Subset of Metric Space is Subset of its Closure
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Then:
- $H \subseteq H^-$
where $H^-$ denotes the closure of $H$.
Proof
By definition of closure:
- $H^- = H' \cup H^i$
where:
- $H'$ denotes the set of limit points of $H$
- $H^i$ denotes the set of isolated points of $H$.
Let $a \in H$.
If $a$ is a limit point of $H$ then $a \in H'$.
Suppose $a \notin H'$.
Then by definition of limit point:
- $\neg \forall \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} \ne \O$
That is:
- $\exists \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} = \O$
and so as $\map d {a, a} = 0$:
- $\exists \epsilon \in \R_{>0}: \set {x \in A: \map d {x, a} < \epsilon} = \set a$
Thus, by definition, $a$ is an isolated point of $H$.
So $a \in H'$ or $a \in H^i$.
Hence by definition of set union:
- $a \in H \subseteq H^-$
By definition of subset:
- $H \subseteq H \subseteq H^-$
and hence the result by definition of closure.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$