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