n:n\in\mathbb{N}\}$ in $\mathbb{R}$. [2] Find the closure of $(0,1)$ in $\mathbb{R}$. [3] Give an example of a set with no limit points. ::: :::answer [1] The derived set is $\{0\}$. [2] The closure is $[0,1]$. [3] Any subset of a discrete metric space has no limit points. ::: :::faq[Frequently Asked Questions] Q: Must a limit point belong to the set? A: No. The point $0$ is a limit point of $\{1/n:n\in\mathbb{N}\}$ but is not in the set. Q: Is every point of a set a limit point? A: No. Isolated points are not limit points. Q: Why are sequences useful here? A: Metric balls of radius $1/n$ naturally produce sequences converging to the point. ::: :::call-to-action[Continue to Closed Sets] subtitle: Use limit points and closure to study closed sets and complements. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/closed-sets-and-complementation button-ghost: Previous Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/open-sets-and-interior :::