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My question is,"When is a singleton set in a metric space open"?

I think the answer will be reached if we can know what metrics induce the discrete topology.

We know that the discrete metric induces the discrete topology.Is there any other metric which induces the same topology?

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  • $\begingroup$ Any metric bounded away from zero will do. But also, a singleton could be open even if not all other singletons are open, as in $\{ 0 \} \cup \{ 1/n : n \in \mathbb{N} \}$. $\endgroup$ – Ian May 29 '18 at 7:51
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It seems that the question that you are interested in is: “When is every singleton set in a metric space open?” And the answer is indeed that that's when the metric induces the discrete topology. But you can't deduce from this that the metric is the discrete metric. For instance, in $\mathbb Z$ the usual metric induces the discrete topology too.

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  • $\begingroup$ Yes,so can we define a metric(other than the discrete metric) on an arbitrary space which induces the discrete topology? $\endgroup$ – timotheechalamet May 29 '18 at 10:01
  • $\begingroup$ @Anwi Sure. If $d$ is the discrete metric, take $2d$, for instance. If your space has more than one point, $2d\neq d$. $\endgroup$ – José Carlos Santos May 29 '18 at 10:03
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Let $X$ be a set and $d$ a metric on $X$:

the topology induced by $d$ on $X$ is the discrete topology iff $\forall x \in X, d(x,\cdot): X \setminus \{x\} \rightarrow \mathbb{R}$ is bounded below by a strictly positive real number.

Brief proof:

If $d$ induces the discrete topology then $\forall x \in X, \{x\}$ is open so exists a basis element $B(x,\epsilon)$ such that $x \in B(x,\epsilon) \subset \{x\}$, so $B(x,\epsilon)=\{x\}$. It follows that $\forall y \in X, y \neq x, d(x,y)\geqslant\epsilon$.

Conversely, if $\exists a$ strictly positive real such that $\forall y \in X, y \neq x, d(x,y)\geqslant a$ we take the open ball $B(x,\frac a2)$ and observe that $B(x,\frac a2)=\{x\}$.

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