Finite Subspace of Dense-in-itself Metric Space is not Open

Theorem

Let $M = \struct {A, d}$ be a metric space that is dense-in-itself.

Let $U$ be a finite subset of $A$.

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

Proof

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

Aiming for a contradiction, suppose $U$ is open.

Let:

$\ds D = \min_{i \mathop \ne j} \map d {x_i, x_j}$

Let $x_j \in U$.

Consider the open ball $\map {B_{D/2} } {x_j}$ of $x_j$.

From Open Ball of Metric Space is Open Set, $\map {B_{D/2} } {x_j}$ is an open set.

Then from Finite Intersection of Open Sets of Metric Space is Open, we can see that $U \cap \map {B_{D/2} } {x_j}$ is also an open set.

However, we have that:

$\dfrac D 2 < \ds \min_{i \mathop \ne j} \map d {x_i, x_j}$

Thus the open ball $\map {B_{D / 2} } {x_j}$ is simply the singleton $\set {x_j}$.

But then $x_j$ is an isolated point of $M$.

This contradicts the fact that $M$ is dense-in-itself.

So in fact, $U$ cannot be open.

The result follows.

$\blacksquare$