Closure of Subset of Metric Space is Closed
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Then $H^-$ is a closed set of $M$.
Proof
Let $\overline {\paren {H^-} }$ denote the complement of $H^-$.
Let $x \in \overline {\paren {H^-} }$.
By definition of the closure of $H$:
- $x$ is not a limit point of $H$.
So:
- $\exists \epsilon \in \R_{> 0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$
From Intersection with Set Difference is Set Difference with Intersection:
- $\paren {\map {B_\epsilon} x \cap H} \setminus \set x = \O$
From Set Difference with Superset is Empty Set:
- $\map {B_\epsilon} x \cap H \subseteq \set x$
By definition of the closure of $H$:
- $x$ is not an isolated point of $H$.
So:
- $\map {B_\epsilon} x \cap H \ne \set x$
Thus:
- $\map {B_\epsilon} x \cap H = \O$
From Empty Intersection iff Subset of Complement:
- $\map {B_\epsilon} x \subseteq \overline {\paren{H^-} }$
It follows that $\overline {\paren{H^-} }$ is open in $M$.
Thus $H^-$ is closed in $M$ by definition.
$\blacksquare$