Filter on a set

In mathematics, a filter on a set is a family of subsets which is closed under supersets and finite intersections. The concept originates in topology, where the neighborhoods of a point form a filter on the space. Filters were introduced by Henri Cartan in 1937^[1][2] and, as described in the article dedicated to filters in topology, they were subsequently used by Nicolas Bourbaki in their book Topologie Générale as an alternative to the related notion of a net developed in 1922 by E. H. Moore and Herman L. Smith. They have also found applications in model theory and set theory.

For filters on a poset, see Filter (mathematics).

For other uses, see Filter (disambiguation).

Filters on a set were later generalized to order filters. Specifically, a filter on a set ${\displaystyle X}$ is a order filter on the power set of ${\displaystyle X}$ ordered by inclusion.

The notion dual to a filter is an ideal. Ultrafilters are a particularly important subclass of filters.

Definition

Given a set ${\displaystyle X}$, a filter ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle X}$ is a set of subsets of ${\displaystyle X}$ such that:

• ${\displaystyle F}$ is upwards-closed: If ${\displaystyle A,B\subseteq X}$ are such that ${\displaystyle A\in {\mathcal {F}}}$ and ${\displaystyle A\subseteq B}$ then ${\displaystyle B\in {\mathcal {F}}}$,
• ${\displaystyle F}$ is closed under finite intersections: ${\displaystyle X\in {\mathcal {F}}}$,^{[note 1]}, and if ${\displaystyle A\in {\mathcal {F}}}$ and ${\displaystyle B\in {\mathcal {F}}}$ then ${\displaystyle A\cap B\in {\mathcal {F}}}$.

A proper (or non-degenerate) filter is a filter which is proper as a subset of the powerset ${\displaystyle {\mathcal {P}}(X)}$ (i.e., the only improper filter is ${\displaystyle {\mathcal {P}}(X)}$, consisting of all possible subsets). By upwards-closure, a filter is proper if and only if it does not contain the empty set. Many authors adopt the convention that a filter must be proper by definition.

Examples

• The singleton set ${\displaystyle {\mathcal {F}}=\{X\}}$ is called the trivial or indiscrete filter on ${\displaystyle X}$.^[3][4]
• If ${\displaystyle Y}$ is a subset of ${\displaystyle X}$, the subsets of ${\displaystyle X}$ which are supersets of ${\displaystyle Y}$ form a principal filter.
• If ${\displaystyle X}$ is a topological space and ${\displaystyle x\in X}$, then the set of neighborhoods of ${\displaystyle x}$ is a filter on ${\displaystyle X}$, the neighborhood filter of ${\displaystyle x}$.
• Many examples arise from various "largeness" conditions:
  • If ${\displaystyle X}$ is a set, the set of all cofinite subsets of ${\displaystyle X}$ (i.e., those sets whose complement in ${\displaystyle X}$ is finite) is a filter on ${\displaystyle X}$, the Fréchet filter.^[4][3]
  • Similarly, if ${\displaystyle X}$ is a set, the cocountable subsets of ${\displaystyle X}$ (those whose complement is countable) form a filter. More generally, for any infinite cardinal ${\displaystyle \kappa }$, the subsets whose complement has cardinal at most ${\displaystyle \kappa }$ form a filter.
  • If ${\displaystyle X}$ is a complete measure space (e.g., ${\displaystyle \mathbb {R} ^{n}}$ with the Lebesgue measure), the conull subsets of ${\displaystyle X}$, i.e., the subsets whose complement has measure zero, form a filter on ${\displaystyle X}$. (For a non-complete measure space, one can take the subsets which, while not necessarily measurable, are contained in a measurable subset of measure zero.)
  • Similarly, if ${\displaystyle X}$ is a measure space, the subsets whose complement is contained in a measurable subset of finite measure form a filter on ${\displaystyle X}$.
  • If ${\displaystyle X}$ is a topological space, the comeager subsets of ${\displaystyle X}$, i.e., those whose complement is meager, form a filter on ${\displaystyle X}$.
  • The subsets of ${\displaystyle \mathbb {N} }$ which have a natural density of 1 form a filter on ${\displaystyle \mathbb {N} }$.^[5]
• The club filter of a regular uncountable cardinal ${\displaystyle \kappa }$ is the filter of all sets containing a club subset of ${\displaystyle \kappa }$.
• If ${\displaystyle ({\mathcal {F}}_{i})_{i\in I}}$ be a family of filters on ${\displaystyle X}$ and ${\displaystyle {\mathcal {J}}}$ is a filter on ${\displaystyle I}$ then ${\displaystyle \bigcup _{A\in {\mathcal {J}}}\bigcap _{i\in A}{\mathcal {F}}_{i}}$ is a filter on ${\displaystyle X}$ called Kowalsky's filter.^[6]

Principal and free filters

The kernel of a filter ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle X}$ is the intersection of all the subsets of ${\displaystyle X}$ in ${\displaystyle F}$.

A principal filter on ${\displaystyle X}$ is a filter ${\displaystyle {\mathcal {F}}}$ which has a particularly simple form: it contains exactly the supersets of ${\displaystyle Y}$, for some fixed subset ${\displaystyle Y\subseteq X}$. When ${\displaystyle Y=\varnothing }$, this yields the improper filter. In the particular case where ${\displaystyle Y}$ is a singleton, the filter is said to be discrete, i.e., a discrete filter on ${\displaystyle X}$ is the set of subsets of ${\displaystyle X}$ which contain ${\displaystyle y}$, for some fixed element ${\displaystyle y\in X}$ (the filter is also said to be "principal at ${\displaystyle y}$").^[3]

A filter ${\displaystyle {\mathcal {F}}}$ is principal if and only if the kernel of ${\displaystyle {\mathcal {F}}}$ is an element of ${\displaystyle {\mathcal {F}}}$, and when this is the case, ${\displaystyle {\mathcal {F}}}$ consists of the supersets of its kernel. On a finite set, every filter is principal (since the intersection defining the kernel is finite).

A filter is said to be free when it has empty kernel, otherwise it is fixed (and if ${\displaystyle x}$ is an element of the kernel, it is fixed by ${\displaystyle x}$). A filter on a set ${\displaystyle X}$ is free if and only if it contains the Fréchet filter on ${\displaystyle X}$.^[7]

A filter ${\displaystyle {\mathcal {F}}}$ is countably deep if the kernel of any countable subset of ${\displaystyle {\mathcal {F}}}$ belongs to ${\displaystyle {\mathcal {F}}}$.^[8]

Correspondence with order filters

The concept of a filter on a set is a special case of the more general concept of filter on a partially ordered set. By definition, a filter on a partially ordered set ${\displaystyle P}$ is a subset ${\displaystyle F}$ of ${\displaystyle P}$ which is upwards-closed (if ${\displaystyle x\in F}$ and ${\displaystyle x\leq y}$ then ${\displaystyle y\in F}$) and downwards-directed (every finite subset of ${\displaystyle F}$ has a lower bound in ${\displaystyle F}$). A filter on a set ${\displaystyle X}$ is the same as a filter on the powerset ${\displaystyle {\mathcal {P}}(X)}$ ordered by inclusion.^{[proof 1]}

Filter bases and subbases

The intersection of a family of filters on ${\displaystyle X}$ is a filter. Hence, for all set ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X)}$, the intersection of all filters containing ${\displaystyle {\mathcal {B}}}$ is a minimum filter (in the sense of inclusion) containing ${\displaystyle {\mathcal {B}}}$. This filter ${\displaystyle {\mathcal {F}}}$ is called the filter generated by ${\displaystyle {\mathcal {B}}}$, and ${\displaystyle {\mathcal {B}}}$ is said to be a filter subbase of ${\displaystyle {\mathcal {F}}}$. Furthermore, ${\displaystyle {\mathcal {F}}}$ has a more explicit description: it is obtained by closing ${\displaystyle {\mathcal {B}}}$ under finite intersections, then upwards, i.e., ${\displaystyle {\mathcal {F}}}$ consists of the subsets ${\displaystyle Y\subseteq X}$ such that ${\displaystyle A_{0}\cap \dots \cap A_{n-1}\subseteq Y}$ for some natural number ${\displaystyle n}$ and some ${\displaystyle A_{0},\dots ,A_{n-1}\in {\mathcal {B}}}$. Since these operations preserve the kernel, it follows that a set ${\displaystyle {\mathcal {F}}}$ is a proper filter if and only if ${\displaystyle {\mathcal {B}}}$ has the finite intersection property: the intersection of a finite subfamily of ${\displaystyle {\mathcal {B}}}$ is non-empty.

A filter base of a filter ${\displaystyle {\mathcal {F}}}$ is a set ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X)}$ such that ${\displaystyle {\mathcal {F}}}$ is obtained only by applying the second step of closing upwards, i.e., when ${\displaystyle {\mathcal {F}}}$ consists of those subsets ${\displaystyle Y\subseteq X}$ for which ${\displaystyle A\subseteq Y}$ for some ${\displaystyle A\in {\mathcal {B}}}$. This upwards closure is a filter if and only if ${\displaystyle {\mathcal {B}}}$ is downwards-directed, i.e., ${\displaystyle {\mathcal {B}}}$ is non-empty and for all ${\displaystyle A,B\in {\mathcal {B}}}$ there exists ${\displaystyle C\in {\mathcal {B}}}$ such that ${\displaystyle C\subseteq A\cap B}$. When this is the case, ${\displaystyle {\mathcal {B}}}$ is also called a prefilter. The generated filter is proper if and only if ${\displaystyle {\mathcal {B}}}$ does not contain the empty set.

Examples

• When ${\displaystyle X}$ is a topological space and ${\displaystyle x\in X}$, a filter base of the neighborhood filter of ${\displaystyle x}$ is known as a neighborhood base for ${\displaystyle x}$, and similarly, a filter subbase of the neighborhood filter of ${\displaystyle x}$ is known as a neighborhood subbase for ${\displaystyle x}$. The open neighborhoods of ${\displaystyle x}$ always form a neighborhood base of ${\displaystyle x}$, by definition of the neighborhood filter. In ${\displaystyle X=\mathbb {R} ^{n}}$, the closed balls of positive radius around ${\displaystyle x}$ also form a neighborhood base for ${\displaystyle x}$.

• Let ${\displaystyle X}$ be an infinite set and let ${\displaystyle {\mathcal {F}}}$ consist of the subsets of ${\displaystyle X}$ which contain all points but one. Then ${\displaystyle {\mathcal {F}}}$ is a filter subbase of the Fréchet filter on ${\displaystyle X}$, which consists of the cofinite subsets. Its closure under finite intersections is the entire Fréchet filter, but there are smaller bases of the Fréchet filter which contain the subbase ${\displaystyle {\mathcal {F}}}$, such as the one formed by the subsets of ${\displaystyle X}$ which contain all points but a finite odd number. In fact, for every base of the Fréchet filter, removing any subset yields another base of the Fréchet filter.

• If ${\displaystyle X}$ is a topological space, the dense open subsets of ${\displaystyle X}$ form a filter base on ${\displaystyle X}$, because they are closed under finite intersection. The filter they generate consists of the complements of nowhere dense subsets. On ${\displaystyle X=\mathbb {R} ^{n}}$, restricting to the null dense open subsets yields another filter base for the same filter.^{[citation needed]}

• Similarly, if ${\displaystyle X}$ is nonmeager then the countable intersections of dense open subsets form a filter base which generates the filter of comeager subsets.

• Let ${\displaystyle X}$ be a set and let ${\displaystyle (x_{i})_{i\in I}}$ be a net with values in ${\displaystyle X}$, i.e., a family whose domain ${\displaystyle I}$ is a directed set. The filter base of tails of ${\displaystyle (x_{i})}$ consists of the sets ${\displaystyle \{x_{j},j\geq i\}}$ for ${\displaystyle i\in I}$; it is downwards-closed by directedness of ${\displaystyle I}$. A elementary filter is a filter which has a filter base of this form. This example is fundamental in the application of filters in topology.^[9]

• Every π-system is a filter base.

Constructions of filters

Trace

If ${\displaystyle {\mathcal {F}}}$ is a filter on ${\displaystyle X}$ and ${\displaystyle Y\subseteq X}$, the trace of ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle Y}$ is ${\displaystyle \{A\cap Y,A\in {\mathcal {F}}\}}$, which is a filter.

Product

Given a family of sets ${\displaystyle (X_{i})_{i\in I}}$ and a filter ${\displaystyle {\mathcal {F}}_{i}}$ on each ${\displaystyle X_{i}}$, the product filter ${\displaystyle \prod _{i\in I}{\mathcal {F}}_{i}}$ on the product set ${\displaystyle \prod _{i\in I}X_{i}}$ is defined as the filter generated by the sets ${\displaystyle \pi _{i}^{-1}(A)}$ for ${\displaystyle i\in I}$ and ${\displaystyle A\in {\mathcal {F}}_{i}}$, where ${\displaystyle \pi _{i}:\left(\prod _{j\in I}X_{j}\right)\to X_{i}}$ is the projection from the product set onto the ${\displaystyle i}$-th component. This construction is similar to the product topology.^[4]

If each ${\displaystyle {\mathcal {B}}_{i}}$ is a filter base on ${\displaystyle {\mathcal {F}}_{i}}$, a filter base of ${\displaystyle \prod _{i\in I}{\mathcal {F}}_{i}}$ is given by the sets ${\displaystyle \prod _{i\in I}A_{i}}$ where ${\displaystyle (A_{i})}$ is a family such that ${\displaystyle A_{i}\in {\mathcal {F}}_{i}}$ for all ${\displaystyle i\in I}$ and ${\displaystyle A_{i}=X_{i}}$ for all but finitely many ${\displaystyle i\in I}$.^[4]

Image and preimage

Let ${\displaystyle f:X\to Y}$ be a function.

When ${\displaystyle {\mathcal {F}}}$ is a family of subsets of ${\displaystyle X}$, its image by ${\displaystyle f}$ is defined as $${\displaystyle f({\mathcal {F}})=\{\{f(x),x\in A\},A\in {\mathcal {F}}\}}$$ If ${\displaystyle {\mathcal {F}}}$ is a filter on ${\displaystyle X}$ and ${\displaystyle f}$ is additionally surjective, then ${\displaystyle f({\mathcal {F}})}$ is a filter on ${\displaystyle Y}$; furthermore, if ${\displaystyle {\mathcal {B}}}$ is a filter base of ${\displaystyle {\mathcal {F}}}$ then ${\displaystyle f({\mathcal {B}})}$ is a filter base of ${\displaystyle f({\mathcal {F}})}$.

Similarly, when ${\displaystyle {\mathcal {F}}}$ is a family of subsets of ${\displaystyle Y}$, its preimage by ${\displaystyle f}$ is defined as $${\displaystyle f^{-1}({\mathcal {F}})=\{\{x\in X\mid f(x)\in A\},A\in {\mathcal {F}}\}}$$ If ${\displaystyle {\mathcal {F}}}$ is a filter on ${\displaystyle Y}$ and ${\displaystyle f}$ is additionally injective, then ${\displaystyle f^{-1}({\mathcal {F}})}$ is a filter on ${\displaystyle X}$; furthermore, if ${\displaystyle {\mathcal {B}}}$ is a filter base of ${\displaystyle {\mathcal {F}}}$ then ${\displaystyle f^{-1}({\mathcal {B}})}$ is a filter base of ${\displaystyle f^{-1}({\mathcal {F}})}$, and unlike the image case, it also holds that if ${\displaystyle {\mathcal {S}}}$ is a filter subbase of ${\displaystyle {\mathcal {F}}}$ then ${\displaystyle f^{-1}({\mathcal {S}})}$ is a filter subbase of ${\displaystyle f^{-1}({\mathcal {F}})}$.

Preliminaries, notation, and basic notions

From now on, all filters are assumed to be proper and possibly improper filters will be called "dual ideals".

Notation and Definition	Name
${\displaystyle \ker {\mathcal {B}}=\bigcap _{B\in {\mathcal {B}}}B}$	Kernel of ${\displaystyle {\mathcal {B}}}$[7]
${\displaystyle S\setminus {\mathcal {B}}:=\{S\setminus B~:~B\in {\mathcal {B}}\}=\{S\}\,(\setminus )\,{\mathcal {B}}}$	Dual of ${\displaystyle {\mathcal {B}}{\text{ in }}S}$ where ${\displaystyle S}$ is a set.[10]
${\displaystyle {\mathcal {B}}{\big \vert }_{S}:=\{B\cap S~:~B\in {\mathcal {B}}\}={\mathcal {B}}\,(\cap )\,\{S\}}$	Trace of ${\displaystyle {\mathcal {B}}{\text{ on }}S}$[10] or the restriction of ${\displaystyle {\mathcal {B}}{\text{ to }}S}$ where ${\displaystyle S}$ is a set; sometimes denoted by ${\displaystyle {\mathcal {B}}\cap S}$
${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {C}}=\{B\cap C~:~B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$[11]	Elementwise (set) intersection (${\displaystyle {\mathcal {B}}\cap {\mathcal {C}}}$ will denote the usual intersection)
${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {C}}=\{B\cup C~:~B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$[11]	Elementwise (set) union (${\displaystyle {\mathcal {B}}\cup {\mathcal {C}}}$ will denote the usual union)
${\displaystyle {\mathcal {B}}\,(\setminus )\,{\mathcal {C}}=\{B\setminus C~:~B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}\}}$	Elementwise (set) subtraction (${\displaystyle {\mathcal {B}}\setminus {\mathcal {C}}}$ will denote the usual set subtraction)
${\displaystyle {\mathcal {P}}(X)=\{S~:~S\subseteq X\}}$	Power set of a set ${\displaystyle X}$[7]

For any two families ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}},}$ declare that ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$ if and only if for every ${\displaystyle C\in {\mathcal {C}}}$ there exists some ${\displaystyle F\in {\mathcal {F}}{\text{ such that }}F\subseteq C,}$ in which case it is said that ${\displaystyle {\mathcal {C}}}$ is coarser than ${\displaystyle {\mathcal {F}}}$ and that ${\displaystyle {\mathcal {F}}}$ is finer than (or subordinate to) ${\displaystyle {\mathcal {C}}.}$^[4][12][13] The notation ${\displaystyle {\mathcal {F}}\vdash {\mathcal {C}}{\text{ or }}{\mathcal {F}}\geq {\mathcal {C}}}$ may also be used in place of ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}.}$

Two families ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh,^[10] written ${\displaystyle {\mathcal {B}}\#{\mathcal {C}},}$ if ${\displaystyle B\cap C\neq \varnothing {\text{ for all }}B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}.}$

Throughout, ${\displaystyle f}$ is a map and ${\displaystyle S}$ is a set.

Notation and Definition	Name
${\displaystyle f^{-1}({\mathcal {B}})=\left\{f^{-1}(B)~:~B\in {\mathcal {B}}\right\}}$[14]	Image of ${\displaystyle {\mathcal {B}}{\text{ under }}f^{-1},}$ or the preimage of ${\displaystyle {\mathcal {B}}}$ under ${\displaystyle f}$
${\displaystyle f^{-1}(S)=\{x\in \operatorname {domain} f~:~f(x)\in S\}}$	Image of ${\displaystyle S{\text{ under }}f^{-1},}$ or the preimage of ${\displaystyle S{\text{ under }}f}$
${\displaystyle f({\mathcal {B}})=\{f(B)~:~B\in {\mathcal {B}}\}}$[15]	Image of ${\displaystyle {\mathcal {B}}}$ under ${\displaystyle f}$
${\displaystyle f(S)=\{f(s)~:~s\in S\cap \operatorname {domain} f\}}$	Image of ${\displaystyle S{\text{ under }}f}$

A filter ${\displaystyle {\mathcal {B}}}$ is a subfilter of a filter ${\displaystyle {\mathcal {F}}}$^[6][16] if ${\displaystyle {\mathcal {B}}}$ is a filter and ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}}$ where for filters, ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}{\text{ if and only if }}{\mathcal {B}}\leq {\mathcal {F}}.}$ Importantly, the expression "is a superfilter of" is for filters the analog of "is a subsequence of". So despite having the prefix "sub" in common, "is a subfilter of" is actually the reverse of "is a subsequence of." However, ${\displaystyle {\mathcal {B}}\leq {\mathcal {F}}}$ can also be written ${\displaystyle {\mathcal {F}}\vdash {\mathcal {B}}}$ which is described by saying "${\displaystyle {\mathcal {F}}}$ is subordinate to ${\displaystyle {\mathcal {B}}.}$" With this terminology, "is subordinate to" becomes for filters (and also for prefilters) the analog of "is a subsequence of,"^[17] which makes this one situation where using the term "subordinate" and symbol ${\displaystyle \,\vdash \,}$ may be helpful.

Basic examples

Named examples

The intersection of all elements in any non–empty family ${\displaystyle \mathbb {F} \subseteq \operatorname {Filters} (X)}$ is itself a filter on ${\displaystyle X}$ called the infimum or greatest lower bound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X),}$ which is why it may be denoted by ${\displaystyle \bigwedge _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}.}$ Said differently, ${\displaystyle \ker \mathbb {F} =\bigcap _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}\in \operatorname {Filters} (X).}$ Because every filter on ${\displaystyle X}$ has ${\displaystyle \{X\}}$ as a subset, this intersection is never empty. By definition, the infimum is the finest/largest (relative to ${\displaystyle \,\subseteq \,{\text{ and }}\,\leq \,}$) filter contained as a subset of each member of ${\displaystyle \mathbb {F} .}$^[4]

• If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are filters then their infimum in ${\displaystyle \operatorname {Filters} (X)}$ is the filter ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}.}$^[11] If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are prefilters then ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}}$ is a prefilter that is coarser (with respect to ${\displaystyle \,\leq }$) than both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ (that is, ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {B}}\,(\cup )\,{\mathcal {F}}\leq {\mathcal {F}}}$); indeed, it is one of the finest such prefilters, meaning that if ${\displaystyle {\mathcal {S}}}$ is a prefilter such that ${\displaystyle {\mathcal {S}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {S}}\leq {\mathcal {F}}}$ then necessarily ${\displaystyle {\mathcal {S}}\leq {\mathcal {B}}\,(\cup )\,{\mathcal {F}}.}$^[11] More generally, if ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are non−empty families and if ${\displaystyle \mathbb {S}$ :=\{{\mathcal {S}}\subseteq {\mathcal {P}}(X)~:~{\mathcal {S}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {S}}\leq {\mathcal {F}}\}} then ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}\in \mathbb {S} }$ and ${\displaystyle {\mathcal {B}}\,(\cup )\,{\mathcal {F}}}$ is a greatest element (with respect to ${\displaystyle \leq }$) of ${\displaystyle \mathbb {S} .}$^[11]

Let ${\displaystyle \varnothing \neq \mathbb {F} \subseteq \operatorname {DualIdeals} (X)}$ and let ${\displaystyle \cup \mathbb {F} =\bigcup _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}.}$ The supremum or least upper bound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {DualIdeals} (X),}$ denoted by ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}},}$ is the smallest (relative to ${\displaystyle \subseteq }$) dual ideal on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset; that is, it is the smallest (relative to ${\displaystyle \subseteq }$) dual ideal on ${\displaystyle X}$ containing ${\displaystyle \cup \mathbb {F} }$ as a subset. This dual ideal is ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X},}$ where ${\displaystyle \pi \left(\cup \mathbb {F} \right):=\left\{F_{1}\cap \cdots \cap F_{n}~:~n\in \mathbb {N} {\text{ and every }}F_{i}{\text{ belongs to some }}{\mathcal {F}}\in \mathbb {F} \right\}}$ is the π–system generated by ${\displaystyle \cup \mathbb {F} .}$ As with any non–empty family of sets, ${\displaystyle \cup \mathbb {F} }$ is contained in some filter on ${\displaystyle X}$ if and only if it is a filter subbase, or equivalently, if and only if ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$ is a filter on ${\displaystyle X,}$ in which case this family is the smallest (relative to ${\displaystyle \subseteq }$) filter on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset and necessarily ${\displaystyle \mathbb {F} \subseteq \operatorname {Filters} (X).}$

Let ${\displaystyle \varnothing \neq \mathbb {F} \subseteq \operatorname {Filters} (X)}$ and let ${\displaystyle \cup \mathbb {F} =\bigcup _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}.}$ The supremum or least upper bound of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X),}$ denoted by ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}}$ if it exists, is by definition the smallest (relative to ${\displaystyle \subseteq }$) filter on ${\displaystyle X}$ containing every element of ${\displaystyle \mathbb {F} }$ as a subset. If it exists then necessarily ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}=\pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$^[4] (as defined above) and ${\displaystyle \bigvee _{{\mathcal {F}}\in \mathbb {F} }{\mathcal {F}}}$ will also be equal to the intersection of all filters on ${\displaystyle X}$ containing ${\displaystyle \cup \mathbb {F} .}$ This supremum of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X)}$ exists if and only if the dual ideal ${\displaystyle \pi \left(\cup \mathbb {F} \right)^{\uparrow X}}$ is a filter on ${\displaystyle X.}$ The least upper bound of a family of filters ${\displaystyle \mathbb {F} }$ may fail to be a filter.^[4] Indeed, if ${\displaystyle X}$ contains at least 2 distinct elements then there exist filters ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}{\text{ on }}X}$ for which there does not exist a filter ${\displaystyle {\mathcal {F}}{\text{ on }}X}$ that contains both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}.}$ If ${\displaystyle \cup \mathbb {F} }$ is not a filter subbase then the supremum of ${\displaystyle \mathbb {F} {\text{ in }}\operatorname {Filters} (X)}$ does not exist and the same is true of its supremum in ${\displaystyle \operatorname {Prefilters} (X)}$ but their supremum in the set of all dual ideals on ${\displaystyle X}$ will exist (it being the degenerate filter ${\displaystyle {\mathcal {P}}(X)}$).^[8]

• If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ are prefilters (resp. filters on ${\displaystyle X}$) then ${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {F}}}$ is a prefilter (resp. a filter) if and only if it is non–degenerate (or said differently, if and only if ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}}}$ mesh), in which case it is one of the coarsest prefilters (resp. the coarsest filter) on ${\displaystyle X}$ (with respect to ${\displaystyle \,\leq }$) that is finer (with respect to ${\displaystyle \,\leq }$) than both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {F}};}$ this means that if ${\displaystyle {\mathcal {S}}}$ is any prefilter (resp. any filter) such that ${\displaystyle {\mathcal {B}}\leq {\mathcal {S}}{\text{ and }}{\mathcal {F}}\leq {\mathcal {S}}}$ then necessarily ${\displaystyle {\mathcal {B}}\,(\cap )\,{\mathcal {F}}\leq {\mathcal {S}},}$^[11] in which case it is denoted by ${\displaystyle {\mathcal {B}}\vee {\mathcal {F}}.}$^[8]

Kernels

If ${\displaystyle f}$ is a map then ${\displaystyle f(\ker {\mathcal {B}})\subseteq \ker f({\mathcal {B}})}$ and ${\displaystyle f^{-1}(\ker {\mathcal {B}})=\ker f^{-1}({\mathcal {B}}).}$ If ${\displaystyle {\mathcal {B}}\leq {\mathcal {C}}}$ then ${\displaystyle \ker {\mathcal {C}}\subseteq \ker {\mathcal {B}}}$ while if ${\displaystyle {\mathcal {B}}}$ and ${\displaystyle {\mathcal {C}}}$ are equivalent then ${\displaystyle \ker {\mathcal {B}}=\ker {\mathcal {C}}.}$ Equivalent families have equal kernels. Two principal families are equivalent if and only if their kernels are equal; that is, if ${\displaystyle {\mathcal {B}}}$ and ${\displaystyle {\mathcal {C}}}$ are principal then they are equivalent if and only if ${\displaystyle \ker {\mathcal {B}}=\ker {\mathcal {C}}.}$

Classifying families by their kernels

For every filter ${\displaystyle {\mathcal {F}}{\text{ on }}X}$ there exists a unique pair of dual ideals ${\displaystyle {\mathcal {F}}^{*}{\text{ and }}{\mathcal {F}}^{\bullet }{\text{ on }}X}$ such that ${\displaystyle {\mathcal {F}}^{*}}$ is free, ${\displaystyle {\mathcal {F}}^{\bullet }}$ is principal, and ${\displaystyle {\mathcal {F}}^{*}\wedge {\mathcal {F}}^{\bullet }={\mathcal {F}},}$ and ${\displaystyle {\mathcal {F}}^{*}{\text{ and }}{\mathcal {F}}^{\bullet }}$ do not mesh (that is, ${\displaystyle {\mathcal {F}}^{*}\vee {\mathcal {F}}^{\bullet }={\mathcal {P}}(X)}$). The dual ideal ${\displaystyle {\mathcal {F}}^{*}}$ is called the free part of ${\displaystyle {\mathcal {F}}}$ while ${\displaystyle {\mathcal {F}}^{\bullet }}$ is called the principal part^[8] where at least one of these dual ideals is filter. If ${\displaystyle {\mathcal {F}}}$ is principal then ${\displaystyle {\mathcal {F}}^{\bullet }:={\mathcal {F}}{\text{ and }}{\mathcal {F}}^{*}:={\mathcal {P}}(X);}$ otherwise, ${\displaystyle {\mathcal {F}}^{\bullet }:=\{\ker {\mathcal {F}}\}^{\uparrow X}}$ and ${\displaystyle {\mathcal {F}}^{*}:={\mathcal {F}}\vee \{X\setminus \left(\ker {\mathcal {F}}\right)\}^{\uparrow X}}$ is a free (non–degenerate) filter.^[8]

Finer/coarser, subordination, and meshing

The preorder ${\displaystyle \,\leq \,}$ that is defined below is of fundamental importance for the use of prefilters (and filters) in topology. For instance, this preorder is used to define the prefilter equivalent of "subsequence",^[17] where "${\displaystyle {\mathcal {F}}\geq {\mathcal {C}}}$" can be interpreted as "${\displaystyle {\mathcal {F}}}$ is a subsequence of ${\displaystyle {\mathcal {C}}}$" (so "subordinate to" is the prefilter equivalent of "subsequence of"). It is also used to define prefilter convergence in a topological space. The definition of ${\displaystyle {\mathcal {B}}}$ meshes with ${\displaystyle {\mathcal {C}},}$ which is closely related to the preorder ${\displaystyle \,\leq ,}$ is used in Topology to define cluster points.

Two families of sets ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh^[10] and are compatible, indicated by writing ${\displaystyle {\mathcal {B}}\#{\mathcal {C}},}$ if ${\displaystyle B\cap C\neq \varnothing {\text{ for all }}B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}.}$ If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ do not mesh then they are dissociated. If ${\displaystyle S\subseteq X{\text{ and }}{\mathcal {B}}\subseteq {\mathcal {P}}(X)}$ then ${\displaystyle {\mathcal {B}}{\text{ and }}S}$ are said to mesh if ${\displaystyle {\mathcal {B}}{\text{ and }}\{S\}}$ mesh, or equivalently, if the trace of ${\displaystyle {\mathcal {B}}{\text{ on }}S,}$ which is the family $${\displaystyle {\mathcal {B}}{\big \vert }_{S}=\{B\cap S~:~B\in {\mathcal {B}}\},}$$ does not contain the empty set, where the trace is also called the restriction of ${\displaystyle {\mathcal {B}}{\text{ to }}S.}$

Declare that ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}},{\mathcal {F}}\geq {\mathcal {C}},{\text{ and }}{\mathcal {F}}\vdash {\mathcal {C}},}$ stated as ${\displaystyle {\mathcal {C}}}$ is coarser than ${\displaystyle {\mathcal {F}}}$ and ${\displaystyle {\mathcal {F}}}$ is finer than (or subordinate to) ${\displaystyle {\mathcal {C}},}$^{[4][12][13][11][8]} if any of the following equivalent conditions hold:

1. Definition: Every ${\displaystyle C\in {\mathcal {C}}}$ contains some ${\displaystyle F\in {\mathcal {F}}.}$ Explicitly, this means that for every ${\displaystyle C\in {\mathcal {C}},}$ there is some ${\displaystyle F\in {\mathcal {F}}}$ such that ${\displaystyle F\subseteq C.}$
   • Said more briefly in plain English, ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$ if every set in ${\displaystyle {\mathcal {C}}}$ is larger than some set in ${\displaystyle {\mathcal {F}}.}$ Here, a "larger set" means a superset.
2. ${\displaystyle \{C\}\leq {\mathcal {F}}{\text{ for every }}C\in {\mathcal {C}}.}$
   • In words, ${\displaystyle \{C\}\leq {\mathcal {F}}}$ states exactly that ${\displaystyle C}$ is larger than some set in ${\displaystyle {\mathcal {F}}.}$ The equivalence of (a) and (b) follows immediately.
   • From this characterization, it follows that if ${\displaystyle \left({\mathcal {C}}_{i}\right)_{i\in I}}$ are families of sets, then ${\displaystyle \bigcup _{i\in I}{\mathcal {C}}_{i}\leq {\mathcal {F}}{\text{ if and only if }}{\mathcal {C}}_{i}\leq {\mathcal {F}}{\text{ for all }}i\in I.}$
3. ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}^{\uparrow X},}$ which is equivalent to ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {F}}^{\uparrow X}}$;
4. ${\displaystyle {\mathcal {C}}^{\uparrow X}\leq {\mathcal {F}}}$;
5. ${\displaystyle {\mathcal {C}}^{\uparrow X}\leq {\mathcal {F}}^{\uparrow X},}$ which is equivalent to ${\displaystyle {\mathcal {C}}^{\uparrow X}\subseteq {\mathcal {F}}^{\uparrow X}}$;

and if in addition ${\displaystyle {\mathcal {F}}}$ is upward closed, which means that ${\displaystyle {\mathcal {F}}={\mathcal {F}}^{\uparrow X},}$ then this list can be extended to include:

6. ${\displaystyle {\mathcal {C}}\subseteq {\mathcal {F}}.}$^[18]
   • So in this case, this definition of "${\displaystyle {\mathcal {F}}}$ is finer than ${\displaystyle {\mathcal {C}}}$" would be identical to the topological definition of "finer" had ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ been topologies on ${\displaystyle X.}$

If an upward closed family ${\displaystyle {\mathcal {F}}}$ is finer than ${\displaystyle {\mathcal {C}}}$ (that is, ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$) but ${\displaystyle {\mathcal {C}}\neq {\mathcal {F}}}$ then ${\displaystyle {\mathcal {F}}}$ is said to be strictly finer than ${\displaystyle {\mathcal {C}}}$ and ${\displaystyle {\mathcal {C}}}$ is strictly coarser than ${\displaystyle {\mathcal {F}}.}$

Two families are comparable if one of these sets is finer than the other.^[4]

Example: If ${\displaystyle x_{i_{\bullet }}=\left(x_{i_{n}}\right)_{n=1}^{\infty }}$ is a subsequence of ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }}$ then ${\displaystyle \operatorname {Tails} \left(x_{i_{\bullet }}\right)}$ is subordinate to ${\displaystyle \operatorname {Tails} \left(x_{\bullet }\right);}$ in symbols: ${\displaystyle \operatorname {Tails} \left(x_{i_{\bullet }}\right)\vdash \operatorname {Tails} \left(x_{\bullet }\right)}$ and also ${\displaystyle \operatorname {Tails} \left(x_{\bullet }\right)\leq \operatorname {Tails} \left(x_{i_{\bullet }}\right).}$ Stated in plain English, the prefilter of tails of a subsequence is always subordinate to that of the original sequence. To see this, let ${\displaystyle C:=x_{\geq i}\in \operatorname {Tails} \left(x_{\bullet }\right)}$ be arbitrary (or equivalently, let ${\displaystyle i\in \mathbb {N} }$ be arbitrary) and it remains to show that this set contains some ${\displaystyle F:=x_{i_{\geq n}}\in \operatorname {Tails} \left(x_{i_{\bullet }}\right).}$ For the set ${\displaystyle x_{\geq i}=\left\{x_{i},x_{i+1},\ldots \right\}}$ to contain ${\displaystyle x_{i_{\geq n}}=\left\{x_{i_{n}},x_{i_{n+1}},\ldots \right\},}$ it is sufficient to have ${\displaystyle i\leq i_{n}.}$ Since ${\displaystyle i_{1}<i_{2}<\cdots }$ are strictly increasing integers, there exists ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle i_{n}\geq i,}$ and so ${\displaystyle x_{\geq i}\supseteq x_{i_{\geq n}}}$ holds, as desired. Consequently, ${\displaystyle \operatorname {TailsFilter} \left(x_{\bullet }\right)\subseteq \operatorname {TailsFilter} \left(x_{i_{\bullet }}\right).}$ The left hand side will be a strict/proper subset of the right hand side if (for instance) every point of ${\displaystyle x_{\bullet }}$ is unique (that is, when ${\displaystyle x_{\bullet }:\mathbb {N} \to X}$ is injective) and ${\displaystyle x_{i_{\bullet }}}$ is the even-indexed subsequence ${\displaystyle \left(x_{2},x_{4},x_{6},\ldots \right)}$ because under these conditions, every tail ${\displaystyle x_{i_{\geq n}}=\left\{x_{2n},x_{2n+2},x_{2n+4},\ldots \right\}}$ (for every ${\displaystyle n\in \mathbb {N} }$) of the subsequence will belong to the right hand side filter but not to the left hand side filter.

For another example, if ${\displaystyle {\mathcal {B}}}$ is any family then ${\displaystyle \varnothing \leq {\mathcal {B}}\leq {\mathcal {B}}\leq \{\varnothing \}}$ always holds and furthermore, ${\displaystyle \{\varnothing \}\leq {\mathcal {B}}{\text{ if and only if }}\varnothing \in {\mathcal {B}}.}$

Assume that ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ are families of sets that satisfy ${\displaystyle {\mathcal {B}}\leq {\mathcal {F}}{\text{ and }}{\mathcal {C}}\leq {\mathcal {F}}.}$ Then ${\displaystyle \ker {\mathcal {F}}\subseteq \ker {\mathcal {C}},}$ and ${\displaystyle {\mathcal {C}}\neq \varnothing {\text{ implies }}{\mathcal {F}}\neq \varnothing ,}$ and also ${\displaystyle \varnothing \in {\mathcal {C}}{\text{ implies }}\varnothing \in {\mathcal {F}}.}$ If in addition to ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}},{\mathcal {F}}}$ is a filter subbase and ${\displaystyle {\mathcal {C}}\neq \varnothing ,}$ then ${\displaystyle {\mathcal {C}}}$ is a filter subbase^[11] and also ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ mesh.^{[19][proof 2]} More generally, if both ${\displaystyle \varnothing \neq {\mathcal {B}}\leq {\mathcal {F}}{\text{ and }}\varnothing \neq {\mathcal {C}}\leq {\mathcal {F}}}$ and if the intersection of any two elements of ${\displaystyle {\mathcal {F}}}$ is non–empty, then ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh.^{[proof 2]} Every filter subbase is coarser than both the π–system that it generates and the filter that it generates.^[11]

If ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ are families such that ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}},}$ the family ${\displaystyle {\mathcal {C}}}$ is ultra, and ${\displaystyle \varnothing \not \in {\mathcal {F}},}$ then ${\displaystyle {\mathcal {F}}}$ is necessarily ultra. It follows that any family that is equivalent to an ultra family will necessarily be ultra. In particular, if ${\displaystyle {\mathcal {C}}}$ is a prefilter then either both ${\displaystyle {\mathcal {C}}}$ and the filter ${\displaystyle {\mathcal {C}}^{\uparrow X}}$ it generates are ultra or neither one is ultra. If a filter subbase is ultra then it is necessarily a prefilter, in which case the filter that it generates will also be ultra. A filter subbase ${\displaystyle {\mathcal {B}}}$ that is not a prefilter cannot be ultra; but it is nevertheless still possible for the prefilter and filter generated by ${\displaystyle {\mathcal {B}}}$ to be ultra. If ${\displaystyle S\subseteq X{\text{ and }}{\mathcal {B}}\subseteq {\mathcal {P}}(X)}$ is upward closed in ${\displaystyle X}$ then ${\displaystyle S\not \in {\mathcal {B}}{\text{ if and only if }}(X\setminus S)\#{\mathcal {B}}.}$^[8]

Relational properties of subordination

The relation ${\displaystyle \,\leq \,}$ is reflexive and transitive, which makes it into a preorder on ${\displaystyle {\mathcal {P}}({\mathcal {P}}(X)).}$^[20] The relation ${\displaystyle \,\leq \,{\text{ on }}\operatorname {Filters} (X)}$ is antisymmetric but if ${\displaystyle X}$ has more than one point then it is not symmetric.

Symmetry: For any ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X),{\mathcal {B}}\leq \{X\}{\text{ if and only if }}\{X\}={\mathcal {B}}.}$ So the set ${\displaystyle X}$ has more than one point if and only if the relation ${\displaystyle \,\leq \,{\text{ on }}\operatorname {Filters} (X)}$ is not symmetric.

Antisymmetry: If ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {C}}{\text{ then }}{\mathcal {B}}\leq {\mathcal {C}}}$ but while the converse does not hold in general, it does hold if ${\displaystyle {\mathcal {C}}}$ is upward closed (such as if ${\displaystyle {\mathcal {C}}}$ is a filter). Two filters are equivalent if and only if they are equal, which makes the restriction of ${\displaystyle \,\leq \,}$ to ${\displaystyle \operatorname {Filters} (X)}$ antisymmetric. But in general, ${\displaystyle \,\leq \,}$ is not antisymmetric on ${\displaystyle \operatorname {Prefilters} (X)}$ nor on ${\displaystyle {\mathcal {P}}({\mathcal {P}}(X))}$; that is, ${\displaystyle {\mathcal {C}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {B}}\leq {\mathcal {C}}}$ does not necessarily imply ${\displaystyle {\mathcal {B}}={\mathcal {C}}}$; not even if both ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {B}}}$ are prefilters.^[13] For instance, if ${\displaystyle {\mathcal {B}}}$ is a prefilter but not a filter then ${\displaystyle {\mathcal {B}}\leq {\mathcal {B}}^{\uparrow X}{\text{ and }}{\mathcal {B}}^{\uparrow X}\leq {\mathcal {B}}{\text{ but }}{\mathcal {B}}\neq {\mathcal {B}}^{\uparrow X}.}$

Equivalent families of sets

The preorder ${\displaystyle \,\leq \,}$ induces its canonical equivalence relation on ${\displaystyle {\mathcal {P}}({\mathcal {P}}(X)),}$ where for all ${\displaystyle {\mathcal {B}},{\mathcal {C}}\in {\mathcal {P}}({\mathcal {P}}(X)),}$ ${\displaystyle {\mathcal {B}}}$ is equivalent to ${\displaystyle {\mathcal {C}}}$ if any of the following equivalent conditions hold:^[11][18]

1. ${\displaystyle {\mathcal {C}}\leq {\mathcal {B}}{\text{ and }}{\mathcal {B}}\leq {\mathcal {C}}.}$
2. The upward closures of ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {B}}}$ are equal.

Two upward closed (in ${\displaystyle X}$) subsets of ${\displaystyle {\mathcal {P}}(X)}$ are equivalent if and only if they are equal.^[11] If ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X)}$ then necessarily ${\displaystyle \varnothing \leq {\mathcal {B}}\leq {\mathcal {P}}(X)}$ and ${\displaystyle {\mathcal {B}}}$ is equivalent to ${\displaystyle {\mathcal {B}}^{\uparrow X}.}$ Every equivalence class other than ${\displaystyle \{\varnothing \}}$ contains a unique representative (that is, element of the equivalence class) that is upward closed in ${\displaystyle X.}$^[11]

Properties preserved between equivalent families

Let ${\displaystyle {\mathcal {B}},{\mathcal {C}}\in {\mathcal {P}}({\mathcal {P}}(X))}$ be arbitrary and let ${\displaystyle {\mathcal {F}}}$ be any family of sets. If ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ are equivalent (which implies that ${\displaystyle \ker {\mathcal {B}}=\ker {\mathcal {C}}}$) then for each of the statements/properties listed below, either it is true of both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ or else it is false of both ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$:^[20]

1. Not empty
2. Proper (that is, ${\displaystyle \varnothing }$ is not an element)
   • Moreover, any two degenerate families are necessarily equivalent.
3. Filter subbase
4. Prefilter
   • In which case ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ generate the same filter on ${\displaystyle X}$ (that is, their upward closures in ${\displaystyle X}$ are equal).
5. Free
6. Principal
7. Ultra
8. Is equal to the trivial filter ${\displaystyle \{X\}}$
   • In words, this means that the only subset of ${\displaystyle {\mathcal {P}}(X)}$ that is equivalent to the trivial filter is the trivial filter. In general, this conclusion of equality does not extend to non−trivial filters (one exception is when both families are filters).
9. Meshes with ${\displaystyle {\mathcal {F}}}$
10. Is finer than ${\displaystyle {\mathcal {F}}}$
11. Is coarser than ${\displaystyle {\mathcal {F}}}$
12. Is equivalent to ${\displaystyle {\mathcal {F}}}$

Missing from the above list is the word "filter" because this property is not preserved by equivalence. However, if ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ are filters on ${\displaystyle X,}$ then they are equivalent if and only if they are equal; this characterization does not extend to prefilters.

Equivalence of prefilters and filter subbases

If ${\displaystyle {\mathcal {B}}}$ is a prefilter on ${\displaystyle X}$ then the following families are always equivalent to each other:

1. ${\displaystyle {\mathcal {B}}}$;
2. the π–system generated by ${\displaystyle {\mathcal {B}}}$;
3. the filter on ${\displaystyle X}$ generated by ${\displaystyle {\mathcal {B}}}$;

and moreover, these three families all generate the same filter on ${\displaystyle X}$ (that is, the upward closures in ${\displaystyle X}$ of these families are equal).

In particular, every prefilter is equivalent to the filter that it generates. By transitivity, two prefilters are equivalent if and only if they generate the same filter.^{[11][proof 3]} Every prefilter is equivalent to exactly one filter on ${\displaystyle X,}$ which is the filter that it generates (that is, the prefilter's upward closure). Said differently, every equivalence class of prefilters contains exactly one representative that is a filter. In this way, filters can be considered as just being distinguished elements of these equivalence classes of prefilters.^[11]

A filter subbase that is not also a prefilter cannot be equivalent to the prefilter (or filter) that it generates. In contrast, every prefilter is equivalent to the filter that it generates. This is why prefilters can, by and large, be used interchangeably with the filters that they generate while filter subbases cannot. Every filter is both a π–system and a ring of sets.

Set theoretic properties and constructions

Subordination is preserved by images and preimages

The relation ${\displaystyle \,\leq \,}$ is preserved under both images and preimages of families of sets.^[4] This means that for any families ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}},}$^[21] $${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}\quad {\text{ implies }}\quad g({\mathcal {C}})\leq g({\mathcal {F}})\quad {\text{ and }}\quad f^{-1}({\mathcal {C}})\leq f^{-1}({\mathcal {F}}).}$$

Moreover, the following relations always hold for any family of sets ${\displaystyle {\mathcal {C}}}$:^[21] $${\displaystyle {\mathcal {C}}\leq f\left(f^{-1}({\mathcal {C}})\right)}$$ where equality will hold if ${\displaystyle f}$ is surjective.^[21] Furthermore, $${\displaystyle f^{-1}({\mathcal {C}})=f^{-1}\left(f\left(f^{-1}({\mathcal {C}})\right)\right)\quad {\text{ and }}\quad g({\mathcal {C}})=g\left(g^{-1}(g({\mathcal {C}}))\right).}$$

If ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X){\text{ and }}{\mathcal {C}}\subseteq {\mathcal {P}}(Y)}$ then^[8] $${\displaystyle f({\mathcal {B}})\leq {\mathcal {C}}\quad {\text{ if and only if }}\quad {\mathcal {B}}\leq f^{-1}({\mathcal {C}})}$$ and ${\displaystyle g^{-1}(g({\mathcal {C}}))\leq {\mathcal {C}}}$^[21] where equality will hold if ${\displaystyle g}$ is injective.^[21]

Set subtraction and some examples

Using duality between ideals and dual ideals

There is a dual relation ${\displaystyle {\mathcal {B}}\vartriangleleft {\mathcal {C}}}$ or ${\displaystyle {\mathcal {C}}\vartriangleright {\mathcal {B}},}$ which is defined to mean that every ${\displaystyle B\in {\mathcal {B}}}$ is contained in some ${\displaystyle C\in {\mathcal {C}}.}$ Explicitly, this means that for every ${\displaystyle B\in {\mathcal {B}}}$ , there is some ${\displaystyle C\in {\mathcal {C}}}$ such that ${\displaystyle B\subseteq C.}$ This relation is dual to ${\displaystyle \,\leq \,}$ in sense that ${\displaystyle {\mathcal {B}}\vartriangleleft {\mathcal {C}}}$ if and only if ${\displaystyle (X\setminus {\mathcal {B}})\leq (X\setminus {\mathcal {C}}).}$^[18] The relation ${\displaystyle {\mathcal {B}}\vartriangleleft {\mathcal {C}}}$ is closely related to the downward closure of a family in a manner similar to how ${\displaystyle \,\leq \,}$ is related to the upward closure family.

For an example that uses this duality, suppose ${\displaystyle f:X\to Y}$ is a map and ${\displaystyle \Xi \subseteq {\mathcal {P}}(Y).}$ Define $${\displaystyle \Xi _{f}:=\{I\subseteq X~:~f(I)\in \Xi \}}$$ which contains the empty set if and only if ${\displaystyle \Xi }$ does. It is possible for ${\displaystyle \Xi }$ to be an ultrafilter and for ${\displaystyle \Xi _{f}}$ to be empty or not closed under finite intersections (see footnote for example).^{[note 2]} Although ${\displaystyle \Xi _{f}}$ does not preserve properties of filters very well, if ${\displaystyle \Xi }$ is downward closed (resp. closed under finite unions, an ideal) then this will also be true for ${\displaystyle \Xi _{f}.}$ Using the duality between ideals and dual ideals allows for a construction of the following filter.

     Suppose ${\displaystyle {\mathcal {B}}}$ is a filter on ${\displaystyle Y}$ and let ${\displaystyle \Xi$ :=Y\setminus {\mathcal {B}}} be its dual in ${\displaystyle Y.}$ If ${\displaystyle X\not \in \Xi _{f}}$ then ${\displaystyle \Xi _{f}}$'s dual ${\displaystyle X\setminus \Xi _{f}}$ will be a filter.

See also

• Characterizations of the category of topological spaces
• Convergence space – Generalization of the notion of convergence that is found in general topology
• Filter (mathematics) – Special subset of a partially ordered set
• Filter quantifier
• Filters in topology – Use of filters to describe and characterize all basic topological notions and results
• Filtration (mathematics) – Indexed set in mathematics
• Filtration (probability theory) – Model of information available at a given point of a random process
• Filtration (abstract algebra)
• Fréchet filter
• Generic filter
• Ideal (set theory) – Non-empty family of sets that is closed under finite unions and subsets
• Stone–Čech compactification#Construction using ultrafilters – Concept in topology
• The fundamental theorem of ultraproducts – Mathematical constructionPages displaying short descriptions of redirect targets
• Ultrafilter – Maximal proper filter
• Ultrafilter on a set – Maximal proper filter

Notes

1. The intersection of zero subsets of ${\displaystyle X}$ is ${\displaystyle X}$ itself.
2. Suppose ${\displaystyle X}$ has more than one point, ${\displaystyle f:X\to Y}$ is a constant map, and ${\displaystyle \Xi =\{f(X)\}}$ then ${\displaystyle \Xi _{f}}$ will consist of all non–empty subsets of ${\displaystyle Y.}$

Proofs

1. It is immediate that a filter on ${\displaystyle X}$ is an order filter on ${\displaystyle {\mathcal {P}}(X)}$. For the converse, let ${\displaystyle {\mathcal {F}}}$ be an order filter on ${\displaystyle {\mathcal {P}}(X)}$. It is upwards-closed by definition. We check closure under finite intersections. If ${\displaystyle A_{0},\dots ,A_{n-1}}$ is a finite family of subsets from ${\displaystyle {\mathcal {F}}}$, it has a lower bound in ${\displaystyle {\mathcal {F}}}$ by downwards-closure, which is some ${\displaystyle B\in {\mathcal {F}}}$ such that ${\displaystyle B\subseteq A_{0},\dots ,B\subseteq A_{n-1}}$. Then ${\displaystyle B\subseteq (A_{0}\cap \dots \cap A_{n-1})}$, hence ${\displaystyle A_{0}\cap \dots \cap A_{n-1}\in {\mathcal {F}}}$ by upwards-closure.
2. To prove that ${\displaystyle {\mathcal {B}}{\text{ and }}{\mathcal {C}}}$ mesh, let ${\displaystyle B\in {\mathcal {B}}{\text{ and }}C\in {\mathcal {C}}.}$ Because ${\displaystyle {\mathcal {B}}\leq {\mathcal {F}}}$ (resp. because ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}}$), there exists some ${\displaystyle F,G\in {\mathcal {F}}{\text{ such that }}F\subseteq B{\text{ and }}G\subseteq C}$ where by assumption ${\displaystyle F\cap G\neq \varnothing }$ so ${\displaystyle \varnothing \neq G\cap F\subseteq B\cap C.\blacksquare }$ If ${\displaystyle {\mathcal {F}}}$ is a filter subbase and if ${\displaystyle \varnothing \neq {\mathcal {C}}\leq {\mathcal {F}},}$ then taking ${\displaystyle {\mathcal {B}}:={\mathcal {F}}}$ implies that ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}{\text{ mesh. }}\blacksquare }$ If ${\displaystyle C_{1},\ldots ,C_{n}\in {\mathcal {C}}}$ then there are ${\displaystyle F_{1},\ldots ,F_{n}\in {\mathcal {F}}}$ such that ${\displaystyle F_{i}\subseteq C_{i}}$ and now ${\displaystyle \varnothing \neq F_{1}\cap \cdots F_{n}\subseteq C_{1}\cap \cdots C_{n}.}$ This shows that ${\displaystyle {\mathcal {C}}}$ is a filter subbase. ${\displaystyle \blacksquare }$
3. This is because if ${\displaystyle {\mathcal {C}}{\text{ and }}{\mathcal {F}}}$ are prefilters on ${\displaystyle X}$ then ${\displaystyle {\mathcal {C}}\leq {\mathcal {F}}{\text{ if and only if }}{\mathcal {C}}^{\uparrow X}\subseteq {\mathcal {F}}^{\uparrow X}.}$