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One can prove that a metric space is discrete iff each of its elements is an isolated point. Also, a subset of a metric space is closed iff it contains all its limit points. Is this a way to prove that each subset of a discrete metric space is closed, as if there are no limit points, than trivially, each subset contains all its limit points?

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  • $\begingroup$ Sure. But a definition for discrete space is that all its subsets are closed. So there is not much to prove then. If the definition say sthat all its subsets are open then we are not very far from the conclusion that they are all closed as well. Just note that every set is a complement of its complement. $\endgroup$
    – drhab
    Commented May 11, 2018 at 9:57

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Yes, that would be correct. If $S$ is a subset of a discrete metric space, then the set of limit points of $S$ is the empty set, which is, of course, a subset of $S$. Therefore, $S$ is closed.

But it is much more natural to prove that each set is closed by proving that each set is open.

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  • $\begingroup$ Thank you! I know there are other (more straightforward) ways to prove this, I was just curious. $\endgroup$
    – user509037
    Commented May 11, 2018 at 11:31

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