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If $M$ is a discrete metric space then $\forall a\in M$ $B(a,1)=\{x\in M:d(x,a)<1\}=\{a\}$ is open in $M$. Is the converse true?

If $\{a\}$ is open $\forall a\in M$ then $\{a\}$ is both closed and open:

For any metric space $(M,d)$ $\{a\}$ is closed because if $x\in M\backslash\{a\}$ then $B(x,d(x,a)>0)\subset M\backslash\{a\}$ proving $M\backslash\{a\}$ is open.

So $\{a\}$ is either $M$ or $\emptyset$. Does that make sense?

John Cataldo
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1 Answers1

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If the topology is discrete every subset is open, including $\{a\}$.

If every set $\{a\}$ is open, then every subset is open, because unions of open sets are open.

egreg
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