Element is Meet Irreducible iff Complement of Element is Irreducible

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $P = \left({\tau, \preceq}\right)$ be an ordered set where $\mathord\preceq = \mathord\subseteq \cap \left({\tau \times \tau}\right)$

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 $P$ $\complement_S\left({A}\right)$ is irreducible

where $\complement_S\left({A}\right)$ 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:
 * $\complement_S\left({A}\right) \ne \varnothing$

Thus by definition:
 * $\complement_S\left({A}\right)$ is non-empty.

Thus by definition:
 * $\complement_S\left({A}\right)$ is closed.

Let $B, C$ be closed subsets of $S$ such that
 * $\complement_S\left({A}\right) = B \cup C$

By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union and Relative Complement of Relative Complement:
 * $A = \complement_S\left({B}\right) \cap \complement_S\left({C}\right)$

By definition of closed set:
 * $\complement_S\left({B}\right) \in \tau$ and $\complement_S\left({C}\right) \in \tau$

By definition of topological space:
 * $\complement_S\left({B}\right) \cap \complement_S\left({C}\right) \in \tau$

By Meet in Inclusion Ordered Set:
 * $\complement_S\left({B}\right) \wedge \complement_S\left({C}\right) = A$

By definition of meet irreducible:
 * $A = \complement_S\left({B}\right)$ or $A = \complement_S\left({C}\right)$

Thus by Relative Complement of Relative Complement:
 * $\complement_S\left({A}\right) = B$ or $\complement_S\left({A}\right) = C$

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