# Equivalence of Definitions of Extremally Disconnected Space

## Theorem

The following definitions of the concept of Extremally Disconnected Space are equivalent:

### Definition 1: using Closures of Open Sets

A $T_2$ (Hausdorff) topological space $T = \struct {S, \tau}$ is extremally disconnected if and only if the closure of every open set of $T$ is open.

### Definition 2: using Interiors of Closed Sets

A $T_2$ (Hausdorff) topological space $T = \struct {S, \tau}$ is extremally disconnected if and only if the interior of every closed set of $T$ is closed.

### Definition 3: using Disjoint Open Sets

A $T_2$ (Hausdorff) topological space $T = \struct {S, \tau}$ is extremally disconnected if and only if the closures of every pair of open sets which are disjoint are also disjoint.

## Proof

### $(1)$ iff $(2)$

Extremally Disconnected by Interior of Closed Sets

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) topological space such that the closure of every open set of $T$ is open.

Let $V \subseteq S$ be closed in $T$.

Then $S \setminus V$ is open by definition.

Then its closure $\paren {S \setminus V}^-$ is open by hypothesis.

By Complement of Interior equals Closure of Complement we have that:

$\paren {S \setminus V}^- = S \setminus V^\circ$

where $V^\circ$ is the interior of $V$.

As $S \setminus V^\circ$ is open in $T$, it follows that $V^-$ is closed.

So the interior of every closed set of $T$ is closed.

$\Box$

By a similar argument we see that if the interior of every closed set of $T$ is closed in $T$, then the closure of every open set of $T$ is open.

Hence the result.

$\Box$

### $(1)$ iff $(3)$

Extremally Disconnected by Disjoint Open Sets

### Definition $1$ if and only if Definition $2$

Follows directly from Complement of Interior equals Closure of Complement.

$\Box$

### Definition $1$ implies Definition $3$

Let $A, B \subseteq S$ be disjoint open sets.

Then:

 $\displaystyle A \cap B$ $=$ $\displaystyle \O$ Definition of Disjoint Sets $\displaystyle \leadsto \ \$ $\displaystyle A^- \cap B$ $=$ $\displaystyle \O$ Disjoint Open Sets remain Disjoint with one Closure $\displaystyle \leadsto \ \$ $\displaystyle A^- \cap B^-$ $=$ $\displaystyle \O$ Disjoint Open Sets remain Disjoint with one Closure; $A^-$ is open

As $A, B$ are arbitrary:

the closures of every pair of open sets which are disjoint are also disjoint.

$\Box$

### Definition $3$ implies Definition $1$

Let $A \subseteq S$ be an open set.

By Topological Closure is Closed, $A^-$ is closed.

Hence $\relcomp S {A^-}$ is open.

We have:

 $\displaystyle A$ $\subseteq$ $\displaystyle A^-$ Set is Subset of its Topological Closure $\displaystyle \leadsto \ \$ $\displaystyle A \cap \relcomp S {A^-}$ $=$ $\displaystyle \O$ Empty Intersection iff Subset of Complement $\displaystyle \leadsto \ \$ $\displaystyle A^- \cap \paren {\relcomp S {A^-} }^-$ $=$ $\displaystyle \O$ Definition $3$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {\relcomp S {A^-} }^-$ $\subseteq$ $\displaystyle \relcomp S {A^-}$ Empty Intersection iff Subset of Complement $\displaystyle \leadsto \ \$ $\displaystyle \paren {\relcomp S {A^-} }^-$ $=$ $\displaystyle \relcomp S {A^-}$ Set is Subset of its Topological Closure

By Set is Closed iff Equals Topological Closure, $\relcomp S {A^-}$ is closed.

Thus $\relcomp S {\relcomp S {A^-} } = A^-$ is open.

As $A$ is arbitrary:

the closure of every open set is open.

$\blacksquare$