Complement of Element is Irreducible implies Element is Meet 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 $A \in \tau$.

Then $\complement_S\left({A}\right)$ is irreducible implies $A$ is meet irreducible in $P$

where $\complement_S\left({A}\right)$ denotes the relative complement of $A$ relative to $S$.

Proof
Assume that
 * $\complement_S\left({A}\right)$ 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 (Set Theory)/Relative Complement/Complement of Intersection:
 * $\complement_S\left({A}\right) = \complement_S\left({x}\right) \cup \complement_S\left({y}\right)$

By definition:
 * $\complement_S\left({x}\right)$ and $\complement_S\left({y}\right)$ are closed.

By definition of irreducible:
 * $\complement_S\left({A}\right) = \complement_S\left({x}\right)$ or $\complement_S\left({A}\right) = \complement_S\left({y}\right)$

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