# Complement of Irreducible Topological Subset is Prime Element

## Theorem

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

Let $X$ be an irreducible subset of $S$ such that:

$\relcomp S X \in \tau$

Let $L = \struct {\tau, \preceq}$ be an inclusion ordered set of topology $\tau$.

Then $\relcomp S X$ is prime element in $L$.

## Proof

Let $Y, Z \in \tau$ such that

$Y \wedge Z \preceq \relcomp S X$

By definition of topological space:

$Y \cap Z \in \tau$
$Y \cap Z = Y \wedge Z$

By definition of inclusion ordered set:

$Y \cap Z \subseteq \relcomp S X$
$X \subseteq \relcomp S {Y \cap Z}$
$X \subseteq \relcomp S Y \cup \relcomp S Z$
$X = \paren {\relcomp S Y \cup \relcomp S Z} \cap X$
$X = \paren {\relcomp S Y \cap X} \cup \paren {\relcomp S Z \cap X}$

By definition of closed set and Relative Complement of Relative Complement:

$X$, $\relcomp S Y$, and $\relcomp S Z$ are closed sets.
$\relcomp S Y \cap X$, $\relcomp S Z \cap X$ are closed sets.

By definition of irreducible:

$\relcomp S Y \cap X = X$ or $\relcomp S Z \cap X = X$
$X \subseteq \relcomp S Y$ or $X \subseteq \relcomp S Z$
$Y \subseteq \relcomp S X$ or $Z \subseteq \relcomp S X$

Thus by definition of inclusion ordered set:

$Y \preceq \relcomp S X$ or $Z \preceq \relcomp S X$

$\blacksquare$