Condition for Point being in Closure

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Let $H^-$ denote the closure of $H$ in $T$.


Let $x \in S$.

Then $x \in H^-$ if and only if every open neighborhood of $x$ contains a point in $H$.


Metric Space

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 1

From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$.

By definition, an open neighborhood of $x$ in $T$ is an open set of $T$ which contains $x$.


Necessary Condition

Let $x \in H^-$.

Then either:

$(1): \quad x \in H$, in which case every open neighborhood of $x$ in $T$ trivially contains a point in $H$ (that is, $x$ itself);
$(2): \quad x$ is a limit point of $H$ in $T$.


Suppose $(2)$ holds.

Then it follows directly from the definition of limit point that every open neighborhood of $x$ in $T$ contains a point in $H$ other than $x$.


Sufficient Condition

Suppose that every open neighborhood of $x$ in $T$ contains a point in $H$.


If $x \in H$, then $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:

$x \in H^-$


If $x \notin H$ then $x$ must be a limit point of $H$ by definition.

So again, $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:

$x \in H^-$

$\blacksquare$


Proof 2

The condition to be proved is equivalent to showing that $x \in H^-$ if and only if, for every open neighborhood $U$ of $x$, the intersection $H \cap U$ is non-empty.


For $U \in \tau$, let $U^{\complement}$ denote the relative complement of $U$ in $S$.

By definition, $U^{\complement}$ is closed in $T$


We have that:

\(\ds H \cap U\) \(=\) \(\ds \O\)
\(\ds \leadstoandfrom \ \ \) \(\ds H\) \(\subseteq\) \(\ds U^{\complement}\) Empty Intersection iff Subset of Complement
\(\ds \leadstoandfrom \ \ \) \(\ds H^-\) \(\subseteq\) \(\ds U^{\complement}\) Set Closure is Smallest Closed Set in Topological Space
\(\ds \leadstoandfrom \ \ \) \(\ds U\) \(\subseteq\) \(\ds \paren {H^-}^{\complement}\) Relative Complement inverts Subsets and Relative Complement of Relative Complement
\(\ds \leadstoandfrom \ \ \) \(\ds H^- \cap U\) \(=\) \(\ds \O\)

Thus:

$x \in U \iff x \notin H^-$

The result follows from the Rule of Transposition.

$\blacksquare$


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