Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Get an example of a metric on a countable set that not generates the discrete topology.

I think it may be a set in this way $0 \cup\{1/n:n\in\mathbb N\}$ with the metric $d(x,y)= \vert x-y \vert$ but I can not do a rigorous proof of because cannot be the metric discrete, is because there is no open ball that is exactly$\{0\}$ or what would be the example?

share|cite|improve this question

As Lord Farin says, your example is fine. Another example, even further from the discrete topology, is the usual metric on $\Bbb Q$, the space of rational numbers, given by $d(p,q)=|p-q|$.

share|cite|improve this answer

Your example works. To prove that $\{0\}$ is not open, recall the definition of the induced topology, which has "open balls" $B(x; \epsilon) = \{y \in S: d(x,y) < \epsilon\}$ as a basis. (you can take $\epsilon \in \Bbb Q_{>0}$ if you like). We thus have to show that:

$$\forall \epsilon >0 : B(0; \epsilon) \ne \{0\}$$

that is, there is for each $\epsilon > 0$ a point $x$ in the set so that $d(x, 0) < \epsilon$. Can you do this?

share|cite|improve this answer

Another example can be $\mathbb{Q}$ with the $p$-adic metric.

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.