Finite Subset of Space is Closed

Theorem

Metric Space

Let $M = \struct {A, d}$ be a metric space.

Let $S \subseteq A$ be finite.

Then $S$ is closed in $M$.

Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $F \subseteq X$ be finite.

Then $F$ is closed in $M$.