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  • 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?

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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
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3 Answers 3

up vote 2 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.

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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.

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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.

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