Metric balls and neighbourhoods form the first topological language of metric spaces. Once a metric $d$ has been fixed on a set $X$, nearness to a point is expressed by points whose distance from that point is less than a positive number. This idea is the basis for open sets, convergence, continuity, closure, and completeness. In this lesson, students learn how open balls, closed balls, and neighbourhoods are written, interpreted, and used in proofs. :::definition[Open Ball] Let $(X,d)$ be a metric space. For $x_0\in X$ and $r>0$, the set $$\begin{equation*} S(x_0,r)=\{x\in X:d(x_0,x)<r\} \end{equation*}$$ is called the $open ball$ with centre $x_0$ and radius $r$. ::: :::definition[Closed Ball] Let $(X,d)$ be a metric space. For $x_0\in X$ and $r>0$, the set $$\begin{equation*} \overline{S}(x_0,r)=\{x\in X:d(x_0,x)\leq r\} \end{equation*}$$ is called the $closed ball$ with centre $x_0$ and radius $r$. ::: :::definition[Neighbourhood] Let $(X,d)$ be a metric space and let $x_0\in X$. A $neighbourhood$ of $x_0$ is an open ball in $X$ with centre $x_0$. Thus a neighbourhood of $x_0$ has the form $S(x_0,r)$ for some $r>0$. ::: The notation depends completely on the metric. In the usual metric on $\mathbb{R}$, balls are intervals. In a discrete metric, small balls may contain only one point. In a function space with the uniform metric, a ball around a function consists of functions whose graphs remain within a fixed vertical band. Students often make the mistake of drawing Euclidean circles even when the metric is not Euclidean. :::example[Balls in the Real Line] Let $\mathbb{R}$ have the usual metric $d(x,y)=|x-y|$. Then $$\begin{equation*} S(x_0,r)=\{x\in\mathbb{R}:|x-x_0|<r\}=(x_0-r,x_0+r). \end{equation*}$$ Similarly, $$\begin{equation*} \overline{S}(x_0,r)=\{x\in\mathbb{R}:|x-x_0|\leq r\}=[x_0-r,x_0+r]. \end{equation*}$$ Hence open balls in the real line are bounded open intervals, and closed balls are bounded closed intervals. ::: :::example[Balls in the Euclidean Plane] Let $\mathbb{R}^2$ have the Euclidean metric $$\begin{equation*} d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}. \end{equation*}$$ Then $S(x_0,r)$ is the set of all points strictly inside the circle with centre $x_0$ and radius $r$. The closed ball $\overline{S}(x_0,r)$ is the set of all points inside or on that circle. ::: :::example[Balls in a Discrete Metric Space] Let $(X,d)$ be a discrete metric space, where $$\begin{equation*} d(x,y)= \begin{cases} 0,&x=y,\\ 1,&x\neq y. \end{cases} \end{equation*}$$ If $0<r\leq1$, then $S(x,r)=\{x\}$. If $r>1$, then $S(x,r)=X$. Thus in a discrete metric space, sufficiently small open balls isolate their centres. ::: :::example[Balls in a Function Space] Let $C_{\mathbb{R}}[a,b]$ be the metric space of real-valued continuous functions on $[a,b]$ with the uniform metric $$\begin{equation*} d(f,g)=\sup_{t\in[a,b]}|f(t)-g(t)|. \end{equation*}$$ For $f_0\in C_{\mathbb{R}}[a,b]$ and $r>0$, $$\begin{equation*} S(f_0,r)=\{f\in C_{\mathbb{R}}[a,b]:\sup_{t\in[a,b]}|f(t)-f_0(t)|<r\}. \end{equation*}$$ This says that $f$ lies uniformly within distance $r$ of $f_0$ on the whole interval. ::: :::theorem Let $(X,d)$ be a metric space. Then every open ball in $X$ is an open set. ::: :::proof Given that $(X,d)$ is a metric space and $S(x,r)$ is an open ball in $X$. To prove that $S(x,r)$ is open. Let $y\in S(x,r)$. Then $d(x,y)<r$. Define $$\begin{equation*} r_0=r-d(x,y). \end{equation*}$$ Then $r_0>0$. We show that $S(y,r_0)\subset S(x,r)$. Let $z\in S(y,r_0)$. Using the triangle inequality, $$\begin{align*} d(x,z) &\leq d(x,y)+d(y,z)\\ &<d(x,y)+r_0\\ &=d(x,y)+r-d(x,y)\\ &=r. \end{align*}$$ Therefore $z\in S(x,r)$. Hence $S(y,r_0)\subset S(x,r)$. Thus every point of $S(x,r)$ has a neighbourhood contained in $S(x,r)$. $\square$ ::: :::solved-problem Let $(X,d)$ be a metric space. Suppose $z\in S(x,r)$ and $0<\varepsilon<r-d(x,z)$. Show that $$\begin{equation*} S(z,\varepsilon)\subset S(x,r). \end{equation*}$$ ::: :::solution Let $y\in S(z,\varepsilon)$. Then $d(z,y)<\varepsilon$. Since $\varepsilon<r-d(x,z)$, using the triangle inequality gives $$\begin{align*} d(x,y) &\leq d(x,z)+d(z,y)\\ &<d(x,z)+r-d(x,z)\\ &=r. \end{align*}$$ Therefore $y\in S(x,r)$. Hence $S(z,\varepsilon)\subset S(x,r)$. $\square$ ::: :::solved-problem Let $(X,d)$ be a metric space and let $x,y\in X$ with $x\neq y$. Show that there exist disjoint open balls, one centred at $x$ and the other centred at $y$. ::: :::solution Since $x\neq y$, $d(x,y)>0$. Put $$\begin{equation*} r=\frac{d(x,y)}{2}. \end{equation*}$$ If possible let $z\in S(x,r)\cap S(y,r)$. Then $d(x,z)<r$ and $d(z,y)<r$. Therefore $$\begin{align*} d(x,y) &\leq d(x,z)+d(z,y)\\ &<r+r\\ &=d(x,y). \end{align*}$$ A contradiction. Hence $S(x,r)\cap S(y,r)=\varnothing$. $\square$ ::: :::exercise [1] Define open ball and closed ball in a metric space. [2] Find $S(2,3)$ in $\mathbb{R}$ with the usual metric. [3] In a discrete metric space, find $S(x,1)$ and $S(x,2)$. [4] Prove that $S(x,\varepsilon)\subset\{y\in X:d(x,y)\leq\varepsilon\}$. ::: :::answer [1] $S(x_0,r)=\{x\in X:d(x_0,x)<r\}$ and $\overline{S}(x_0,r)=\{x\in X:d(x_0,x)\leq r\}$. [2] $S(2,3)=(-1,5)$. [3] $S(x,1)=\{x\}$ and $S(x,2)=X$. [4] If $y\in S(x,\varepsilon)$, then $d(x,y)<\varepsilon$, hence $d(x,y)\leq\varepsilon$. ::: :::faq[Frequently Asked Questions] Q: Is every neighbourhood an open set? A: Yes. A neighbourhood is an open ball, and every open ball is open. Q: Does the shape of a ball depend on the metric? A: Yes. The same underlying set can have different ball shapes under different metrics. Q: Why is the triangle inequality used so often? A: It transfers control from one centre to another nearby centre. ::: :::call-to-action[Continue to Open Sets] subtitle: Review ball inclusion carefully, because it becomes the main proof method for open sets and interior. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/open-sets-and-interior button-ghost: Section Start | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/metric-balls-and-neighbourhoods :::