Finite Subspace of Metric Space is Not Open

Theorem

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

Let $U \subseteq M$ be a finite subset of $M$.

Then $U$ is not an open set of $M$.

Proof

Let $U = \left\{{x_0, x_1, \ldots, x_n}\right\}$.

Let $x_j \in U$.

Let $\displaystyle D = \min_{i \ne j} d \left({x_i, x_j}\right)$.

Then $N_{D/2} \left({x_j}\right)$ is a neighborhood of $x_j$ containing only $x_j$.

The result follows.

$\blacksquare$