After closure has been defined, closed sets become natural. A set is closed when it already contains all limit points forced by approximation. Closedness is equivalent to openness of the complement, so many metric-space proofs pass between open and closed statements by taking complements. This lesson develops closed sets, closed balls, and the algebra of closed sets. :::definition[Closed Set] Let $(X,d)$ be a metric space. A subset $F$ of $X$ is called $closed$ if it contains all of its limit points. Equivalently, $$\begin{equation*} F'\subset F. \end{equation*}$$ ::: :::example[A Nonclosed Set] Let $A=\{1,1/2,1/3,\ldots,1/n,\ldots\}\subset\mathbb{R}$. Then $0\in A'$ but $0\notin A$. Therefore $A$ is not closed. ::: :::example[Closed Sets in a Discrete Metric Space] Every subset of a discrete metric space is closed, because no subset has a limit point. Thus every subset contains all of its limit points. ::: :::theorem[Closed Set and Open Complement] Let $(X,d)$ be a metric space and let $F\subset X$. Then $F$ is closed in $X$ if and only if $X\setminus F$ is open in $X$. ::: :::proof Given that $(X,d)$ is a metric space and $F\subset X$. To prove that $F$ is closed if and only if $X\setminus F$ is open. Suppose $F$ is closed. Let $x\in X\setminus F$. Then $x$ is not a limit point of $F$, so there exists $r>0$ such that $S(x,r)\cap F=\varnothing$. Hence $S(x,r)\subset X\setminus F$, and $X\setminus F$ is open. Conversely, suppose $X\setminus F$ is open. Let $x$ be a limit point of $F$. If $x\notin F$, then $x\in X\setminus F$, so some ball centred at $x$ lies in $X\setminus F$. This contradicts the definition of limit point. Therefore $x\in F$, so $F$ is closed. $\square$ ::: :::theorem[Closed Balls Are Closed] Let $(X,d)$ be a metric space. For $x\in X$ and $r>0$, the closed ball $$\begin{equation*} \overline{S}(x,r)=\{y\in X:d(x,y)\leq r\} \end{equation*}$$ is closed. ::: :::proof Given that $(X,d)$ is a metric space. To prove that $\overline{S}(x,r)$ is closed. Let $F=\overline{S}(x,r)$ and let $y\in X\setminus F$. Then $d(x,y)>r$. Put $\rho=d(x,y)-r>0$. If $z\in S(y,\rho)$, then $$\begin{align*} d(x,z) &\geq d(x,y)-d(y,z)\\ &>d(x,y)-\rho\\ &=r. \end{align*}$$ Thus $z\notin F$, so $S(y,\rho)\subset X\setminus F$. Hence $X\setminus F$ is open and $F$ is closed. $\square$ ::: :::theorem[Algebra of Closed Sets] In a metric space, $\varnothing$ and $X$ are closed, arbitrary intersections of closed sets are closed, and finite unions of closed sets are closed. ::: :::proof Given that closedness is equivalent to openness of complements. To prove the algebra of closed sets. The complements of $\varnothing$ and $X$ are $X$ and $\varnothing$, both open. Hence both $\varnothing$ and $X$ are closed. For closed sets $F_\alpha$, $$\begin{equation*} X\setminus\bigcap_{\alpha}F_\alpha=\bigcup_{\alpha}(X\setminus F_\alpha), \end{equation*}$$ which is open. Therefore the intersection is closed. For finitely many closed sets, $$\begin{equation*} X\setminus\bigcup_{i=1}^{n}F_i=\bigcap_{i=1}^{n}(X\setminus F_i), \end{equation*}$$ which is open because it is a finite intersection of open sets. Hence the finite union is closed. $\square$ ::: :::exercise [1] Prove that $\{0\}$ is closed in $\mathbb{R}$. [2] Give an infinite union of closed subsets of $\mathbb{R}$ that is not closed. [3] Explain why a closed ball is closed. ::: :::answer [1] Its complement $(-\infty,0)\cup(0,\infty)$ is open. [2] $\bigcup_{n=1}^{\infty}[-1+1/n,1-1/n]=(-1,1)$. [3] Its complement is open by the reverse triangle inequality argument. ::: :::faq[Frequently Asked Questions] Q: Is every closed set open? A: No. The interval $[0,1]$ is closed but not open in $\mathbb{R}$. Q: Can a set be both open and closed? A: Yes. $\varnothing$ and $X$ are always both open and closed. Q: Why are complements important? A: They convert closed-set statements into open-set statements. ::: :::call-to-action[Continue to Bounded Sets] subtitle: Move from topological containment to metric size through bounded sets and diameters. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/bounded-sets-and-diameters button-ghost: Previous Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/limit-points-and-closure :::