Open and closed balls are the local measuring devices of analysis. In a [Metric Space](/page/Metric%20Space), they turn the numerical distance between two points into neighbourhoods, convergence tests, continuity tests, and compactness arguments in finite-dimensional Euclidean spaces. In Euclidean analysis they are the standard regions on which estimates are localized, while in [Topology](/page/Topology) they are the model examples behind [Open Set](/page/Open%20Set), [Closed Set](/page/Closed%20Set), [Closure](/page/Closure), and [Interior](/page/Interior). A ball answers a simple question: which points lie within a prescribed distance of a centre? The open version records strict nearness and is built for perturbation arguments. The closed version records bounded nearness and is built for limiting arguments. Many basic statements in analysis use both at once: a function may be controlled on an open ball, extended to a closed ball, and then studied through the boundary between them. ## Definition The most useful neighbourhoods in a metric space are those determined by the metric itself. If $x_0$ is the point of interest and $r>0$ is the scale of observation, the open ball consists of all points that can move a positive amount before reaching distance $r$ from $x_0$. [definition: Open Ball] Let $(X,d)$ be a metric space, let $x_0 \in X$, and let $r>0$. The open ball of radius $r$ centred at $x_0$ is \begin{align*} B(x_0,r) := \{x \in X : d(x,x_0)<r\}. \end{align*} [/definition] Limit arguments often produce points only as limits of approximating points. A strict inequality can be lost at the limit, so one needs the version of a ball that permits points exactly at distance $r$ from the centre. [definition: Closed Ball] Let $(X,d)$ be a metric space, let $x_0 \in X$, and let $r>0$. The closed ball of radius $r$ centred at $x_0$ is \begin{align*} \overline{B}(x_0,r) := \{x \in X : d(x,x_0)\le r\}. \end{align*} [/definition] In Euclidean estimates, the notation $B(x_0,r)$ is usually used without naming the metric each time. To make that convention unambiguous, we need a separate Euclidean definition in which distance is measured by the norm $|x-x_0|$. [definition: Euclidean Open Ball] Let $x_0 \in \mathbb{R}^n$ and let $r>0$. The Euclidean open ball of radius $r$ centred at $x_0$ is \begin{align*} B(x_0,r) := \{x \in \mathbb{R}^n : |x-x_0|<r\}. \end{align*} [/definition] When estimates include limiting points or boundary values, the strict Euclidean inequality is too small. To handle compactness and boundary arguments in Euclidean space, we need the version that keeps all points whose Euclidean distance from the centre is at most the chosen radius. [definition: Euclidean Closed Ball] Let $x_0 \in \mathbb{R}^n$ and let $r>0$. The Euclidean closed ball of radius $r$ centred at $x_0$ is \begin{align*} \overline{B}(x_0,r) := \{x \in \mathbb{R}^n : |x-x_0|\le r\}. \end{align*} [/definition] The points added when one passes from the open Euclidean ball to the closed Euclidean ball form the boundary at exactly one fixed distance from the centre. Boundary-value problems and surface integrals need this set as an object in its own right. [definition: Sphere] Let $x_0 \in \mathbb{R}^n$ and let $r>0$. The sphere of radius $r$ centred at $x_0$ is \begin{align*} \partial B(x_0,r) := \{x \in \mathbb{R}^n : |x-x_0|=r\}. \end{align*} [/definition] A radius of zero cannot describe an open neighbourhood, but closed-ball arguments sometimes allow radii to decrease to zero. To handle that endpoint case cleanly, we use a separate degenerate closed ball whose only possible point is the centre itself. [definition: Degenerate Closed Ball] Let $(X,d)$ be a metric space and let $x_0 \in X$. The degenerate closed ball centred at $x_0$ is \begin{align*} \overline{B}(x_0,0) := \{x_0\}. \end{align*} [/definition] [example: Centre as a Zero-Radius Closed Ball] For any metric space $(X,d)$ and any $x_0\in X$, the condition $d(x,x_0)\le 0$ forces $d(x,x_0)=0$, hence $x=x_0$. Thus the zero-radius closed ball records the centre alone. [/example] Most analysis uses only radii $r>0$ for open balls, because neighbourhoods must contain room around their centre. Closed balls of radius zero appear naturally when a sequence of closed balls shrinks to a single point. The real line gives the quickest calibration for the notation. In one dimension, the metric definition recovers the intervals used in limits, so the symbols $B(x_0,r)$ and $\overline{B}(x_0,r)$ should be read as distance-based versions of familiar open and closed intervals. [example: Basic Real-Line Balls] In $\mathbb{R}$ with the Euclidean metric $d(x,y)=|x-y|$, the open ball centred at $x_0\in\mathbb{R}$ with radius $r>0$ is computed from the inequality $|x-x_0|<r$. For any $x\in\mathbb{R}$, \begin{align*} x\in B(x_0,r) \Longleftrightarrow |x-x_0|<r. \end{align*} By the defining property of absolute value on $\mathbb{R}$, this is equivalent to \begin{align*} -r<x-x_0<r. \end{align*} Adding $x_0$ to all three parts gives \begin{align*} x_0-r<x<x_0+r. \end{align*} Therefore \begin{align*} B(x_0,r)=(x_0-r,x_0+r). \end{align*} The closed ball is obtained by replacing strict inequality with non-strict inequality. For any $x\in\mathbb{R}$, \begin{align*} x\in \overline{B}(x_0,r) \Longleftrightarrow |x-x_0|\le r. \end{align*} Again using the absolute value inequality on $\mathbb{R}$, this is equivalent to \begin{align*} -r\le x-x_0\le r. \end{align*} Adding $x_0$ to all three parts gives \begin{align*} x_0-r\le x\le x_0+r. \end{align*} Hence \begin{align*} \overline{B}(x_0,r)=[x_0-r,x_0+r]. \end{align*} Thus, on the real line, metric open and closed balls are exactly the familiar open and closed intervals centred at $x_0$. [/example] ## Equivalent Characterisations Open balls are the local atoms from which all open sets in a metric topology are built. This is the bridge between the metric definition and the topological definition of openness. [quotetheorem:7780] This criterion is the practical test for openness in metric spaces. It says that openness is not a global mystery about the whole set: every point of an open set has some positive amount of room around it. The radius may depend on the point, and that dependence is important near the boundary of the set. Limit arguments with balls require a complementary stability principle. If every approximating point satisfies $d(x,x_0)\le r$, continuity of the distance function should prevent the limit from escaping that same inequality. [quotetheorem:7781] The point of this closedness result is that closed balls are safe containers for convergent sequences. When estimates put all approximating objects inside one closed ball, the limit remains inside the same bound. This is why closed balls appear naturally in compactness and existence arguments, while open balls are better suited to perturbations. There is a tempting identification between the closed ball and the closure of the open ball. To compare these two constructions in Euclidean space, we need the fact that no distance levels are missing and every boundary point can be approached from strictly smaller radii. [quotetheorem:7782] In Euclidean space, the equality tells us that the boundary sphere is exactly what is added when the open ball is closed. The result depends on the ability to move radially inward from a boundary point, so it uses the linear and norm structure of $\mathbb{R}^n$, not just the formal inequality $|x-x_0|\le r$. That Euclidean equality hides a real metric-space obstruction: some spaces have gaps in the possible distances from a centre. Containment can be forced by closedness, but equality needs enough nearby points inside the open ball. [quotetheorem:7783] This containment is the part that survives in every metric space: limits of points with distance strictly less than $r$ can only land at distance at most $r$. Equality can fail because a boundary point at distance $r$ may have no points of smaller distance approaching it. The theorem is therefore a warning to distinguish the closure of an open ball from the closed ball notation. To describe the topology generated by a metric, individual balls are not enough; we need the whole family of possible centres and positive radii. This collection is the basic supply from which metric-open sets are assembled. [definition: Basis of Open Balls] Let $(X,d)$ be a metric space. The basis of open balls associated to $d$ is the collection \begin{align*} \mathcal{B}_d := \{B(x,r) : x \in X,\ r>0\}. \end{align*} [/definition] Once the family $\mathcal{B}_d$ is named, the remaining question is whether it really behaves like a basis. To prove that it generates the metric topology, every metric-open set should be recoverable by taking unions of these balls, and each point of an open set should lie in a ball still contained in that set. [quotetheorem:7784] This theorem is one reason metric spaces are more concrete than arbitrary topological spaces. Instead of checking all open sets at once, many arguments reduce to checking what happens on balls. ## Computations in Common Metrics ### Subspace Effects Balls are always computed inside the ambient metric space. A subset of the real line inherits the same distance formula as $\mathbb{R}$, but points outside the subset are no longer available, so the visible shape of a ball can change near the edge of the space. [example: Balls in a Subspace] Let $X=(0,1)$ with the metric inherited from $\mathbb{R}$, so $d(x,y)=|x-y|$ for $x,y\in X$. We compute the ball in the subspace $X$, which means that the variable $x$ is required to lie in $(0,1)$ throughout. For the open ball centred at $1/4$ with radius $1/2$, the definition gives \begin{align*} B_X(1/4,1/2)=\{x\in(0,1): |x-1/4|<1/2\}. \end{align*} For $x\in(0,1)$, the absolute value inequality is equivalent to \begin{align*} |x-1/4|<1/2 \Longleftrightarrow -1/2<x-1/4<1/2. \end{align*} Adding $1/4$ to all three parts gives \begin{align*} -1/4<x<3/4. \end{align*} Together with the ambient condition $x\in(0,1)$, this becomes \begin{align*} 0<x<3/4. \end{align*} Hence \begin{align*} B_X(1/4,1/2)=(0,3/4). \end{align*} For the closed ball, the same computation uses the non-strict inequality: \begin{align*} \overline{B}_X(1/4,1/2)=\{x\in(0,1): |x-1/4|\le 1/2\}. \end{align*} For $x\in(0,1)$, \begin{align*} |x-1/4|\le 1/2 \Longleftrightarrow -1/2\le x-1/4\le 1/2. \end{align*} Adding $1/4$ to all three parts gives \begin{align*} -1/4\le x\le 3/4. \end{align*} Intersecting this interval with the ambient space $(0,1)$ gives \begin{align*} (0,1)\cap[-1/4,3/4]=(0,3/4]. \end{align*} Therefore \begin{align*} \overline{B}_X(1/4,1/2)=(0,3/4]. \end{align*} The endpoint $-1/4$ would occur in the real-line computation, but it is not a point of $X$, so it is not available in the subspace ball; the ambient metric space is part of the data of the ball. [/example] ### Euclidean and Coordinate Shapes Intervals explain the one-dimensional case, but higher-dimensional balls also encode geometry. To connect the formal definition with the shapes used in multivariable calculus, we compute the unit ball and its boundary in the plane. [example: Euclidean Balls in the Plane] In $\mathbb{R}^2$ with the Euclidean metric, the centre $0$ means $(0,0)$, and for $x=(x_1,x_2)$ the Euclidean distance from $x$ to $0$ is \begin{align*} |x-0|=\sqrt{(x_1-0)^2+(x_2-0)^2}. \end{align*} Thus \begin{align*} |x-0|=\sqrt{x_1^2+x_2^2}. \end{align*} Since both sides are nonnegative, the inequality $|x-0|<1$ is equivalent to its square: \begin{align*} \sqrt{x_1^2+x_2^2}<1 \Longleftrightarrow x_1^2+x_2^2<1. \end{align*} Therefore the open ball is the open unit disk \begin{align*} B(0,1)=\{(x_1,x_2)\in \mathbb{R}^2 : x_1^2+x_2^2<1\}. \end{align*} The closed ball uses the same distance calculation with a non-strict inequality. Since $\sqrt{x_1^2+x_2^2}\ge 0$ and $1\ge 0$, \begin{align*} \sqrt{x_1^2+x_2^2}\le 1 \Longleftrightarrow x_1^2+x_2^2\le 1. \end{align*} Hence \begin{align*} \overline{B}(0,1)=\{(x_1,x_2)\in \mathbb{R}^2 : x_1^2+x_2^2\le 1\}. \end{align*} The sphere definition gives the boundary circle by replacing the inequality with equality: \begin{align*} \partial B(0,1)=\{(x_1,x_2)\in\mathbb{R}^2:x_1^2+x_2^2=1\}. \end{align*} Thus Euclidean balls in the plane are disks, and their spheres are the usual circles that appear as boundaries in multivariable calculus and planar analysis. [/example] The Euclidean metric is not the only meaningful way to measure distance on $\mathbb{R}^n$. To model coordinatewise control rather than rotational symmetry, we need the maximum coordinate difference metric and its different-looking balls. [definition: Supremum Metric] On $\mathbb{R}^n$, the supremum metric is the map \begin{align*} d_\infty: \mathbb{R}^n \times \mathbb{R}^n \to [0,\infty), \qquad (x,y)\mapsto \max_{1\le i\le n}|x_i-y_i|. \end{align*} [/definition] The supremum metric is useful because its balls are coordinate boxes. This makes it natural in estimates where each coordinate is controlled separately. [example: Supremum Metric Balls] In $(\mathbb{R}^2,d_\infty)$, the centre $0$ means $(0,0)$. For $x=(x_1,x_2)$, the supremum metric gives \begin{align*} d_\infty(x,0)=\max(|x_1-0|,|x_2-0|). \end{align*} Since $|x_1-0|=|x_1|$ and $|x_2-0|=|x_2|$, this becomes \begin{align*} d_\infty(x,0)=\max(|x_1|,|x_2|). \end{align*} Thus $x\in B(0,1)$ exactly when \begin{align*} \max(|x_1|,|x_2|)<1. \end{align*} For two [real numbers](/page/Real%20Numbers) $a,b$, the inequality $\max(a,b)<1$ is equivalent to $a<1$ and $b<1$. Applying this with $a=|x_1|$ and $b=|x_2|$ gives \begin{align*} \max(|x_1|,|x_2|)<1 \Longleftrightarrow |x_1|<1 \text{ and } |x_2|<1. \end{align*} The absolute value inequalities are equivalent to \begin{align*} |x_1|<1 \Longleftrightarrow -1<x_1<1. \end{align*} and \begin{align*} |x_2|<1 \Longleftrightarrow -1<x_2<1. \end{align*} Therefore \begin{align*} B(0,1)=\{(x_1,x_2)\in\mathbb{R}^2:-1<x_1<1 \text{ and } -1<x_2<1\}. \end{align*} This is exactly the Cartesian product \begin{align*} B(0,1)=(-1,1)\times(-1,1). \end{align*} For the closed ball, the definition gives \begin{align*} \overline{B}(0,1)=\{(x_1,x_2)\in\mathbb{R}^2:\max(|x_1|,|x_2|)\le 1\}. \end{align*} The inequality $\max(|x_1|,|x_2|)\le 1$ is equivalent to $|x_1|\le 1$ and $|x_2|\le 1$, so \begin{align*} |x_1|\le 1 \Longleftrightarrow -1\le x_1\le 1. \end{align*} and \begin{align*} |x_2|\le 1 \Longleftrightarrow -1\le x_2\le 1. \end{align*} Hence \begin{align*} \overline{B}(0,1)=\{(x_1,x_2)\in\mathbb{R}^2:-1\le x_1\le 1 \text{ and } -1\le x_2\le 1\}. \end{align*} Equivalently, \begin{align*} \overline{B}(0,1)=[-1,1]\times[-1,1]. \end{align*} Thus, in the supremum metric, balls are coordinate squares rather than Euclidean disks; the word ball refers to the chosen distance, not to visual roundness. [/example] To see why Euclidean intuition can fail, it helps to use a metric with the largest possible gap in distances. In the discrete metric, distinct points are all exactly distance $1$ apart, so there are no points at intermediate distances from a centre. [definition: Discrete Metric] Let $X$ be a set. The discrete metric on $X$ is the map $d: X \times X \to [0,\infty)$ given by $d(x,y)=0$ when $x=y$ and $d(x,y)=1$ when $x\ne y$. [/definition] ### Discrete Gaps This metric is a useful warning: closed balls and closures of open balls can diverge when the space has no points at intermediate distances. [example: Closed Ball Not Equal to Closure of Open Ball] Let $X$ be a set with at least two points, equipped with the discrete metric, and fix $x_0\in X$. For any $x\in X$, the definition of the discrete metric gives \begin{align*} d(x,x_0)=0 \text{ if } x=x_0,\quad \text{and}\quad d(x,x_0)=1 \text{ if } x\ne x_0. \end{align*} Therefore \begin{align*} d(x,x_0)<1 \Longleftrightarrow x=x_0. \end{align*} Using the definition of an open ball, \begin{align*} B(x_0,1)=\{x\in X:d(x,x_0)<1\}=\{x_0\}. \end{align*} The singleton $\{x_0\}$ is closed in the discrete metric because its complement is open: if $y\in X\setminus\{x_0\}$, then $B(y,1)=\{y\}\subset X\setminus\{x_0\}$. Since $\{x_0\}$ is already closed, its closure is itself: \begin{align*} \overline{B(x_0,1)}=\overline{\{x_0\}}=\{x_0\}. \end{align*} For the closed ball, the defining inequality is non-strict. For every $x\in X$, either $d(x,x_0)=0$ or $d(x,x_0)=1$, so in both cases \begin{align*} d(x,x_0)\le 1. \end{align*} Hence \begin{align*} \overline{B}(x_0,1)=\{x\in X:d(x,x_0)\le 1\}=X. \end{align*} Because $X$ has at least two points, $\{x_0\}\subsetneq X$, and therefore \begin{align*} \overline{B(x_0,1)}\subsetneq \overline{B}(x_0,1). \end{align*} This shows that, in a general metric space, the closure of an open ball need not equal the closed ball with the same centre and radius. [/example] The example explains why metric notation must be read carefully. The same symbol with a bar placed outside the parentheses has a different meaning from a closed ball symbol. ## Local and Limiting Behavior The fundamental local property of an open ball is that every point inside it is itself the centre of a smaller open ball still contained inside the original one. This is the metric expression of openness. [quotetheorem:7785] This result is the working form of openness used in epsilon-delta arguments. The number $\rho$ measures how much room remains before the point reaches the radius $r$. Closed balls are stable under limits in metric spaces. The next statement is the sequential form, which is often the most practical version in analysis. [quotetheorem:7786] The statement is frequently used without naming the closed ball: if every approximating point satisfies a uniform distance bound from $x_0$, then the limit satisfies the same bound. In finite-dimensional Euclidean analysis, this limiting stability combines with boundedness to give compactness. [quotetheorem:7787] This compactness statement is finite-dimensional. In infinite-dimensional normed spaces, closed bounded balls need not be compact, so compactness cannot be inferred from closedness and boundedness alone. Another finite-dimensional feature is that every Euclidean ball can be translated and rescaled to the unit ball. [quotetheorem:7788] Volume should respond predictably to translations and changes of scale. In $\mathbb{R}^n$, every ball is obtained from the unit ball by translating its centre and multiplying all distances by its radius. This normalization is often enough for a first analysis page. It explains why estimates are commonly proved on a unit ball and then transported to an arbitrary ball by translation and rescaling. Exact volume and surface-area formulas belong to measure theory; here the important point is that Euclidean balls have a stable shape under the basic geometric operations used in local analysis. ## Beyond and Connected Topics Open balls are the starting point for [Continuity](/page/Continuity) in metric spaces. A function is continuous at a point when sufficiently small balls around the input point are sent inside prescribed balls around the output value. [quotetheorem:7789] This theorem is the ball-language form of the $\varepsilon$-$\delta$ definition. The output ball is prescribed first, and continuity says that some input ball can be chosen small enough to map inside it. It is useful because it separates the target tolerance from the local scale needed in the domain. Convergence of sequences has the same neighbourhood structure as continuity. To use balls as convergence tests, one must know that every prescribed radius around the limit eventually contains all terms of the sequence. [quotetheorem:7790] The sequence criterion turns convergence into eventual membership in every ball around the proposed limit. It is local in radius but global in time: for each radius, all sufficiently late terms must remain inside the corresponding ball. This is the version used whenever estimates show that tails of a sequence are trapped in smaller and smaller neighbourhoods. Vector spaces with norms combine metric neighbourhoods with linear operations. To use ball language in functional analysis, we need the version where the distance from $v$ to $v_0$ is measured by the norm of the difference $v-v_0$. [definition: Norm Ball] Let $(V,\|\cdot\|_V)$ be a [normed vector space](/page/Normed%20Vector%20Space), let $v_0\in V$, and let $r>0$. The open norm ball centred at $v_0$ with radius $r$ is \begin{align*} B(v_0,r):=\{v\in V:\|v-v_0\|_V<r\}, \end{align*} and the closed norm ball centred at $v_0$ with radius $r$ is \begin{align*} \overline{B}(v_0,r):=\{v\in V:\|v-v_0\|_V\le r\}. \end{align*} [/definition] Norm balls are central in [Banach Space](/page/Banach%20Space) theory. Boundedness, operator norms, compactness failures, weak compactness, and fixed-point theorems are all expressed in terms of behaviour on balls. In measure theory and PDE, balls are the standard localization domains. For example, an estimate may hold on $B(x_0,r)$ and then be improved on $B(x_0,r/2)$ to avoid boundary effects. [remark: Localisation by Balls] For $U\subset\mathbb{R}^n$ open and $x_0\in U$, the openness of $U$ implies that there exists $r>0$ such that $B(x_0,r)\subset U$. This permits local arguments inside $U$ to be reduced to arguments on Euclidean balls. [/remark] Questions about nearness to a set ask whether some ball around a point reaches that set. Instead of testing every radius separately, one records the infimum of all distances from the point to points of the set. [definition: Distance to a Set] Let $(X,d)$ be a metric space and let $A\subset X$ be nonempty. The distance-to-$A$ function is the map \begin{align*} \operatorname{dist}(\cdot,A):X\to[0,\infty), \qquad x\mapsto \inf_{a\in A} d(x,a). \end{align*} [/definition] The inequality $\operatorname{dist}(x,A)<r$ says that the open ball $B(x,r)$ intersects $A$. The inequality $\operatorname{dist}(x,A)\le r$ is more subtle unless the infimum is attained, which is another place where compactness enters. ## References [Metric Space](/page/Metric%20Space). [Open Set](/page/Open%20Set). [Closed Set](/page/Closed%20Set). [Closure](/page/Closure). [Continuity](/page/Continuity). [Hausdorff Measure](/page/Hausdorff%20Measure). Walter Rudin, *Principles of Mathematical Analysis* (1976). James R. Munkres, *Topology* (2000). Gerald B. Folland, *Real Analysis* (1999).