Complement of Element is Irreducible implies Element is Meet Irreducible

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

Let $P = \struct {\tau, \preceq}$ be an ordered set where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$

Let $A \in \tau$.

Then $\relcomp S A$ is irreducible implies $A$ is meet irreducible in $P$

where $\relcomp S A$ denotes the relative complement of $A$ relative to $S$.

Proof
Assume that
 * $\relcomp S A$ is irreducible.

Let $x, y \in \tau$ such that
 * $A = x \wedge y$

By definition of topological space:
 * $x \cap y \in \tau$

By Meet in Inclusion Ordered Set:
 * $x \wedge y = x \cap y$

By De Morgan's Laws: Complement of Intersection:
 * $\relcomp S A = \relcomp S x \cup \relcomp S y$

By definition:
 * $\relcomp S x$ and $\relcomp S y$ are closed.

By definition of irreducible:
 * $\relcomp S A = \relcomp S x$ or $\relcomp S A = \relcomp S y$

Thus by Relative Complement of Relative Complement:
 * $A = x$ or $A = y$