Metric Space is Open in Itself

Theorem

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

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

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.

$\blacksquare$

Let $\R$ be the real number line considered as an Euclidean space.

Let $\closedint a b \subset \R$ be a closed interval of $\R$.

Then from Closed Real Interval is not Open Set, $\closedint a b$ is not an open set of $\R$.

However, if $\closedint a b$ is considered as a subspace of $\R$, then it is seen that $\closedint a b$ is an open set of $\closedint a b$.

1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.10$