It is well known that the discrete metric induces the discrete topology. One can tinker with this and get a great deal of metrics which induce the discrete topology.

However, given a set equipped with the discrete topology, is there some general characterization of metrics which induce the topology?

I suspect there's something along the lines of the metric being separated from $0$ as follows: "For a metric space $(X,d)$, there exists a constant $c>0$ such that $d(x,y)\geq c$ for all $x,y\in X$ if and only if $d$ induces the discrete topology."

One direction (necessity) is trivial. However, when I try to prove the other direction I always end up needing something like compactness somewhere, but that forces $X$ to have finite cardinality, and of course any metric on a set of finite cardinality induces the discrete topology.

In any case, I'm not sure is what I put in quotes above is true or not, but does anyone know of a way to characterize metrics which induce the discrete topology?


2 Answers 2


This isn't true. Consider the metric space $\{1/n:n\in\mathbb N\}$ with the usual Euclidean metric. This is discrete as for any $n\in\mathbb N$ we have $$d\left(\frac{1}{n+1},\frac1n\right)=\frac{1}{n(n+1)}\text{ and }d\left(\frac{1}{n-1},\frac1n\right)=\frac{1}{n(n-1)}$$ and so the open ball of radius $\frac{1}{n(n+1)}$ is just $\left\{\frac1n\right\}$. Yet clearly $d(\frac{1}{n+1},\frac1n)\to 0$.

  • $\begingroup$ Hmm, thanks for the counterexample! $\endgroup$
    – danzibr
    Mar 13, 2013 at 2:01

The correct statement is that a metric induces the discrete topology if and only if $d(x, -)$ is bounded away from $0$ as a function of the second variable for all $x$. (It need not be uniformly bounded away from $0$ in $x$.) But this is practically a tautology: it's equivalent to the claim that there always exists an open ball around each point $x$ consisting only of $x$.

  • $\begingroup$ Excellent. Thanks. I'll sit down and verify it for myself. $\endgroup$
    – danzibr
    Mar 13, 2013 at 2:02
  • $\begingroup$ This was incredibly helpful , thanks ! $\endgroup$
    – nerdy
    Apr 1, 2015 at 0:26

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