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:
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:
- $x \vee y \in N$
$\blacksquare$