$X\in {\mathcal {B))$ and $\emptyset \notin {\mathcal {B))$
if $A\in {\mathcal {B))$ and $B\in {\mathcal {B))$,then $A\cap B\in {\mathcal {B))$
If $A,B\subset X,A\in {\mathcal {B))$,and $A\subset B$,then $B\in {\mathcal {B))$
A filter on a set may be thought of as representing a "collection of large subsets".^{[2]} Filters appear in order, model theory, set theory, but can also be found in topology, from which they originate. The dual notion of a filter is an ideal.
In this article, upper case Roman letters like $S{\text{ and ))X$ denote sets (but not families unless indicated otherwise) and $\wp (X)$ will denote the power set of $X.$ A subset of a power set is called a family of sets (or simply, a family) where it is over $X$ if it is a subset of $\wp (X).$ Families of sets will be denoted by upper case calligraphy letters such as ${\mathcal {B)),{\mathcal {C)),{\text{ and )){\mathcal {F)).$
Whenever these assumptions are needed, then it should be assumed that $X$ is non–empty and that ${\mathcal {B)),{\mathcal {F)),$ etc. are families of sets over $X.$
The terms "prefilter" and "filter base" are synonyms and will be used interchangeably.
Warning about competing definitions and notation
There are unfortunately several terms in the theory of filters that are defined differently by different authors.
These include some of the most important terms such as "filter."
While different definitions of the same term usually have significant overlap, due to the very technical nature of filters (and point–set topology), these differences in definitions nevertheless often have important consequences.
When reading mathematical literature, it is recommended that readers check how the terminology related to filters is defined by the author.
For this reason, this article will clearly state all definitions that are used in this article.
Unfortunately, not all notation related to filters is well established and some notation varies greatly across the literature (for example, the notation for the set of all prefilters on a set) so in such cases this article uses whatever notation is most self describing or easily remembered.
The theory of filters and prefilters is well developed and has a plethora of definitions and notations, many of which are now unceremoniously listed to prevent this article from becoming prolix and to allow for the easy look up of notation and definitions.
Their important properties are described later.
${\mathcal {B))^{\uparrow X}:=\{S\subseteq X~:~B\subseteq S{\text{ for some ))B\in {\mathcal {B))\,\}=\bigcup _{B\in {\mathcal {B))}\{S~:~B\subseteq S\subseteq X\))$
and similarly the downward closure of ${\mathcal {B))$ is ${\mathcal {B))^{\downarrow }:=\{S\subseteq B~:~B\in {\mathcal {B))\,\}=\bigcup _{B\in {\mathcal {B))}\wp (B).$
Trace of ${\mathcal {B)){\text{ on ))S$^{[7]} or the restriction of ${\mathcal {B)){\text{ to ))S$ where $S$ is a set; sometimes denoted by ${\mathcal {B))\cap S$
For any two families ${\mathcal {C)){\text{ and )){\mathcal {F)),$ declare that ${\mathcal {C))\leq {\mathcal {F))$ if and only if for every $C\in {\mathcal {C))$ there exists some $F\in {\mathcal {F)){\text{ such that ))F\subseteq C,$ in which case it is said that ${\mathcal {C))$ is coarser than${\mathcal {F))$ and that ${\mathcal {F))$ is finer than (or subordinate to) ${\mathcal {C)).$^{[10]}^{[11]}^{[12]} The notation ${\mathcal {F))\vdash {\mathcal {C)){\text{ or )){\mathcal {F))\geq {\mathcal {C))$ may also be used in place of ${\mathcal {C))\leq {\mathcal {F)).$
Two families ${\mathcal {B)){\text{ and )){\mathcal {C))$mesh,^{[7]} written ${\mathcal {B))\#{\mathcal {C)),$ if $B\cap C\neq \varnothing {\text{ for all ))B\in {\mathcal {B)){\text{ and ))C\in {\mathcal {C)).$
A directed set is a set $I$ together with a preorder, which will be denoted by $\,\leq \,$ (unless explicitly indicated otherwise), that makes $(I,\leq )$ into an (upward) directed set;^{[15]} this means that for all $i,j\in I,$ there exists some $k\in I$ such that $i\leq k{\text{ and ))j\leq k.$ For any indices $i{\text{ and ))j,$ the notation $j\geq i$ is defined to mean $i\leq j$ while $i<j$ is defined to mean that $i\leq j$ holds but it is not true that $j\leq i$ (if $\,\leq \,$ is antisymmetric then this is equivalent to $i\leq j{\text{ and ))i\neq j$).
A net in $X$^{[15]} is a map from a non–empty directed set into $X.$
The notation $x_{\bullet }=\left(x_{i}\right)_{i\in I))$ will be used to denote a net with domain $I.$
Notation and Definition
Name
$I_{\geq i}=\{j\in I~:~j\geq i\))$
Tail or section of $I$ starting at $i\in I$ where $(I,\leq )$ is a directed set.
$x_{\geq i}=\left\{x_{j}~:~j\geq i{\text{ and ))j\in I\right\))$
Tail or section of $x_{\bullet }=\left(x_{i}\right)_{i\in I))$ starting at $i\in I$
Set or prefilter of tails/sections of $x_{\bullet }.$ Also called the eventuality filter base generated by (the tails of) $x_{\bullet }=\left(x_{i}\right)_{i\in I}.$ If $x_{\bullet ))$ is a sequence then $\operatorname {Tails} \left(x_{\bullet }\right)$ is also called the sequential filter base.^{[16]}
(Eventuality) filter of/generated by (tails of) $x_{\bullet }=\left(x_{i}\right)_{i\in I))$^{[16]}
$f\left(I_{\geq i}\right)=\{f(j)~:~j\geq i{\text{ and ))j\in I\))$
Tail or section of a net $f:I\to X$ starting at $i\in I$^{[16]} where $(I,\leq )$ is a directed set.
Warning about using strict comparison
If $x_{\bullet }=\left(x_{i}\right)_{i\in I))$ is a net and $i\in I$ then it is possible for the set $x_{>i}=\left\{x_{j}~:~j>i{\text{ and ))j\in I\right\},$ which is called the tail of $x_{\bullet ))$after$i$, to be empty (for example, this happens if $i$ is an upper bound of the directed set$I$).
In this case, the family $\left\{x_{>i}~:~i\in I\right\))$ would contain the empty set, which would prevent it from being a prefilter (defined later).
This is the (important) reason for defining $\operatorname {Tails} \left(x_{\bullet }\right)$ as $\left\{x_{\geq i}~:~i\in I\right\))$ rather than $\left\{x_{>i}~:~i\in I\right\))$ or even $\left\{x_{>i}~:~i\in I\right\}\cup \left\{x_{\geq i}~:~i\in I\right\))$ and it is for this reason that in general, when dealing with the prefilter of tails of a net, the strict inequality $\,<\,$ may not be used interchangeably with the inequality $\,\leq .$
Additionally, a semiring is a π-system where every complement $B\setminus A$ is equal to a finite disjoint union of sets in ${\mathcal {F)).$
A semialgebra is a semiring that contains $\Omega .$ $A,B,A_{1},A_{2},\ldots$ are arbitrary elements of ${\mathcal {F))$ and it is assumed that ${\mathcal {F))\neq \varnothing .$
The following is a list of properties that a family ${\mathcal {B))$ of sets may possess and they form the defining properties of filters, prefilters, and filter subbases. Whenever it is necessary, it should be assumed that ${\mathcal {B))\subseteq \wp (X).$
The family of sets ${\mathcal {B))$ is:
Proper or nondegenerate if $\varnothing \not \in {\mathcal {B)).$ Otherwise, if $\varnothing \in {\mathcal {B)),$ then it is called improper^{[17]} or degenerate.
Directed downward^{[15]} if whenever $A,B\in {\mathcal {B))$ then there exists some $C\in {\mathcal {B))$ such that $C\subseteq A\cap B.$
This property can be characterized in terms of directedness, which explains the word "directed": A binary relation$\,\preceq \,$ on ${\mathcal {B))$ is called (upward) directed if for any two $A{\text{ and ))B,$ there is some $C$ satisfying $A\preceq C{\text{ and ))B\preceq C.$ Using $\,\supseteq \,$ in place of $\,\preceq \,$ gives the definition of directed downward whereas using $\,\subseteq \,$ instead gives the definition of directed upward. Explicitly, ${\mathcal {B))$ is directed downward (resp. directed upward) if and only if for all $A,B\in {\mathcal {B)),$ there exists some "greater" $C\in {\mathcal {B))$ such that $A\supseteq C{\text{ and ))B\supseteq C$ (resp. such that $A\subseteq C{\text{ and ))B\subseteq C$) − where the "greater" element is always on the right hand side,^{[note 1]} − which can be rewritten as $A\cap B\supseteq C$ (resp. as $A\cup B\subseteq C$).
If a family ${\mathcal {B))$ has a greatest element with respect to $\,\supseteq \,$ (for example, if $\varnothing \in {\mathcal {B))$) then it is necessarily directed downward.
Closed under finite intersections (resp. unions) if the intersection (resp. union) of any two elements of ${\mathcal {B))$ is an element of ${\mathcal {B)).$
If ${\mathcal {B))$ is closed under finite intersections then ${\mathcal {B))$ is necessarily directed downward. The converse is generally false.
Upward closed or Isotone in $X$^{[5]} if ${\mathcal {B))\subseteq \wp (X){\text{ and )){\mathcal {B))={\mathcal {B))^{\uparrow X},$ or equivalently, if whenever $B\in {\mathcal {B))$ and some set $C$ satisfies $B\subseteq C\subseteq X,{\text{ then ))C\in {\mathcal {B)).$ Similarly, ${\mathcal {B))$ is downward closed if ${\mathcal {B))={\mathcal {B))^{\downarrow }.$ An upward (respectively, downward) closed set is also called an upper set or upset (resp. a lower set or down set).
The family ${\mathcal {B))^{\uparrow X},$ which is the upward closure of ${\mathcal {B)){\text{ in ))X,$ is the unique smallest (with respect to $\,\subseteq$) isotone family of sets over $X$ having ${\mathcal {B))$ as a subset.
Many of the properties of ${\mathcal {B))$ defined above and below, such as "proper" and "directed downward," do not depend on $X,$ so mentioning the set $X$ is optional when using such terms. Definitions involving being "upward closed in $X,$" such as that of "filter on $X,$" do depend on $X$ so the set $X$ should be mentioned if it is not clear from context.
Ideal^{[17]}^{[18]} if ${\mathcal {B))\neq \varnothing$ is downward closed and closed under finite unions.
Dual ideal on $X$^{[19]} if ${\mathcal {B))\neq \varnothing$ is upward closed in $X$ and also closed under finite intersections. Equivalently, ${\mathcal {B))\neq \varnothing$ is a dual ideal if for all $R,S\subseteq X,$$R\cap S\in {\mathcal {B))\;{\text{ if and only if ))\;R,S\in {\mathcal {B)).$^{[9]}
Explanation of the word "dual": A family ${\mathcal {B))$ is a dual ideal (resp. an ideal) on $X$ if and only if the dual of ${\mathcal {B)){\text{ in ))X,$ which is the family
is an ideal (resp. a dual ideal) on $X.$ In other words, dual ideal means "dual of an ideal". The family $X\setminus {\mathcal {B))$ should not be confused with $\wp (X)\setminus {\mathcal {B))=\{S\subseteq X~:~S\notin {\mathcal {B))\))$ because these two sets are not equal in general; for instance, $X\setminus {\mathcal {B))=\wp (X){\text{ if and only if )){\mathcal {B))=\wp (X).$ The dual of the dual is the original family, meaning $X\setminus (X\setminus {\mathcal {B)))={\mathcal {B)).$ The set $X$ belongs to the dual of ${\mathcal {B))$ if and only if $\varnothing \in {\mathcal {B)).$^{[17]}
Filter on $X$^{[19]}^{[7]} if ${\mathcal {B))$ is a properdual ideal on $X.$ That is, a filter on $X$ is a non−empty subset of $\wp (X)\setminus \{\varnothing \))$ that is closed under finite intersections and upward closed in $X.$ Equivalently, it is a prefilter that is upward closed in $X.$ In words, a filter on $X$ is a family of sets over $X$ that (1) is not empty (or equivalently, it contains $X$), (2) is closed under finite intersections, (3) is upward closed in $X,$ and (4) does not have the empty set as an element.
Warning: Some authors, particularly algebrists, use "filter" to mean a dual ideal; others, particularly topologists, use "filter" to mean a proper/non–degenerate dual ideal.^{[20]} It is recommended that readers always check how "filter" is defined when reading mathematical literature. However, the definitions of "ultrafilter," "prefilter," and "filter subbase" always require non-degeneracy. This article uses Henri Cartan's original definition of "filter",^{[3]}^{[4]} which required non–degeneracy.
A dual filter on $X$ is a family ${\mathcal {B))$ whose dual $X\setminus {\mathcal {B))$ is a filter on $X.$ Equivalently, it is an ideal on $X$ that does not contain $X$ as an element.
The power set $\wp (X)$ is the one and only dual ideal on $X$ that is not also a filter. Excluding $\wp (X)$ from the definition of "filter" in topology has the same benefit as excluding $1$ from the definition of "prime number": it obviates the need to specify "non-degenerate" (the analog of "non-unital" or "non-$1$") in many important results, thereby making their statements less awkward.
Prefilter or filter base^{[7]}^{[21]} if ${\mathcal {B))\neq \varnothing$ is proper and directed downward. Equivalently, ${\mathcal {B))$ is called a prefilter if its upward closure ${\mathcal {B))^{\uparrow X))$ is a filter. It can also be defined as any family that is equivalent (with respect to $\leq$) to some filter.^{[8]} A proper family ${\mathcal {B))\neq \varnothing$ is a prefilter if and only if ${\mathcal {B))\,(\cap )\,{\mathcal {B))\leq {\mathcal {B)).$^{[8]} A family is a prefilter if and only if the same is true of its upward closure.
If ${\mathcal {B))$ is a prefilter then its upward closure ${\mathcal {B))^{\uparrow X))$ is the unique smallest (relative to $\subseteq$) filter on $X$ containing ${\mathcal {B))$ and it is called the filter generated by${\mathcal {B)).$ A filter ${\mathcal {F))$ is said to be generated by a prefilter ${\mathcal {B))$ if ${\mathcal {F))={\mathcal {B))^{\uparrow X},$ in which ${\mathcal {B))$ is called a filter base for ${\mathcal {F)).$
Unlike a filter, a prefilter is not necessarily closed under finite intersections.
π–system if ${\mathcal {B))\neq \varnothing$ is closed under finite intersections. Every non–empty family ${\mathcal {B))$ is contained in a unique smallest π–system called the π–system generated by${\mathcal {B)),$ which is sometimes denoted by $\pi ({\mathcal {B))).$ It is equal to the intersection of all π–systems containing ${\mathcal {B))$ and also to the set of all possible finite intersections of sets from ${\mathcal {B))$:
A π–system is a prefilter if and only if it is proper. Every filter is a proper π–system and every proper π–system is a prefilter but the converses do not hold in general.
A prefilter is equivalent (with respect to $\leq$) to the π–system generated by it and both of these families generate the same filter on $X.$
Filter subbase^{[7]}^{[22]} and centered^{[8]} if ${\mathcal {B))\neq \varnothing$ and ${\mathcal {B))$ satisfies any of the following equivalent conditions:
${\mathcal {B))$ has the finite intersection property, which means that the intersection of any finite family of (one or more) sets in ${\mathcal {B))$ is not empty; explicitly, this means that whenever $n\geq 1{\text{ and ))B_{1},\ldots ,B_{n}\in {\mathcal {B))$ then $\varnothing \neq B_{1}\cap \cdots \cap B_{n}.$
The π–system generated by ${\mathcal {B))$ is proper; that is, $\varnothing \not \in \pi ({\mathcal {B))).$
The π–system generated by ${\mathcal {B))$ is a prefilter.
${\mathcal {B))$ is a subset of some prefilter.
${\mathcal {B))$ is a subset of some filter.
Assume that ${\mathcal {B))$ is a filter subbase. Then there is a unique smallest (relative to $\subseteq$) filter ${\mathcal {F))_{\mathcal {B)){\text{ on ))X$ containing ${\mathcal {B))$ called the filter generated by${\mathcal {B))$, and ${\mathcal {B))$ is said to be a filter subbase for this filter. This filter is equal to the intersection of all filters on $X$ that are supersets of ${\mathcal {B)).$ The π–system generated by ${\mathcal {B)),$ denoted by $\pi ({\mathcal {B))),$ will be a prefilter and a subset of ${\mathcal {F))_{\mathcal {B)).$ Moreover, the filter generated by ${\mathcal {B))$ is equal to the upward closure of $\pi ({\mathcal {B))),$ meaning $\pi ({\mathcal {B)))^{\uparrow X}={\mathcal {F))_{\mathcal {B)).$^{[8]} However, ${\mathcal {B))^{\uparrow X}={\mathcal {F))_{\mathcal {B))$ if and only if${\mathcal {B))$ is a prefilter (although ${\mathcal {B))^{\uparrow X))$ is always an upward closed filter subbase for ${\mathcal {F))_{\mathcal {B))$).
A $\subseteq$ –smallest (meaning smallest relative to $\subseteq$ ) prefilter containing a filter subbase ${\mathcal {B))$ will exist only under certain circumstances. It exists, for example, if the filter subbase ${\mathcal {B))$ happens to also be a prefilter. It also exists if the filter (or equivalently, the π–system) generated by ${\mathcal {B))$ is principal, in which case ${\mathcal {B))\cup \{\ker {\mathcal {B))\))$ is the unique smallest prefilter containing ${\mathcal {B)).$ Otherwise, in general, a $\subseteq$ –smallest prefilter containing ${\mathcal {B))$ might not exist. For this reason, some authors may refer to the π–system generated by ${\mathcal {B))$ as the prefilter generated by${\mathcal {B)).$ However, as shown in an example below, if a $\subseteq$ –smallest prefilter does exist (say it is denoted by $\operatorname {minPre} {\mathcal {B))$) then contrary to usual expectations, it is not necessarily equal to "the prefilter generated by ${\mathcal {B))$" (that is, $\operatorname {minPre} {\mathcal {B))\neq \pi ({\mathcal {B)))$ is possible). And if the filter subbase ${\mathcal {B))$ happens to also be a prefilter but not a π-system then unfortunately, "the prefilter generated by this prefilter" (meaning $\pi ({\mathcal {B)))$) will not be ${\mathcal {B))=\operatorname {minPre} {\mathcal {B))$ (that is, $\pi ({\mathcal {B)))\neq {\mathcal {B))$ is possible even when ${\mathcal {B))$ is a prefilter), which is why this article will prefer the accurate and unambiguous terminology of "the π–system generated by ${\mathcal {B))$".
Subfilter of a filter ${\mathcal {F))$ and that ${\mathcal {F))$ is a superfilter of ${\mathcal {B))$^{[17]}^{[23]} if ${\mathcal {B))$ is a filter and ${\mathcal {B))\subseteq {\mathcal {F))$ where for filters, ${\mathcal {B))\subseteq {\mathcal {F)){\text{ if and only if )){\mathcal {B))\leq {\mathcal {F)).$
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, ${\mathcal {B))\leq {\mathcal {F))$ can also be written ${\mathcal {F))\vdash {\mathcal {B))$ which is described by saying "${\mathcal {F))$ is subordinate to ${\mathcal {B)).$" With this terminology, "is subordinate to" becomes for filters (and also for prefilters) the analog of "is a subsequence of,"^{[24]} which makes this one situation where using the term "subordinate" and symbol $\,\vdash \,$ may be helpful.
There are no prefilters on $X=\varnothing$ (nor are there any nets valued in $\varnothing$), which is why this article, like most authors, will automatically assume without comment that $X\neq \varnothing$ whenever this assumption is needed.
Basic examples
Named examples
The singleton set ${\mathcal {B))=\{X\))$ is called the indiscrete or trivial filter on $X.$^{[25]}^{[10]} It is the unique minimal filter on $X$ because it is a subset of every filter on $X$; however, it need not be a subset of every prefilter on $X.$
The dual ideal $\wp (X)$ is also called the degenerate filter on $X$^{[9]} (despite not actually being a filter). It is the only dual ideal on $X$ that is not a filter on $X.$
If $(X,\tau )$ is a topological space and $x\in X,$ then the neighborhood filter${\mathcal {N))(x)$ at $x$ is a filter on $X.$ By definition, a family ${\mathcal {B))\subseteq \wp (X)$ is called a neighborhood basis (resp. a neighborhood subbase) at $x{\text{ for ))(X,\tau )$ if and only if ${\mathcal {B))$ is a prefilter (resp. ${\mathcal {B))$ is a filter subbase) and the filter on $X$ that ${\mathcal {B))$ generates is equal to the neighborhood filter ${\mathcal {N))(x).$ The subfamily $\tau (x)\subseteq {\mathcal {N))(x)$ of open neighborhoods is a filter base for ${\mathcal {N))(x).$ Both prefilters ${\mathcal {N))(x){\text{ and ))\tau (x)$ also form a bases for topologies on $X,$ with the topology generated $\tau (x)$ being coarser than $\tau .$ This example immediately generalizes from neighborhoods of points to neighborhoods of non–empty subsets $S\subseteq X.$
${\mathcal {B))$ is an elementary prefilter^{[26]} if ${\mathcal {B))=\operatorname {Tails} \left(x_{\bullet }\right)$ for some sequence $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }{\text{ in ))X.$
${\mathcal {B))$ is an elementary filter or a sequential filter on $X$^{[27]} if ${\mathcal {B))$ is a filter on $X$ generated by some elementary prefilter. The filter of tails generated by a sequence that is not eventually constant is necessarily not an ultrafilter.^{[28]} Every principal filter on a countable set is sequential as is every cofinite filter on a countably infinite set.^{[9]} The intersection of finitely many sequential filters is again sequential.^{[9]}
The set ${\mathcal {F))$ of all cofinite subsets of $X$ (meaning those sets whose complement in $X$ is finite) is proper if and only if ${\mathcal {F))$ is infinite (or equivalently, $X$ is infinite), in which case ${\mathcal {F))$ is a filter on $X$ known as the Fréchet filter or the cofinite filter on $X.$^{[10]}^{[25]} If $X$ is finite then ${\mathcal {F))$ is equal to the dual ideal $\wp (X),$ which is not a filter. If $X$ is infinite then the family $\{X\setminus \{x\}~:~x\in X\))$ of complements of singleton sets is a filter subbase that generates the Fréchet filter on $X.$ As with any family of sets over $X$ that contains $\{X\setminus \{x\}~:~x\in X\},$ the kernel of the Fréchet filter on $X$ is the empty set: $\ker {\mathcal {F))=\varnothing .$
The intersection of all elements in any non–empty family $\mathbb {F} \subseteq \operatorname {Filters} (X)$ is itself a filter on $X$ called the infimum or greatest lower bound of $\mathbb {F} {\text{ in ))\operatorname {Filters} (X),$ which is why it may be denoted by $\bigwedge _((\mathcal {F))\in \mathbb {F} }{\mathcal {F)).$ Said differently, $\ker \mathbb {F} =\bigcap _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))\in \operatorname {Filters} (X).$ Because every filter on $X$ has $\{X\))$ as a subset, this intersection is never empty. By definition, the infimum is the finest/largest (relative to $\,\subseteq \,{\text{ and ))\,\leq \,$) filter contained as a subset of each member of $\mathbb {F} .$^{[10]}
If ${\mathcal {B)){\text{ and )){\mathcal {F))$ are filters then their infimum in $\operatorname {Filters} (X)$ is the filter ${\mathcal {B))\,(\cup )\,{\mathcal {F)).$^{[8]} If ${\mathcal {B)){\text{ and )){\mathcal {F))$ are prefilters then ${\mathcal {B))\,(\cup )\,{\mathcal {F))$ is a prefilter that is coarser (with respect to $\,\leq$) than both ${\mathcal {B)){\text{ and )){\mathcal {F))$ (that is, ${\mathcal {B))\,(\cup )\,{\mathcal {F))\leq {\mathcal {B)){\text{ and )){\mathcal {B))\,(\cup )\,{\mathcal {F))\leq {\mathcal {F))$); indeed, it is one of the finest such prefilters, meaning that if ${\mathcal {S))$ is a prefilter such that ${\mathcal {S))\leq {\mathcal {B)){\text{ and )){\mathcal {S))\leq {\mathcal {F))$ then necessarily ${\mathcal {S))\leq {\mathcal {B))\,(\cup )\,{\mathcal {F)).$^{[8]} More generally, if ${\mathcal {B)){\text{ and )){\mathcal {F))$ are non−empty families and if $\mathbb {S} :=\((\mathcal {S))\subseteq \wp (X)~:~{\mathcal {S))\leq {\mathcal {B)){\text{ and )){\mathcal {S))\leq {\mathcal {F))\))$ then ${\mathcal {B))\,(\cup )\,{\mathcal {F))\in \mathbb {S}$ and ${\mathcal {B))\,(\cup )\,{\mathcal {F))$ is a greatest element (with respect to $\leq$) of $\mathbb {S} .$^{[8]}
Let $\varnothing \neq \mathbb {F} \subseteq \operatorname {DualIdeals} (X)$ and let $\cup \mathbb {F} =\bigcup _((\mathcal {F))\in \mathbb {F} }{\mathcal {F)).$
The supremum or least upper bound of $\mathbb {F} {\text{ in ))\operatorname {DualIdeals} (X),$ denoted by $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F)),$ is the smallest (relative to $\subseteq$) dual ideal on $X$ containing every element of $\mathbb {F}$ as a subset; that is, it is the smallest (relative to $\subseteq$) dual ideal on $X$ containing $\cup \mathbb {F}$ as a subset.
This dual ideal is $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))=\pi \left(\cup \mathbb {F} \right)^{\uparrow X},$ where $\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 $\cup \mathbb {F} .$
As with any non–empty family of sets, $\cup \mathbb {F}$ is contained in some filter on $X$ if and only if it is a filter subbase, or equivalently, if and only if $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))=\pi \left(\cup \mathbb {F} \right)^{\uparrow X))$ is a filter on $X,$ in which case this family is the smallest (relative to $\subseteq$) filter on $X$ containing every element of $\mathbb {F}$ as a subset and necessarily $\mathbb {F} \subseteq \operatorname {Filters} (X).$
Let $\varnothing \neq \mathbb {F} \subseteq \operatorname {Filters} (X)$ and let $\cup \mathbb {F} =\bigcup _((\mathcal {F))\in \mathbb {F} }{\mathcal {F)).$
The supremum or least upper bound of $\mathbb {F} {\text{ in ))\operatorname {Filters} (X),$ denoted by $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))$ if it exists, is by definition the smallest (relative to $\subseteq$) filter on $X$ containing every element of $\mathbb {F}$ as a subset.
If it exists then necessarily $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))=\pi \left(\cup \mathbb {F} \right)^{\uparrow X))$^{[10]} (as defined above) and $\bigvee _((\mathcal {F))\in \mathbb {F} }{\mathcal {F))$ will also be equal to the intersection of all filters on $X$ containing $\cup \mathbb {F} .$
This supremum of $\mathbb {F} {\text{ in ))\operatorname {Filters} (X)$ exists if and only if the dual ideal $\pi \left(\cup \mathbb {F} \right)^{\uparrow X))$ is a filter on $X.$
The least upper bound of a family of filters $\mathbb {F}$ may fail to be a filter.^{[10]} Indeed, if $X$ contains at least 2 distinct elements then there exist filters ${\mathcal {B)){\text{ and )){\mathcal {C)){\text{ on ))X$ for which there does not exist a filter ${\mathcal {F)){\text{ on ))X$ that contains both ${\mathcal {B)){\text{ and )){\mathcal {C)).$
If $\cup \mathbb {F}$ is not a filter subbase then the supremum of $\mathbb {F} {\text{ in ))\operatorname {Filters} (X)$ does not exist and the same is true of its supremum in $\operatorname {Prefilters} (X)$ but their supremum in the set of all dual ideals on $X$ will exist (it being the degenerate filter $\wp (X)$).^{[9]}
If ${\mathcal {B)){\text{ and )){\mathcal {F))$ are prefilters (resp. filters on $X$) then ${\mathcal {B))\,(\cap )\,{\mathcal {F))$ is a prefilter (resp. a filter) if and only if it is non–degenerate (or said differently, if and only if ${\mathcal {B)){\text{ and )){\mathcal {F))$ mesh), in which case it is one of the coarsest prefilters (resp. the coarsest filter) on $X$ (with respect to $\,\leq$) that is finer (with respect to $\,\leq$) than both ${\mathcal {B)){\text{ and )){\mathcal {F));$ this means that if ${\mathcal {S))$ is any prefilter (resp. any filter) such that ${\mathcal {B))\leq {\mathcal {S)){\text{ and )){\mathcal {F))\leq {\mathcal {S))$ then necessarily ${\mathcal {B))\,(\cap )\,{\mathcal {F))\leq {\mathcal {S)),$^{[8]} in which case it is denoted by ${\mathcal {B))\vee {\mathcal {F)).$^{[9]}
Let $I{\text{ and ))X$ be non−empty sets and for every $i\in I$ let ${\mathcal {D))_{i))$ be a dual ideal on $X.$ If ${\mathcal {I))$ is any dual ideal on $I$ then $\bigcup _{\Xi \in {\mathcal {I))}\;\;\bigcap _{i\in \Xi }\;{\mathcal {D))_{i))$ is a dual ideal on $X$ called Kowalsky's dual ideal or Kowalsky's filter.^{[17]}
Other examples
Let $X=\{p,1,2,3\))$ and let ${\mathcal {B))=\{\{p\},\{p,1,2\},\{p,1,3\}\},$ which makes ${\mathcal {B))$ a prefilter and a filter subbase that is not closed under finite intersections. Because ${\mathcal {B))$ is a prefilter, the smallest prefilter containing ${\mathcal {B))$ is ${\mathcal {B)).$ The π–system generated by ${\mathcal {B))$ is $\{\{p,1\}\}\cup {\mathcal {B)).$ In particular, the smallest prefilter containing the filter subbase ${\mathcal {B))$ is not equal to the set of all finite intersections of sets in ${\mathcal {B)).$ The filter on $X$ generated by ${\mathcal {B))$ is ${\mathcal {B))^{\uparrow X}=\{S\subseteq X:p\in S\}=\{\{p\}\cup T~:~T\subseteq \{1,2,3\}\}.$ All three of ${\mathcal {B)),$ the π–system ${\mathcal {B))$ generates, and ${\mathcal {B))^{\uparrow X))$ are examples of fixed, principal, ultra prefilters that are principal at the point $p;{\mathcal {B))^{\uparrow X))$ is also an ultrafilter on $X.$
Let $(X,\tau )$ be a topological space, ${\mathcal {B))\subseteq \wp (X),$ and define ${\overline {\mathcal {B))}:=\left\{\operatorname {cl} _{X}B~:~B\in {\mathcal {B))\right\},$ where ${\mathcal {B))$ is necessarily finer than ${\overline {\mathcal {B))}.$^{[29]} If ${\mathcal {B))$ is non–empty (resp. non–degenerate, a filter subbase, a prefilter, closed under finite unions) then the same is true of ${\overline {\mathcal {B))}.$ If ${\mathcal {B))$ is a filter on $X$ then ${\overline {\mathcal {B))))$ is a prefilter but not necessarily a filter on $X$ although $\left({\overline {\mathcal {B))}\right)^{\uparrow X))$ is a filter on $X$ equivalent to ${\overline {\mathcal {B))}.$
The set ${\mathcal {B))$ of all dense open subsets of a (non–empty) topological space $X$ is a proper π–system and so also a prefilter. If $X=\mathbb {R} ^{n))$ (with $1\leq n\in \mathbb {N}$) then the set ${\mathcal {B))_{\operatorname {LebFinite} ))$ of all $B\in {\mathcal {B))$ such that $B$ has finite Lebesgue measure is a proper π–system and prefilter that is also a proper subset of ${\mathcal {B)).$ The prefilters ${\mathcal {B))_{\operatorname {LebFinite} }{\text{ and )){\mathcal {B))$ generate the same filter on $X.$
A filter subbase with no $\,\subseteq -$smallest prefilter containing it: In general, if a filter subbase ${\mathcal {S))$ is not a π–system then an intersection $S_{1}\cap \cdots \cap S_{n))$ of $n$ sets from ${\mathcal {S))$ will usually require a description involving $n$ variables that cannot be reduced down to only two (consider, for instance $\pi ({\mathcal {S)))$ when ${\mathcal {S))=\{(-\infty ,r)\cup (r,\infty )~:~r\in \mathbb {R} \))$). This example illustrates an atypical class of a filter subbases ${\mathcal {S))_{R))$ where all sets in both ${\mathcal {S))_{R))$ and its generated π–system can be described as sets of the form $B_{r,s},$ so that in particular, no more than two variables (specifically, $r{\text{ and ))s$) are needed to describe the generated π–system.
For all $r,s\in \mathbb {R} ,$ let
$B_{r,s}=(r,0)\cup (s,\infty ),$
where $B_{r,s}=B_{\min(r,s),s))$ always holds so no generality is lost by adding the assumption $r\leq s.$
For all real $r\leq s{\text{ and ))u\leq v,$ if $s{\text{ or ))v$ is non-negative then $B_{-r,s}\cap B_{-u,v}=B_{-\min(r,u),\max(s,v)}.$^{[note 2]}
For every set $R$ of positive reals, let^{[note 3]}
${\mathcal {S))_{R}:=\left\{B_{-r,r}:r\in R\right\}=\{(-r,0)\cup (r,\infty ):r\in R\}\quad {\text{ and ))\quad {\mathcal {B))_{R}:=\left\{B_{-r,s}:r\leq s{\text{ with ))r,s\in R\right\}=\{(-r,0)\cup (s,\infty ):r\leq s{\text{ in ))R\}.$
Let $X=\mathbb {R}$ and suppose $\varnothing \neq R\subseteq (0,\infty )$ is not a singleton set. Then ${\mathcal {S))_{R))$ is a filter subbase but not a prefilter and ${\mathcal {B))_{R}=\pi \left({\mathcal {S))_{R}\right)$ is the π–system it generates, so that ${\mathcal {B))_{R}^{\uparrow X))$ is the unique smallest filter in $X=\mathbb {R}$ containing ${\mathcal {S))_{R}.$ However, ${\mathcal {S))_{R}^{\uparrow X))$ is not a filter on $X$ (nor is it a prefilter because it is not directed downward, although it is a filter subbase) and ${\mathcal {S))_{R}^{\uparrow X))$ is a proper subset of the filter ${\mathcal {B))_{R}^{\uparrow X}.$
If $R,S\subseteq (0,\infty )$ are non−empty intervals then the filter subbases ${\mathcal {S))_{R}{\text{ and )){\mathcal {S))_{S))$ generate the same filter on $X$ if and only if $R=S.$
If ${\mathcal {C))$ is a prefilter satisfying ${\mathcal {S))_{(0,\infty )}\subseteq {\mathcal {C))\subseteq {\mathcal {B))_{(0,\infty )))$^{[note 4]} then for any $C\in {\mathcal {C))\setminus {\mathcal {S))_{(0,\infty )},$ the family ${\mathcal {C))\setminus \{C\))$ is also a prefilter satisfying ${\mathcal {S))_{(0,\infty )}\subseteq {\mathcal {C))\setminus \{C\}\subseteq {\mathcal {B))_{(0,\infty )}.$ This shows that there cannot exist a minimal/least (with respect to $\subseteq$) prefilter that both contains ${\mathcal {S))_{(0,\infty )))$ and is a subset of the π–system generated by ${\mathcal {S))_{(0,\infty )}.$ This remains true even if the requirement that the prefilter be a subset of ${\mathcal {B))_{(0,\infty )}=\pi \left({\mathcal {S))_{(0,\infty )}\right)$ is removed; that is, (in sharp contrast to filters) there does not exist a minimal/least (with respect to $\subseteq$) prefilter containing the filter subbase ${\mathcal {S))_{(0,\infty )}.$
There are many other characterizations of "ultrafilter" and "ultra prefilter," which are listed in the article on ultrafilters. Important properties of ultrafilters are also described in that article.
A non–empty family ${\mathcal {B))\subseteq \wp (X)$ of sets is/is an:
Ultra^{[7]}^{[30]} if $\varnothing \not \in {\mathcal {B))$ and any of the following equivalent conditions are satisfied:
For every set $S\subseteq X$ there exists some set $B\in {\mathcal {B))$ such that $B\subseteq S{\text{ or ))B\subseteq X\setminus S$ (or equivalently, such that $B\cap S{\text{ equals ))B{\text{ or ))\varnothing$).
For every set $S\subseteq \bigcup _{B\in {\mathcal {B))}B$ there exists some set $B\in {\mathcal {B))$ such that $B\cap S{\text{ equals ))B{\text{ or ))\varnothing .$
This characterization of "${\mathcal {B))$ is ultra" does not depend on the set $X,$ so mentioning the set $X$ is optional when using the term "ultra."
For every set $S$ (not necessarily even a subset of $X$) there exists some set $B\in {\mathcal {B))$ such that $B\cap S{\text{ equals ))B{\text{ or ))\varnothing .$
If ${\mathcal {B))$ satisfies this condition then so does every superset ${\mathcal {F))\supseteq {\mathcal {B)).$ For example, if $T$ is any singleton set then $\{T\))$ is ultra and consequently, any non–degenerate superset of $\{T\))$ (such as its upward closure) is also ultra.
Ultra prefilter^{[7]}^{[30]} if it is a prefilter that is also ultra. Equivalently, it is a filter subbase that is ultra. A prefilter ${\mathcal {B))$ is ultra if and only if it satisfies any of the following equivalent conditions:
${\mathcal {B))$ is maximal in $\operatorname {Prefilters} (X)$ with respect to $\,\leq ,\,$ which means that
Although this statement is identical to that given below for ultrafilters, here ${\mathcal {B))$ is merely assumed to be a prefilter; it need not be a filter.
${\mathcal {B))^{\uparrow X))$ is ultra (and thus an ultrafilter).
${\mathcal {B))$ is equivalent (with respect to $\leq$) to some ultrafilter.
A filter subbase that is ultra is necessarily a prefilter. A filter subbase is ultra if and only if it is a maximal filter subbase with respect to $\,\leq \,$ (as above).^{[17]}
Ultrafilter on $X$^{[7]}^{[30]} if it is a filter on $X$ that is ultra. Equivalently, an ultrafilter on $X$ is a filter ${\mathcal {B)){\text{ on ))X$ that satisfies any of the following equivalent conditions:
${\mathcal {B))$ is generated by an ultra prefilter.
For any $S\subseteq X,S\in {\mathcal {B)){\text{ or ))X\setminus S\in {\mathcal {B)).$^{[17]}
${\mathcal {B))\cup (X\setminus {\mathcal {B)))=\wp (X).$ This condition can be restated as: $\wp (X)$ is partitioned by ${\mathcal {B))$ and its dual $X\setminus {\mathcal {B)).$
The sets ${\mathcal {B)){\text{ and ))X\setminus {\mathcal {B))$ are disjoint whenever ${\mathcal {B))$ is a prefilter.
$\wp (X)\setminus {\mathcal {B))=\{S\in \wp (X):S\not \in {\mathcal {B))\))$ is an ideal.^{[17]}
For any $R,S\subseteq X,$ if $R\cup S=X$ then $R\in {\mathcal {B)){\text{ or ))S\in {\mathcal {B)).$
For any $R,S\subseteq X,$ if $R\cup S\in {\mathcal {B))$ then $R\in {\mathcal {B)){\text{ or ))S\in {\mathcal {B))$ (a filter with this property is called a prime filter).
This property extends to any finite union of two or more sets.
For any $R,S\subseteq X,$ if $R\cup S\in {\mathcal {B)){\text{ and ))R\cap S=\varnothing$ then either$R\in {\mathcal {B)){\text{ or ))S\in {\mathcal {B)).$
${\mathcal {B))$ is a maximal filter on $X$; meaning that if ${\mathcal {C))$ is a filter on $X$ such that ${\mathcal {B))\subseteq {\mathcal {C))$ then necessarily ${\mathcal {C))={\mathcal {B))$ (this equality may be replaced by ${\mathcal {C))\subseteq {\mathcal {B)){\text{ or by )){\mathcal {C))\leq {\mathcal {B))$).
If ${\mathcal {C))$ is upward closed then ${\mathcal {B))\leq {\mathcal {C)){\text{ if and only if )){\mathcal {B))\subseteq {\mathcal {C)).$ So this characterization of ultrafilters as maximal filters can be restated as:
Because subordination $\,\geq \,$ is for filters the analog of "is a subnet/subsequence of" (specifically, "subnet" should mean "AA–subnet," which is defined below), this characterization of an ultrafilter as being a "maximally subordinate filter" suggests that an ultrafilter can be interpreted as being analogous to some sort of "maximally deep net" (which could, for instance, mean that "when viewed only from $X$" in some sense, it is indistinguishable from its subnets, as is the case with any net valued in a singleton set for example),^{[note 5]} which is an idea that is actually made rigorous by ultranets. The ultrafilter lemma is then the statement that every filter ("net") has some subordinate filter ("subnet") that is "maximally subordinate" ("maximally deep").
Any non–degenerate family that has a singleton set as an element is ultra, in which case it will then be an ultra prefilter if and only if it also has the finite intersection property.
The trivial filter $\{X\}{\text{ on ))X$ is ultra if and only if $X$ is a singleton set.
The ultrafilter lemma
The following important theorem is due to Alfred Tarski (1930).^{[31]}
A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.^{[10]}^{[proof 1]}
Assuming the axioms of Zermelo–Fraenkel (ZF), the ultrafilter lemma follows from the Axiom of choice (in particular from Zorn's lemma) but is strictly weaker than it. The ultrafilter lemma implies the Axiom of choice for finite sets. If only dealing with Hausdorff spaces, then most basic results (as encountered in introductory courses) in Topology (such as Tychonoff's theorem for compact Hausdorff spaces and the Alexander subbase theorem) and in functional analysis (such as the Hahn–Banach theorem) can be proven using only the ultrafilter lemma; the full strength of the axiom of choice might not be needed.
Kernels
The kernel is useful in classifying properties of prefilters and other families of sets.
The kernel^{[5]} of a family of sets ${\mathcal {B))$ is the intersection of all sets that are elements of ${\mathcal {B)):$
If ${\mathcal {B))\subseteq \wp (X)$ then for any point $x,x\not \in \ker {\mathcal {B)){\text{ if and only if ))X\setminus \{x\}\in {\mathcal {B))^{\uparrow X}.$
Properties of kernels
If ${\mathcal {B))\subseteq \wp (X)$ then $\ker \left({\mathcal {B))^{\uparrow X}\right)=\ker {\mathcal {B))$ and this set is also equal to the kernel of the π–system that is generated by ${\mathcal {B)).$
In particular, if ${\mathcal {B))$ is a filter subbase then the kernels of all of the following sets are equal:
(1) ${\mathcal {B)),$ (2) the π–system generated by ${\mathcal {B)),$ and (3) the filter generated by ${\mathcal {B)).$
If $f$ is a map then $f(\ker {\mathcal {B)))\subseteq \ker f({\mathcal {B))){\text{ and ))f^{-1}(\ker {\mathcal {B)))=\ker f^{-1}({\mathcal {B))).$
If ${\mathcal {B))\leq {\mathcal {C)){\text{ then ))\ker {\mathcal {C))\subseteq \ker {\mathcal {B))$ while if ${\mathcal {B)){\text{ and )){\mathcal {C))$ are equivalent then $\ker {\mathcal {B))=\ker {\mathcal {C)).$
If ${\mathcal {B)){\text{ and )){\mathcal {C))$ are principal then they are equivalent if and only if $\ker {\mathcal {B))=\ker {\mathcal {C)).$
Classifying families by their kernels
A family ${\mathcal {B))$ of sets is:
Free^{[6]} if $\ker {\mathcal {B))=\varnothing ,$ or equivalently, if $\{X\setminus \{x\}~:~x\in X\}\subseteq {\mathcal {B))^{\uparrow X};$ this can be restated as $\{X\setminus \{x\}~:~x\in X\}\leq {\mathcal {B)).$
A filter ${\mathcal {F)){\text{ on ))X$ is free if and only if $X$ is infinite and ${\mathcal {F))$ contains the Fréchet filter on $X$ as a subset.
Fixed if $\ker {\mathcal {B))\neq \varnothing$ in which case, ${\mathcal {B))$ is said to be fixed by any point $x\in \ker {\mathcal {B)).$
Any fixed family is necessarily a filter subbase.
Principal^{[6]} if $\ker {\mathcal {B))\in {\mathcal {B)).$
A proper principal family of sets is necessarily a prefilter.
Discrete or Principal at$x\in X$^{[25]} if $\{x\}=\ker {\mathcal {B))\in {\mathcal {B)),$ in which case $x$ is called its principal element.
The principal filter at $x{\text{ on ))X$ is the filter $\{x\}^{\uparrow X}.$ A filter ${\mathcal {F))$ is principal at $x$ if and only if ${\mathcal {F))=\{x\}^{\uparrow X}.$
Countably deep if whenever ${\mathcal {C))\subseteq {\mathcal {F))$ is a countable subset then $\ker {\mathcal {C))\in {\mathcal {B)).$^{[9]}
If ${\mathcal {B))$ is a principal filter on $X$ then $\varnothing \neq \ker {\mathcal {B))\in {\mathcal {B))$ and
where $\{\ker {\mathcal {B))\))$ is also the smallest prefilter that generates ${\mathcal {B)).$
Family of examples: For any non–empty $C\subseteq \mathbb {R} ,$ the family ${\mathcal {B))_{C}=\{\mathbb {R} \setminus (r+C)~:~r\in \mathbb {R} \))$ is free but it is a filter subbase if and only if no finite union of the form $\left(r_{1}+C\right)\cup \cdots \cup \left(r_{n}+C\right)$ covers $\mathbb {R} ,$ in which case the filter that it generates will also be free. In particular, ${\mathcal {B))_{C))$ is a filter subbase if $C$ is countable (for example, $C=\mathbb {Q} ,\mathbb {Z} ,$ the primes), a meager set in $\mathbb {R} ,$ a set of finite measure, or a bounded subset of $\mathbb {R} .$ If $C$ is a singleton set then ${\mathcal {B))_{C))$ is a subbase for the Fréchet filter on $\mathbb {R} .$
For every filter ${\mathcal {F)){\text{ on ))X$ there exists a unique pair of dual ideals ${\mathcal {F))^{*}{\text{ and )){\mathcal {F))^{\bullet }{\text{ on ))X$ such that ${\mathcal {F))^{*))$ is free, ${\mathcal {F))^{\bullet ))$ is principal, and ${\mathcal {F))^{*}\wedge {\mathcal {F))^{\bullet }={\mathcal {F)),$ and ${\mathcal {F))^{*}{\text{ and )){\mathcal {F))^{\bullet ))$ do not mesh (that is, $$