# Equivalence of Definitions of Closure of Topological Subspace

## Theorem

The following definitions of the concept of Closure in the context of Topology are equivalent:

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

Let $H \subseteq S$.

### Definition 1

The closure of $H$ (in $T$) is defined as:

$H^- := H \cup H'$

where $H'$ is the derived set of $H$.

### Definition 2

The closure of $H$ (in $T$) is defined as:

$\displaystyle H^- := \bigcap \left\{{K \supseteq H: K}\right.$ is closed in $\left.{T}\right\}$

### Definition 3

The closure of $H$ (in $T$), denoted $H^-$, is defined as the smallest closed set of $T$ that contains $H$.

### Definition 4

The closure of $H$ (in $T$) is defined as the union of $H$ and its boundary in $T$:

$H^- := H \cup \partial H$

### Definition 5

The closure of $H$ (in $T$) is the union of the set of all isolated points of $H$ and the set of all limit points of $H$:

$H^- := H^i \cup H'$

### Definition 6

The closure of $H$ (in $T$), denoted $H^-$, is the set of all adherent points of $H$.

## Proof

### Definition $1$ is equivalent to Definition $2$

This is proved in Set Closure as Intersection of Closed Sets.

$\Box$

### Definition $2$ is equivalent to Definition $3$

This is proved in Set Closure is Smallest Closed Set.

$\Box$

### Definition $1$ is equivalent to Definition $4$

By the definition of the interior, and Set is Subset of its Topological Closure, it easily follows that

$H^\circ \subseteq H \subseteq H^-$

Then:

 $\displaystyle H \cup \partial H$ $=$ $\displaystyle H \cup \paren {H^- \setminus H^\circ}$ Definition of Boundary (Topology) $\displaystyle$ $=$ $\displaystyle H \cup \paren {H \setminus H^\circ} \cup \paren {H^- \setminus H}$ Union of Relative Complements of Nested Subsets $\displaystyle$ $=$ $\displaystyle H \cup \paren {H^- \setminus H}$ $\displaystyle$ $=$ $\displaystyle H^-$ Union with Relative Complement

$\Box$

### Definition $1$ is equivalent to Definition $5$

Every isolated point of $H$ is a point of $H$.

$(1): \quad H^i \cup H' \subseteq H \cup H'$

If $S \setminus \paren {H^i \cup H'} = \O$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.

Otherwise, let $x \in S \setminus \paren {H^i \cup H'} \ne \O$.

From De Morgan's Laws:

$S \setminus \paren {H^i \cup H'} = \paren {S \setminus H^i} \cap \paren {S \setminus H'}$

Thus $x$ is a point of $S$ that is neither an isolated point of $H$ nor a limit point of $H$.

Then there exists an open set $U$ that contains $x$ such that:

$U \cap H \ne \set x$

and:

$H \cap \paren {U \setminus \set x} = \O$

This implies that

$U \cap H = \O$
$x \notin H \cup H'$

It has been shown that:

$x \notin H^i \cup H' \implies x \notin H \cup H'$
$x \in H \cup H' \implies x \in H^i \cup H'$

And so

$(2): \quad H \cup H' \subseteq H^i \cup H'$

Combining $(1)$ and $(2)$:

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

Hence the result.

$\Box$

### Definition $1$ is equivalent to Definition $6$

By one of the equivalent definitions of an adherent point:

A point $x \in S$ is an adherent point of $H$ if and only if every open neighborhood $U$ of $x$ satisfies $H \cap U \ne \O$
$x \in H^-$ if and only if every open set of $T$ which contains $x$ contains a point in $H$

where:

$H^-$ is the union of $H$ and all the limit points of $H$ in $T$

The equivalence follows.

$\blacksquare$