Condition for Point being in Closure/Metric Space

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Theorem

Let $M = \struct {S, d}$ be a metric space.

Let $H \subseteq S$.

Let $\map \cl H$ denote the closure of $H$ in $M$.


Let $x \in S$.

Then $x \in \map \cl H$ if and only if:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.


Proof

By definition of closure of $H$ in $M$:

$\map \cl H = H^i \cup H'$

where:

$H^i$ denotes the set of isolated points of $H$
$H'$ denotes the set of limit points of $H$.


Sufficient Condition

Let $x \in \map \cl H$.


Suppose $x \in H^i$.

By definition of isolated point:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set a$

But also by definition of isolated point, we do at least have that:

$x \in H$

and from Center is Element of Open Ball, we have that:

$\forall \epsilon \in \R_{>0}: x \in \map {B_\epsilon} x$

and so:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$


Suppose $x \in H'$:

By definition of limit point:

$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$

and so:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$


So in both cases:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

$\Box$


Necessary Condition

Let $x$ be such that:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$


To show that $x \in \map \cl H$, it is sufficient to show that either:

$x \in H^i$

or that:

$x \in H'$


There are $2$ cases to consider.


$(1): \quad$ Suppose that:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$

Then $x$ is by definition an isolated point of $H$.

That is:

$x \in H^i$


$(2): \quad$ Suppose that:

$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \set x$

Let $\epsilon$ be arbitrary.

As $\map {B_\epsilon} x \cap H \ne \O$, it must follow that:

$\exists y \in \map {B_\epsilon} x \cap H: y \ne x$

and so:

$y \in \paren {\map {B_\epsilon} x \setminus \set x} \cap H$

Hence:

$\paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$

Then $x$ is by definition a limit point of $H$.

That is:

$x \in H'$


Hence the result.

$\blacksquare$


Sources

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