Metric Space is Open in Itself
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Theorem
Let $M = \struct {A, d}$ 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.
$\blacksquare$
Examples
Closed Real Interval
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 $\closedint a b$.
Sources
- 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 $2.3.10$