Topology induced by metric

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?

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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 $d(p,q)=|p-q|$.

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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?

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

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