Set of Isolated Points of Metric Space is Disjoint from Limit Points
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Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H'$ be the set of limit points of $H$.
Let $H^i$ be the set of isolated points of $H$.
Then:
- $H' \cap H^i = \O$
Proof
Let $a \in H_i$.
Then by definition of isolated point:
- $\exists \epsilon \in \R_{>0}: \set {x \in H: \map d {x, a} < \epsilon} = \set a$
But by Metric Space Axiom $(\text M 1)$:
- $\map d {a, a} = 0$
and so:
- $\set {x \in H: 0 < \map d {x, a} < \epsilon} = \O$
So by definition $a$ is not a limit point of $H$.
That is:
- $a \notin H'$
or:
- $a \in \relcomp A {H'}$
It follows from Intersection with Complement is Empty iff Subset that:
- $H' \cap H^i = \O$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$