Maximal Element of Complement of Filter is Meet Irreducible

From ProofWiki
Jump to navigation Jump to search


Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

Let $F$ be a filter in $L$.

Let $p \in S$.

Let $p = \max \complement_S\left({F}\right)$.

Then $p$ is meet irreducible.


Let $x, y \in S$.

Aiming for a contradiction suppose that

$p = x \wedge y$ and $p \ne x$ and $p \ne y$

By Meet Precedes Operands:

$p \preceq x$ and $p \preceq y$

By definition of $\prec$:

$p \prec x$ and $p \prec y$

We will prove that

$x \notin F$ or $y \notin F$

Aiming for a contradiction suppose that

$x \in F \land y \in F$

By definition of filtered:

$\exists z \in F: z \preceq x \land z \preceq y$

By definition of infimum:

$z \preceq p$

By definition of upper set:

$p \in F$

Thus this contradicts $p \in \complement_S\left({F}\right)$ by definition of greatest element.


By definition of relative complement:

$x \in \complement_S\left({F}\right)$ or $y \in \complement_S\left({F}\right)$

Thus by definition of greatest element: this contradicts $p \prec x$ and $p \prec y$