Filter (set theory)

In mathematics, a filter on a set $X$ is a family $\mathcal$ of subsets such that:
# $X \in \mathcal$ and $\emptyset \notin \mathcal$
# if $A\in \mathcal$ and $B \in \mathcal$, then $A\cap B\in \mathcal$
# If $A,B\subset X,A\in \mathcal$, and $A\subset B$, then $B\in \mathcal$

A filter on a set may be thought of as representing a "collection of large subsets". 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.

Filters were introduced by Henri Cartan in 1937 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. Order filters are generalizations of filters from sets to arbitrary partially ordered sets. Specifically, a filter on a set is just a proper order filter in the special case where the partially ordered set consists of the power set ordered by set inclusion.

Preliminaries, notation, and basic notions

In this article, upper case Roman letters like $S \text X$ 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 (or simply, ) where it is if it is a subset of $\wp(X).$ Families of sets will be denoted by upper case calligraphy letters such as $\mathcal, \mathcal, \text \mathcal.$
Whenever these assumptions are needed, then it should be assumed that $X$ is non–empty and that $\mathcal, \mathcal,$ 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 as they are used.
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.

Sets operations

The or in $X$ of a family of sets $\mathcal \subseteq \wp(X)$ is

and similarly the of $\mathcal$ is $\mathcal^ := \ = \bigcup_ \wp(B).$

Throughout, $f$ is a map and $S$ is a set.

Nets and their tails

A is a set $I$ together with a preorder, which will be denoted by $\,\leq\,$ (unless explicitly indicated otherwise), that makes $(I, \leq)$ into an () ; this means that for all $i, j \in I,$ there exists some $k \in I$ such that $i \leq k \text j \leq k.$ For any indices $i \text j,$ 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 true that $j \leq i$ (if $\,\leq\,$ is antisymmetric then this is equivalent to $i \leq j \text i \neq j$).

A is a map from a non–empty directed set into $X.$
The notation $x_ = \left(x_i\right)_$ will be used to denote a net with domain $I.$

Warning about using strict comparison

If $x_ = \left(x_i\right)_$ is a net and $i \in I$ then it is possible for the set $x_ = \left\,$ which is called , to be empty (for example, this happens if $i$ is an upper bound of the directed set $I$).
In this case, the family $\left\$ would contain the empty set, which would prevent it from being a prefilter (defined later).
This is the (important) reason for defining $\operatorname\left(x_\right)$ as $\left\$ rather than $\left\$ or even $\left\\cup \left\$ 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 $\,\leq.$

Filters and prefilters

The following is a list of properties that a family $\mathcal$ 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 $\mathcal \subseteq \wp(X).$

Many of the properties of $\mathcal$ 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.

$\textrm(X) \quad=\quad \textrm(X) \,\setminus\, \ \quad\subseteq\quad \textrm(X) \quad\subseteq\quad \textrm(X).$

\text X containing $\mathcal$ called the , and $\mathcal$ is said to this filter. This filter is equal to the intersection of all filters on $X$ that are supersets of $\mathcal.$ The –system generated by $\mathcal,$ denoted by $\pi(\mathcal),$ will be a prefilter and a subset of $\mathcal_.$ Moreover, the filter generated by $\mathcal$ is equal to the upward closure of $\pi(\mathcal),$ meaning $\pi(\mathcal)^ = \mathcal_.$ However, $\mathcal^ = \mathcal_$ if $\mathcal$ is a prefilter (although $\mathcal^$ is always an upward closed filter base for $\mathcal_$).
* A $\subseteq$–smallest (meaning smallest relative to $\subseteq$) filter containing a filter subbase $\mathcal$ will exist only under certain circumstances. It exists, for example, if the filter subbase $\mathcal$ happens to also be a prefilter. It also exists if the filter (or equivalently, the –system) generated by $\mathcal$ is principal, in which case $\mathcal \cup \$ is the unique smallest prefilter containing $\mathcal.$ Otherwise, in general, a $\subseteq$–smallest filter containing $\mathcal$ might not exist. For this reason, some authors may refer to the –system generated by $\mathcal$ as However, if a $\subseteq$–smallest prefilter does exist (say it is denoted by $\operatorname \mathcal$) then contrary to usual expectations, it is necessarily equal to " the prefilter generated by $\mathcal$" (that is, $\operatorname \mathcal \neq \pi(\mathcal)$ is possible). And if the filter subbase $\mathcal$ happens to also be a prefilter but not a -system then unfortunately, " the prefilter generated by this prefilter" (meaning $\pi(\mathcal)$) will not be $\mathcal = \operatorname \mathcal$ (that is, $\pi(\mathcal) \neq \mathcal$ is possible even when $\mathcal$ is a prefilter), which is why this article will prefer the accurate and unambiguous terminology of "the –system generated by $\mathcal$".

of a filter $\mathcal$ and that $\mathcal$ is a of $\mathcal$ if $\mathcal$ is a filter and $\mathcal \subseteq \mathcal$ where for filters, $\mathcal \subseteq \mathcal \text \mathcal \leq \mathcal.$
* Importantly, the expression "is a filter of" is for filters the analog of "is a sequence of". So despite having the prefix "sub" in common, "is a filter of" is actually the of "is a sequence of." However, $\mathcal \leq \mathcal$ can also be written $\mathcal \vdash \mathcal$ which is described by saying "$\mathcal$ is subordinate to $\mathcal.$" With this terminology, "is ordinate to" becomes for filters (and also for prefilters) the analog of "is a sequence of," 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 = \$ is called the or It is the unique 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 (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 $\mathcal(x)$ at $x$ is a filter on $X.$ By definition, a family $\mathcal \subseteq \wp(X)$ is called a (resp. a ) at $x \text (X, \tau)$ if and only if $\mathcal$ is a prefilter (resp. $\mathcal$ is a filter subbase) and the filter on $X$ that $\mathcal$ generates is equal to the neighborhood filter $\mathcal(x).$ The subfamily $\tau(x) \subseteq \mathcal(x)$ of open neighborhoods is a filter base for $\mathcal(x).$ Both prefilters $\mathcal(x) \text \tau(x)$ 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 $S \subseteq X.$
• $\mathcal$ is an if $\mathcal = \operatorname\left(x_\right)$ for some sequence $x_ = \left(x_i\right)_^ \text X.$
• $\mathcal$ is an or a on $X$ if $\mathcal$ 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 an ultrafilter. Every principal filter on a countable set is sequential as is every cofinite filter on a countably infinite set. The intersection of finitely many sequential filters is again sequential.
• The set $\mathcal$ of all cofinite subsets of $X$ (meaning those sets whose complement in $X$ is finite) is proper if and only if $\mathcal$ is infinite (or equivalently, $X$ is infinite), in which case $\mathcal$ is a filter on $X$ known as the or the on $X.$ If $X$ is finite then $\mathcal$ is equal to the dual ideal $\wp(X),$ which is not a filter. If $X$ is infinite then the family $\$ 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 $\,$ the kernel of the Fréchet filter on $X$ is the empty set: $\ker \mathcal = \varnothing.$
• The intersection of all elements in any non–empty family $\mathbb \subseteq \operatorname(X)$ is itself a filter on $X$ called the or of $\mathbb \text \operatorname(X),$ which is why it may be denoted by $\bigwedge_ \mathcal.$ Said differently, $\ker \mathbb = \bigcap_ \mathcal \in \operatorname(X).$ Because every filter on $X$ has $\$ as a subset, this intersection is never empty. By definition, the infimum is the finest/largest (relative to $\,\subseteq\, \text \,\leq\,$) filter contained as a subset of each member of $\mathbb.$
  * If $\mathcal \text \mathcal$ are filters then their infimum in $\operatorname(X)$ is the filter $\mathcal \,(\cup)\, \mathcal.$ If $\mathcal \text \mathcal$ are prefilters then $\mathcal \,(\cup)\, \mathcal$ is a prefilter that is coarser (with respect to $\,\leq$) than both $\mathcal \text \mathcal$ (that is, $\mathcal \,(\cup)\, \mathcal \leq \mathcal \text \mathcal \,(\cup)\, \mathcal \leq \mathcal$); indeed, it is one of the finest such prefilters, meaning that if $\mathcal$ is a prefilter such that $\mathcal \leq \mathcal \text \mathcal \leq \mathcal$ then necessarily $\mathcal \leq \mathcal \,(\cup)\, \mathcal.$ More generally, if $\mathcal \text \mathcal$ are non−empty families and if $\mathbb := \$ then $\mathcal \,(\cup)\, \mathcal \in \mathbb$ and $\mathcal \,(\cup)\, \mathcal$ is a greatest element (with respect to $\leq$) of $\mathbb.$
• Let $\varnothing \neq \mathbb \subseteq \operatorname(X)$ and let $\cup \mathbb = \bigcup_ \mathcal.$
  The or of $\mathbb \text \operatorname(X),$ denoted by $\bigvee_ \mathcal,$ is the smallest (relative to $\subseteq$) dual ideal on $X$ containing every element of $\mathbb$ as a subset; that is, it is the smallest (relative to $\subseteq$) dual ideal on $X$ containing $\cup \mathbb$ as a subset.
  This dual ideal is $\bigvee_ \mathcal = \pi\left(\cup \mathbb\right)^,$ where $\pi\left(\cup \mathbb\right) := \left\$ is the –system generated by $\cup \mathbb.$
  As with any non–empty family of sets, $\cup \mathbb$ is contained in filter on $X$ if and only if it is a filter subbase, or equivalently, if and only if $\bigvee_ \mathcal = \pi\left(\cup \mathbb\right)^$ is a filter on $X,$ in which case this family is the smallest (relative to $\subseteq$) filter on $X$ containing every element of $\mathbb$ as a subset and necessarily $\mathbb \subseteq \operatorname(X).$
  
• Let $\varnothing \neq \mathbb \subseteq \operatorname(X)$ and let $\cup \mathbb = \bigcup_ \mathcal.$
  The or of $\mathbb \text \operatorname(X),$ denoted by $\bigvee_ \mathcal$ if it exists, is by definition the smallest (relative to $\subseteq$) filter on $X$ containing every element of $\mathbb$ as a subset.
  If it exists then necessarily $\bigvee_ \mathcal = \pi\left(\cup \mathbb\right)^$ (as defined above) and $\bigvee_ \mathcal$ will also be equal to the intersection of all filters on $X$ containing $\cup \mathbb.$
  This supremum of $\mathbb \text \operatorname(X)$ exists if and only if the dual ideal $\pi\left(\cup \mathbb\right)^$ is a filter on $X.$
  The least upper bound of a family of filters $\mathbb$ may fail to be a filter. Indeed, if $X$ contains at least 2 distinct elements then there exist filters $\mathcal \text \mathcal \text X$ for which there does exist a filter $\mathcal \text X$ that contains both $\mathcal \text \mathcal.$
  If $\cup \mathbb$ is not a filter subbase then the supremum of $\mathbb \text \operatorname(X)$ does not exist and the same is true of its supremum in $\operatorname(X)$ but their supremum in the set of all dual ideals on $X$ will exist (it being the degenerate filter $\wp(X)$).
  * If $\mathcal \text \mathcal$ are prefilters (resp. filters on $X$) then $\mathcal \,(\cap)\, \mathcal$ is a prefilter (resp. a filter) if and only if it is non–degenerate (or said differently, if and only if $\mathcal \text \mathcal$ mesh), in which case it is coarsest prefilters (resp. coarsest filter) on $X$ (with respect to $\,\leq$) that is finer (with respect to $\,\leq$) than both $\mathcal \text \mathcal;$ this means that if $\mathcal$ is any prefilter (resp. any filter) such that $\mathcal \leq \mathcal \text \mathcal \leq \mathcal$ then necessarily $\mathcal \,(\cap)\, \mathcal \leq \mathcal,$ in which case it is denoted by $\mathcal \vee \mathcal.$
  
• Let $I \text X$ be non−empty sets and for every $i \in I$ let $\mathcal_i$ be a dual ideal on $X.$ If $\mathcal$ is any dual ideal on $I$ then $\bigcup_ \;\;\bigcap_ \;\mathcal_i$ is a dual ideal on $X$ called or .
• The club filter of a regular uncountable cardinal is the filter of all sets containing a club subset of $\kappa.$ It is a $\kappa$-complete filter closed under diagonal intersection.

Other examples

• Let $X = \$ and let $\mathcal = \,$ which makes $\mathcal$ a prefilter and a filter subbase that is not closed under finite intersections. Because $\mathcal$ is a prefilter, the smallest prefilter containing $\mathcal$ is $\mathcal.$ The –system generated by $\mathcal$ is $\ \cup \mathcal.$ In particular, the smallest prefilter containing the filter subbase $\mathcal$ is equal to the set of all finite intersections of sets in $\mathcal.$ The filter on $X$ generated by $\mathcal$ is $\mathcal^ = \ = \.$ All three of $\mathcal,$ the –system $\mathcal$ generates, and $\mathcal^$ are examples of fixed, principal, ultra prefilters that are principal at the point $p; \mathcal^$ is also an ultrafilter on $X.$
• Let $(X, \tau)$ be a topological space, $\mathcal \subseteq \wp(X),$ and define $\overline := \left\,$ where $\mathcal$ is necessarily finer than $\overline.$ If $\mathcal$ is non–empty (resp. non–degenerate, a filter subbase, a prefilter, closed under finite unions) then the same is true of $\overline.$ If $\mathcal$ is a filter on $X$ then $\overline$ is a prefilter but not necessarily a filter on $X$ although $\left(\overline\right)^$ is a filter on $X$ equivalent to $\overline.$
• The set $\mathcal$ of all dense open subsets of a (non–empty) topological space $X$ is a proper –system and so also a prefilter. If the space is a Baire space, then the set of all countable intersections of dense open subsets is a –system and a prefilter that is finer than $\mathcal.$ If $X = \R^n$ (with $1 \leq n \in \N$) then the set $\mathcal_$ of all $B \in \mathcal$ such that $B$ has finite Lebesgue measure is a proper –system and free prefilter that is also a proper subset of $\mathcal.$ The prefilters $\mathcal_$ and $\mathcal$ are equivalent and so generate the same filter on $X.$
  The prefilter $\mathcal_$ is properly contained in, and not equivalent to, the prefilter consisting of all dense subsets of $\R.$ Since $X$ is a Baire space, every countable intersection of sets in $\mathcal_$ is dense in $X$ (and also comeagre and non–meager) so the set of all countable intersections of elements of $\mathcal_$ is a prefilter and –system; it is also finer than, and not equivalent to, $\mathcal_.$
• ''A filter subbase with no $\,\subseteq-$smallest prefilter containing it'': In general, if a filter subbase $\mathcal$ is not a –system then an intersection $S_1 \cap \cdots \cap S_n$ of $n$ sets from $\mathcal$ will usually require a description involving $n$ variables that cannot be reduced down to only two (consider, for instance $\pi(\mathcal)$ when $\mathcal = \$). This example illustrates an atypical class of a filter subbases $\mathcal_R$ where all sets in both $\mathcal_R$ and its generated –system can be described as sets of the form $B_,$ so that in particular, no more than two variables (specifically, $r \text s$) are needed to describe the generated –system.

  For all $r, s \in \R,$ let
  $B_ = (r, 0) \cup (s, \infty),$
  where $B_ = B_$ always holds so no generality is lost by adding the assumption $r \leq s.$
  For all real $r \leq s \text u \leq v,$ if $s \text v$ is non-negative then $B_ \cap B_ = B_.$More generally, for any real numbers satisfying $r \leq s \text u \leq v, B_ \cap B_ = B_$ where $m := \min(s,v,\max(r,u)).$
  For every set $R$ of positive reals, letIf $R, S \subseteq \R \text \mathcal_R \cap \mathcal_S = \mathcal_.$ This property and the fact that $\mathcal_R$ is nonempty and proper if and only if $R \neq \varnothing$ actually allows for the construction of even more examples of prefilters, because if $\mathcal \subseteq \wp(\R)$ is any prefilter (resp. filter subbase, –system) then so is $\left\.$
  $\mathcal_R := \left\ = \ \quad \text \quad \mathcal_R := \left\ = \.$

  Let $X = \R$ and suppose $\varnothing \neq R \subseteq (0, \infty)$ is not a singleton set. Then $\mathcal_R$ is a filter subbase but not a prefilter and $\mathcal_R = \pi\left(\mathcal_R\right)$ is the –system it generates, so that $\mathcal_R^$ is the unique smallest filter in $X = \R$ containing $\mathcal_R.$ However, $\mathcal_R^$ is a filter on $X$ (nor is it a prefilter because it is not directed downward, although it is a filter subbase) and $\mathcal_R^$ is a proper subset of the filter $\mathcal_R^.$
  If $R, S \subseteq (0, \infty)$ are non−empty intervals then the filter subbases $\mathcal_R \text \mathcal_S$ generate the same filter on $X$ if and only if $R = S.$

  If $\mathcal$ is a prefilter satisfying $\mathcal_ \subseteq \mathcal \subseteq \mathcal_$It may be shown that if $\mathcal$ is any family such that $\mathcal_ \subseteq \mathcal \subseteq \mathcal_$ then $\mathcal$ is a prefilter if and only if for all real $0 < r \leq s$ there exist real $0 < u \leq v$ such that $u \leq r \leq s \leq v \text B_ \in \mathcal.$ then for any $C \in \mathcal \setminus \mathcal_,$ the family $\mathcal \setminus \$ is also a prefilter satisfying $\mathcal_ \subseteq \mathcal \setminus \ \subseteq \mathcal_.$ This shows that there cannot exist a minimal/ least (with respect to $\subseteq$) prefilter that both contains $\mathcal_$ and is a subset of the –system generated by $\mathcal_.$ This remains true even if the requirement that the prefilter be a subset of $\mathcal_ = \pi\left(\mathcal_\right)$ is removed; that is, (in sharp contrast to filters) there does exist a minimal/least (with respect to $\subseteq$) filter containing the filter subbase $\mathcal_.$
  

Ultrafilters

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.

$\begin \textrm(X)\; &=\; \textrm(X) \,\cap\, \textrm(X)\\ &\subseteq\; \textrm(X) = \textrm(X)\\ &\subseteq\; \textrm(X) \\ \end$

B there exists some set $B \in \mathcal$ such that $B \cap S \text B \text \varnothing.$
* This characterization of "$\mathcal$ is ultra" does not depend on the set $X,$ so mentioning the set $X$ is optional when using the term "ultra."

For set $S$ (not necessarily even a subset of $X$) there exists some set $B \in \mathcal$ such that $B \cap S \text B \text \varnothing.$
* If $\mathcal$ satisfies this condition then so does superset $\mathcal \supseteq \mathcal.$ For example, if $T$ is any singleton set then $\$ is ultra and consequently, any non–degenerate superset of $\$ (such as its upward closure) is also ultra.

if it is a prefilter that is also ultra. Equivalently, it is a filter subbase that is ultra. A prefilter $\mathcal$ is ultra if and only if it satisfies any of the following equivalent conditions:

1. $\mathcal$ is in $\operatorname(X)$ with respect to $\,\leq,\,$ which means that $\text \mathcal \in \operatorname(X), \; \mathcal \leq \mathcal \; \text \; \mathcal \leq \mathcal.$


2. $\text \mathcal \in \operatorname(X), \; \mathcal \leq \mathcal \; \text \; \mathcal \leq \mathcal.$
   * Although this statement is identical to that given below for ultrafilters, here $\mathcal$ is merely assumed to be a prefilter; it need not be a filter.


3. $\mathcal^$ is ultra (and thus an ultrafilter).


4. $\mathcal$ 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).

if it is a filter on $X$ that is ultra. Equivalently, an ultrafilter on $X$ is a filter $\mathcal \text X$ that satisfies any of the following equivalent conditions:

1. $\mathcal$ is generated by an ultra prefilter.


2. For any $S \subseteq X, S \in \mathcal \text X \setminus S \in \mathcal.$


3. $\mathcal \cup (X \setminus \mathcal) = \wp(X).$ This condition can be restated as: $\wp(X)$ is partitioned by $\mathcal$ and its dual $X \setminus \mathcal.$
   * The sets $\mathcal \text X \setminus \mathcal$ are disjoint whenever $\mathcal$ is a prefilter.


4. $\wp(X) \setminus \mathcal = \$ is an ideal.


5. For any $R, S \subseteq X,$ if $R \cup S = X$ then $R \in \mathcal \text S \in \mathcal.$


6. For any $R, S \subseteq X,$ if $R \cup S \in \mathcal$ then $R \in \mathcal \text S \in \mathcal$ (a filter with this property is called a ).
   * This property extends to any finite union of two or more sets.


7. For any $R, S \subseteq X,$ if $R \cup S \in \mathcal \text R \cap S = \varnothing$ then $R \in \mathcal \text S \in \mathcal.$


8. $\mathcal$ is a filter on $X$; meaning that if $\mathcal$ is a filter on $X$ such that $\mathcal \subseteq \mathcal$ then necessarily $\mathcal = \mathcal$ (this equality may be replaced by $\mathcal \subseteq \mathcal \text \mathcal \leq \mathcal$).
   * If $\mathcal$ is upward closed then $\mathcal \leq \mathcal \text \mathcal \subseteq \mathcal.$ So this characterization of ultrafilters as maximal filters can be restated as: $\text \mathcal \in \operatorname(X), \; \mathcal \leq \mathcal \; \text \; \mathcal \leq \mathcal.$
   * 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),For instance, one sense in which a net $u_ \text X$ could be interpreted as being "maximally deep" is if all important properties related to $X$ (such as convergence for example) of any subnet is completely determined by $u_$ in all topologies on $X.$ In this case $u_$ and its subnet become effectively indistinguishable (at least topologically) if one's information about them is limited to only that which can be described in solely in terms of $X$ and directly related sets (such as its subsets). 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 $\ \text X$ is ultra if and only if $X$ is a singleton set.

The ultrafilter lemma

The following important theorem is due to Alfred Tarski (1930).

A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.Let $\mathcal$ be a filter on $X$ that is not an ultrafilter. If $S \subseteq X$ is such that $S \not\in \mathcal \text \ \cup \mathcal$ has the finite intersection property (because if $F \in \mathcal \text F \cap (X \setminus S) = \varnothing \text F \subseteq S$) so that by the ultrafilter lemma, there exists some ultrafilter $\mathcal_S \text X$ such that $\ \cup \mathcal \subseteq \mathcal_S$ (so in particular, $S \not\in \mathcal_S$). Intersecting all such $\mathcal_S$ proves that $\mathcal = \bigcap_ \mathcal_S. \blacksquare$
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 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.

B

If $\mathcal \subseteq \wp(X)$ then for any point $x, x \not\in \ker \mathcal \text X \setminus \ \in \mathcal^.$

Properties of kernels

If $\mathcal \subseteq \wp(X)$ then $\ker \left(\mathcal^\right) = \ker \mathcal$ and this set is also equal to the kernel of the –system that is generated by $\mathcal.$
In particular, if $\mathcal$ is a filter subbase then the kernels of all of the following sets are equal:
:(1) $\mathcal,$ (2) the –system generated by $\mathcal,$ and (3) the filter generated by $\mathcal.$

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

Classifying families by their kernels

If $\mathcal$ is a principal filter on $X$ then $\varnothing \neq \ker \mathcal \in \mathcal$ and
$\mathcal = \^ = \ = \wp(X \setminus \ker \mathcal) \,(\cup)\, \$
where $\$ is also the smallest prefilter that generates $\mathcal.$

Family of examples: For any non–empty $C \subseteq \R,$ the family $\mathcal_C = \$ 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 $\R,$ in which case the filter that it generates will also be free. In particular, $\mathcal_C$ is a filter subbase if $C$ is countable (for example, $C = \Q, \Z,$ the primes), a meager set in $\R,$ a set of finite measure, or a bounded subset of $\R.$ If $C$ is a singleton set then $\mathcal_C$ is a subbase for the Fréchet filter on $\R.$

For every filter $\mathcal \text X$ there exists a unique pair of dual ideals $\mathcal^* \text \mathcal^ \text X$ such that $\mathcal^*$ is free, $\mathcal^$ is principal, and $\mathcal^* \wedge \mathcal^ = \mathcal,$ and $\mathcal^* \text \mathcal^$ do not mesh (that is, $\mathcal^* \vee \mathcal^ = \wp(X)$). The dual ideal $\mathcal^*$ is called of $\mathcal$ while $\mathcal^$ is called where at least one of these dual ideals is filter. If $\mathcal$ is principal then $\mathcal^ := \mathcal \text \mathcal^* := \wp(X);$ otherwise, $\mathcal^ := \^$ and $\mathcal^* := \mathcal \vee \^$ is a free (non–degenerate) filter.

Finite prefilters and finite sets

If a filter subbase $\mathcal$ is finite then it is fixed (that is, not free);
this is because $\ker \mathcal = \bigcap_ B$ is a finite intersection and the filter subbase $\mathcal$ has the finite intersection property.
A finite prefilter is necessarily principal, although it does not have to be closed under finite intersections.

If $X$ is finite then all of the conclusions above hold for any $\mathcal \subseteq \wp(X).$
In particular, on a finite set $X,$ there are no free filter subbases (and so no free prefilters), all prefilters are principal, and all filters on $X$ are principal filters generated by their (non–empty) kernels.

The trivial filter $\$ is always a finite filter on $X$ and if $X$ is infinite then it is the only finite filter because a non–trivial finite filter on a set $X$ is possible if and only if $X$ is finite.
However, on any infinite set there are non–trivial filter subbases and prefilters that are finite (although they cannot be filters).
If $X$ is a singleton set then the trivial filter $\$ is the only proper subset of $\wp(X)$ and moreover, this set $\$ is a principal ultra prefilter and any superset $\mathcal \supseteq \mathcal$ (where $\mathcal \subseteq \wp(Y) \text X \subseteq Y$) with the finite intersection property will also be a principal ultra prefilter (even if $Y$ is infinite).

Characterizing fixed ultra prefilters

If a family of sets $\mathcal$ is fixed (that is, $\ker \mathcal \neq \varnothing$) then $\mathcal$ is ultra if and only if some element of $\mathcal$ is a singleton set, in which case $\mathcal$ will necessarily be a prefilter. Every principal prefilter is fixed, so a principal prefilter $\mathcal$ is ultra if and only if $\ker \mathcal$ is a singleton set.

Every filter on $X$ that is principal at a single point is an ultrafilter, and if in addition $X$ is finite, then there are no ultrafilters on $X$ other than these.

The next theorem shows that every ultrafilter falls into one of two categories: either it is free or else it is a principal filter generated by a single point.

Finer/coarser, subordination, and meshing

The preorder $\,\leq\,$ 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", where "$\mathcal \geq \mathcal$" can be interpreted as "$\mathcal$ is a subsequence of $\mathcal$" (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 $\mathcal$ meshes with $\mathcal,$ which is closely related to the preorder $\,\leq,$ is used in Topology to define cluster points.

Two families of sets $\mathcal \text \mathcal$ and are , indicated by writing $\mathcal \# \mathcal,$ if $B \cap C \neq \varnothing \text B \in \mathcal \text C \in \mathcal.$ If $\mathcal \text \mathcal$ do not mesh then they are . If $S \subseteq X \text \mathcal \subseteq \wp(X)$ then $\mathcal \text S$ are said to if $\mathcal \text \$ mesh, or equivalently, if the of $\mathcal \text S,$ which is the family
$\mathcal\big\vert_S = \,$
does not contain the empty set, where the trace is also called the of $\mathcal \text S.$

''Example'': If $x_ = \left(x_\right)_^\infty$ is a subsequence of $x_ = \left(x_i\right)_^\infty$ then $\operatorname\left(x_\right)$ is subordinate to $\operatorname\left(x_\right);$ in symbols: $\operatorname\left(x_\right) \vdash \operatorname\left(x_\right)$ and also $\operatorname\left(x_\right) \leq \operatorname\left(x_\right).$
Stated in plain English, the prefilter of tails of a subsequence is always subordinate to that of the original sequence.
To see this, let $C := x_ \in \operatorname\left(x_\right)$ be arbitrary (or equivalently, let $i \in \N$ be arbitrary) and it remains to show that this set contains some $F := x_ \in \operatorname\left(x_\right).$
For the set $x_ = \left\$ to contain $x_ = \left\,$ it is sufficient to have $i \leq i_n.$
Since $i_1 < i_2 < \cdots$ are strictly increasing integers, there exists $n \in \N$ such that $i_n \geq i,$ and so $x_ \supseteq x_$ holds, as desired.
Consequently, $\operatorname\left(x_\right) \subseteq \operatorname\left(x_\right).$
The left hand side will be a subset of the right hand side if (for instance) every point of $x_$ is unique (that is, when $x_ : \N \to X$ is injective) and $x_$ is the even-indexed subsequence $\left(x_2, x_4, x_6, \ldots\right)$ because under these conditions, every tail $x_ = \left\$ (for every $n \in \N$) of the subsequence will belong to the right hand side filter but not to the left hand side filter.

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

Assume that $\mathcal \text \mathcal$ are families of sets that satisfy $\mathcal \leq \mathcal \text \mathcal \leq \mathcal.$ Then $\ker \mathcal \subseteq \ker \mathcal,$ and $\mathcal \neq \varnothing \text \mathcal \neq \varnothing,$ and also $\varnothing \in \mathcal \text \varnothing \in \mathcal.$
If in addition to $\mathcal \leq \mathcal, \mathcal$ is a filter base and $\mathcal \neq \varnothing,$ then $\mathcal$ is a filter subbase and also $\mathcal \text \mathcal$ mesh.To prove that $\mathcal \text \mathcal$ mesh, let $B \in \mathcal \text C \in \mathcal.$ Because $\mathcal \leq \mathcal$ (resp. because $\mathcal \leq \mathcal$), there exists some $F, G \in \mathcal \text F \subseteq B \text G \subseteq C$ where by assumption $F \cap G \neq \varnothing$ so $\varnothing \neq G \cap F \subseteq B \cap C. \blacksquare$ If $\mathcal$ is a filter subbase and if $\varnothing \neq \mathcal \leq \mathcal,$ then taking $\mathcal := \mathcal$ implies that $\mathcal \text \mathcal \text \blacksquare$ If $C_1, \ldots, C_n \in \mathcal$ then there are $F_1, \ldots, F_n \in \mathcal$ such that $F_i \subseteq C_i$ and now $\varnothing \neq F_1 \cap \cdots F_n \subseteq C_1 \cap \cdots C_n.$ This shows that $\mathcal$ is a filter subbase. $\blacksquare$
More generally, if both $\varnothing \neq \mathcal \leq \mathcal \text \varnothing \neq \mathcal \leq \mathcal$ and if the intersection of any two elements of $\mathcal$ is non–empty, then $\mathcal \text \mathcal$ mesh.
Every filter subbase is coarser than both the –system that it generates and the filter that it generates.

If $\mathcal \text \mathcal$ are families such that $\mathcal \leq \mathcal,$ the family $\mathcal$ is ultra, and $\varnothing \not\in \mathcal,$ then $\mathcal$ is necessarily ultra. It follows that any family that is equivalent to an ultra family will necessarily ultra. In particular, if $\mathcal$ is a prefilter then either both $\mathcal$ and the filter $\mathcal^$ 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 $\mathcal$ that is not a prefilter cannot be ultra; but it is nevertheless still possible for the prefilter and filter generated by $\mathcal$ to be ultra. If $S \subseteq X \text \mathcal \subseteq \wp(X)$ is upward closed in $X$ then $S \not\in \mathcal \text (X \setminus S) \# \mathcal.$

Relational properties of subordination

The relation $\,\leq\,$ is reflexive and transitive, which makes it into a preorder on $\wp(\wp(X)).$
The relation $\,\leq\, \text \operatorname(X)$ is antisymmetric but if $X$ has more than one point then it is symmetric.

:
For any $\mathcal \subseteq \wp(X), \mathcal \leq \ \text \ = \mathcal.$
So the set $X$ has more than one point if and only if the relation $\,\leq\, \text \operatorname(X)$ is symmetric.

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

Equivalent families of sets

The preorder $\,\leq\,$ induces its canonical equivalence relation on $\wp(\wp(X)),$ where for all $\mathcal, \mathcal \in \wp(\wp(X)),$ $\mathcal$ is to $\mathcal$ if any of the following equivalent conditions hold:

1. $\mathcal \leq \mathcal \text \mathcal \leq \mathcal.$


2. The upward closures of $\mathcal \text \mathcal$ are equal.

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

Properties preserved between equivalent families

Let $\mathcal, \mathcal \in \wp(\wp(X))$ be arbitrary and let $\mathcal$ be any family of sets. If $\mathcal \text \mathcal$ are equivalent (which implies that $\ker \mathcal = \ker \mathcal$) then for each of the statements/properties listed below, either it is true of $\mathcal \text \mathcal$ or else it is false of $\mathcal \text \mathcal$:

1. Not empty


2. Proper (that is, $\varnothing$ is not an element)
   * Moreover, any two degenerate families are necessarily equivalent.


3. Filter subbase


4. Prefilter
   * In which case $\mathcal \text \mathcal$ generate the same filter on $X$ (that is, their upward closures in $X$ are equal).


5. Free


6. Principal


7. Ultra


8. Is equal to the trivial filter $\$
   * In words, this means that the only subset of $\wp(X)$ that is equivalent to the trivial filter 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 $\mathcal$


10. Is finer than $\mathcal$


11. Is coarser than $\mathcal$


12. Is equivalent to $\mathcal$

Missing from the above list is the word "filter" because this property is preserved by equivalence.
However, if $\mathcal \text \mathcal$ are filters on $X,$ then they are equivalent if and only if they are equal; this characterization does extend to prefilters.

Equivalence of prefilters and filter subbases

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

1. $\mathcal$;


2. the –system generated by $\mathcal$;


3. the filter on $X$ generated by $\mathcal$;

and moreover, these three families all generate the same filter on $X$ (that is, the upward closures in $X$ 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.This is because if $\mathcal \text \mathcal$ are prefilters on $X$ then $\mathcal \leq \mathcal \text \mathcal^ \subseteq \mathcal^.$
Every prefilter is equivalent to exactly one filter on $X,$ 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.

A filter subbase that is also a prefilter can 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 –system and a ring of sets.

Examples of determining equivalence/non–equivalence

Examples: Let $X = \R$ and let $E$ be the set $\Z$ of integers (or the set $\N$). Define the sets
$\mathcal = \ \qquad \text \qquad \mathcal_ = \ \qquad \text \qquad \mathcal_ = \.$

All three sets are filter subbases but none are filters on $X$ and only $\mathcal$ is prefilter (in fact, $\mathcal$ is even free and closed under finite intersections). The set $\mathcal_$ is fixed while $\mathcal_$ is free (unless $E = \N$). They satisfy $\mathcal_ \leq \mathcal_ \leq \mathcal,$ but no two of these families are equivalent; moreover, no two of the filters generated by these three filter subbases are equivalent/equal. This conclusion can be reached by showing that the –systems that they generate are not equivalent. Unlike with $\mathcal_,$ every set in the –system generated by $\mathcal_$ contains $\Z$ as a subset,The –system generated by $\mathcal_$ (resp. by $\mathcal_$) is a prefilter whose elements are finite unions of open (resp. closed) intervals having endpoints in $E \cup \$ with two of these intervals being of the forms $(-\infty, e_1) \text (e_2, \infty)$ (resp. $(-\infty, e_1] \text [e_2, \infty)$) where $e_1 \leq 1 + e_2$; in the case of $\mathcal_,$ it is possible for one or more of these closed intervals to be singleton sets (that is, degenerate closed intervals). which is what prevents their generated –systems (and hence their generated filters) from being equivalent. If $E$ was instead $\Q \text \R$ then all three families would be free and although the sets $\mathcal_ \text \mathcal_$ would remain equivalent to each other, their generated –systems would be equivalent and consequently, they would generate the same filter on $X$; however, this common filter would still be strictly coarser than the filter generated by $\mathcal.$

Set theoretic properties and constructions

Trace and meshing

If $\mathcal$ is a prefilter (resp. filter) on $X \text S \subseteq X$ then the trace of $\mathcal \text S,$ which is the family $\mathcal\big\vert_S := \mathcal (\cap) \,$ is a prefilter (resp. a filter) if and only if $\mathcal \text S$ mesh (that is, $\varnothing \not\in \mathcal (\cap) \$), in which case the trace of $\mathcal \text S$ is said to be .
If $\mathcal$ is ultra and if $\mathcal \text S$ mesh then the trace $\mathcal\big\vert_S$ is ultra.
If $\mathcal$ is an ultrafilter on $X$ then the trace of $\mathcal \text S$ is a filter on $S$ if and only if $S \in \mathcal.$

For example, suppose that $\mathcal$ is a filter on $X \text S \subseteq X$ is such that $S \neq X \text X \setminus S \not\in \mathcal.$ Then $\mathcal \text S$ mesh and $\mathcal \cup \$ generates a filter on $X$ that is strictly finer than $\mathcal.$

When prefilters mesh

Given non–empty families $\mathcal \text \mathcal,$ the family
$\mathcal (\cap) \mathcal := \$
satisfies $\mathcal \leq \mathcal (\cap) \mathcal$ and $\mathcal \leq \mathcal (\cap) \mathcal.$
If $\mathcal (\cap) \mathcal$ is proper (resp. a prefilter, a filter subbase) then this is also true of both $\mathcal \text \mathcal.$
In order to make any meaningful deductions about $\mathcal (\cap) \mathcal$ from $\mathcal \text \mathcal, \mathcal (\cap) \mathcal$ needs to be proper (that is, $\varnothing \not\in \mathcal (\cap) \mathcal,$ which is the motivation for the definition of "mesh".
In this case, $\mathcal (\cap) \mathcal$ is a prefilter (resp. filter subbase) if and only if this is true of both $\mathcal \text \mathcal.$
Said differently, if $\mathcal \text \mathcal$ are prefilters then they mesh if and only if $\mathcal (\cap) \mathcal$ is a prefilter.
Generalizing gives a well known characterization of "mesh" entirely in terms of subordination (that is, $\,\leq\,$):

Two prefilters (resp. filter subbases) $\mathcal \text \mathcal$ mesh if and only if there exists a prefilter (resp. filter subbase) $\mathcal$ such that $\mathcal \leq \mathcal$ and $\mathcal \leq \mathcal.$

If the least upper bound of two filters $\mathcal \text \mathcal$ exists in $\operatorname(X)$ then this least upper bound is equal to $\mathcal (\cap) \mathcal.$

Images and preimages under functions

Throughout, $f : X \to Y \text g : Y \to Z$ will be maps between non–empty sets.

Images of prefilters

Let $\mathcal \subseteq \wp(Y).$ Many of the properties that $\mathcal$ may have are preserved under images of maps; notable exceptions include being upward closed, being closed under finite intersections, and being a filter, which are not necessarily preserved.

Explicitly, if one of the following properties is true of $\mathcal \text Y,$ then it will necessarily also be true of $g(\mathcal) \text g(Y)$ (although possibly not on the codomain $Z$ unless $g$ is surjective):

• Filter properties: ultra, ultrafilter, filter, prefilter, filter subbase, dual ideal, upward closed, proper/non–degenerate.


• Ideal properties: ideal, closed under finite unions, downward closed, directed upward.

Moreover, if $\mathcal \subseteq \wp(Y)$ is a prefilter then so are both $g(\mathcal) \text g^(g(\mathcal)).$
The image under a map $f : X \to Y$ of an ultra set $\mathcal \subseteq \wp(X)$ is again ultra and if $\mathcal$ is an ultra prefilter then so is $f(\mathcal).$

If $\mathcal$ is a filter then $g(\mathcal)$ is a filter on the range $g(Y),$ but it is a filter on the codomain $Z$ if and only if $g$ is surjective.
Otherwise it is just a prefilter on $Z$ and its upward closure must be taken in $Z$ to obtain a filter.
The upward closure of $g(\mathcal) \text Z$ is
$g(\mathcal)^ = \left\$
where if $\mathcal$ is upward closed in $Y$ (that is, a filter) then this simplifies to:
$g(\mathcal)^ = \left\.$

If $X \subseteq Y$ then taking $g$ to be the inclusion map $X \to Y$ shows that any prefilter (resp. ultra prefilter, filter subbase) on $X$ is also a prefilter (resp. ultra prefilter, filter subbase) on $Y.$

Preimages of prefilters

Let $\mathcal \subseteq \wp(Y).$
Under the assumption that $f : X \to Y$ is Surjective function, surjective:

$f^(\mathcal)$ is a prefilter (resp. filter subbase, –system, closed under finite unions, proper) if and only if this is true of $\mathcal.$

However, if $\mathcal$ is an ultrafilter on $Y$ then even if $f$ is surjective (which would make $f^(\mathcal)$ a prefilter), it is nevertheless still possible for the prefilter $f^(\mathcal)$ to be neither ultra nor a filter on $X$ (see thisFor an example of how this failure can happen, consider the case where there exists some $B \in \mathcal \text y \in Y \setminus B$ such that both $f^(y)$ and its complement in $X$ contains at least two distinct points. footnote for an example).

If $f : X \to Y$ is not surjective then denote the trace of $\mathcal \text f(X)$ by $\mathcal\big\vert_,$ where in this case particular case the trace satisfies:
$\mathcal\big\vert_ = f\left(f^(\mathcal)\right)$
and consequently also:
$f^(\mathcal) = f^\left(\mathcal\big\vert_\right).$

This last equality and the fact that the trace $\mathcal\big\vert_$ is a family of sets over $f(X)$ means that to draw conclusions about $f^(\mathcal),$ the trace $\mathcal\big\vert_$ can be used in place of $\mathcal$ and the $f : X \to f(X)$ can be used in place of $f : X \to Y.$
For example:

$f^(\mathcal)$ is a prefilter (resp. filter subbase, –system, proper) if and only if this is true of $\mathcal\big\vert_.$

In this way, the case where $f$ is not (necessarily) surjective can be reduced down to the case of a surjective function (which is a case that was described at the start of this subsection).

Even if $\mathcal$ is an ultrafilter on $Y,$ if $f$ is not surjective then it is nevertheless possible that $\varnothing \in \mathcal\big\vert_,$ which would make $f^(\mathcal)$ degenerate as well. The next characterization shows that degeneracy is the only obstacle. If $\mathcal$ is a prefilter then the following are equivalent:

1. $f^(\mathcal)$ is a prefilter;


2. $\mathcal\big\vert_$ is a prefilter;


3. $\varnothing \not\in \mathcal\big\vert_$;


4. $\mathcal$ meshes with $f(X)$

and moreover, if $f^(\mathcal)$ is a prefilter then so is $f\left(f^(\mathcal)\right).$

If $S \subseteq Y$ and if $\operatorname : S \to Y$ denotes the inclusion map then the trace of $\mathcal \text S$ is equal to $\operatorname^(\mathcal).$ This observation allows the results in this subsection to be applied to investigating the trace on a set.

Bijections, injections, and surjections

All properties involving filters are preserved under bijections. This means that if $\mathcal \subseteq \wp(Y) \text g : Y \to Z$ is a bijection, then $\mathcal$ is a prefilter (resp. ultra, ultra prefilter, filter on $X,$ ultrafilter on $X,$ filter subbase, –system, ideal on $X,$ etc.) if and only if the same is true of $g(\mathcal) \text Z.$

A map $g : Y \to Z$ is injective if and only if for all prefilters $\mathcal \text Y, \mathcal$ is equivalent to $g^(g(\mathcal)).$ The image of an ultra family of sets under an injection is again ultra.

The map $f : X \to Y$ is a surjection if and only if whenever $\mathcal$ is a prefilter on $Y$ then the same is true of $f^(\mathcal) \text X$ (this result does not require the ultrafilter lemma).

Subordination is preserved by images and preimages

The relation $\,\leq\,$ is preserved under both images and preimages of families of sets.
This means that for families $\mathcal \text \mathcal,$
$\mathcal \leq \mathcal \quad \text \quad g(\mathcal) \leq g(\mathcal) \quad \text \quad f^(\mathcal) \leq f^(\mathcal).$

Moreover, the following relations always hold for family of sets $\mathcal$:

$\mathcal \leq f\left(f^(\mathcal)\right)$
where equality will hold if $f$ is surjective.
Furthermore,
$f^(\mathcal) = f^\left(f\left(f^(\mathcal)\right)\right) \quad \text \quad g(\mathcal) = g\left(g^(g(\mathcal))\right).$

If $\mathcal \subseteq \wp(X) \text \mathcal \subseteq \wp(Y)$ then
$f(\mathcal) \leq \mathcal \quad \text \quad \mathcal \leq f^(\mathcal)$
and $g^(g(\mathcal)) \leq \mathcal$ where equality will hold if $g$ is injective.

Products of prefilters

Suppose $X_ = \left(X_i\right)_$ is a family of one or more non–empty sets, whose product will be denoted by $\prod X_ := \prod_ X_i,$ and for every index $i \in I,$ let
$\Pr_ : \prod X_ \to X_i$
denote the canonical projection.
Let $\mathcal_ := \left(\mathcal_i\right)_$ be non−empty families, also indexed by $I,$ such that $\mathcal_i \subseteq \wp\left(X_i\right)$ for each $i \in I.$
The of the families $\mathcal_$ is defined identically to how the basic open subsets of the product topology are defined (had all of these $\mathcal_i$ been topologies). That is, both the notations
$\prod_ \mathcal_ = \prod_ \mathcal_i$
denote the family of all subsets $\prod_ S_i \subseteq \prod_ X_$ such that $S_i = X_i$ for all but finitely many $i \in I$ and where $S_i \in \mathcal_i$ for any one of these finitely many exceptions (that is, for any $i$ such that $S_i \neq X_i,$ necessarily $S_i \in \mathcal_i$).
When every $\mathcal_i$ is a filter subbase then the family $\bigcup_ \Pr_^ \left(\mathcal_i\right)$ is a filter subbase for the filter on $\prod X_$ generated by $\mathcal_.$
If $\prod \mathcal_$ is a filter subbase then the filter on $\prod X_$ that it generates is called the .
If every $\mathcal_i$ is a prefilter on $X_i$ then $\prod \mathcal_$ will be a prefilter on $\prod X_$ and moreover, this prefilter is equal to the coarsest prefilter $\mathcal \text \prod X_$ such that
$\Pr_ (\mathcal) = \mathcal_i$
for every $i \in I.$
However, $\prod \mathcal_$ may fail to be a filter on $\prod X_$ even if every $\mathcal_i$ is a filter on $X_i.$

Set subtraction and some examples

Set subtracting away a subset of the kernel

If $\mathcal$ is a prefilter on $X, S \subseteq \ker \mathcal, \text S \not\in \mathcal$ then $\$ is a prefilter, where this latter set is a filter if and only if $\mathcal$ is a filter and $S = \varnothing.$ In particular, if $\mathcal$ is a neighborhood basis at a point $x$ in a topological space $X$ having at least 2 points, then $\$ is a prefilter on $X.$ This construction is used to define $\lim_ f(x) \to y$ in terms of prefilter convergence.

Using duality between ideals and dual ideals

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

For an example that uses this duality, suppose $f : X \to Y$ is a map and $\Xi \subseteq \wp(Y).$ Define
$\Xi_f := \$
which contains the empty set if and only if $\Xi$ does. It is possible for $\Xi$ to be an ultrafilter and for $\Xi_f$ to be empty or not closed under finite intersections (see footnote for example).Suppose $X$ has more than one point, $f : X \to Y$ is a constant map, and $\Xi = \$ then $\Xi_f$ will consist of all non–empty subsets of $Y.$ Although $\Xi_f$ does not preserve properties of filters very well, if $\Xi$ is downward closed (resp. closed under finite unions, an ideal) then this will also be true for $\Xi_f.$ Using the duality between ideals and dual ideals allows for a construction of the following filter.

Suppose $\mathcal$ is a filter on $Y$ and let $\Xi := Y \setminus \mathcal$ be its dual in $Y.$ If $X \not\in \Xi_f$ then $\Xi_f$'s dual $X \setminus \Xi_f$ will be a filter.

Other examples

Example: The set $\mathcal$ of all dense open subsets of a topological space is a proper –system and a prefilter. If the space is a Baire space, then the set of all countable intersections of dense open subsets is a –system and a prefilter that is finer than $\mathcal.$

Example: The family $\mathcal_$ of all dense open sets of $X = \R$ having finite Lebesgue measure is a proper –system and a free prefilter. The prefilter $\mathcal_$ is properly contained in, and not equivalent to, the prefilter consisting of all dense open subsets of $\R.$ Since $X$ is a Baire space, every countable intersection of sets in $\mathcal_$ is dense in $X$ (and also comeagre and non–meager) so the set of all countable intersections of elements of $\mathcal_$ is a prefilter and –system; it is also finer than, and not equivalent to, $\mathcal_.$

Filters and nets

This section will describe the relationships between prefilters and nets in great detail because of how important these details are applying filters to topology − particularly in switching from utilizing nets to utilizing filters and vice verse − and because it to make it easier to understand later why subnets (with their most commonly used definitions) are not generally equivalent with "sub–prefilters".

Nets to prefilters

A net $x_ = \left(x_i\right)_ \text X$ is canonically associated with its prefilter of tails $\operatorname\left(x_\right).$
If $f : X \to Y$ is a map and $x_$ is a net in $X$ then $\operatorname\left(f\left(x_\right)\right) = f\left(\operatorname\left(x_\right)\right).$

Prefilters to nets

A is a pair $(S, s)$ consisting of a non–empty set $S$ and an element $s \in S.$
For any family $\mathcal,$ let
$\operatorname(\mathcal) := \left\.$

Define a canonical preorder $\,\leq\,$ on pointed sets by declaring
$(R, r) \leq (S, s) \quad \text \quad R \supseteq S.$

If $s_0, s_1 \in S \text \left(S, s_0\right) \leq \left(S, s_1\right) \text \left(S, s_1\right) \leq \left(S, s_0\right)$ even if $s_0 \neq s_1,$ so this preorder is not antisymmetric and given any family of sets $\mathcal,$ $(\operatorname(\mathcal), \leq)$ is partially ordered if and only if $\mathcal \neq \varnothing$ consists entirely of singleton sets.
If $\ \in \mathcal \text (\, x)$ is a maximal element of $\operatorname(\mathcal)$; moreover, all maximal elements are of this form.
If $\left(B, b_0\right) \in \operatorname(\mathcal) \text \left(B, b_0\right)$ is a greatest element if and only if $B = \ker \mathcal,$ in which case $\$ is the set of all greatest elements. However, a greatest element $(B, b)$ is a maximal element if and only if $B = \ = \ker \mathcal,$ so there is at most one element that is both maximal and greatest.
There is a canonical map $\operatorname_ ~:~ \operatorname(\mathcal) \to X$ defined by $(B, b) \mapsto b.$
If $i_0 = \left(B_0, b_0\right) \in \operatorname(\mathcal)$ then the tail of the assignment $\operatorname_$ starting at $i_0$ is $\left\ = B_0.$

Although $(\operatorname(\mathcal), \leq)$ is not, in general, a partially ordered set, it is a directed set if (and only if) $\mathcal$ is a prefilter.
So the most immediate choice for the definition of "the net in $X$ induced by a prefilter $\mathcal$" is the assignment $(B, b) \mapsto b$ from $\operatorname(\mathcal)$ into $X.$

:\;&& (\operatorname(\mathcal), \leq) &&\,\to \;& X \\
&& (B, b) &&\,\mapsto\;& b \\
\end
that is, $\operatorname_(B, b) := b.$

If $\mathcal$ is a prefilter on $X \text \operatorname_$ is a net in $X$ and the prefilter associated with $\operatorname_$ is $\mathcal$; that is:The set equality $\operatorname\left(\operatorname_\right) = \mathcal$ holds more generally: if the family of sets $\mathcal \neq \varnothing \text \varnothing \not\in \mathcal$ then the family of tails of the map $\operatorname(\mathcal) \to X$ (defined by $(B, b) \mapsto b$) is equal to $\mathcal.$
$\operatorname\left(\operatorname_\right) = \mathcal.$

This would not necessarily be true had $\operatorname_$ been defined on a proper subset of $\operatorname(\mathcal).$
For example, suppose $X$ has at least two distinct elements, $\mathcal := \$ is the indiscrete filter, and $x \in X$ is arbitrary. Had $\operatorname_$ instead been defined on the singleton set $D := \,$ where the restriction of $\operatorname_$ to $D$ will temporarily be denote by $\operatorname_D : D \to X,$ then the prefilter of tails associated with $\operatorname_D : D \to X$ would be the principal prefilter $\$ rather than the original filter $\mathcal = \$;
this means that the equality $\operatorname\left(\operatorname_D\right) = \mathcal$ is , so unlike $\operatorname_,$ the prefilter $\mathcal$ can be recovered from $\operatorname_D.$
Worse still, while $\mathcal$ is the unique filter on $X,$ the prefilter $\operatorname\left(\operatorname_D\right) = \$ instead generates a filter (that is, an ultrafilter) on $X.$

However, if $x_ = \left(x_i\right)_$ is a net in $X$ then it is in general true that $\operatorname_$ is equal to $x_$ because, for example, the domain of $x_$ may be of a completely different cardinality than that of $\operatorname_$ (since unlike the domain of $\operatorname_,$ the domain of an arbitrary net in $X$ could have cardinality).

Ultranets and ultra prefilters

A net $x_ \text X$ is called an or in $X$ if for every subset $S \subseteq X, x_$ is eventually in $S$ or it is eventually in $X \setminus S$;
this happens if and only if $\operatorname\left(x_\right)$ is an ultra prefilter.
A prefilter $\mathcal \text X$ is an ultra prefilter if and only if $\operatorname_$ is an ultranet in $X.$

Partially ordered net

The domain of the canonical net $\operatorname_$ is in general not partially ordered. However, in 1955 Bruns and Schmidt discoveredBruns G., Schmidt J.,Zur Aquivalenz von Moore-Smith-Folgen und Filtern, Math. Nachr. 13 (1955), 169-186. a construction that allows for the canonical net to have a domain that is both partially ordered and directed; this was independently rediscovered by Albert Wilansky in 1970.
It begins with the construction of a strict partial order (meaning a transitive and irreflexive relation) $\,<\,$ on a subset of $\mathcal \times \N \times X$ that is similar to the lexicographical order on $\mathcal \times \N$ of the strict partial orders $(\mathcal, \supsetneq) \text (\N, <).$
For any $i = (B, m, b) \text j = (C, n, c)$ in $\mathcal \times \N \times X,$ declare that $i < j$ if and only if
$B \supseteq C \text \text B \neq C \text B = C \text m < n,$
or equivalently, if and only if $\text B \supseteq C, \text B = C \text m < n.$

The non−strict partial order associated with $\,<,$ denoted by $\,\leq,$ is defined by declaring that $i \leq j\, \text i < j \text i = j.$
Unwinding these definitions gives the following characterization:

which shows that $\,\leq\,$ is just the lexicographical order on $\mathcal \times \N \times X$ induced by $(\mathcal, \supseteq), \,(\N, \leq), \text (X, =),$ where $X$ is partially ordered by equality $\,=.\,$Explicitly, the partial order on $X$ induced by equality $\,=\,$ refers to the diagonal $\Delta := \,$ which is a homogeneous relation on $X$ that makes $(X, \Delta)$ into a partially ordered set. If this partial order $\Delta$ is denoted by the more familiar symbol $\,\leq\,$ (that is, define $\leq \;:=\; \Delta$) then for any $b, c \in X,$ $\;b \leq c\, \text \,b = c,$ which shows that $\,\leq\,$ (and thus also $\Delta$) is nothing more than a new symbol for equality on $X;$ that is, $(X, \Delta) \ =\ (X, =).$ The notation $(X, =)$ is used because it avoids the unnecessary introduction of a new symbol for the diagonal.
Both $\,< \text \leq\,$ are serial and neither possesses a greatest element or a maximal element; this remains true if they are each restricted to the subset of $\mathcal \times \N \times X$ defined by
$\begin \operatorname_ \;&:=\; \, \\ \end$
where it will henceforth be assumed that they are.
Denote the assignment $i = (B, m, b) \mapsto b$ from this subset by:
\begin
\operatorname_\ :\ &&\ \operatorname_\ &&\,\to \;& X \\ .5ex &&\ (B, m, b) \ &&\,\mapsto\;& b \\ .5ex\end
If $i_0 = \left(B_0, m_0, b_0\right) \in \operatorname_$ then just as with $\operatorname_$ before, the tail of the $\operatorname_$ starting at $i_0$ is equal to $B_0.$
If $\mathcal$ is a prefilter on $X$ then $\operatorname_$ is a net in $X$ whose domain $\operatorname_$ is a partially ordered set and moreover, $\operatorname\left(\operatorname_\right) = \mathcal.$
Because the tails of $\operatorname_ \text \operatorname_$ are identical (since both are equal to the prefilter $\mathcal$), there is typically nothing lost by assuming that the domain of the net associated with a prefilter is both directed partially ordered. If the set $\N$ is replaced with the positive rational numbers then the strict partial order $<$ will also be a dense order.

Subordinate filters and subnets

The notion of "$\mathcal$ is subordinate to $\mathcal$" (written $\mathcal \vdash \mathcal$) is for filters and prefilters what "$x_ = \left(x_\right)_^$ is a subsequence of $x_ = \left(x_i\right)_^$" is for sequences.
For example, if $\operatorname\left(x_\right) = \left\$ denotes the set of tails of $x_$ and if $\operatorname\left(x_\right) = \left\$ denotes the set of tails of the subsequence $x_$ (where $x_ := \left\$) then $\operatorname\left(x_\right) ~\vdash~ \operatorname\left(x_\right)$ (that is, $\operatorname\left(x_\right) \leq \operatorname\left(x_\right)$) is true but $\operatorname\left(x_\right) ~\vdash~ \operatorname\left(x_\right)$ is in general false.

Non–equivalence of subnets and subordinate filters

A subset $R \subseteq I$ of a preordered space $(I, \leq)$ is or in $I$ if for every $i \in I$ there exists some $r \in R \text i \leq r.$ If $R \subseteq I$ contains a tail of $I$ then $R$ is said to be or ; explicitly, this means that there exists some $i \in I \text I_ \subseteq R$ (that is, $j \in R \text j \in I \text i \leq j$). An eventual set is necessarily not empty. A subset is eventual if and only if its complement is not frequent (which is termed ).
A map $h : A \to I$ between two preordered sets is if whenever $a, b \in A \text a \leq b, \text h(a) \leq h(b).$

Subnets in the sense of Willard and subnets in the sense of Kelley are the most commonly used definitions of "subnet."
The first definition of a subnet was introduced by John L. Kelley in 1955.
Stephen Willard introduced his own variant of Kelley's definition of subnet in 1970.
AA–subnets were introduced independently by Smiley (1957), Aarnes and Andenaes (1972), and Murdeshwar (1983); AA–subnets were studied in great detail by Aarnes and Andenaes but they are not often used.

Kelley did not require the map $h$ to be order preserving while the definition of an AA–subnet does away entirely with any map between the two nets' domains and instead focuses entirely on $X$ − the nets' common codomain.
Every Willard–subnet is a Kelley–subnet and both are AA–subnets.
In particular, if $y_ = \left(y_a\right)_$ is a Willard–subnet or a Kelley–subnet of $x_ = \left(x_i\right)_$ then $\operatorname\left(x_\right) \leq \operatorname\left(y_\right).$

• Example: Let $I = \N$ and let $x_$ be a constant sequence, say $x_ = \left(0\right)_.$ Let $s_1 = 0$ and $A = \$ so that $s_ = \left(s_a\right)_ = \left(s_1\right)$ is a net on $A.$ Then $s_$ is an AA-subnet of $x_$ because $\operatorname\left(x_\right) = \ = \operatorname\left(s_\right).$ But $s_$ is not a Willard-subnet of $x_$ because there does not exist any map $h : A \to I$ whose image is a cofinal subset of $I = \N.$ Nor is $s_$ a Kelley-subnet of $x_$ because if $h : A \to I$ is any map then $E := I \setminus \$ is a cofinal subset of $I = \N$ but $h^(E) = \varnothing$ is not eventually in $A.$
  

AA–subnets have a defining characterization that immediately shows that they are fully interchangeable with sub(ordinate)filters.
Explicitly, what is meant is that the following statement is true for AA–subnets:

If $\mathcal \text \mathcal$ are prefilters then $\mathcal \leq \mathcal \text \operatorname_$ is an AA–subnet of $\;\operatorname_.$

If "AA–subnet" is replaced by "Willard–subnet" or "Kelley–subnet" then the above statement becomes . In particular, the problem is that the following statement is in general false:

statement: If $\mathcal \text \mathcal$ are prefilters such that $\mathcal \leq \mathcal \text \operatorname_$ is a Kelley–subnet of $\;\operatorname_.$

Since every Willard–subnet is a Kelley–subnet, this statement remains false if the word "Kelley–subnet" is replaced with "Willard–subnet".

• : For all $n \in \N,$ let $B_n = \ \cup \N_.$ Let $\mathcal = \,$ which is a proper –system, and let $\mathcal = \ \cup \mathcal,$ where both families are prefilters on the natural numbers $X := \N = \.$
  Because $\mathcal \leq \mathcal, \mathcal$ is to $\mathcal$ as a subsequence is to a sequence.
  So ideally, $S = \operatorname_$ should be a subnet of $B = \operatorname_.$
  Let $I := \operatorname(\mathcal)$ be the domain of $\operatorname_,$ so $I$ contains a cofinal subset that is order isomorphic to $\N$ and consequently contains neither a maximal nor greatest element.
  Let $A := \operatorname(\mathcal) = \ \cup I, \text M := (1, \)$ is both a maximal and greatest element of $A.$
  The directed set $A$ also contains a subset that is order isomorphic to $\N$ (because it contains $I,$ which contains such a subset) but no such subset can be cofinal in $A$ because of the maximal element $M.$
  Consequently, any order–preserving map $h : A \to I$ must be eventually constant (with value $h(M)$) where $h(M)$ is then a greatest element of the range $h(A).$
  Because of this, there can be no order preserving map $h : A \to I$ that satisfies the conditions required for $\operatorname_$ to be a Willard–subnet of $\operatorname_$ (because the range of such a map $h$ cannot be cofinal in $I$).
  Suppose for the sake of contradiction that there exists a map $h : A \to I$ such that $h^\left(I_\right)$ is eventually in $A$ for all $i \in I.$
  Because $h(M) \in I,$ there exist $n, n_0 \in \N$ such that $h(M) = \left(n_0, B_n\right) \text n_0 \in B_n.$
  For every $i \in I,$ because $h^\left(I_\right)$ is eventually in $A,$ it is necessary that $h(M) \in I_.$
  In particular, if $i := \left(n + 2, B_\right)$ then $h(M) \geq i = \left(n + 2, B_\right),$ which by definition is equivalent to $B_n \subseteq B_,$ which is false.
  Consequently, $\operatorname_$ is not a Kelley–subnet of $\operatorname_.$
  

If "subnet" is defined to mean Willard–subnet or Kelley–subnet then nets and filters are not completely interchangeable because there exists a filter–sub(ordinate)filter relationships that cannot be expressed in terms of a net–subnet relationship between the two induced nets. In particular, the problem is that Kelley–subnets and Willard–subnets are fully interchangeable with subordinate filters. If the notion of "subnet" is not used or if "subnet" is defined to mean AA–subnet, then this ceases to be a problem and so it becomes correct to say that nets and filters are interchangeable. Despite the fact that AA–subnets do not have the problem that Willard and Kelley subnets have, they are not widely used or known about.

See also

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Notes

Proofs

Citations

References

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{{Set theory

General topology
Order theory
Set theory