# 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 $T = \left({S, \tau}\right)$ 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 $H$:

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

## 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:

 $\displaystyle H \cup \partial H$ $=$ $\displaystyle H \cup \left({H^- \setminus H^\circ}\right)$ Definition of Boundary (Topology) $\displaystyle$ $=$ $\displaystyle H \cup \left({H \setminus H^\circ}\right) \cup \left({H^- \setminus H}\right)$ Union of Relative Complements of Nested Subsets $\displaystyle$ $=$ $\displaystyle H \cup \left({H^- \setminus H}\right)$ $\displaystyle$ $=$ $\displaystyle H^-$ 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 $S \setminus \left({H^i \cup H'}\right) = \varnothing$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.

Otherwise, let $x \in S \setminus \left({H^i \cup H'}\right) \ne \varnothing$.

From De Morgan's Laws:

$S \setminus \left({H^i \cup H'}\right) = \left({S \setminus H^i}\right) \cap \left({S \setminus H'}\right)$

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:

$U \cap H \ne \left\{{x}\right\}$

and:

$H \cap \left({U \setminus \left\{{x}\right\}}\right) = \varnothing$

This implies that

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

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.

$\blacksquare$