Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  • Is it true that any metric on a finite set is the discrete metric?

    I can see that it's at least equivalent with the discrete metric since $B(x,\delta)=\{x\}$ where $X=\{a_i\}_{i=1}^n,$ $\delta=\min\{d(a_i,a_j):i\ne j\}, d$ being the metric on $X.$

Added: Does for $X=\{a,b,c\},d:X\to\mathbb R$ such that $d(a,b)=d(b,c)=d(c,a)=2,$$d(a,a)=d(b,b)=d(c,c)=0$ work as a counterexample?

share|cite|improve this question
The usual definition of the discrete metric is distance between distinct points is $1$. Then the answer is clearly no, Make the distances $2$. Or $1$ plus little bits. – André Nicolas Aug 7 '13 at 3:13
up vote 3 down vote accepted

You are quite correct. The usual definition of the discrete metric is $$d(x,y):=\begin{cases}0 & x=y\\1 & x\ne y.\end{cases}$$ Rather, we can say that every metric on a finite set induces the discrete topology.

share|cite|improve this answer

Consider the set $S=\{a,b\}$ endowed with the metric $$d(x,y)=\begin{cases} 0 & \text{if }x=y,\\ 2 & \text{if }x\neq y \end{cases}$$ (in other words, $2\times$ the discrete metric on $S$). This is a valid metric on the finite set $S$ that is different from the discrete metric.

The most you can say is that, for any finite set $S$ endowed with a metric, the induced topology is discrete (in other words, every subset of $S$ is open). See the relevant Wikipedia article.

share|cite|improve this answer

It's been pointed out that two metrics on a finite set are not necessary equal, or isometric. However, they are Lipschitz equivalent! Or to put it another way: Yes, any metric on a finite set is Lipschitz-equivalent to the standard discrete metric.

share|cite|improve this answer

Sorry to bump this thread up but I couldn't help myself seeing that all the counterexamples are scaled discrete metrics only. However, you can define many more metrics on finite sets.

You can create metrics such as $d(a,b)=1$, $d(b,c)=1.5$ and $d(c,a)=2$. More generally, just plot the 3 points on the plane $\mathbb{R}^2$ and construct "triangles" of your liking (the 3 points can also be on the same line). The length of the edges will define your metric which is a scaling of the standard discrete metric iff you have an equilateral triangle.

Of course, they all generate the same discrete topology on the finite set because they are Lipschitz equivalent to the discrete metric as per the previous answer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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