Filters in topology

Filters in topology, a subfield of mathematics, can be used to study topological spaces and define all basic topological notions such a convergence, continuity, compactness, and more. Filters, which are special families of subsets of some given set, also provide a common framework for defining various types of limits of functions such as limits from the left/right, to infinity, to a point or a set, and many others. Special types of filters called have many useful technical properties and they may often be used in place of arbitrary filters.

Filters have generalizations called (also known as ) and , all of which appear naturally and repeatedly throughout topology. Examples include neighborhood filters/ bases/subbases and uniformities. Every filter is a prefilter and both are filter subbases. Every prefilter and filter subbase is contained in a unique smallest filter, which they are said to . This establishes a relationship between filters and prefilters that may often be exploited to allow one to use whichever of these two notions is more technically convenient. There is a certain preorder on families of sets, denoted by $\,\leq,\,$ that helps to determine exactly when and how one notion (filter, prefilter, etc.) can or cannot be used in place of another. This preorder's importance is amplified by the fact that it also defines the notion of filter convergence, where by definition, a filter (or prefilter) $\mathcal$ to a point if and only if $\mathcal \leq \mathcal,$ where $\mathcal$ is that point's neighborhood filter. Consequently, subordination also plays an important role in many concepts that are related to convergence, such as cluster points and limits of functions. In addition, the relation $\mathcal \geq \mathcal,$ which denotes $\mathcal \leq \mathcal$ and is expressed by saying that $\mathcal$ $\mathcal,$ also establishes a relationship in which $\mathcal$ is to $\mathcal$ as a subsequence is to a sequence (that is, the relation $\geq,$ which is called , is for filters the analog of "is a subsequence of").

Filters were introduced by Henri Cartan in 1937 and subsequently used by Bourbaki in their book as an alternative to the similar notion of a net developed in 1922 by E. H. Moore and H. L. Smith.
Filters can also be used to characterize the notions of sequence and net convergence. But unlikeSequences and nets in a space $X$ are maps from directed sets like the natural number, which in general maybe entirely unrelated to the set $X$ and so they, and consequently also their notions of convergence, are not intrinsic to $X.$ sequence and net convergence, filter convergence is defined in terms of subsets of the topological space $X$ and so it provides a notion of convergence that is completely intrinsic to the topological space; indeed, the category of topological spaces can be equivalently defined entirely in terms of filters. Every net induces a canonical filter and dually, every filter induces a canonical net, where this induced net (resp. induced filter) converges to a point if and only if the same is true of the original filter (resp. net). This characterization also holds for many other definitions such as cluster points. These relationships make it possible to switch between filters and nets, and they often also allow one to choose whichever of these two notions (filter or net) is more convenient for the problem at hand.
However, assuming that "subnet" is defined using either of its most popular definitions (which are those given by Willard and by Kelley), then in general, this relationship does extend to subordinate filters and subnets because as detailed below, there exist subordinate filters whose filter/subordinate–filter relationship cannot be described in terms of the corresponding net/subnet relationship; this issue can however be resolved by using a less commonly encountered definition of "subnet", which is that of an AA–subnet.

Thus filters/prefilters and this single preorder $\,\leq\,$ provide a framework that seamlessly ties together fundamental topological concepts such as topological spaces ( via neighborhood filters), neighborhood bases, convergence, various limits of functions, continuity, compactness, sequences (via sequential filters), the filter equivalent of "subsequence" (subordination), uniform spaces, and more; concepts that otherwise seem relatively disparate and whose relationships are less clear.

Motivation

Archetypical example of a filter

The archetypical example of a filter is the $\mathcal(x)$ at a point $x$ in a topological space $(X, \tau),$ which is the family of sets consisting of all neighborhoods of $x.$
By definition, a neighborhood of some given point $x$ is any subset $B \subseteq X$ whose topological interior contains this point; that is, such that $x \in \operatorname_X B.$ Importantly, neighborhoods are required to be open sets; those are called .
The fundamental properties shared by neighborhood filters, which are listed below, ultimately became the definition of a "filter."
A is a set $\mathcal$ of subsets of $X$ that satisfies all of the following conditions:

1. :   $X \in \mathcal$  –  just as $X \in \mathcal(x),$ since $X$ is always a neighborhood of $x$ (and of anything else that it contains);


2. :   $\varnothing \not\in \mathcal$  –  just as no neighborhood of $x$ is empty;


3. :   If $B, C \in \mathcal \text B \cap C \in \mathcal$  –  just as the intersection of any two neighborhoods of $x$ is again a neighborhood of $x$;


4. :   If $B \in \mathcal \text B \subseteq S \subseteq X$ then $S \in \mathcal$  –  just as any subset of $X$ that contains a neighborhood of $x$ will necessarily a neighborhood of $x$ (this follows from $\operatorname_X B \subseteq \operatorname_X S$ and the definition of "a neighborhood of $x$").

Generalizing sequence convergence by using sets − determining sequence convergence without the sequence

A is by definition a map $\N \to X$ from the natural numbers into the space $X.$
The original notion of convergence in a topological space was that of a sequence converging to some given point in a space, such as a metric space.
With metrizable spaces (or more generally first–countable spaces or Fréchet–Urysohn spaces), sequences usually suffices to characterize, or "describe", most topological properties, such as the closures of subsets or continuity of functions.
But there are many spaces where sequences can be used to describe even basic topological properties like closure or continuity.
This failure of sequences was the motivation for defining notions such as nets and filters, which fail to characterize topological properties.

Nets directly generalize the notion of a sequence since nets are, by definition, maps $I \to X$ from an arbitrary directed set $(I, \leq)$ into the space $X.$ A sequence is just a net whose domain is $I = \N$ with the natural ordering. Nets have their own notion of convergence, which is a direct generalization of sequence convergence.

Filters generalize sequence convergence in a different way by considering the values of a sequence.
To see how this is done, consider a sequence $x_ = \left(x_i\right)_^ \text X,$ which is by definition just a function $x_ : \N \to X$ whose value at $i \in \N$ is denoted by $x_i$ rather than by the usual parentheses notation $x_(i)$ that is commonly used for arbitrary functions.
Knowing only the image (sometimes called "the range") $\operatorname x_ := \left\ = \left\$ of the sequence is not enough to characterize its convergence; multiple sets are needed.
It turns out that the needed sets are the following,Technically, any infinite subfamily of this set of tails is enough to characterize this sequence's convergence. But in general, unless indicated otherwise, the set of tails is taken unless there is some reason to do otherwise. which are called the of the sequence $x_$:
\begin
&\ \\ .3ex&\ \\ .3ex&\ \\ .3ex& && && &&\;\,\vdots && && && \\ .3ex&\ \\ .3ex& && && &&\;\,\vdots && && && \\ .3ex\end

These sets completely determine this sequence's convergence (or non–convergence) because given any point, this sequence converges to it if and only if for every neighborhood $U$ (of this point), there is some integer $n$ such that $U$ contains all of the points $x_n, x_, \ldots .$ This can be reworded as:

every neighborhood $U$ must contain some set of the form $\$ as a subset.

It is the above characterization that can be used with the above family of tails to determine convergence (or non–convergence) of the sequence $x_ : \N \to X.$
Specifically, with these in hand, the $x_ : \N \to X$ is no longer needed to determine convergence of this sequence (no matter what topology is placed on $X$).
By generalizing this observation, the notion of "convergence" can be extended from functions/sequences to families of sets.

The above set of tails of a sequence is in general not a filter but it does "" a filter via taking its (which consists of all supersets of all tails). The same is true of other important families of sets such as any neighborhood basis at a given point, which in general is also not a filter but does generate a filter via its upward closure (in particular, it generates the neighborhood filter at that point). The properties that these families share led to the notion of a , also called a , which by definition is any family having the minimal properties necessary and sufficient for it to generate a filter via taking its upward closure.

Nets vs. filters − advantages and disadvantages

Filters and nets each have their own advantages and drawbacks and there's no reason to use one notion exclusively over the other.Indeed, net convergence is defined using neighborhood filters while (pre)filters are directed sets with respect to $\,\supseteq\,,$ so it is difficult to keep these notions completely separate.
Depending on what is being proved, a proof may be made significantly easier by using one of these notions instead of the other. Both filters and nets can be used to completely characterize any given topology.
Nets are direct generalizations of sequences and can often be used similarly to sequences, so the learning curve for nets is typically much less steep than that for filters.
However, filters, and especially ultrafilters, have many more uses outside of topology, such as in set theory, mathematical logic, model theory (ultraproducts, for example), abstract algebra, order theory, generalized convergence spaces, Cauchy spaces, and in the definition and use of hyperreal numbers.

Like sequences, nets are and so they have the .
For example, like sequences, nets can be "plugged into" other functions, where "plugging in" is just function composition.
Theorems related to functions and function composition may then be applied to nets.
One example is the universal property of inverse limits, which is defined in terms of composition of functions rather than sets and it is more readily applied to functions like nets than to sets like filters (a prominent example of an inverse limit is the Cartesian product).
Filters may be awkward to use in certain situations, such as when switching between a filter on a space $X$ and a filter on a dense subspace $S \subseteq X.$

In contrast to nets, filters (and prefilters) are families of and so they have the .
For example, if $f$ is surjective then the $f^(\mathcal) := \left\$ under $f^$ of an arbitrary filter or prefilter $\mathcal$ is both easily defined and guaranteed to be a prefilter on $f$'s domain, whereas it is less clear how to pullback (unambiguously/without choice) an arbitrary sequence (or net) $y_$ so as to obtain a sequence or net in the domain (unless $f$ is also injective and consequently a bijection, which is a stringent requirement). Similarly, the intersection of any collection of filters is once again a filter whereas it is not clear what this could mean for sequences or nets.
Because filters are composed of subsets of the very topological space $X$ that is under consideration, topological set operations (such as closure or interior) may be applied to the sets that constitute the filter.
Taking the closure of all the sets in a filter is sometimes useful in functional analysis for instance.
Theorems and results about images or preimages of sets under a function may also be applied to the sets that constitute a filter; an example of such a result might be one of continuity's characterizations in terms of preimages of open/closed sets or in terms of the interior/closure operators.
Special types of filters called have many useful properties that can significantly help in proving results.
One downside of nets is their dependence on the directed sets that constitute their domains, which in general may be entirely unrelated to the space $X.$ In fact, the class of nets in a given set $X$ is too large to even be a set (it is a proper class); this is because nets in $X$ can have domains of cardinality.
In contrast, the collection of all filters (and of all prefilters) on $X$ is a set whose cardinality is no larger than that of $\wp(\wp(X)).$
Similar to a topology on $X,$ a filter on $X$ is "intrinsic to $X$" in the sense that both structures consist of subsets of $X$ and neither definition requires any set that cannot be constructed from $X$ (such as $\N$ or other directed sets, which sequences and nets require).

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.

Topology notation

Denote the set of all topologies on a set $X \text \operatorname(X).$
Suppose $\tau \in \operatorname(X),$ $S \subseteq X$ is any subset, and $x \in X$ is any point.

If $\varnothing \neq S \subseteq X$ then $\tau(S) = \bigcap_ \tau(s) \text \mathcal_(S) = \bigcap_ \mathcal_(s).$

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.

\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 $\mathcal \vee \mathcal.$
  

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, $\mathcal_.$

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.

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 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 $X \setminus \mathcal.$


4. 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.


5. $\mathcal$ 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").

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.
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 $\ker \left(\mathcal^\right) = \ker \mathcal$ 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 $f(\ker \mathcal) \subseteq \ker f(\mathcal) \text f^(\ker \mathcal) = \ker f^(\mathcal).$
Equivalent families have equal kernels.
Two principal families are equivalent if and only if their kernels are equal.

=Classifying families by their kernels

=

If $\mathcal$ is a principal filter on $X$ then $\varnothing \neq \ker \mathcal \in \mathcal$ and
$\mathcal = \^$
and $\$ 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 $\R.$

=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, $\ \leq \mathcal \text \varnothing \in \mathcal.$

A non-empty family that is coarser than a filter subbase must itself be a filter subbase.
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.

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.

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.
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.

Set theoretic properties and constructions relevant to topology

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 .
The trace is always finer than the original family; that is, $\mathcal \leq \mathcal\big\vert_S.$
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):
ultra, ultrafilter, filter, prefilter, filter subbase, dual ideal, upward closed, proper/non–degenerate, 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:

$f^(\mathcal)$ 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 $X.$

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.

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 $X_i.$

Convergence, limits, and cluster points

Throughout, $(X, \tau)$ is a topological space.

Prefilters vs. filters

With respect to maps and subsets, the property of being a prefilter is in general more well behaved and better preserved than the property of being a filter. For instance, the image of a prefilter under some map is again a prefilter; but the image of a filter under a non–surjective map is a filter on the codomain, although it will be a prefilter. The situation is the same with preimages under non–injective maps (even if the map is surjective). If $S \subseteq X$ is a proper subset then any filter on $S$ will not be a filter on $X,$ although it will be a prefilter.

One advantage that filters have is that they are distinguished representatives of their equivalence class (relative to $\,\leq$), meaning that any equivalence class of prefilters contains a unique filter. This property may be useful when dealing with equivalence classes of prefilters (for instance, they are useful in the construction of completions of uniform spaces via Cauchy filters). The many properties that characterize ultrafilters are also often useful. They are used to, for example, construct the Stone–Čech compactification. The use of ultrafilters generally requires that the ultrafilter lemma be assumed. But in the many fields where the axiom of choice (or the Hahn–Banach theorem) is assumed, the ultrafilter lemma necessarily holds and does not require an addition assumption.

A note on intuition

Suppose that $\mathcal$ is a non–principal filter on an infinite set $X.$ $\mathcal$ has one "upward" property (that of being closed upward) and one "downward" property (that of being directed downward).
Starting with any $F_0 \in \mathcal,$ there always exists some $F_1 \in \mathcal$ that is a subset of $F_0$; this may be continued ad infinitum to get a sequence $F_0 \supset F_1 \supset \cdots$ of sets in $\mathcal$ with each $F_$ being a subset of $F_i.$ The same is true going "upward", for if $F_0 = X \in \mathcal$ then there is no set in $\mathcal$ that contains $X$ as a proper subset.
Thus when it comes to limiting behavior (which is a topic central to the field of topology), going "upward" leads to a dead end, while going "downward" is typically fruitful. So to gain understanding and intuition about how filters (and prefilter) relate to concepts in topology, the "downward" property is usually the one to concentrate on. This is also why so many topological properties can be described by using only prefilters, rather than requiring filters (which only differ from prefilters in that they are also upward closed).
The "upward" property of filters is less important for topological intuition but it is sometimes useful to have for technical reasons. For example, with respect to $\,\subseteq,$ every filter subbase is contained in a unique smallest filter but there may not exist a unique smallest prefilter containing it.

Limits and convergence

The following well known definition will be generalized to prefilters.
A point $x \in X$ is called a , , or of a subset $S \subseteq X$ if every neighborhood of $x \text X$ contains a point of $S$ different from $x,$ or equivalently, if $x \in \operatorname_(S \setminus \).$ The set of all limit points of $S$ is called the derived set of $S \text X.$
The closure of a set $S \subseteq X$ is equal to the union of $S$ together with the set of all limit points of $S.$

A family $\mathcal$ is said to to a point or subset $x$ of $X$ written $\mathcal \to x \text \lim \mathcal \to x \text X,$ if $\mathcal \geq \mathcal(x),$ in which case $x$ is said to be a (or if $x$ is a point, also ) of $\mathcal \text X.$ Denote the set of all these limit points by $\lim _X \mathcal \text \lim \mathcal.$ As usual, $\lim \mathcal = x$ is defined to mean that $\mathcal \to x \text X$ and $x \in X$ is the limit point of $\mathcal \text X;$ that is, if also $\mathcal \to z \text X \text z = x.$ (If the notation "$\lim \mathcal = x$" did not also require that the limit point $x$ be unique then the equals sign would no longer be guaranteed to be transitive).

In words, $\mathcal$ converges to a point if and only if $\mathcal$ is than the neighborhood filter at that point. Explicitly, $\mathcal(x) \leq \mathcal$ means that every neighborhood $N \text x$ contains some $B \in \mathcal$ as a subset (that is, $B \subseteq N$); thus the following then holds: $\mathcal \ni N \supseteq B \in \mathcal.$
In the above definitions, it suffices to check that $\mathcal$ is finer than some (or equivalently, finer than every) neighborhood base in $(X, \tau)$ of the point or set (for example, such as $\tau(x) = \$ or $\tau(S) = \bigcap_ \tau(s)$).

For example, if $x \in X$ then $x$ is a limit point of the principle ultra prefilter $\,$ the ultrafilter that it generates, the neighborhood filter $\mathcal(x),$ and of any neighborhood basis at $x.$
The one and only limit point in $X := \R$ of the free prefilter $\mathcal := \$ is $0.$

If $\mathcal$ converges to a point or subset then the same is true of any family finer than $\mathcal$ (such as $\mathcal$'s restriction to any given subset of $X,$ for example).
Consequently, the limit points of a family $\mathcal$ are the same as the limit points of its upward closure: $\operatorname_X \mathcal ~=~ \operatorname_X \left(\mathcal^\right).$
In particular, the limit points of a prefilter are the same as the limit points of the filter that it generates.
If a filter subbase converges to a point or subset then so does the filter that it generates, although the converse is not guaranteed. For example, the filter subbase $\$ does not converge to $0$ in $X := \R$ although the (principle ultra) filter that it generates does.
If $\varnothing \neq S \subseteq X$ then because $\mathcal(S) = \bigcap_ \mathcal(s),$ if $\mathcal \to s \text s \in S$ then $\mathcal \to S.$
Because $\varnothing$ is an open set, a family $\mathcal$ converges to $\varnothing$ if and only if $\varnothing \in \mathcal;$ so in particular, no filter or other non-degenerate family can converge to the empty set, which is why when dealing with convergent prefilters (or filter subbases), it is typically assumed (often without mention) that $S \neq \varnothing.$

Given $x \in X,$ the following are equivalent for a prefilter $\mathcal:$

1. $\mathcal$ converges to $x.$


2. $\mathcal$ converges to the set $\.$


3. $\mathcal^$ converges to $x.$


4. There exists a family equivalent to $\mathcal$ that converges to $x.$

If $\mathcal$ is a prefilter and $B \in \mathcal$ then $\mathcal$ converges to a point (or subset) of $X$ if and only if this is true of the trace $\mathcal\big\vert_B.$ If $\mathcal$ is a filter subbase that converges to $x \text S$ then this is also true of the filter that it generates (and also of any prefilter equivalent to this filter, such as the -system generated by $\mathcal$).

Because subordination is transitive, if $\mathcal \leq \mathcal \text \lim _ \mathcal \subseteq \lim _ \mathcal$ and moreover, for every $x \in X,$ both $\$ and the maximal/ultrafilter $\^$ converge to $x \text X.$ Thus every topological space $(X, \tau)$ induces a canonical convergence $\xi \subseteq X \times \operatorname(X)$ defined by $(x, \mathcal) \in \xi \text x \in \lim _ \mathcal.$
At the other extreme, the neighborhood filter $\mathcal(x)$ is the smallest (that is, coarsest) filter on $X$ that converges to $x \text X;$ that is, any filter converging to $x$ must contain $\mathcal(x)$ as a subset. Said differently, the family of filters that converge to $x$ consists exactly of those filter on $X$ that contain $\mathcal(x)$ as a subset.
Consequently, the finer the topology on $X$ then the prefilters exist that have any limit points in $X.$

Cluster points

Say that $x \in X$ is a or an of a family $\mathcal$ if $\mathcal$ meshes with the neighborhood filter at $x$; that is, if $\mathcal \# \mathcal(x).$ The set of all cluster points of $\mathcal$ is denoted by $\operatorname_X \mathcal \text \operatorname \mathcal.$

Explicitly, this means that $B \cap N \neq \varnothing \text B \in \mathcal$ and every neighborhood $N$ of $x.$ When $\mathcal$ is a prefilter then the definition of "$\mathcal \text \mathcal$ mesh" can be characterized entirely in terms of the preorder $\,\leq\,.$

More generally, given $S \subseteq X,$ say that

$\mathcal$ $S$ if $\mathcal$ meshes with the neighborhood filter of $S$; that is, if $\mathcal \# \mathcal(S).$

In the above definitions, it suffices to check that $\mathcal$ meshes with some (or equivalently, meshes with every) neighborhood base in $X$ of $x \text S.$ Two equivalent families of sets have the exact same limit points and also the same cluster points. No matter the topology, for every $x \in X,$ both $\$ and the principal ultrafilter $\^$ cluster at $x.$
For any $S \subseteq X,$ if $\mathcal$ clusters at some $s \in S$ then $\mathcal$ clusters at $S.$ No family clusters at $S := \varnothing$ and if $\varnothing \in \mathcal \text \varnothing = \operatorname \mathcal.$
If $\mathcal$ clusters to a point or subset then the same is true of any family coarser than $\mathcal.$ Consequently, the cluster points of a family $\mathcal$ are the same as the cluster points of its upward closure: $\operatorname_X \mathcal ~=~ \operatorname_X \left(\mathcal^\right).$
In particular, the cluster points of a prefilter are the same as the cluster points of the filter that it generates.

Given $x \in X,$ the following are equivalent for a prefilter $\mathcal \text X$:

1. $\mathcal$ clusters at $x.$


2. $\mathcal$ clusters at the set $\.$


3. The family $\mathcal^$ generated by $\mathcal$ clusters at $x.$


4. There exists a family equivalent to $\mathcal$ that clusters at $x.$


5. $x \in \bigcap_ \operatorname_X F.$


6. $X \setminus N \not\in \mathcal^$ for every neighborhood $N$ of $x.$
   * If $\mathcal$ is a filter on $X$ then $x \in \operatorname_X \mathcal \text X \setminus N \not\in \mathcal$ for every neighborhood $N \text x.$


7. There exists a prefilter $\mathcal$ subordinate to $\mathcal$ (that is, $\mathcal \geq \mathcal$) such that $\mathcal \to x.$
   * This is the filter equivalent of "$x$ is a cluster point of a sequence if and only if there exists a subsequence converging to $x.$
   * In particular, if $x$ is a cluster point of a prefilter $\mathcal$ then $\mathcal (\cap) \mathcal(x)$ is a prefilter subordinate to $\mathcal$ that converges to $x \text X.$

The set $\operatorname_X \mathcal$ of all cluster points of a prefilter $\mathcal \text X$ satisfies
$\operatorname_X \mathcal = \bigcap_ \operatorname_X B.$
Consequently, the set $\operatorname_X \mathcal$ of all cluster points of prefilter $\mathcal$ is a closed subset of $X.$ This also justifies the notation $\operatorname_X \mathcal$ for the set of cluster points.
In particular, if $K \subseteq X$ is non-empty (so that $\mathcal := \$ is a prefilter) then $\operatorname_X \ = \operatorname_X K$ since both sides are equal to $\bigcap_ \operatorname_X B.$

Properties and relationships

Just like sequences and nets, it is possible for a prefilter on a topological space of infinite cardinality to not have cluster points or limit points.

If $x$ is a limit point of $\mathcal$ then $x$ is necessarily a limit point of any family $\mathcal$ than $\mathcal$ (that is, if $\mathcal(x) \leq \mathcal \text \mathcal \leq \mathcal$ then $\mathcal(x) \leq \mathcal$).
In contrast, if $x$ is a cluster point of $\mathcal$ then $x$ is necessarily a cluster point of any family $\mathcal$ than $\mathcal$ (that is, if $\mathcal(x) \text \mathcal$ mesh and $\mathcal \leq \mathcal$ then $\mathcal(x) \text \mathcal$ mesh).

Equivalent families and subordination

Any two equivalent families $\mathcal \text \mathcal$ can be used in the definitions of "limit of" and "cluster at" because their equivalency guarantees that $\mathcal \leq \mathcal$ if and only if $\mathcal \leq \mathcal,$ and also that $\mathcal \# \mathcal$ if and only if $\mathcal \# \mathcal.$
In essence, the preorder $\,\leq\,$ is incapable of distinguishing between equivalent families.
Given two prefilters, whether or not they mesh can be characterized entirely in terms of subordination.
Thus the two most fundamental concepts related to (pre)filters to Topology (that is, limit and cluster points) can both be defined in terms of the subordination relation. This is why the preorder $\,\leq\,$ is of such great importance in applying (pre)filters to Topology.

Limit and cluster point relationships and sufficient conditions

Every limit point of a non-degenerate family $\mathcal$ is also a cluster point; in symbols:
$\operatorname_X \mathcal ~\subseteq~ \operatorname_X \mathcal.$
This is because if $x$ is a limit point of $\mathcal$ then $\mathcal(x) \text \mathcal$ mesh, which makes $x$ a cluster point of $\mathcal.$ But in general, a cluster point need not be a limit point. For instance, every point in any given non-empty subset $K \subseteq X$ is a cluster point of the principle prefilter $\mathcal := \$ (no matter what topology is on $X$) but if $X$ is Hausdorff and $K$ has more than one point then this prefilter has no limit points; the same is true of the filter $\^$ that this prefilter generates.

However, every cluster point of an prefilter is a limit point. Consequently, if $\mathcal$ is an prefilter then $\operatorname_X \mathcal = \operatorname_X \mathcal;$ that is to say, a point $x \in X$ will be a cluster point of an ultra prefilter $\mathcal$ if and only if it is a limit point of $\mathcal.$
Although a cluster point of a filter need not be a limit point, there will always exist a finer filter that does converge to it; in particular, if $\mathcal$ clusters at $x$ then $\mathcal \,(\cap)\, \mathcal(x) = \$ is a filter subbase whose generated filter converges to $x.$

If $\varnothing \neq \mathcal \subseteq \wp(X) \text \mathcal \geq \mathcal$ is a filter subbase such that $\mathcal \to x \text X$ then $x \in \operatorname_X \mathcal.$ In particular, any limit point of a filter subbase subordinate to $\mathcal \neq \varnothing$ is necessarily also a cluster point of $\mathcal.$
If $x$ is a cluster point of a prefilter $\mathcal$ then $\mathcal (\cap) \mathcal(x)$ is a prefilter subordinate to $\mathcal$ that converges to $x \text X.$

If $S \subseteq X$ and if $\mathcal$ is a prefilter on $S$ then every cluster point of $\mathcal \text X$ belongs to $\operatorname_X S$ and any point in $\operatorname_X S$ is a limit point of a filter on $S.$

Primitive sets

A subset $P \subseteq X$ is called if it is the set of limit points of some ultrafilter (or equivalently, some ultra prefilter). That is, if there exists an ultrafilter $\mathcal \text X$ such that $P$ is equal to $\operatorname_X \mathcal,$ which recall denotes the set of limit points of $\mathcal \text X.$ Since limit points are the same as cluster points for ultra prefilters, a subset is primitive if and only if it is equal to the set $\operatorname_X \mathcal$ of cluster points of some ultra prefilter $\mathcal.$
For example, every closed singleton subset is primitive. The image of a primitive subset of $X$ under a continuous map $f : X \to Y$ is contained in a primitive subset of $Y.$

Assume that $P, Q \subseteq X$ are two primitive subset of $X.$
If $U$ is an open subset of $X$ that intersects $P$ then $U \in \mathcal$ for any ultrafilter $\mathcal \text X$ such that $P = \operatorname_X \mathcal.$
In addition, if $P \text Q$ are distinct then there exists some $S \subseteq X$ and some ultrafilters $\mathcal_P \text \mathcal_Q \text X$ such that $P = \operatorname_X \mathcal_P, Q = \operatorname_X \mathcal_Q, S \in \mathcal_P,$ and $X \setminus S \in \mathcal_Q.$

Other results

If $X$ is a complete lattice then:
* The limit inferior of $B$ is the infimum of the set of all cluster points of $B.$
* The limit superior of $B$ is the supremum of the set of all cluster points of $B.$
* $B$ is a convergent prefilter if and only if its limit inferior and limit superior agree; in this case, the value on which they agree is the limit of the prefilter.

Limits of functions defined as limits of prefilters

Suppose $f : X \to Y$ is a map from a set into a topological space $Y,$ $\mathcal \subseteq \wp(X),$ and $y \in Y.$ If $y$ is a limit point (respectively, a cluster point) of $f(\mathcal) \text Y$ then $y$ is called a or (respectively, a )
Explicitly, $y$ is a limit of $f$ with respect to $\mathcal$ if and only if $\mathcal(y) \leq f(\mathcal),$ which can be written as $f(\mathcal) \to y \text \lim f(\mathcal) \to y \text Y$ (by definition of this notation) and stated as $f$ If the limit $y$ is unique then the arrow $\to$ may be replaced with an equals sign $=.$ The neighborhood filter $\mathcal(y)$ can be replaced with any family equivalent to it and the same is true of $\mathcal.$

The definition of a convergent net is a special case of the above definition of a limit of a function.
Specifically, if $x \in X \text \chi : (I, \leq) \to X$ is a net then
$\chi \to x \text X \quad \text \quad \chi(\operatorname(I, \leq)) \to x \text X,$
where the left hand side states that $x$ is a Convergent net, limit $\chi$ while the right hand side states that $x$ is a limit $\chi$ with respect to $\mathcal := \operatorname(I, \leq)$ (as just defined above).

The table below shows how various types of limits encountered in analysis and topology can be defined in terms of the convergence of images (under $f$) of particular prefilters on the domain $X.$
This shows that prefilters provide a general framework into which many of the various definitions of limits fit.
The limits in the left–most column are defined in their usual way with their obvious definitions.

Throughout, let $f : X \to Y$ be a map between topological spaces, $x_0 \in X, \text y \in Y.$
If $Y$ is Hausdorff then all arrows in the table may be replaced with equal signs and may be replaced with

, $X = \N \text f : \N \to Y$ is a sequence in $Y$
, -

, $\lim_ f(x) \to y$
, ⇔
, style='text-align:left;', $f(\mathcal) \to y \text \mathcal\,:=\,(\R, \infty) \,:=\,\$
, $X = \R$
, -

, $\lim_ f(x) \to y$
, ⇔
, style='text-align:left;', $f(\mathcal) \to y \text \mathcal\,:=\,(-\infty,\R) \,:=\,\$
, $X = \R$
, -

, $\lim_ f(x) \to y$
, ⇔
, style='text-align:left;', $f(\mathcal) \to y \text \mathcal\,:=\,\$
, $X = \R \text X = \Z$ for a double-ended sequence
, -

, $\lim_ f(x) \to y$
, ⇔
, style='text-align:left;', $f(\mathcal) \to y \text \mathcal\,:=\,\$
, style="padding-left:2em; padding-right:2em;", $(X, \, \cdot\, ) \text$ a seminormed space; $\text X = \Complex$

By defining different prefilters, many other notions of limits can be defined; for example, $\lim_ f(x) \to y.$

Divergence to infinity

Divergence of a real-valued function to infinity can be defined/characterized by using the prefilters
$(\R, \infty) := \ ~~ \text ~~ (-\infty, \R) := \,$
where $f \to \infty$ along $\mathcal$ if and only if $(\R, \infty) \leq f(\mathcal)$ and similarly, $f \to -\infty$ along $\mathcal$ if and only if $(-\infty, \R) \leq f(\mathcal).$ The family $(\R, \infty)$ can be replaced by any family equivalent to it, such as $[\R, \infty) := \$ for instance (in real analysis, this would correspond to replacing the strict inequality in the definition with and the same is true of $\mathcal$ and $(-\infty, \R).$

So for example, if $\mathcal\,:=\,\mathcal\left(x_0\right)$ then $\lim_ f(x) \to \infty$ if and only if $(\R, \infty) \leq f(\mathcal)$ holds. Similarly, $\lim_ f(x) \to - \infty$ if and only if $(-\infty, \R) \leq f\left(\mathcal\left(x_0\right)\right),$ or equivalently, if and only if $(-\infty, \R] \leq f\left(\mathcal\left(x_0\right)\right).$

More generally, if $f$ is valued in $Y = \R^n \text Y = \Complex^n$ (or some other seminormed vector space) and if $B_ := \ = Y \setminus B_$ then $\lim_ , f(x), \to \infty$ if and only if $B_ \leq f\left(\mathcal\left(x_0\right)\right)$ holds, where $B_ := \left\.$

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.

Nets to prefilters

In the definitions below, the first statement is the standard definition of a limit point of a net (respectively, a cluster point of a net) and it is gradually reworded until the corresponding filter concept is reached.

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
$(R, r) \leq (S, s) \quad \text \quad R \supseteq S.$

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 $\operatorname(\mathcal).$

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).

\to x.

$x$ is a cluster point of $\mathcal$ if and only if $x$ is a cluster point of $\operatorname_.$

\right) and that if $x_$ is a net in $X$ then (1) $x_ \to x \text \operatorname\left(x_\right) \to x,$ and (2) $x$ is a cluster point of $x_$ if and only if $x$ is a cluster point of $\operatorname\left(x_\right).$
By using $x_ := \operatorname_ \text \mathcal = \operatorname\left(\operatorname_\right),$ it follows that
$\mathcal \to x \quad \text \quad \operatorname\left(\operatorname_\right) \to x \quad \text \quad \operatorname_ \to x.$
It also follows that $x$ is a cluster point of $\mathcal$ if and only if $x$ is a cluster point of $\operatorname\left(\operatorname_\right)$ if and only if $x$ is a cluster point of $\operatorname_.$

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 (detailed here: Filter (set theory)#Partially ordered net) 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.
Because the tails of this partially ordered net are identical to the tails of $\operatorname_$ (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 can further be assumed that the partially ordered domain is also 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 $\operatorname\left(x_\right) ~\vdash~ \operatorname\left(x_\right)$ (which by definition means $\operatorname\left(x_\right) \leq \operatorname\left(x_\right)$) is true but $\operatorname\left(x_\right) ~\vdash~ \operatorname\left(x_\right)$ is in general false.
If $x_ = \left(x_i\right)_$ is a net in a topological space $X$ and if $\mathcal(x)$ is the neighborhood filter at a point $x \in X,$ then $x_ \to x \text X \text \mathcal(x) \leq \operatorname\left(x_\right).$

If $f : X \to Y$ is an surjective open map, $x \in X,$ and $\mathcal$ is a prefilter on $Y$ that converges to $f(x),$ then there exist a prefilter $\mathcal$ on $X$ such that $\mathcal \to x$ and $f(\mathcal)$ is equivalent to $\mathcal$ (that is, $\mathcal \leq f(\mathcal) \leq \mathcal$).

Subordination analogs of results involving subsequences

The following results are the prefilter analogs of statements involving subsequences. The condition "$\mathcal \geq \mathcal,$" which is also written $\mathcal \vdash \mathcal,$ is the analog of "$\mathcal$ is a subsequence of $\mathcal.$" So "finer than" and "subordinate to" is the prefilter analog of "subsequence of." Some people prefer saying "subordinate to" instead of "finer than" because it is more reminiscent of "subsequence of."

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 $r \in R$ such that $i \leq r.$ If $R \subseteq I$ contains a tail of $I$ then $R$ is said to be or ; explicitly, this means that there exists some $i \in I$ such that $I_ \subseteq R$ (that is, $j \in R$ for all $j \in I$ satisfying $i \leq j$). 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 $a, b \in A$ satisfy $a \leq b,$ then $h(a) \leq h(b).$

#Willard–subnet, Subnets in the sense of Willard and #Kelley–subnet, 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: If $I = \N$ and $x_ = \left(x_i\right)_$ is a constant sequence and if $A = \$ and $s_1 := x_1$ then $s_ = \left(s_a\right)_$ is an AA-subnet of $x_$ but it is neither a Willard-subnet nor a Kelley-subnet of $x_.$

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 $\mathcal \leq \mathcal$ if and only if $\operatorname_$ is an AA–subnet of $\operatorname_.$

If "AA–subnet" is replaced by "Willard–subnet" or "Kelley–subnet" then the above statement becomes . In particular, as Filter (set theory)#Example of subordination that Kelley subnets can not express, this counter-example demonstrates, 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".

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.

Topologies and prefilters

Throughout, $(X, \tau)$ is a topological space.

Examples of relationships between filters and topologies

Bases and prefilters

Let $\mathcal \neq \varnothing$ be a family of sets that covers $X$ and define $\mathcal_x = \$ for every $x \in X.$ The definition of a base for some topology can be immediately reworded as: $\mathcal$ is a base for some topology on $X$ if and only if $\mathcal_x$ is a filter base for every $x \in X.$
If $\tau$ is a topology on $X$ and $\mathcal \subseteq \tau$ then the definitions of $\mathcal$ is a Base (topology), basis (resp. subbase) for $\tau$ can be reworded as:

$\mathcal$ is a base (resp. subbase) for $\tau$ if and only if for every $x \in X, \mathcal_x$ is a filter base (resp. filter subbase) that generates the neighborhood filter of $(X, \tau)$ at $x.$

Neighborhood filters

The archetypical example of a filter is the set of all neighborhoods of a point in a topological space.
Any neighborhood basis of a point in (or of a subset of) a topological space is a prefilter. In fact, the definition of a neighborhood base can be equivalently restated as: "a neighborhood base is any prefilter that is equivalent the neighborhood filter."

Neighborhood bases at points are examples of prefilters that are fixed but may or may not be principal.
If $X = \R$ has its usual topology and if $x \in X,$ then any neighborhood filter base $\mathcal$ of $x$ is fixed by $x$ (in fact, it is even true that $\ker \mathcal = \$) but $\mathcal$ is principal since $\ \not\in \mathcal.$
In contrast, a topological space has the discrete topology if and only if the neighborhood filter of every point is a principal filter generated by exactly one point.
This shows that a non–principal filter on an infinite set is not necessarily free.

The neighborhood filter of every point $x$ in topological space $X$ is fixed since its kernel contains $x$ (and possibly other points if, for instance, $X$ is not a T1 space, T₁ space). This is also true of any neighborhood basis at $x.$
For any point $x$ in a T1 space, T₁ space (for example, a Hausdorff space), the kernel of the neighborhood filter of $x$ is equal to the singleton set $\.$

However, it is possible for a neighborhood filter at a point to be principal but discrete (that is, not principal at a point).
A neighborhood basis $\mathcal$ of a point $x$ in a topological space is principal if and only if the kernel of $\mathcal$ is an open set. If in addition the space is T1 space, T₁ then $\ker \mathcal = \$ so that this basis $\mathcal$ is principal if and only if $\$ is an open set.

Generating topologies from filters and prefilters

Suppose $\mathcal \subseteq \wp(X)$ is not empty (and $X \neq \varnothing$). If $\mathcal$ is a filter on $X$ then $\ \cup \mathcal$ is a topology on $X$ but the converse is in general false. This shows that in a sense, filters are topologies. Topologies of the form $\ \cup \mathcal$ where $\mathcal$ is an filter on $X$ are an even more specialized subclass of such topologies; they have the property that proper subset $\varnothing \neq S \subseteq X$ is open or closed, but (unlike the discrete topology) never both. These spaces are, in particular, examples of door spaces.

If $\mathcal$ is a prefilter (resp. filter subbase, –system, proper) on $X$ then the same is true of both $\ \cup \mathcal$ and the set $\mathcal_$ of all possible unions of one or more elements of $\mathcal.$ If $\mathcal$ is closed under finite intersections then the set $\tau_ = \ \cup \mathcal_$ is a topology on $X$ with both $\ \cup \mathcal_ \text \ \cup \mathcal$ being Base (topology), bases for it. If the –system $\mathcal$ covers $X$ then both $\mathcal_ \text \mathcal$ are also bases for $\tau_.$ If $\tau$ is a topology on $X$ then $\tau \setminus \$ is a prefilter (or equivalently, a –system) if and only if it has the finite intersection property (that is, it is a filter subbase), in which case a subset $\mathcal \subseteq \tau$ will be a basis for $\tau$ if and only if $\mathcal \setminus \$ is equivalent to $\tau \setminus \,$ in which case $\mathcal \setminus \$ will be a prefilter.

Topologies on directed sets and net convergence

Let $(I, \leq)$ be a non–empty directed set and let $\operatorname(I) = \left\,$ where $I_ = \.$ Then $\operatorname(I)$ is a prefilter that Cover (topology), covers $I$ and if $I$ is totally ordered then $\operatorname(I)$ is also closed under finite intersections. This particular prefilter $\operatorname(I)$ forms a base for a topology on $I$ in which all sets of the form $I_ = \$ are also open.
The same is true of the topology $\tau_I := \ \cup \operatorname(I) \text I,$ where $\operatorname(I)$ is the filter on $I$ generated by $\operatorname(I).$ With this topology, convergent nets can be viewed as continuous functions in the following way.
Let $(X, \tau)$ be a topological space, let $x \in X,$ let $x_ = \left(x_i\right)_ ~:~ I \to X$ be a net in $X,$ and let $\tau(x) \subseteq \tau$ denote the set of all open neighborhoods of $x.$
If the net $x_$ converges to $x \text (X, \tau)$ then $x_ ~:~ \left(I, \tau_I\right) \to \left(X, \ \cup \tau(x)\right)$ is necessarily continuous although in general, the converse is false (for example, consider if $x_$ is constant and not equal to $x$).
But if in addition to continuity, the preimage under $x_$ of every $N \in \tau(x)$ is not empty, then the net $x_$ will necessarily converge to $x \text (X, \tau).$
In this way, the empty set is all that separates net convergence and continuity.

Another way in which a convergent nets can be viewed as continuous functions is, for any given $x \in X$ and net $x_ = \left(x_i\right)_ ~:~ I \to X,$ to first extend the net to a new net $\hat_ := \left(\hat_i\right)_ : I \cup \ \to X,$ where $\infty \not\in I$ is a new symbol, by defining $\hat_ := x \text \hat_i := x_i$ for every $i \in I.$ If $I \cup \$ is endowed with the topology $\tau_ ~:=~ \wp(I) ~\cup~ \left\ ~=~ \wp(I) ~\cup~ \left(\,\ \,(\cup)\, \operatorname(I)\,\right)$ then $x_ \to x \text X$ (that is, the net $x_$ Convergent net, converges to $x$) if and only if $\hat_ : \left(I \cup \, \tau_\right) \to (X, \tau)$ is a continuous function. Moreover, $I$ is always a Dense set, dense subset of $I \cup \.$

Topological properties and prefilters

Neighborhoods and topologies

The neighborhood filter of a nonempty subset $S \subseteq X$ in a topological space $X$ is equal to the intersection of all neighborhood filters of all points in $S.$
If $S \subseteq X$ then $S$ is open in $X$ if and only if whenever $\mathcal$ is a filter on $X$ and $s \in S,$ then $\mathcal \to s \text X \text S \in \mathcal.$

Suppose $\sigma \text \tau$ are topologies on $X.$
Then $\tau$ is finer than $\sigma$ (that is, $\sigma \subseteq \tau$) if and only if whenever $x \in X \text \mathcal$ is a filter on $X,$ if $\mathcal \to x \text (X, \tau)$ then $\mathcal \to x \text (X, \sigma).$ Consequently, $\sigma = \tau$ if and only if for every filter $\mathcal \text X$ and every $x \in X, \mathcal \to x \text (X, \sigma)$ if and only if $\mathcal \to x \text (X, \tau).$
However, it is possible that $\sigma \neq \tau$ while also for every filter $\mathcal \text X, \mathcal$ converges to point of $X \text (X, \sigma)$ if and only if $\mathcal$ converges to point of $X \text (X, \tau).$

Closure

If $x \in X \text S \subseteq X \text S \neq \varnothing$ then the following are equivalent:

1. $x \in \operatorname_X S$


2. $x$ is a limit point of the prefilter $\;$ that is, $\ \to x \text X.$


3. There exists a prefilter $\mathcal \subseteq \wp(X) \text X$ such that $S \in \mathcal \text \mathcal \to x \text X.$


4. There exists a prefilter $\mathcal \subseteq \wp(S) \text S$ such that $\mathcal \to x \text X.$


5. $x$ is a cluster point of the prefilter $\.$


6. The prefilter $\$ meshes with the neighborhood filter $\mathcal(x).$


7. The prefilter $\$ meshes with some (or equivalently, with every) prefilter of $\mathcal(x).$

The following are equivalent:

1. $x$ is a limit points of $S \text X.$


2. There exists a prefilter $\mathcal \subseteq \wp(S) \text \ \setminus \$ such that $\mathcal \to x \text X.$

Closed sets

If $S \subseteq X$ is not empty then the following are equivalent:

1. $S$ is a closed subset of $X.$


2. If $x \in X \text \mathcal \subseteq \wp(S)$ is a prefilter on $S$ such that $\mathcal \to x \text X,$ then $x \in S.$


3. If $x \in X \text \mathcal \subseteq \wp(S)$ is a prefilter on $S$ such that $x$ is an accumulation points of $\mathcal \text X,$ then $x \in S.$


4. If $x \in X$ is such that the neighborhood filter $\mathcal(x)$ meshes with $\$ then $x \in S.$
   * The proof of this characterization depends the ultrafilter lemma, which depends on the axiom of choice.

Hausdorffness

The following are equivalent:

1. $X$ is a Hausdorff space.


2. Every prefilter on $X$ converges to at most one point in $X.$


3. The above statement but with the word "prefilter" replaced by any one of the following: filter, ultra prefilter, ultrafilter.

Compactness

As discussed Ultrafilter lemma, in this article, the Ultrafilter Lemma is closely related to many important theorems involving compactness.

The following are equivalent:

1. $(X, \tau)$ is a compact space.


2. Every ultrafilter on $X$ converges to at least one point in $X.$
   * That this condition implies compactness can be proven by using only the ultrafilter lemma. That compactness implies this condition can be proven without the ultrafilter lemma (or even the axiom of choice).


3. The above statement but with the word "ultrafilter" replaced by any one of the following: ultra prefilter, filter, prefilter.


4. For every filter $\mathcal \text X$ there exists a filter $\mathcal \text X$ such that $\mathcal \leq \mathcal$ and $\mathcal$ converges to some point of $X.$


5. The above statement but with each instance of the word "filter" replaced by: prefilter.


6. Every filter on $X$ has at least one cluster point in $X.$
   * That this condition is equivalent to compactness can be proven by using only the ultrafilter lemma.


7. The above statement but with the word "prefilter" replaced by any one of the following: prefilter.


8. Alexander subbase theorem: There exists a subbase $\mathcal \text \tau$ such that every cover of $X$ by sets in $\mathcal$ has a finite subcover.
   * That this condition is equivalent to compactness can be proven by using only the ultrafilter lemma.

If $\mathcal$ is the set of all complements of compact subsets of a given topological space $X,$ then $\mathcal$ is a filter on $X$ if and only if $X$ is compact.

Continuity

Let $f : X \to Y$ is a map between topological spaces $(X, \tau) \text (Y, \upsilon).$

Given $x \in X,$ the following are equivalent:

1. $f : X \to Y$ is Continuous function (topology), continuous at $x.$


2. Definition: For every neighborhood $V$ of $f(x) \text Y$ there exists some neighborhood $N$ of $x \text X$ such that $f(N) \subseteq V.$


3. $f(\mathcal(x)) \to f(x) \text Y.$


4. If $\mathcal$ is a filter on $X$ such that $\mathcal \to x \text X$ then $f(\mathcal) \to f(x) \text Y.$


5. The above statement but with the word "filter" replaced by "prefilter".

The following are equivalent:

1. $f : X \to Y$ is continuous.


2. If $x \in X \text \mathcal$ is a prefilter on $X$ such that $\mathcal \to x \text X$ then $f(\mathcal) \to f(x) \text Y.$


3. If $x \in X$ is a limit point of a prefilter $\mathcal \text X$ then $f(x)$ is a limit point of $f(\mathcal) \text Y.$


4. Any one of the above two statements but with the word "prefilter" replaced by any one of the following: filter.

If $\mathcal$ is a prefilter on $X, x \in X$ is a cluster point of $\mathcal, \text f : X \to Y$ is continuous, then $f(x)$ is a cluster point in $Y$ of the prefilter $f(\mathcal).$

A subset $D$ of a topological space $X$ is Dense set, dense in $X$ if and only if for every $x \in X,$ the trace $\mathcal_X(x)\big\vert_D$ of the neighborhood filter $\mathcal_X(x)$ along $D$ does not contain the empty set (in which case it will be a filter on $D$).

Suppose $f : D \to Y$ is a continuous map into a Hausdorff regular space $Y$ and that $D$ is a dense subset of a topological space $X.$ Then $f$ has a continuous extension $F : X \to Y$ if and only if for every $x \in X,$ the prefilter $f\left(\mathcal_X(x)\big\vert_D\right)$ converges to some point in $Y.$ Furthermore, this continuous extension will be unique whenever it exists.

Products

Suppose $X_ := \left(X_i\right)_$ is a non–empty family of non–empty topological spaces and that is a family of prefilters where each $\mathcal_i$ is a prefilter on $X_i.$
Then the product $\mathcal_$ of these prefilters (defined above) is a prefilter on the product space $\prod X_,$ which as usual, is endowed with the product topology.

If $x_ := \left(x_i\right)_ \in \prod X_,$ then $\mathcal_ \to x_ \text \prod X_$ if and only if $\mathcal_i \to x_i \text X_i \text i \in I.$

Suppose $X \text Y$ are topological spaces, $\mathcal$ is a prefilter on $X$ having $x \in X$ as a cluster point, and $\mathcal$ is a prefilter on $Y$ having $y \in Y$ as a cluster point.
Then $(x, y)$ is a cluster point of $\mathcal \times \mathcal$ in the product space $X \times Y.$
However, if $X = Y = \Q$ then there exist sequences $x_ := \left(x_i\right)_^ \subseteq X \text y_ := \left(y_i\right)_^ \subseteq Y$ such that both of these sequences have a cluster point in $\Q$ but the sequence $\left(x_i, y_i\right)_^ \subseteq X \times Y$ does have a cluster point in $X \times Y.$

Example application: The ultrafilter lemma along with the axioms of Zermelo–Fraenkel set theory, ZF imply Tychonoff's theorem for compact Hausdorff spaces:

Let $X_ := \left(X_i\right)_$ be compact Hausdorff space, topological spaces.
Assume that the ultrafilter lemma holds (because of the Hausdorff assumption, this proof does need the full strength of the axiom of choice; the ultrafilter lemma suffices).
Let $X := \prod X_$ be given the product topology (which makes $X$ a Hausdorff space) and for every $i,$ let $\Pr_i : X \to X_i$ denote this product's projections.
If $X = \varnothing$ then $X$ is compact and the proof is complete so assume $X \neq \varnothing.$
Despite the fact that $X \neq \varnothing,$ because the axiom of choice is not assumed, the projection maps $\Pr_i : X \to X_i$ are not guaranteed to be surjective.

Let $\mathcal$ be an ultrafilter on $X$ and for every $i,$ let $\mathcal_i$ denote the ultrafilter on $X_i$ generated by the ultra prefilter $\Pr_i(\mathcal).$
Because $X_i$ is compact and Hausdorff, the ultrafilter $\mathcal_i$ converges to a unique limit point $x_i \in X_i$ (because of $x_i$'s uniqueness, this definition does not require the axiom of choice).
Let $x := \left(x_i\right)_$ where $x$ satisfies $\Pr_i(x) = x_i$ for every $i.$
The characterization of convergence in the product topology that was given above implies that $\mathcal \to x \text X.$
Thus every ultrafilter on $X$ converges to some point of $X,$ which implies that $X$ is compact (recall that this implication's proof only required the ultrafilter lemma). $\blacksquare$

Examples of applications of prefilters

Uniformities and Cauchy prefilters

A uniform space is a set $X$ equipped with a filter on $X \times X$ that has certain properties. A or is a prefilter on $X \times X$ whose upward closure is a uniform space.
A prefilter $\mathcal$ on a uniform space $X$ with uniformity $\mathcal$ is called a if for every entourage $N \in \mathcal,$ there exists some $B \in \mathcal$ that is , which means that $B \times B \subseteq N.$
A is a minimal element (with respect to $\,\leq\,$ or equivalently, to $\,\subseteq$) of the set of all Cauchy filters on $X.$
Examples of minimal Cauchy filters include the neighborhood filter $\mathcal_X(x)$ of any point $x \in X.$
Every convergent filter on a uniform space is Cauchy. Moreover, every cluster point of a Cauchy filter is a limit point.

A uniform space $(X, \mathcal)$ is called (resp. ) if every Cauchy prefilter (resp. every elementary Cauchy prefilter) on $X$ converges to at least one point of $X$ (replacing all instance of the word "prefilter" with "filter" results in equivalent statement).
Every compact uniform space is complete because any Cauchy filter has a cluster point (by compactness), which is necessarily also a limit point (since the filter is Cauchy).

Uniform spaces were the result of attempts to generalize notions such as "uniform continuity" and "uniform convergence" that are present in metric spaces. Every topological vector space, and more generally, every topological group can be made into a uniform space in a canonical way. Every uniformity also generates a canonical induced topology. Filters and prefilters play an important role in the theory of uniform spaces. For example, the completion of a Hausdorff uniform space (even if it is not Metrizable space, metrizable) is typically constructed by using minimal Cauchy filters.
Nets are less ideal for this construction because their domains are extremely varied (for example, the class of all Cauchy nets is not a set); sequences cannot be used in the general case because the topology might not be metrizable, First-countable space, first–countable, or even Sequential space, sequential.
The set of all on a Hausdorff topological vector space (TVS) $X$ can made into a vector space and topologized in such a way that it becomes a Complete topological vector space, completion of $X$ (with the assignment $x \mapsto \mathcal_X(x)$ becoming a TVS-embedding, linear topological embedding that identifies $X$ as a dense vector subspace of this completion).

More generally, a Cauchy space, is a pair $(X, \mathfrak)$ consisting of a set $X$ together a family $\mathfrak \subseteq \wp(\wp(X))$ of (proper) filters, whose members are declared to be "", having all of the following properties:
# For each $x \in X,$ the discrete ultrafilter at $x$ is an element of $\mathfrak.$
# If $F \in \mathfrak$ is a subset of a proper filter $G,$ then $G \in \mathfrak.$
# If $F, G \in \mathfrak$ and if each member of $F$ intersects each member of $G,$ then $F \cap G \in \mathfrak.$
The set of all Cauchy filters on a uniform space forms a Cauchy space. Every Cauchy space is also a convergence space.
A map $f : X \to Y$ between two Cauchy spaces is called if the image of every Cauchy filter in $X$ is a Cauchy filter in $Y.$
Unlike the category of topological spaces, the Category (mathematics), category of Cauchy spaces and Cauchy continuous maps is Cartesian closed, and contains the category of proximity spaces.

Convergence of nets of sets

There is often a personal preference of nets over filters or filters over nets. This example shows that the choice between nets and filters is not a dichotomy by combining them together.

A or a refers to a net in the power set $\wp(X)$ of $X;$ that is, a net of sets in $X$ is a function from a non–empty directed set into $\wp(X).$
However, a "net in $X$" will always refer to a net valued in $X$ and never to a net valued in $\wp(X)$ although for emphasis or contrast, a net in $X$ may also be referred to as a .
A net $S_ = \left(S_i\right)_$ of sets in $X$ is called a (resp. , , , etc.) if every $S_i$ has this property.
Similarly, $S_$ is called (resp. , , , etc.) if there is some index $i$ such that this is true of $S_j$ for every index $j \geq i.$

The following definition generalizes that of a #Tail of a net, tail of a net of points.

Suppose $S_ = \left(S_i\right)_$ is a net of sets in $X.$ Define for every index $i$ the to be the set
$S_ := \bigcup_ S_j$
and define the or generated by $S_$ to be the family
$\operatorname\left(S_\right) := \left\.$
The family $\operatorname\left(S_\right)$ is a prefilter if and only if it does not contain the empty set, which is equivalent to $S_$ not being eventually empty; in this case the upward closure in $X$ of this prefilter of tails is called the or in $X$ generated by $S_.$
A net $y_$ (of sets or points) is eventually contained in a set $C$ if and only if $\ \leq \operatorname\left(y_\right);$ so $S_$ is eventually empty if and only if $\ \leq \operatorname\left(S_\right).$

Nets of sets arise naturally when pulling back nets in a function's codomain.
If $f : X \to Y$ is a map and $y_ = \left(y_i\right)_$ is a net of sets (or points) then let $f^\left(y_\right) := \left(f^\left(y_i\right)\right)_$ and $f\left(y_\right) := \left(f\left(y_i\right)\right)_;$ that is, $f^\left(y_\right)$ denotes the net of sets $I \to \wp(X)$ defined by $i \mapsto f^\left(y_i\right).$
The tail of $f\left(y_\right)$ starting at an index $i$ is equal to $f\left(y_\right)$ and similarly, the tail of $f^\left(y_\right)$ starting at $i$ is $f^\left(y_\right).$
Consequently, $\operatorname\left(f\left(y_\right)\right) = f\left(\operatorname\left(y_\right)\right)$ where this family is a prefilter if and only if $\operatorname\left(y_\right)$ is a prefilter; similarly, $\operatorname\left(f^\left(y_\right)\right) = f^\left(\operatorname\left(y_\right)\right).$
One useful consequence of this definition is that $f^\left(y_\right)$ is a prefilter if and only if $y_$ (or for points, ) $f(X),$ meaning that for every index $i,$ there is some $j \geq i$ such that $y_j \cap f(X) \neq \varnothing$ (where this intersection means $y_j \in f(X)$ if $y_j$ is a point instead of a set).
In particular, $y_$ (meaning that $y_ \subseteq f(X)$ for some $i$) is a necessary condition for $\operatorname\left(f^\left(y_\right)\right)$ to be a prefilter. So even if a net $\left(y_i\right)_$ of points in $Y$ cannot be pulled back by $f$ to a net $\left(x_i\right)_$ of in $X$ (say because it is not entirely/eventually in the image of $f$), it is nevertheless still possible to talk about the net of $f^\left(y_\right)$ and its properties (such as convergence or clustering).

Convergence and clustering

Consideration of the following bijective correspondence leads naturally to the definitions of convergence and clustering for a net of sets, which are defined analogously to the original definitions given for a net of points.

(Nets of points $\leftrightarrow$ Nets of singleton sets): Every net $x_ = \left(x_i\right)_$ of points in $X$ can be uniquely associated with the $\left(\left\\right)_$ and conversely, every net of singleton sets in $X$ is uniquely associated with a (defined in the obvious way).
The tail of $\left(\left\\right)_$ starting at an index $i$ is equal to that of $x_$ (that is, to $x_$); consequently, $\operatorname\left(x_\right) = \operatorname\left(\left(\left\\right)_\right).$
This makes it apparent that the following definition of "#Convergent net of sets, convergence of a net of sets" in $X$ is indeed a generalization of #Convergent net, the original definition of "convergence of a net of points" in $X$ (because $x_ \to R$ if and only if $\left(\left\\right)_ \to R$); similarly, a net of points clusters at a given point or subset $x$ (according to #Cluster point of a net, the original definition) if and only if its associated net of singleton sets clusters at $x$ (according to #Cluster point of a net of sets, the definition below).

A net of sets $S_$ is said to to a given point or subset $x$ of $X,$ written $S_ \to x \text (X, \tau),$ if $\,\operatorname\left(S_\right) \to x \text (X, \tau),$ which recall was defined to mean that $\mathcal_(x) \leq \operatorname\left(S_\right).$ Explicitly, this happens if and only if for every neighborhood $U$ of $x,$ there exists some index $i$ such that $S_ \subseteq U.$ Similarly, $S_$ is said to a given point or subset $x$ of $X$ if $\operatorname\left(S_\right)$ meshes with $\mathcal_(x)$ (written $\mathcal_(x) \;\#\; \operatorname\left(S_\right)$); explicitly, this means that $\varnothing \neq N \cap S_$ for every index $i \in I$ and neighborhood $N$ of $x.$

Every net of sets that is eventually empty converges to every point/subset. However, a net of sets converges to $\varnothing$ if and only if it is eventually empty. No net of sets clusters at $\varnothing.$ If a net of sets converges to $x$ then it will cluster at $x$ if and only if it is not eventually empty (which implies $x \neq \varnothing$).

If $x_ = \left(x_i\right)_$ is a net in $X$ then $\left(x_\right)_$ is a net of sets in $X$ and for any point or subset $x$ of $X,$ $x_$ converges to (respectively, clusters at) $x$ in $X$ if and only if this is true of $\left(x_\right)_.$ This statement remains true if $x_$ is instead a net of sets.
If $f : X \to Y$ is a map and $y_ = \left(y_i\right)_$ is a net (of points or of sets) then $f^\left(y_\right)$ converges to (respectively, clusters at) some given point or subset of $X$ if and only if every neighborhood of it contains (respectively, intersects) some set of the form $f^\left(y_\right).$
Moreover, the net $f^\left(y_\right)$ converges in $X$ to some given point or subset if and only if this is true of $f^\left(\operatorname\left(y_\right)\right).$

Prefilters and nets of sets

If $\mathcal$ is a prefilter on $X$ then $(\mathcal, \supseteq)$ is a (partially ordered) directed set, so that the identity map $\operatorname_ : (\mathcal, \supseteq) \to \mathcal$ is a net of sets in $X.$
Every prefilter can be canonically identified with this net of sets (that is, with its identity map when the prefilter/domain is directed by $\supseteq$).
Thus it is significantly easier to canonically associate every prefilter with a net of than with a net of (as was #Prefilters to nets, done above), and because the relationship is also much simpler, it is easier utilize.
For instance, it is readily seen that the tail of the net $\operatorname_$ starting at a given index $B \in B$ is equal to $B$ (in other words, the tail starting at an index is the index itself) so that $\operatorname\left(\operatorname_\right) = \mathcal$ (that is, this net's tails are its indices) and the prefilter $\mathcal$ converges to (respectively, clusters at) a given point or subset if and only if the same is true of its canonical net of sets $\operatorname_.$
In particular, information (including intuition and visualizations) about how or why a prefilter $\mathcal$ converges to (or doesn't converge to, or clusters at, etc.) a point or set can almost immediately be obtained from information about how/why the net of sets $\operatorname_$ does the same (or vice versa).

Applications

Some applications are now given showing how nets of sets can be used to characterize various properties.
In the statements below, unless indicated otherwise, $y$ and the net $y_$ are in $Y$ (not sets) and the map $f : X \to Y$ is not necessarily surjective.

• A map $f : X \to Y$ is Closed map, closed (meaning it sends closed sets to closed subset of $Y$) if and only if whenever $y_ \to y \text Y$ then $f^\left(y_\right) \to f^(y) \text X.$
  * In comparison, $f : X \to Y$ is Continuity (topology), continuous if and only if whenever $x_ \to x \text X$ then $f\left(x_\right) \to f(x) \text Y.$
  This characterization remains true if $y_ \text y$ are allowed to be sets (instead of restricted to being points) such that $y_ \to y \text Y.$

  Assume $f : X \to Y$ is closed and $y_ \to y \text Y.$
  If $y \not\in f(X)$ then $y$ is in the open set $Y \setminus f(X)$ so that $y_ \to y \text Y$ implies that $f^\left(y_\right)$ is eventually empty and thus that $f^\left(y_\right) \to \varnothing = f^(y)$ in $X.$
  So assume $f^(y) \neq \varnothing$ and let $U$ be an open neighborhood of $f^(y)$ in $X.$
  It remains to show that $f^\left(y_\right) \subseteq U$ for some index $i.$
  Since $f$ is closed, $Y \setminus f(X \setminus U)$ is an open neighborhood of $y$ in $Y$ so there must exists some index $i$ such that $y_ \subseteq Y \setminus f(X \setminus U).$
  This implies $f^\left(y_\right) \subseteq f^(Y \setminus f(X \setminus U))$ where the right hand side is a subset of $U,$ as desired.

  For the converse, assume that $y_ \to y \text Y$ implies $f^\left(y_\right) \to f^(y) \text X.$ Let $C \subseteq X$ be closed and assume it is not empty. Let $y_ = \left(y_i\right)_$ be a net in $f(C)$ (meaning $y_i \in f(C)$ for all $i$) and let $y \in Y$ be such that $y_ \to y \text Y.$ It remains to show that $y \in f(C).$ The hypotheses guarantee that $f^\left(y_\right) \to f^(y) \text X.$ The fact that every fiber $f^\left(y_i\right)$ is not empty and that these fibers converge to $f^(y)$ imply that $f^(y) \neq \varnothing.$
  Since $X \setminus C$ is open, were it true that $f^(y) \subseteq X \setminus C$ then there would exist some index $i$ such that $f^\left(y_\right) \subseteq X \setminus C,$ which is impossible since $y_j \in f(C)$ for every index $j.$
  Thus $f^(y) \not\subseteq X \setminus C$ so there is some $c \in C \cap f^(y),$ which proves that $y = f(c) \in f(C).$ $\blacksquare$

• A map $f : X \to Y$ is Open map, open (meaning it sends open sets to open subset of $Y$) if and only if whenever $x \in X$ and $y_$ is a net that clusters at $f(x) \text Y,$ then $f^\left(y_\right)$ clusters at $x \text X.$
  * In comparison, $f : X \to Y$ is Continuity (topology), continuous if and only if whenever $x_$ is a net that Net (mathematics)#cluster point, clusters at a point $x \text X,$ then $f\left(x_\right)$ clusters at $f(x) \text Y.$
  This characterization remains true if $y_ \text x$ are allowed to be sets.

  For the non-trivial direction, suppose that $f : X \to Y$ is not an open map. Pick an open subset $U \subseteq X$ such that $f(U)$ is not open in $Y,$ where non-openness means that there is some point $y \in f(U)$ such that $f(U)$ is not a neighborhood of $y$ in $Y.$
  Explicitly, this means that $N \not\subseteq f(U)$ for every neighborhood $N$ of $y$ in $Y,$ which guarantees the existence of some $y_N \in N \setminus f(U).$
  Let $I := \mathcal_Y(y)$ denote the neighborhood filter of $y$ in $Y$ and direct it by $\,\supseteq\,$ to make $y_ := \left(y_N\right)_$ into a net that converges to $y$ in $Y,$ which implies that $y_$ clusters at $y$ in $Y.$
  Because $y \in f(U),$ there is some $x \in U \cap f^(y).$
  But $f^\left(y_\right)$ does not clusters at $x$ since $f^\left(y_N\right) \cap U = \varnothing$ for every $N \in I.$
  $\blacksquare$

• A map $f : X \to Y$ is Open map, open if and only if whenever $y_ \to y \text Y$ then any closed subset of $X$ that contains $f^\left(y_\right)$A set $C$ $f^\left(y_\right)$ means that $f^\left(y_\right) \subseteq C$ for every index $i.$ will necessarily also contain $f^(y).$
  * In comparison, by the Continuity (topology)#Closure operator and interior operator definitions, closure characterization of continuity, $f : X \to Y$ is continuous if and only if whenever $x_ \to x \text X$ then any closed subset of $Y$ that contains $f\left(x_\right)$ will necessarily also contain $f(x).$
  This characterization remains true if $y_$ is allowed to be a net of sets that is not eventually empty (instead of being a net of points) while $y \in Y$ continues to be a point (such that $y_ \to y \text Y$); the same is true of the quotient map characterization below.

  If $R \subseteq X$ is any subset then it is readily verified that $f(X \setminus R) = Y \setminus \left\.$
  This implies that a map $f : X \to Y$ is open if and only if whenever $C \subseteq X$ is closed in $X$ then $\left\$ is closed in $Y.$
  This characterization of "open map" combined with the convergent net characterization of closed sets produces the desired conclusion: $f : X \to Y$ is open if and only if whenever $y_ \to y \text Y$ and $C \subseteq X$ is a closed subset of $X$ that contains $f^\left(y_\right),$ then necessarily $f^(y) \subseteq C.$ $\blacksquare$

• A continuous surjection $f : X \to Y$ is a quotient map if and only if whenever $y_ \to y \text Y$ then any closed subset of $X$ that contains $f^\left(y_\right)$ will necessarily also contain $f^(y).$ (A set $C \subseteq X$ is saturated if $f^(f(C)) = C.$)

• A subset $C \subseteq X$ is closed in $X$ if and only if for every point $x \in X$ and every net $c_$ of subsets of $C$ that is not eventually empty, if $c_ \to x \text X$ then $x \in C.$
  
• A map $f : X \to Y$ is continuous if and only if whenever $S$ and $S_$ are sets or points in $X$ such that $S_ \to S \text X,$ then $f\left(S_\right) \to f(S) \text Y.$

  The proof is essentially identical to the usual proof involving only nets of points. One direction (that whose conclusion is that $f$ is continuous) only requires consideration of nets of points and so it is omitted. So suppose that the map is continuous and that $S_ \to S \text X.$ Let $V \subseteq Y$ be an open neighborhood of $f(S)$ in $Y.$ Then $f^(V)$ is an open neighborhood of $S$ in $X$ so there exists some index $i$ such that $S_ \subseteq f^(V).$ Thus $f\left(S_\right) \subseteq V,$ as desired.
  $\blacksquare$



• A map $f : X \to Y$ is continuous if and only if whenever $S_$ is a net of sets or points in $X$ that clusters at (respectively, converges to) some given point or subset $S$ of $X,$ then $f\left(S_\right) := \left(f\left(S_i\right)\right)_$ clusters at (respectively, converges to) $f(S)$ in $Y.$
  

Topologizing the set of prefilters

Starting with nothing more than a set $X,$ it is possible to topologize the set
$\mathbb := \operatorname(X)$
of all filter bases on $X$ with the , which is named after Marshall Harvey Stone.

To reduce confusion, this article will adhere to the following notational conventions:

• Lower case letters for elements $x \in X.$


• Upper case letters for subsets $S \subseteq X.$


• Upper case calligraphy letters for subsets $\mathcal \subseteq \wp(X)$ (or equivalently, for elements $\mathcal \in \wp(\wp(X)),$ such as prefilters).


• Upper case double–struck letters for subsets $\mathbb \subseteq \wp(\wp(X)).$

For every $S \subseteq X,$ let
$\mathbb(S) := \left\$
where $\mathbb(X) = \mathbb \text \mathbb(\varnothing) = \varnothing.$As a side note, had the definitions of "filter" and "prefilter" not required propriety then the degenerate dual ideal $\wp(X)$ would have been a prefilter on $X$ so that in particular, $\mathbb(\varnothing) = \ \neq \varnothing$ with $\wp(X) \in \mathbb(S) \text S \subseteq X.$ These sets will be the basic open subsets of the Stone topology.
If $R \subseteq S \subseteq X$ then
$\left\ ~\subseteq~ \left\.$

From this inclusion, it is possible to deduce all of the subset inclusions displayed below with the exception of $\mathbb(R \cap S) ~\supseteq~ \mathbb(R) \cap \mathbb(S).$This is because the inclusion $\mathbb(R \cap S) ~\supseteq~ \mathbb(R) \cap \mathbb(S)$ is the only one in the sequence below whose proof uses the defining assumption that $\mathbb(S) \subseteq \mathbb.$
For all $R \subseteq S \subseteq X,$
$\mathbb(R \cap S) ~=~ \mathbb(R) \cap \mathbb(S) ~\subseteq~ \mathbb(R) \cup \mathbb(S) ~\subseteq~ \mathbb(R \cup S)$
where in particular, the equality $\mathbb(R \cap S) = \mathbb(R) \cap \mathbb(S)$ shows that the family $\$ is a Pi-system, $\pi$–system that forms a Basis (topology), basis for a topology on $\mathbb$ called the . It is henceforth assumed that $\mathbb$ carries this topology and that any subset of $\mathbb$ carries the induced subspace topology.

In contrast to most other general constructions of topologies (for example, the Product topology, product, Quotient topology, quotient, Subspace topology, subspace topologies, etc.), this topology on $\mathbb$ was defined with using anything other than the set $X;$ there were preexisting Mathematical structure, structures or assumptions on $X$ so this topology is completely independent of everything other than $X$ (and its subsets).

The following criteria can be used for checking for Point of closure, points of closure and neighborhoods.
If $\mathbb \subseteq \mathbb \text \mathcal \in \mathbb$ then:

• : $\ \mathcal$ belongs to the closure of $\mathbb \text \mathbb$ if and only if $\mathcal \subseteq \bigcup_ \mathcal^.$


• : $\ \mathbb$ is a neighborhood of $\mathcal \text \mathbb$ if and only if there exists some $F \in \mathcal$ such that $\mathbb(F) = \left\ \subseteq \mathbb$ (that is, such that for all $\mathcal \in \mathbb, \text F \in \mathcal^ \text \mathcal \in \mathbb$).

It will be henceforth assumed that $X \neq \varnothing$ because otherwise $\mathbb = \varnothing$ and the topology is $\,$ which is uninteresting.

Subspace of ultrafilters

The set of ultrafilters on $X$ (with the subspace topology) is a Stone space, meaning that it is compact, Hausdorff, and Totally disconnected space, totally disconnected.
If $X$ has the discrete topology then the map $\beta : X \to \operatorname(X),$ defined by sending $x \in X$ to the principal ultrafilter at $x,$ is a topological embedding whose image is a dense subset of $\operatorname(X)$ (see the article Stone–Čech compactification#Construction using ultrafilters, Stone–Čech compactification for more details).

Relationships between topologies on $X$ and the Stone topology on $\mathbb$

Every $\tau \in \operatorname(X)$ induces a canonical map $\mathcal_ : X \to \operatorname(X)$ defined by $x \mapsto \mathcal_(x),$ which sends $x \in X$ to the neighborhood filter of $x \text (X, \tau).$
The map $\mathcal_ : X \to \operatorname(X)$ is injective if and only if $\tau \text T_0$ (that is, a Kolmogorov space) and moreover, if $\tau, \sigma \in \operatorname(X)$ then $\tau = \sigma \text \mathcal_ = \mathcal_.$
Thus every $\tau \in \operatorname(X)$ can be identified with the canonical map $\mathcal_,$ which allows $\operatorname(X)$ to be canonically identified as a subset of $\operatorname(X; \mathbb)$ (as a side note, it is now possible to place on $\operatorname(X; \mathbb),$ and thus also on $\operatorname(X),$ the topology of pointwise convergence on $X$ so that it now makes sense to talk about things such as sequences of topologies on $X$ converging pointwise).
For every $\tau \in \operatorname(X),$ the surjection $\mathcal_ : (X, \tau) \to \operatorname \mathcal_$ is continuous, Open and closed maps, closed, and open.
In particular, for every $T_0$ topology $\tau \text X,$ the map $\mathcal_ : (X, \tau) \to \mathbb$ is a topological embedding.

In addition, if $\mathfrak : X \to \operatorname(X)$ is a map such that $x \in \ker \mathfrak(x) := \bigcap_ F \text x \in X$ (which is true of $\mathfrak := \mathcal_,$ for instance), then for every $x \in X \text F \in \mathfrak(x),$ the set $\mathfrak(F) := \$ is a neighborhood (in the subspace topology) of $\mathfrak(x) \text \operatorname \mathfrak.$

See also

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Notes

Proofs

Citations

References

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* (Provides an introductory review of filters in topology and in metric spaces.)
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{{Topology

Filters
General topology