Let $A$ be a Boolean algebra and $F\subseteq A$ be a filter on $A$. Why are the following properties equivalent?

Question

Let $\mathcal{A}$ be a Boolean algebra and $F\subseteq \mathcal{A}$ be a filter on $\mathcal{A}$. Why are the following properties equivalent?

$$(1)\,\,\,A\land B\in F\Rightarrow A,B\in F$$ $$(2)\,\,\,A\in F, A\leq B\Rightarrow B\in F$$ $$(3)\,\,\,A\in F\Rightarrow A\lor B\in F$$

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At least I assume they are (this isn't a textbook exercise). I have found 3 different definitions of filters and they all differ only in these 3 properties, sharing the property that $$A,B\in F\Rightarrow A\land B\in F.$$

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I also know of filters that $$(4)\,\,\,F\neq\emptyset \Leftrightarrow I\in F$$ $$(5)\,\,\,F\neq\mathcal{A}\Leftrightarrow O\notin F$$

are necessary properties. Am I right in that (4) and (5) taken together are not equivalences to (1)–(3)?

Answer 1

A filter $\mathcal{F}\subseteq\mathscr{P}(X)$ is

• closed under binary meets/infs/$\cap$/$\wedge$/$\curlywedge$
• upward closed: i.e if $A\in \mathcal{F}$ then anything bigger than $A$ is also in $\mathcal{F}$
• nonempty: i.e. the entire set $X$ or the top element is an element of $\mathcal{F}$

With that in mind observe your (1-3) are equivalent ways of stating the upward closed condition. Indeed, recall the following are equivalent: $s\preceq t\iff \sup\{s,t\}=t\iff \inf\{s,t\}=s$.

The property shared that you mention is the closure under binary meets condition.

Observe your (4) is the nonempty condition.

Your (5) is the statement that the filter is "proper" i.e. $\mathcal{F}\ne \mathscr{P}(X)$ the powerset. Which is equivalent to stating that the bottom element is not in the filter.

Edit: $\mathscr{P}(X)$ is a quick example of a Boolean algebra $\mathcal{A}$.

Answer 2

(2) implies (1) because $A \land B \le A$ and $A \land B \le B$, so if $A \land B$ is in the filter so are $A$ and $B$ when (2) (closed under enlargements) holds. (2) also implies (3) as $A \le A \lor B$, for all $A,B$.

(1) implies (2) because if $A \in \mathcal{F}$ and $A \le B$ then $A = A \land B \in \mathcal{F}$ and then (1) implies $A,B \in \mathcal{F}$, in particular $B \in \mathcal{F}$ as required.

Also (3) implies (2) because if $A \in \mathcal{F}$ and $A \le B$ then $B = A \lor B \in \mathcal{F}$ by (3).

So indeed the three properties are equivalent for subset $\mathcal{F}$ of a Boolean Algebra. The most common is (2) in combination with the closedness under meets $\land$) you already mentioned separately.

Condition (5) for filters only ensures that the trivial case $\mathcal{F}=\mathcal{A}$ does not occur. If $0 \in \mathcal{F}$ any $A$ would be in the filter, as always $0 \le A$. If any $A$ at all is in $\mathcal{F}$ the same enlargement condition (2) implies $1 \in \mathcal{F}$.

So for filters (4) is saying that $\mathcal{F}$ is non-empty, and (5) is saying it's not the whole algebra, both of which are commonly assumed; we typically only consider non-empty non-trivial filters.