Finite Subset of Metric Space is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $S \subseteq A$ be finite.


Then $S$ is closed in $M$.


Proof

From Metric Space is Hausdorff, $M$ is Hausdorff.

From Finite Subspace of Hausdorff Space is Closed, $S$ is closed.

$\blacksquare$