My question is regarding the notion of balls in metric spaces, and specifically about their diameters. If $(X,d)$ is a metric space and $A \subset X$, then the diameter of $A$ is defined by $$ d(A) = \sup \{ d(a_1,a_2) : a_1 \text{ and } a_2 \in A \}. $$ I wanted to get a "feel" for the definition; so, I tried to verify that the diameter of a ball of radius $r > 0$ in $\mathbb{R}^n$ is exactly $2r$ as per the above definition, and I managed to do this after some effort.

Then, I wondered whether this necessarily happens in every metric space.

Is it possible to give an example of metric space where the diameter of a ball is strictly smaller than the radius?

I am not sure how to go about finding a set $X$ with a metric $d$ such that this condition holds. I guess I am stuck mainly because I am gathering all my intuition from the case of $\mathbb{R}^n$ with the standard metric, and admittedly haven't got a feel for how abstract metric spaces behave. Any help is appreciated.

  • 1
    $\begingroup$ Is it correct to consider the discrete metric? $\endgroup$
    – Jim Art
    Oct 22, 2018 at 8:08

7 Answers 7


Consider the discrete metric $d$ on a set $X$:

$$d(x,y)=\begin{cases} 0,&\text{if }x=y\\ 1,&\text{if }x\ne y\;. \end{cases}$$

Consider the ball of radius $r=1/2$ centered at $x$

Then $B(x,r)=\{x\}$

Now by definition, $\operatorname{diam} A = \sup\{ d(a,b) : a, b \in A \}$

Applying it to our case where $A=B(x,r)$, we have diameter of $A$ equal to $0$

  • 1
    $\begingroup$ But is it even meaningful to say the radius is 1/2 when no two points can be distance 1/2 from each other? $\endgroup$ Oct 22, 2018 at 14:04
  • 1
    $\begingroup$ @MackTuesday yes, it's perfectly meaningful as you can precisely define what the set $B(x,1/2) is. $\endgroup$
    – Ister
    Oct 22, 2018 at 14:06
  • $\begingroup$ @MackTuesday Yes it is meaningful as by definition $B(x_0,r)$ is set of all those points whose distance from $x_0$ is less than $r$. Key word is "Less than". It may not look meaningful if we ask what are elements whose distance from $x_0$ is exactly $1/2$ because we know either two points are at distance zero or a distance one from each other. $\endgroup$ Oct 22, 2018 at 14:24
  • $\begingroup$ You could also define radius as $\inf_x\sup_y d(x,y)$, in which case every singleton would have $r=d=0$, and everything else would have $r=d=1$. $\endgroup$
    – Teepeemm
    Oct 22, 2018 at 17:58
  • $\begingroup$ Ahh sphere v. ball. Right, right. Thanks all. $\endgroup$ Oct 22, 2018 at 21:05

How about the metric space $[0,\infty)$ with the standard metric? The diameter of the ball $B(0,1)$ is $1$, not $2$ (which would be twice its radius)

  • 2
    $\begingroup$ A problem with the question is that in general, in a metric space we can have $B(x,r)=B(x',r')$ with $x\ne x'$ or $r\ne r'$ or both, so an open ball may not have a unique radius or center. Your example is good, but if we ask about closed balls, then $\overline {B(0,1) }=\overline {B(1/2,1/2)}=[0,1]$. $\endgroup$ Oct 22, 2018 at 10:32

This is an elaboration of @user10354138's answer.

Consider a function $d: X \times X \to [0,\infty)$ which is definite, symmetric, and satisfies the ultrametric inequality $$ d(x,y) \leq \max\{ d(x,z),d(z,y) \}\qquad \text{for all } x,y,z \in X \tag{$*$} $$ instead of the usual triangle inequality. Since $\max\{ d(x,z) , d(z,y) \} \leq d(x,z) + d(z,y)$, we see that that $d$ is indeed a metric.

A space $X$ equipped with a metric satisfying $(*)$ has a topology that behaves counter-intuitively in many ways. For instance, every point in a ball is its centre. Precisely, if $x \in X$ and $r > 0$, then for every $y \in B(x,r)$, we have $B(x,r) = B(y,r)$.

To prove this, suppose $y \in B(x,r)$ as above. Suppose $z \in X$ such that $z \in B(y,r)$. We want to show that $z \in B(x,r)$. By the ultrametric inequality and the definition of an open ball, we have $$d(x,z) \leq \max\{ d(x,y), d(y,z) \} < \max\{ r, r \} = r.$$ So, $B(y,r) \subseteq B(x,r)$. Since $x$ and $y$ were arbitrary, the other containment is also true, so $B(x,r) = B(y,r)$ as was to be shown.

What does this mean in the context of diameters? Well, $diam(B(x,r)) = \sup\{ d(x,y) : x,y \in B(x,r) \}$. Since every point of $B(x,r)$ is its centre, $d(x,y) < r$ for all $x,y \in B(x,r)$. Hence, $diam(B(x,r)) \leq r$, so the diameter is no more than the radius!

In fact, there does not exist any open ball in this metric space whose diameter is twice its radius!

Of course, all this theory would be pointless if it turned out that there is no metric satisfying $(*)$. Thankfully, there is indeed a bunch of metrics that satisfy $(*)$, and I will produce some of them below.

Let $X = \mathbb{Q}$, and let $p \in \mathbb{N}$ be prime. Write $m/n \in \mathbb{Q}$ as $$ \frac mn = p^k \frac{m'}{n'}, $$ where $k \in \mathbb{Z}$ and $\gcd(m',p) = 1 = \gcd(n',p)$. This can be done because $\mathbb{Q}$ is the quotient field of $\mathbb{Z}$, which is a unique factorisation domain.

By an absolute value on a field $\mathbb{F}$ we shall mean a function $| \cdot | : \mathbb{F} \to [0,\infty)$ that is definite, multiplicative, and satisfies the triangle inequality. We define the $p$-adic absolute value $| \cdot |_p$ on $\mathbb{Q}$ by $$ \left| \frac mn \right|_p = p^{-k}, $$ where $m/n = p^k (m'n')$ as above. It is easy to check that this satisfies all the properties of an absolute value. In fact, it satisfies an inequality stronger than the triangle inequality, very reminiscent of $(*)$: $$ | x + y |_p \leq \max\{ |x|_p,|y|_p \} \qquad \text{for all } x,y\in \mathbb{Q}.\tag{$\dagger$} $$ This is also called the ultrametric inequality.

Now, any absolute value induces a metric by $d(x,y) = |x - y|$. The metric induced by the $p$-adic absolute value is called the $p$-adic metric, and thanks to $(\dagger)$ the $p$-adic metric satisfies $(*)$. So, for every prime $p \in \mathbb{N}$, you have a metric on $\mathbb{Q}$ for which the diameter of a ball is not equal to twice the radius.

The $p$-adic metric is actually an important object in number theory, so $(\mathbb{Q},| \cdot |_p)$ is an example of a space which arises naturally and is a prototypical answer to your question.


Any ultrametric space would do, for example, the discrete metric on any set, or $\mathbb{Z}$ with the $p$-adic metric.


A fairly common example that isn't completely trivial is the British Rail metric on $\mathbb{R}$, where $d(x,y) = |x| + |y|$ (or $0$ for $x=y$), so $B(x, r)$ has diameter 0 for $x<r$, diameter $r$ for $x<r<2x$ and diameter $2(r-x)$ for $r>2x$.


Consider a plane restricted to two rays from the origin, with a "ball" centered at the origin:

Metric space restricted to two rays with ball at the origin

Nowhere in this ball can you find two points farther apart than the radius of the ball.

  • 1
    $\begingroup$ In fact, any bounded metric space has the property that there is a ball whose diameter is not twice its radius. $\endgroup$
    – user279515
    Oct 23, 2018 at 3:32
  • $\begingroup$ @Brahadeesh That's a good point... in fact if it's bounded, the ball that covers the whole space can have a diameter arbitrarily smaller than its radius... $\endgroup$
    – Owen
    Oct 23, 2018 at 3:41

A bit (very) late answer but I was wondering why this simple example isn't mention here.

Choose $\mathbb{Z}$ with the metric induced by the absolute value $|\cdot |$.

Then, for example, for any $0<r< 1$ we have $\text{diam}(B(0,r))=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.