Example of a metric space where diameter of a ball is not equal twice the radius

Question

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.

Answer 1

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)

Answer 2

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

Answer 3

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 $0$

Answer 4

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!

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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.

Answer 5

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 $r>2x$.

Answer 6

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

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

Answer 7

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 $\text{diam}(B(0,r))=0$.