Metric Space is Open in Itself

Theorem

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

Then the set $A$ is an open set of $M$.

Proof

By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.

Let $x \in A$.

An open ball of $x$ in $M$ is by definition a subset of $A$.

Hence the result.

$\blacksquare$