Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 2

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Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

Let $\struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.

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

Let $M$ be an ideal in $L$ which is disjoint from $F$ such that:

no ideal in $L$ larger than $M$ is disjoint from $F$.

Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.

$N$ is an ideal in $L$.

Proof

Let:

$x \in N$

$y \in L$

$y \le x$

Then by the definition of $N$ there exists an $m \in M$ such that:

$x \le m \vee a$

Since $y \le x$, it follows that:

$y \le m \vee a$

so $y \in N$.

Let:

$x \in N$

$y \in N$

Then there exist $m_x$ and $m_y$ in $M$ such that:

$x \le m_x \vee a$

$y \le m_y \vee a$

Then:

$x \vee y \le \paren {m_x \vee a} \vee \paren {m_y \vee a} = \paren {m_x \vee m_y} \vee a$

But $m_x \vee m_y \in M$, so:

$x \vee y \in N$

$\blacksquare$