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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 $x \land y \in F$ ?

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Say we're working with a definition 1 filter.

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 $x \land y \in F$.


I hope this helps ^_^

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    $\begingroup$ It does, thanks! $\endgroup$ – user428953 Dec 12 '20 at 5:53

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