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 $X=\{a,b,c\},d:X\times X\to\mathbb R$ such that $d(a,b)=d(b,c)=d(c,a)=2,$$d(a,a)=d(b,b)=d(c,c)=0$ work as a counterexample?
Sorry to bump this thread up but I couldn't help myself seeing that all the counterexamples are scaled discrete metrics only. However, you can define many more metrics on finite sets.
You can create metrics such as $d(a,b)=1$, $d(b,c)=1.5$ and $d(c,a)=2$. More generally, just plot the 3 points on the plane $\mathbb{R}^2$ and construct "triangles" of your liking (the 3 points can also be on the same line). The length of the edges will define your metric which is a scaling of the standard discrete metric iff you have an equilateral triangle.
Of course, they all generate the same discrete topology on the finite set because they are Lipschitz equivalent to the discrete metric as per the previous answer.
You are quite correct. The usual definition of the discrete metric is $$d(x,y):=\begin{cases}0 & x=y\\1 & x\ne y.\end{cases}$$ Rather, we can say that every metric on a finite set induces the discrete topology.
Consider the set $S=\{a,b\}$ endowed with the metric $$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.
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.
