# Condition for Point being in Closure

## 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.

$\blacksquare$