Closedness and openness describe topological position, while boundedness and diameter describe metric size. A set may be open and bounded, closed and unbounded, or bounded without being closed. Diameter records the largest possible separation among points of a set, while distance from a point to a set describes approximation. This lesson studies these metric quantities and their behaviour under closure. :::definition[Bounded Subset] Let $(X,d)$ be a metric space and let $A\subset X$ be nonempty. The set $A$ is called $bounded$ if there exists $M>0$ such that $d(x,y)\leq M$ for all $x,y\in A$. ::: :::definition[Diameter] Let $(X,d)$ be a metric space and let $A\subset X$ be nonempty. If $A$ is bounded, its $diameter$ is $$\begin{equation*} \operatorname{diam}(A)=\sup\{d(x,y):x,y\in A\}. \end{equation*}$$ If $A$ is not bounded, then $\operatorname{diam}(A)=\infty$. ::: :::definition[Distance from a Point to a Set] Let $(X,d)$ be a metric space, let $x\in X$, and let $A\subset X$ be nonempty. The $distance from x to A$ is $$\begin{equation*} d(x,A)=\inf\{d(x,a):a\in A\}. \end{equation*}$$ ::: :::definition[Distance Between Two Sets] Let $A$ and $B$ be nonempty subsets of $(X,d)$. The $distance between A and B$ is $$\begin{equation*} d(A,B)=\inf\{d(a,b):a\in A,\ b\in B\}. \end{equation*}$$ ::: :::example[Bounded Sets in a Discrete Metric Space] In a discrete metric space, $d(x,y)\leq1$ for all $x,y$. Therefore every subset is bounded. If $A$ has more than one point, then $\operatorname{diam}(A)=1$. ::: :::example[A Bounded Metric on the Real Line] On $\mathbb{R}$, the metric $d(x,y)=|x-y|/(1+|x-y|)$ satisfies $0\leq d(x,y)<1$. Hence every subset of $\mathbb{R}$ is bounded in this metric. ::: :::theorem[Diameter of a Set and Its Closure] Let $(X,d)$ be a metric space and let $A\subset X$ be nonempty. Then $$\begin{equation*} \operatorname{diam}(A)=\operatorname{diam}(\overline{A}). \end{equation*}$$ ::: :::proof Given that $A\subset X$. To prove that $\operatorname{diam}(A)=\operatorname{diam}(\overline{A})$. Since $A\subset\overline{A}$, $\operatorname{diam}(A)\leq\operatorname{diam}(\overline{A})$. Let $x,y\in\overline{A}$. Choose sequences $\{x_n\}$ and $\{y_n\}$ in $A$ such that $x_n\to x$ and $y_n\to y$. Then $d(x_n,y_n)\to d(x,y)$. Since $d(x_n,y_n)\leq\operatorname{diam}(A)$ for all $n$, we get $d(x,y)\leq\operatorname{diam}(A)$. Taking the supremum over $x,y\in\overline{A}$ gives $\operatorname{diam}(\overline{A})\leq\operatorname{diam}(A)$. Hence the diameters are equal. $\square$ ::: :::solved-problem Let $S(x,r)$ be an open ball in $(X,d)$ and let $A$ be a closed subset with $\operatorname{diam}(A)\leq r$ and $A\cap S(x,r)\neq\varnothing$. Show that $A\subset S(x,2r)$. ::: :::solution Choose $a_0\in A\cap S(x,r)$. Let $a\in A$. Then $d(x,a_0)<r$ and $d(a_0,a)\leq\operatorname{diam}(A)\leq r$. Hence $$\begin{align*} d(x,a) &\leq d(x,a_0)+d(a_0,a)\\ &<r+r\\ &=2r. \end{align*}$$ Thus $a\in S(x,2r)$, and $A\subset S(x,2r)$. $\square$ ::: :::remark Two disjoint nonempty sets may have distance zero. In $\mathbb{R}$, the sets $(0,1)$ and $(1,2)$ are disjoint, but their distance is $0$. ::: :::exercise [1] Find $\operatorname{diam}((0,1))$ in $\mathbb{R}$. [2] Find $d(0,(1,2))$. [3] Give a bounded infinite subset of $\mathbb{R}$. ::: :::answer [1] The diameter is $1$. [2] The distance is $1$. [3] The set $\{1/n:n\in\mathbb{N}\}$ is bounded and infinite. ::: :::faq[Frequently Asked Questions] Q: Does bounded imply closed? A: No. The interval $(0,1)$ is bounded but not closed. Q: Is a diameter always attained? A: No. The diameter of $(0,1)$ is $1$, but no two points in the interval are distance $1$ apart. Q: Can every subset of a metric space be bounded? A: Yes, if the metric itself is bounded. ::: :::call-to-action[Continue to Nested Closed Sets] subtitle: Use shrinking diameters to study nested closed sets in complete spaces. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/nested-closed-sets-in-complete-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/closed-sets-and-complementation :::