# Boolean algebra's filter as a partially order's filter

This is a basic question I'm trying to figure out: why the Boolean's filter definition corresponds to the order-theoretic definition of filter ?

Here follows the relevant definitions.

Definition 1 (Filter in a partially ordered-set) A nonempty subset $$F$$ of a partially ordered-set $$(P, \le)$$ is a filter on $$P$$ if the following conditions hold:

1. $$F$$ is downward directed: for every $$x, y \in F, \exists z \in F$$ s.t. $$z \le x$$ and $$z \le y$$.
2. $$F$$ is upward-closed: for every $$x \in F$$ and $$y \in P$$, if $$x \le y$$ then $$y \in F$$.

Definition 2 (Filter in a Boolean algebra) A nonempty subset $$F$$ of a Boolean algebra $$(\mathbb{B}, \vee, \wedge, \bot, \top, \neg)$$ is a filter if it is

1. downward directed: $$\forall x, y \in F, x \wedge y \in F$$;
2. upward-closed: $$\forall x \in F, y \in \mathbb{B}, x \vee y \in F$$.

Boolean algebra seen as a lattice can be presented with a relation order $$\le$$ with the following equivalence: $$x \le y \iff x \vee y = y \iff x \wedge y = x.$$

My question is, under a Boolean algebra, why the two definitions of filter aforementioned are equivalent ?

It's pretty clear to me that the property of upward closure is equivalent as $$x \vee y = \sup\{x, y\}$$. But I have doubt about the first property (downward directed), especially the direction (Definition 1 $$\implies$$ Definition 2). For the reverse direction, we can take $$z = x \wedge y$$.

Draft:

$$z \le x \textrm{ and } z \le y \iff z = z \wedge x \textrm{ and } z = z \wedge y$$, and by substitution we have $$z = z \land (x \land y)$$. How can I prove that $$z = x \land y$$ and thus by hypothesis $$x \land y \in F$$ ?

Let's let $$x,y \in F$$. We want to show $$x \land y \in F$$.
By definition 1, we know there's some $$z \in F$$ with $$z \leq x$$ and $$z \leq y$$. Then by the definition of $$\land$$, we must have $$z \leq x \land y$$ too. Lastly, by upwards closure, we see $$x \land y \in F$$.