Element is Meet Irreducible iff Complement of Element is 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 $A \in \tau$ such that
 * $A \ne \top_P$

where $\top_P$ denotes the greatest element in $P$.

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

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

Sufficient Condition
Let $A$ be meet irreducible in $P$.

By Top in Ordered Set of Topology:
 * $A \ne S$

By Relative Complement of Empty Set and Relative Complement of Relative Complement:
 * $\relcomp S A \ne \O$

Thus by definition:
 * $\relcomp S A$ is non-empty.

Thus by definition:
 * $\relcomp S A$ is closed.

Let $B, C$ be closed subsets of $S$ such that
 * $\relcomp S A = B \cup C$

By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union and Relative Complement of Relative Complement:
 * $A = \relcomp S B \cap \relcomp S C$

By definition of closed set:
 * $\relcomp S B \in \tau$ and $\relcomp S C \in \tau$

By definition of topological space:
 * $\relcomp S B \cap \relcomp S C \in \tau$

By Meet in Inclusion Ordered Set:
 * $\relcomp S B \wedge \relcomp S C = A$

By definition of meet irreducible:
 * $A = \relcomp S B$ or $A = \relcomp S C$

Thus by Relative Complement of Relative Complement:
 * $\relcomp S A = B$ or $\relcomp S A = C$

Necessary Condition
This follows from Complement of Element is Irreducible implies Element is Meet Irreducible.