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

Definition: 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

Definition: 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.

It is trivial to check that a filter is always a filter base.

Filter generated by a filter base

If $\mathcal B$ is a filter base and $E$ a superset of all the elements of $\mathcal B$ , let $\mathcal F$ be the set of all subsets of $E$ that are superset of some element of $\mathcal B$ :

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

It is trivial to check that $\mathcal F$ is then a filter on $E$ , and is a superset of $\mathcal B$ .

By definition, $\mathcal F$ is the filter generated on $E$ by $\mathcal B$ .

Fineness of filter bases

Definition: If $\mathcal B_1$ and $\mathcal B_2$ are any two filter bases, then $\mathcal B_1$ is finer than $\mathcal B_2$ iff every element of $\mathcal B_2$ is a superset of some element of $\mathcal B_1$ .

Property: If $\mathcal F_1$ and $\mathcal F_2$ are filters on set $E$ and $\mathcal B_1$ and $\mathcal B_2$ filter bases that generate respectively $\mathcal F_1$ and $\mathcal F_2$ , then $\mathcal F_1$ is finer than $\mathcal F_2$ iff $\mathcal B_1$ is finer than $\mathcal B_2$ .

Proof:

Assume that $\mathcal F_1$ is finer than $\mathcal F_2$ . Let $X$ be an element of $\mathcal B_2$ . It is then also an element of $\mathcal F_2$ . Since $\mathcal F_1 \supseteq \mathcal F_2$ , we have $X \in \mathcal F_1$ , and hence $X$ is a superset of some element of $\mathcal B_1$ . This is true for all $X \in \mathcal B_2$ , hence $\mathcal B_1$ is finer than $\mathcal B_2$ .
Assume that $\mathcal B_1$ is finer than $\mathcal B_2$ . Let $X$ be an element of $\mathcal F_2$ . Then $X$ is superset of some element of $\mathcal B_2$ (since $\mathcal B_2$ generates $\mathcal F_2$ ), which itself (since $\mathcal B_1$ is finer than $\mathcal B_2$ ) is a superset of some element of $\mathcal B_1$ , and thus is an element of $\mathcal F_1$ . This is true for all $X \in \mathcal F_2$ , hence $\mathcal F_1$ is finer than $\mathcal F_2$ .

Equivalent filter bases

Definition: Two filter bases are equivalent if each is finer than the other.

Property: Two filter bases generate the same filter on a common enclosing set iff they are equivalent.

Proof: The generated filters are equal, that is, each finer than the other, iff their generating filter bases are each finer than the other, that is, equivalent.

Property: Equivalence is an equivalence relation among filter bases (trivial).

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 $\cap_{X \in \mathcal X} X \neq \emptyset$ .

If $\mathcal B$ is a filter prebase, then together with all the finite intersections of its elements it forms a filter base.

The image of a filter

Let $E$ and $F$ be sets, $\mathcal F$ a filter on $E$ and $f: E \to F$ .

Traditional definition: The filter on $F$ image of $\mathcal F$ by $f$ is $(f^{\rightarrow \rightarrow}(\mathcal F))^{\uparrow F}$ .

Alternative definition: The filter on $F$ image of $\mathcal F$ by $f$ is $\{\ B \subseteq F\ |\ f^\leftarrow(B) \in \mathcal F\ \}$ .

Again, the two definitions are equivalent.