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?
2026-03-31 22:03:59.1774994639
Prove each subset using isolated points.
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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.