Complement of Element is Irreducible implies Element is Meet Irreducible

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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 $A = y$

$\blacksquare$


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