Definitions concerning filters

olivierd

Mar 11, 2018 5 min read

Definitions concerning filters

Definition of a filter

Let $E$ be a set.

Traditional definition of a filter on $E$ : A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff:

$\mathcal F$ is non-empty (equivalently in context: $E \in \mathcal F$ );
$\emptyset \notin \mathcal F$ ;
$\forall A \in \mathcal F, \forall B \subseteq E, (B \supseteq A \implies B \in \mathcal F)$ ;
$\forall A, B \in \mathcal F, A \cap B \in \mathcal F$ .

It follows from the definition that there can be no filters on the empty set.

If we waive the second rule and allow $\emptyset$ among elements of a filter, we obtain a hyperfilter on $E$ . The only hyperfilter on $E$ that is not a filter is $\mathcal P(E)$ , since if $\emptyset$ is an element, all supersets of $\emptyset$ must also belong.

Alternative equivalent definition

A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff $\mathcal F$ is non-empty; and $\emptyset \notin \mathcal F$ ; and $\forall A, B \subseteq E, ((A, B) \in \mathcal F^2 \iff A \cap B \in \mathcal F)$ .

Dispensing with the specification of the enclosing set

One can dispense with the specification of the enclosing set $E$ , retrieving it as $E = \cup_{X \in \mathcal F} X$ . A filter is then a collection $\mathcal F$ of sets such that:

$\mathcal F$ is non-empty;
$\emptyset \notin \mathcal F$ ;
$\forall A \in \mathcal F, \forall B \subseteq \cup_{X \in \mathcal F} X, (B \supseteq A \implies B \in \mathcal F)$ ;
$\forall A, B \in \mathcal F, A \cap B \in \mathcal F$ .

Fineness of filters

If $\mathcal F_1$ and $\mathcal F_2$ are filters on a same set $E$ , then $\mathcal F_1$ is finer than $\mathcal F_2$ iff $\mathcal F_1 \supseteq \mathcal F_2$ .

Filter bases and filter prebases

A collection $\mathcal B$ of sets is a filter base iff:

$\mathcal B$ is non-empty;
$\emptyset \notin \mathcal B$ ;
$\forall A, B \in \mathcal B, \exists C \in \mathcal B, A \cap B \supseteq C$ .

No enclosing set is specified. Indeed, a filter base $\mathcal B$ generates a filter $\mathcal F$ on any superset $E$ of all sets in $\mathcal B$ (that is, on any $E \supseteq \cup_{X \in \mathcal B} X$ ) by:

$\mathcal F = \{ X \subseteq E\ |\ \exists A \in \mathcal B, A \subseteq X \}$

Obviously, the filter that is generated is a superset of the filter base.

Equivalent filter bases

If $\mathcal B_1$ and $\mathcal B_2$ are any two filter bases and $E$ a common enclosing set, let $\mathcal F_1$ and $\mathcal F_2$ be the respective filters generated on $E$ .

Then $\mathcal F_1$ is finer than $\mathcal F_2$ iff every element of $\mathcal B_2$ is superset of some element of $\mathcal B_1$ .

Proof:

Assume that $\mathcal F_1$ is finer than $\mathcal F_2$ . Let $A$ be an element of $\mathcal B_2$ . Then $A$ is also an element of $\mathcal F_2$ , and hence of $\mathcal F_1$ . Since the latter is generated by $\mathcal B_1$ , $A$ is a superset of some element of $\mathcal B_1$ . Hence every element of $\mathcal B_2$ is a superset of some element of $\mathcal B_1$ .
Assume that every element of $\mathcal B_2$ is a superset of some element of $\mathcal B_1$ . Let $A$ be an element of $\mathcal F_2$ . Then it is a superset of some element $B$ of $\mathcal B_2$ . This in turn is a superset of some element $C$ of $\mathcal B_1$ . We have $A \supseteq B \supseteq C$ , hence $A \supseteq C$ with $C \in \mathcal B_1$ , hence $A \in \mathcal F_1$ . Hence $\mathcal F_2 \subseteq \mathcal F_1$ , that is, $\mathcal F_1$ is finer than $\mathcal F_2$ .

It follows that $\mathcal B_1$ and $\mathcal B_2$ will generate the same filter on $E$ iff both every element of $\mathcal B_1$ is superset of some element of $\mathcal B_2$ and every element of $\mathcal B_2$ is superset of some element of $\mathcal B_1$ . Then $\mathcal B_1$ and $\mathcal B2$ are said to be equivalent filter bases (a property that does not depend on the choice of the common enclosing set).

It is trivial to check that equivalence is an equivalence relation among filter bases.

Filter prebases

A nonempty collection $\mathcal B$ of sets is a filter prebase iff all finite intersections of elements of $\mathcal B$ are nonempty: for any finite and nonempty $\mathcal X \subseteq \mathcal B$ , we have $\cap_{X \in \mathcal X} X \neq \emptyset$ .