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 $(2)$
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 $(3)$
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:
| \(\ds A \cap B\) | \(=\) | \(\ds \O\) | Definition of Disjoint Sets | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A^- \cap B\) | \(=\) | \(\ds \O\) | Open Set Disjoint from Set is Disjoint from Closure | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A^- \cap B^-\) | \(=\) | \(\ds \O\) | Open Set Disjoint from Set is Disjoint from Closure; $A^-$ is open |
As $A, B$ are arbitrary:
$\Box$
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:
| \(\ds A\) | \(\subseteq\) | \(\ds A^-\) | Set is Subset of its Topological Closure | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A \cap \relcomp S {A^-}\) | \(=\) | \(\ds \O\) | Empty Intersection iff Subset of Complement | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A^- \cap \paren {\relcomp S {A^-} }^-\) | \(=\) | \(\ds \O\) | Definition 3 of Extremally Disconnected Space | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\relcomp S {A^-} }^-\) | \(\subseteq\) | \(\ds \relcomp S {A^-}\) | Empty Intersection iff Subset of Complement | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {\relcomp S {A^-} }^-\) | \(=\) | \(\ds \relcomp S {A^-}\) | 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:
$\blacksquare$