Open Balls and Their Centers and Radii

Reading the Notation for an Open Ball

A metric assigns a number to each pair of points, measuring their distance. An open ball collects the points whose distance from one chosen point is less than a chosen positive threshold. The chosen point is the center; the threshold is the radius. These labels tell us how to test membership, even though the resulting set does not always determine the labels uniquely.

Throughout, $(X,d)$ is a metric space: $X$ is a set and $d:X\times X\to[0,\infty)$ is a metric. In particular, $d(x,y)\ge 0$, $d(x,y)=0$ exactly when $x=y$, and $d(x,x)=0$. When a topology is relevant, $X$ is equipped with the topology induced by $d$. We use $B_d(x,r)$ to make the metric, center, and radius explicit; when the metric is fixed, we may write $B(x,r)$.

Definition: Given a metric space $(X,d)$, a center $x\in X$, and a radius $r>0$, the open ball with center $x$ and radius $r$ is the subset $$ B_d(x,r)=\{y\in X:d(x,y)<r\}. $$ The center is the point written first, and the radius is the positive number written second. A point $y\in X$ belongs to this ball exactly when its distance from $x$ is strictly less than $r$.

The inequality is strict. Thus $d(x,y)=r$ does not qualify: a point exactly at the radius is not in the ball. The condition is about distance in the chosen metric, not about coordinates, appearance, or a separate notion of “inside.” Also, the ball is a subset of $X$, so only points of $X$ can be its members.

The Membership Test and the Center

Proposition: For $x,y\in X$ and $r>0$, $y\in B_d(x,r)$ if and only if $d(x,y)<r$. In particular, the center $x$ belongs to $B_d(x,r)$.

Proof. By definition, $B_d(x,r)$ consists precisely of the points $y\in X$ satisfying $d(x,y)<r$. This proves both directions of the membership equivalence. For the center, the metric axiom gives $d(x,x)=0$. Since $r>0$, $0<r$, and hence $d(x,x)<r$. Therefore $x\in B_d(x,r)$. $\square$

This proposition is a direct but useful guide to reading the notation. To test a candidate point, first make sure it belongs to the underlying set $X$, then compute its distance from the stated center, and finally compare that distance with the stated radius. If the distance equals the radius, the point is excluded; if it is smaller, the point is included.

Read the ambient space and metric.
The distance formula and the set of available points both matter.

Identify the labels.
In $B_d(x,r)$, $x$ is the center and $r$ is the radius; the definition requires $r>0$.

Test the candidate.
Compute $d(x,y)$. Include $y$ exactly when $d(x,y)<r$, not when $d(x,y)\le r$.

Worked Example: An Open Ball in the Real Line

Give a membership test and describe $B(3,2)$ in $\mathbb{R}$ with its standard topology and usual metric $d(x,y)=|x-y|$. Here the center is $3$ and the radius is $2$.

A real number $y$ belongs exactly when $$ d(3,y)=|3-y|<2. $$ The inequality $|3-y|<2$ is equivalent to $-2<3-y<2$. Subtracting $3$ gives $-5<-y<-1$; multiplying by $-1$ reverses both inequalities, so $1<y<5$. Thus $B(3,2)=(1,5)$.

For example, $y=4$ is included because $d(3,4)=|3-4|=1<2$. The point $y=1$ is excluded because $d(3,1)=|3-1|=2$, which equals the radius rather than being less than it. The point $y=5$ is also excluded because $d(3,5)=|3-5|=2$. The point $y=5.5$ is excluded because $d(3,5.5)=|3-5.5|=2.5>2$.

Check: if $1<y<5$, then subtracting $3$ gives $-2<y-3<2$, so $|3-y|=|y-3|<2$. Conversely, $|3-y|<2$ gives $1<y<5$ by the rearrangement above. The interval and the distance test therefore describe exactly the same set; its endpoints are excluded.

Distances in Other Metric Spaces

The notation for a ball is unchanged when the metric or the underlying set changes. What changes is the distance calculation and, in a subspace, which points are available as members. The following examples use the membership test without relying on a drawing.

Worked Example: A Ball in a Finite Metric Space

Let $X=\{a,b,c\}$, equipped with the metric $d$ given by $d(a,b)=2$, $d(b,c)=3$, and $d(a,c)=4$, with $d(x,x)=0$ and symmetry. Use the metric topology. Find $B_d(a,3)$, and test whether $b$ and $c$ belong.

First, these values do define a metric. Distances are nonnegative, symmetric, and zero exactly on equal points. For the triangle inequality, the only three distinct points have side lengths $2,3,4$. The three required comparisons are $4\le 2+3=5$, $3\le 2+4=6$, and $2\le 3+4=7$. If two points in a triangle coincide, the triangle inequality reduces either to $d(x,y)\le d(x,y)+0$ or to $0\le 2d(x,y)$, both of which hold. Thus the stated function is a metric.

The center is $a$, and the radius is $3$. The underlying set has only $a,b,c$, so test each point. For $a$, $d(a,a)=0<3$, so $a$ is included. For $b$, $d(a,b)=2<3$, so $b$ is included. For $c$, $d(a,c)=4>3$, so $c$ is excluded. Therefore $B_d(a,3)=\{a,b\}$.

Check: the distances from $a$ to the members $a,b$ are $0,2$, and both are strictly less than $3$. The only remaining point $c$ has distance $4$, which is not less than $3$. The computed set satisfies the defining test for every point of $X$.

Worked Example: A Ball in a Restricted Metric Space

Take $X=[0,4]$ with the metric $d(x,y)=|x-y|$, restricted from the usual metric on $\mathbb{R}$, and give $X$ its metric topology. Find $B_d(1,2)$ as a subset of $X$.

The center is $1$ and the radius is $2$. A point $y$ must first belong to $X=[0,4]$, and it must satisfy $$ |1-y|<2. $$ This inequality is equivalent to $-2<1-y<2$. Subtracting $1$ gives $-3<-y<1$. Reversing the inequalities after multiplying by $-1$ gives $-1<y<3$. Intersecting with the available points $0\le y\le4$ yields $B_d(1,2)=[0,3)$.

For direct checks, $y=0$ belongs to $X$ and $d(1,0)=|1-0|=1<2$, so it is included. The point $y=2.5$ belongs to $X$ and $d(1,2.5)=|1-2.5|=1.5<2$, so it is included. The point $y=3$ belongs to $X$, but $d(1,3)=|1-3|=2$, so it is excluded. Points of $\mathbb{R}$ outside $X$ are not candidates for membership, regardless of their distance.

Check: every $y\in[0,3)$ satisfies $y\in[0,4]$ and $-1<y<3$, which gives $|1-y|<2$. Any $y\in[0,4]$ satisfying $|1-y|<2$ must have $-1<y<3$, hence belongs to $[0,3)$. Both the metric inequality and the restriction to $X$ are accounted for.

A Set Need Not Reveal Its Labels

The center and radius in the notation specify how a ball is formed. They are not always uniquely recoverable from the resulting subset. This does not make the membership test ambiguous: for any chosen center and radius, membership is still tested using that center, that radius, and the stated metric.

Worked Example: Different Labels, the Same Ball

Let $X=\{a,b\}$ have the discrete metric $d(x,y)=0$ when $x=y$ and $d(x,y)=1$ when $x\ne y$, with its metric topology. Compare $B_d(a,2)$ and $B_d(b,2)$.

For $B_d(a,2)$, the point $a$ has distance $d(a,a)=0<2$, and $b$ has distance $d(a,b)=1<2$. Thus $B_d(a,2)=\{a,b\}=X$. For $B_d(b,2)$, the point $b$ has distance $d(b,b)=0<2$, and $a$ has distance $d(b,a)=1<2$. Thus $B_d(b,2)=\{a,b\}=X$ as well. The same set is also $B_d(a,3)$, because its two distances from $a$ are still $0<3$ and $1<3$.

Check: each of the two points is tested in each ball, and all four distances are either $0$ or $1$, both strictly below the indicated radius $2$ or $3$. Hence the equalities of sets follow directly from the definition. In particular, the set $X$ alone does not specify one unique center or one unique radius.

Common Pitfall: Replacing “Less Than” with “Less Than or Equal To”

A frequent error is to treat the radius as an included boundary distance. For the real-line ball $B(3,2)$, a mistaken calculation might write $d(3,1)=|3-1|=2\le2$, and conclude that $1$ belongs. The arithmetic is correct, but the membership condition is not: the definition requires $d(3,1)<2$, and $2<2$ is false. Therefore $1\notin B(3,2)$. The same applies to $5$, since $d(3,5)=2$ as well. The correct ball is $(1,5)$, not $[1,5]$.

There is a second point to keep separate: the radius is a threshold in the definition, not necessarily the distance from the center to some point in the space. In a finite metric space there may be no point at distance exactly $r$; even when such points exist, the strict test excludes them. Always apply the defining inequality to candidate points rather than assuming the radius is attained.

Takeaway: In $B_d(x,r)$, the center is $x$, the positive radius is $r$, and a candidate $y\in X$ belongs precisely when $d(x,y)<r$. Equality with the radius means exclusion. Distinct center-radius choices can nevertheless yield the same subset.

Check Your Understanding

For each question, state the ambient metric and use the strict membership test.

In $\mathbb{R}$ with $d(x,y)=|x-y|$, identify the center and radius of $B(2,4)$, and determine whether $y=6$ belongs.
In a metric space, explain why the center of $B_d(x,r)$ belongs whenever $r>0$.
Let $X=\{p,q\}$ have the discrete metric. Find $B_d(p,1)$ and $B_d(p,1.1)$, checking the distance to each point.
For $X=[0,5]$ with the restricted usual metric, solve the membership inequality for $B_d(4,2)$, taking account of the underlying set.
Give a concrete example in which two different centers produce the same open ball, and verify membership point by point.