# Ultrafilter (set theory)

(Redirected from Ultrafilter lemma)

In the mathematical field of set theory, an ultrafilter is a maximal proper filter: it is a filter ${\displaystyle U}$ on a given non-empty set ${\displaystyle X}$ which is a certain type of non-empty family of subsets of ${\displaystyle X,}$ that is not equal to the power set ${\displaystyle \wp (X)}$ of ${\displaystyle X}$ (such filters are called proper) and that is also "maximal" in that there does not exist any other proper filter on ${\displaystyle X}$ that contains it as a proper subset. Said differently, a proper filter ${\displaystyle U}$ is called an ultrafilter if there exists exactly one proper filter that contains it as a subset, that proper filter (necessarily) being ${\displaystyle U}$ itself.

The powerset lattice of the set {1,2,3,4}, with the upper set ↑{1,4} colored dark green. It is a principal filter, but not an ultrafilter, as it can be extended to the larger nontrivial filter ↑{1}, by including also the light green elements. Since ↑{1} cannot be extended any further, it is an ultrafilter.

More formally, an ultrafilter ${\displaystyle U}$ on ${\displaystyle X}$ is a proper filter that is also a maximal filter on ${\displaystyle X}$ with respect to set inclusion, meaning that there does not exist any proper filter on ${\displaystyle X}$ that contains ${\displaystyle U}$ as a proper subset. Ultrafilters on sets are an important special instance of ultrafilters on partially ordered sets, where the partially ordered set consists of the power set ${\displaystyle \wp (X)}$ and the partial order is subset inclusion ${\displaystyle \,\subseteq .}$

Ultrafilters have many applications in set theory, model theory, and topology.[1]:186

## Definitions

Given an arbitrary set ${\displaystyle X,}$  an ultrafilter on ${\displaystyle X}$  is a non-empty family ${\displaystyle U}$  of subsets of ${\displaystyle X}$  such that:

1. Proper or non-degenerate: The empty set is not an element of ${\displaystyle U.}$
2. Upward closed in ${\displaystyle X}$ : If ${\displaystyle A\in U}$  and if ${\displaystyle B\subseteq X}$  is any superset of ${\displaystyle A}$  (that is, if ${\displaystyle A\subseteq B\subseteq X}$ ) then ${\displaystyle B\in U.}$
3. π−system: If ${\displaystyle A}$  and ${\displaystyle B}$  are elements of ${\displaystyle U}$  then so is their intersection ${\displaystyle A\cap B.}$
4. If ${\displaystyle A\subseteq X}$  then either[note 1] ${\displaystyle A}$  or its relative complement ${\displaystyle X\setminus A}$  is an element of ${\displaystyle U.}$

Properties (1), (2), and (3) are the defining properties of a filter on ${\displaystyle X.}$  Some authors do not include non-degeneracy (which is property (1) above) in their definition of "filter". However, the definition of "ultrafilter" (and also of "prefilter" and "filter subbase") always includes non-degeneracy as a defining condition. This article requires that all filters be proper although a filter might be described as "proper" for emphasis.

For a filter ${\displaystyle F}$  that is not an ultrafilter, one would say ${\displaystyle m(A)=1}$  if ${\displaystyle A\in F}$  and ${\displaystyle m(A)=0}$  if ${\displaystyle X\setminus A\in F,}$  leaving ${\displaystyle m}$  undefined elsewhere.[citation needed][clarification needed]

A filter subbase is a non-empty family of sets that has the finite intersection property (i.e. all finite intersections are non-empty). Equivalently, a filter subbase is a non-empty family of sets that is contained in some (proper) filter. The smallest (relative to ${\displaystyle \subseteq }$ ) filter containing a given filter subbase is said to be generated by the filter subbase.

The upward closure in ${\displaystyle X}$  of a family of sets ${\displaystyle P}$  is the set

${\displaystyle P^{\uparrow X}:=\left\{S:A\subseteq S\subseteq X{\text{ for some }}A\in P\right\}.}$

A prefilter or filter base is a non-empty and proper (i.e. ${\displaystyle \varnothing \not \in P}$ ) family of sets ${\displaystyle P}$  that is downward directed, which means that if ${\displaystyle B,C\in P}$  then there exists some ${\displaystyle A\in P}$  such that ${\displaystyle A\subseteq B\cap C.}$  Equivalently, a prefilter is any family of sets ${\displaystyle P}$  whose upward closure ${\displaystyle P^{\uparrow X}}$  is a filter, in which case this filter is called the filter generated by ${\displaystyle P}$  and ${\displaystyle P}$  is said to be a filter base for ${\displaystyle P^{\uparrow X}.}$

The dual in ${\displaystyle X}$ [2] of a family of sets ${\displaystyle P}$  is the set ${\displaystyle X\setminus P:=\left\{X\setminus B:B\in P\right\}.}$

## Generalization to ultra prefilters

A family ${\displaystyle U\neq \varnothing }$  of subsets of ${\displaystyle X}$  is called ultra if ${\displaystyle \varnothing \not \in U}$  and any of the following equivalent conditions are satisfied:[2][3]

1. For every set ${\displaystyle S\subseteq X}$  there exists some set ${\displaystyle B\in U}$  such that ${\displaystyle B\subseteq S}$  or ${\displaystyle B\subseteq X\setminus S}$  (or equivalently, such that ${\displaystyle B\cap S}$  equals ${\displaystyle B}$  or ${\displaystyle \varnothing }$ ).
2. For every set ${\displaystyle S\subseteq \bigcup _{B\in U}B}$  there exists some set ${\displaystyle B\in U}$ } such that ${\displaystyle B\cap S}$  equals ${\displaystyle B}$  or ${\displaystyle \varnothing .}$
• Here, ${\displaystyle \bigcup _{B\in U}B}$  is defined to be the union of all sets in ${\displaystyle U.}$
• This characterization of "${\displaystyle U}$  is ultra" does not depend on the set ${\displaystyle X,}$  so mentioning the set ${\displaystyle X}$  is optional when using the term "ultra."
3. For every set ${\displaystyle S}$  (not necessarily even a subset of ${\displaystyle X}$ ) there exists some set ${\displaystyle B\in U}$  such that ${\displaystyle B\cap S}$  equals ${\displaystyle B}$  or ${\displaystyle \varnothing .}$
• If ${\displaystyle U}$  satisfies this condition then so does every superset ${\displaystyle V\supseteq U.}$  In particular, a set ${\displaystyle V}$  is ultra if and only if ${\displaystyle \varnothing \not \in V}$  and ${\displaystyle V}$  contains as a subset some ultra family of sets.

A filter subbase that is ultra is necessarily a prefilter.[proof 1]

The ultra property can now be used to define both ultrafilters and ultra prefilters:

An ultra prefilter[2][3] is a prefilter that is ultra. Equivalently, it is a filter subbase that is ultra.
An ultrafilter[2][3] on ${\displaystyle X}$  is a (proper) filter on ${\displaystyle X}$  that is ultra. Equivalently, it is any filter on ${\displaystyle X}$  that is generated by an ultra prefilter.
Interpretation as large sets

The elements of a proper filter ${\displaystyle F}$  on ${\displaystyle X}$  may be thought of as being "large sets (relative to ${\displaystyle F}$ )" and the complements in ${\displaystyle X}$  of a large sets can be thought of as being "small" sets[4] (the "small sets" are exactly the elements in the ideal ${\displaystyle X\setminus F}$ ). In general, there may be subsets of ${\displaystyle X}$  that are neither large nor small, or possibly simultaneously large and small. A dual ideal is a filter (i.e. proper) if there is no set that is both large and small, or equivalently, if the ${\displaystyle \varnothing }$  is not large.[4] A filter is ultra if and only if every subset of ${\displaystyle X}$  is either large or else small. With this terminology, the defining properties of a filter can be restarted as: (1) any superset of a large set is large set, (2) the intersection of any two (or finitely many) large sets is large, (3) ${\displaystyle X}$  is a large set (i.e. ${\displaystyle F\neq \varnothing }$ ), (4) the empty set is not large. Different dual ideals give different notions of "large" sets.

Another way of looking at ultrafilters on a power set ${\displaystyle \wp (X)}$  is as follows: for a given ultrafilter ${\displaystyle U}$  define a function ${\displaystyle m}$  on ${\displaystyle \wp (X)}$  by setting ${\displaystyle m(A)=1}$  if ${\displaystyle A}$  is an element of ${\displaystyle U}$  and ${\displaystyle m(A)=0}$  otherwise. Such a function is called a 2-valued morphism. Then ${\displaystyle m}$  is finitely additive, and hence a content on ${\displaystyle \wp (X),}$  and every property of elements of ${\displaystyle X}$  is either true almost everywhere or false almost everywhere. However, ${\displaystyle m}$  is usually not countably additive, and hence does not define a measure in the usual sense.

Ultra prefilters as maximal prefilters

To characterize ultra prefilters in terms of "maximality," the following relation is needed.

Given two families of sets ${\displaystyle M}$  and ${\displaystyle N,}$  the family ${\displaystyle M}$  is said to be coarser[5][6] than ${\displaystyle N,}$  and ${\displaystyle N}$  is finer than and subordinate to ${\displaystyle M,}$  written ${\displaystyle M\leq N}$  or NM, if for every ${\displaystyle X\in M,}$  there is some ${\displaystyle F\in N}$  such that ${\displaystyle F\subseteq C.}$  The families ${\displaystyle M}$  and ${\displaystyle N}$  are called equivalent if ${\displaystyle M\leq N}$  and ${\displaystyle N\leq M.}$  The families ${\displaystyle M}$  and ${\displaystyle N}$  are comparable if one of these sets is finer than the other.[5]

The subordination relationship, i.e. ${\displaystyle \,\geq ,\,}$  is a preorder so the above definition of "equivalent" does form an equivalence relation. If ${\displaystyle M\subseteq N}$  then ${\displaystyle M\leq N}$  but the converse does not hold in general. However, if ${\displaystyle N}$  is upward closed, such as a filter, then ${\displaystyle M\leq N}$  if and only if ${\displaystyle M\subseteq N.}$  Every prefilter is equivalent to the filter that it generates. This shows that it is possible for filters to be equivalent to sets that are not filters.

If two families of sets ${\displaystyle M}$  and ${\displaystyle N}$  are equivalent then either both ${\displaystyle M}$  and ${\displaystyle N}$  are ultra (resp. prefilters, filter subbases) or otherwise neither one of them is ultra (resp. a prefilter, a filter subbase). In particular, if a filter subbase is not also a prefilter, then it is not equivalent to the filter or prefilter that it generates. If ${\displaystyle M}$  and ${\displaystyle N}$  are both filters on ${\displaystyle X}$  then ${\displaystyle M}$  and ${\displaystyle N}$  are equivalent if and only if ${\displaystyle M=N.}$  If a proper filter (resp. ultrafilter) is equivalent to a family of sets ${\displaystyle M}$  then ${\displaystyle M}$  is necessarily a prefilter (resp. ultra prefilter). Using the following characterization, it is possible to define prefilters (resp. ultra prefilters) using only the concept of filters (resp. ultrafilters) and subordination:

An arbitrary family of sets is a prefilter if and only it is equivalent to a (proper) filter.
An arbitrary family of sets is an ultra prefilter if and only it is equivalent to an ultrafilter.
A maximal prefilter on ${\displaystyle X}$ [2][3] is a prefilter ${\displaystyle U\subseteq \wp (X)}$  that satisfies any of the following equivalent conditions:
1. ${\displaystyle U}$  is ultra.
2. ${\displaystyle U}$  is maximal on ${\displaystyle \operatorname {Prefilters} (X)}$  with respect to ${\displaystyle \,\leq ,}$  meaning that if ${\displaystyle P\in \operatorname {Prefilters} (X)}$  satisfies ${\displaystyle U\leq P}$  then ${\displaystyle P\leq U.}$ [3]
3. There is no prefilter properly subordinate to ${\displaystyle U.}$ [3]
4. If a (proper) filter ${\displaystyle F}$  on ${\displaystyle X}$  satisfies ${\displaystyle U\leq F}$  then ${\displaystyle F\leq U.}$
5. The filter on ${\displaystyle X}$  generated by ${\displaystyle U}$  is ultra.

## Characterizations

There are no ultrafilters on ${\displaystyle \wp (\varnothing ),}$  where ${\displaystyle \varnothing }$  is the empty set, so it is henceforth assumed that ${\displaystyle X\neq \varnothing .}$

A filter subbase ${\displaystyle U}$  on ${\displaystyle X}$  is an ultrafilter on ${\displaystyle X}$  if and only if any of the following equivalent conditions hold:[2][3]

1. for any ${\displaystyle S\subseteq X,}$  either ${\displaystyle S\in U}$  or ${\displaystyle X\setminus S\in U.}$
2. ${\displaystyle U}$  is a maximal filter subbase on ${\displaystyle X,}$  meaning that if ${\displaystyle F}$  is any filter subbase on ${\displaystyle X}$  then ${\displaystyle U\subseteq F}$  implies ${\displaystyle U=F.}$ [4]

A (proper) filter ${\displaystyle U}$  on ${\displaystyle X}$  is an ultrafilter on ${\displaystyle X}$  if and only if any of the following equivalent conditions hold:

1. ${\displaystyle U}$  is ultra;
2. ${\displaystyle U}$  is generated by an ultra prefilter;
3. For any subset ${\displaystyle S\subseteq X,}$  ${\displaystyle S\in U}$  or ${\displaystyle X\setminus S\in U.}$ [4]
• So an ultrafilter ${\displaystyle U}$  decides for every ${\displaystyle S\subseteq X}$  whether ${\displaystyle S}$  is "large" (i.e. ${\displaystyle S\in U}$ ) or "small" (i.e. ${\displaystyle X\setminus S\in U}$ ).[7]
4. For each subset ${\displaystyle A\subseteq X,}$  either[note 1] ${\displaystyle A}$  is in ${\displaystyle U}$  or (${\displaystyle X\setminus A}$ ) is.
5. ${\displaystyle U\cup (X\setminus U)=\wp (X).}$  This condition can be restated as: ${\displaystyle \wp (X)}$  is partitioned by ${\displaystyle U}$  and its dual ${\displaystyle X\setminus U.}$
• The sets ${\displaystyle P}$  and ${\displaystyle X\setminus P}$  are disjoint for all prefilters ${\displaystyle P}$  on ${\displaystyle X.}$
6. ${\displaystyle \wp (X)\setminus U=\left\{S\in \wp (X):S\not \in U\right\}}$  is an ideal on ${\displaystyle X.}$ [4]
7. For any finite family ${\displaystyle S_{1},\ldots ,S_{n}}$  of subsets of ${\displaystyle X}$  (where ${\displaystyle n\geq 1}$ ), if ${\displaystyle S_{1}\cup \cdots \cup S_{n}\in U}$  then ${\displaystyle S_{i}\in U}$  for some index ${\displaystyle i.}$
• In words, a "large" set cannot be a finite union of sets that aren't large.[8]
8. For any ${\displaystyle R,S\subseteq X,}$  if ${\displaystyle R\cup S=X}$  then ${\displaystyle R\in U}$  or ${\displaystyle S\in U.}$
9. For any ${\displaystyle R,S\subseteq X,}$  if ${\displaystyle R\cup S\in U}$  then ${\displaystyle R\in U}$  or ${\displaystyle S\in U}$  (a filter with this property is called a prime filter).
10. For any ${\displaystyle R,S\subseteq X,}$  if ${\displaystyle R\cup S\in U}$  and ${\displaystyle R\cap S=\varnothing }$  then either ${\displaystyle R\in U}$  or ${\displaystyle S\in U.}$
11. ${\displaystyle U}$  is a maximal filter; that is, if ${\displaystyle F}$  is a filter on ${\displaystyle X}$  such that ${\displaystyle U\subseteq F}$  then ${\displaystyle U=F.}$  Equivalently, ${\displaystyle U}$  is a maximal filter if there is no filter ${\displaystyle F}$  on ${\displaystyle X}$  that contains ${\displaystyle U}$  as a proper subset (that is, no filter is strictly finer than ${\displaystyle U}$ ).[4]

### Grills and Filter-Grills

If ${\displaystyle {\mathcal {B}}\subseteq \wp (X)}$  then its grill on ${\displaystyle X}$  is the family

${\displaystyle {\mathcal {B}}^{\#X}:=\{S\subseteq X~:~S\cap B\neq \varnothing {\text{ for all }}B\in {\mathcal {B}}\}}$

where ${\displaystyle {\mathcal {B}}^{\#}}$  may be written if ${\displaystyle X}$  is clear from context. For example, ${\displaystyle \varnothing ^{\#}=\wp (X)}$  and if ${\displaystyle \varnothing \in {\mathcal {B}}}$  then ${\displaystyle {\mathcal {B}}^{\#}=\varnothing .}$  If ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {B}}}$  then ${\displaystyle {\mathcal {B}}^{\#}\subseteq {\mathcal {A}}^{\#}}$  and moreover, if ${\displaystyle {\mathcal {B}}}$  is a filter subbase then ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {B}}^{\#}.}$ [9] The grill ${\displaystyle {\mathcal {B}}^{\#X}}$  is upward closed in ${\displaystyle X}$  if and only if ${\displaystyle \varnothing \not \in {\mathcal {B}},}$  which will henceforth be assumed. Moreover, ${\displaystyle {\mathcal {B}}^{\#\#}={\mathcal {B}}^{\uparrow X}}$  so that ${\displaystyle {\mathcal {B}}}$  is upward closed in ${\displaystyle X}$  if and only if ${\displaystyle {\mathcal {B}}^{\#\#}={\mathcal {B}}.}$

The grill of a filter on ${\displaystyle X}$  is called a filter-grill on ${\displaystyle X.}$ [9] For any ${\displaystyle \varnothing \neq {\mathcal {B}}\subseteq \wp (X),}$  ${\displaystyle {\mathcal {B}}}$  is a filter-grill on ${\displaystyle X}$  if and only if (1) ${\displaystyle {\mathcal {B}}}$  is upward closed in ${\displaystyle X}$  and (2) for all sets ${\displaystyle R}$  and ${\displaystyle S,}$  if ${\displaystyle R\cup S\in {\mathcal {B}}}$  then ${\displaystyle R\in {\mathcal {B}}}$  or ${\displaystyle S\in {\mathcal {B}}.}$  The grill operation ${\displaystyle {\mathcal {F}}\mapsto {\mathcal {F}}^{\#X}}$  induces a bijection

${\displaystyle {\bullet }^{\#X}~:~\operatorname {Filters} (X)\to \operatorname {FilterGrills} (X)}$

whose inverse is also given by ${\displaystyle {\mathcal {F}}\mapsto {\mathcal {F}}^{\#X}.}$ [9] If ${\displaystyle {\mathcal {F}}\in \operatorname {Filters} (X)}$  then ${\displaystyle {\mathcal {F}}}$  is a filter-grill on ${\displaystyle X}$  if and only if ${\displaystyle {\mathcal {F}}={\mathcal {F}}^{\#X},}$ [9] or equivalently, if and only if ${\displaystyle {\mathcal {F}}}$  is an ultrafilter on ${\displaystyle X.}$ [9] That is, a filter on ${\displaystyle X}$  is a filter-grill if and only if it is ultra. For any non-empty ${\displaystyle {\mathcal {F}}\subseteq \wp (X),}$  ${\displaystyle {\mathcal {F}}}$  is both a filter on ${\displaystyle X}$  and a filter-grill on ${\displaystyle X}$  if and only if (1) ${\displaystyle \varnothing \not \in {\mathcal {F}}}$  and (2) for all ${\displaystyle R,S\subseteq X,}$  the following equivalences hold:

${\displaystyle R\cup S\in {\mathcal {F}}}$  if and only if ${\displaystyle R,S\in {\mathcal {F}}}$  if and only if ${\displaystyle R\cap S\in {\mathcal {F}}.}$ [9]

### Free or principal

If ${\displaystyle P}$  is any non-empty family of sets then the Kernel of ${\displaystyle P}$  is the intersection of all set in ${\displaystyle P}$ :

${\displaystyle \operatorname {ker} P:=\bigcap _{B\in P}B.}$ [10]

A non-empty family of sets ${\displaystyle P}$  is called:

• free if ${\displaystyle \operatorname {ker} P=\varnothing }$  and fixed otherwise (that is, if ${\displaystyle \operatorname {ker} P\neq \varnothing }$ ),
• principal if ${\displaystyle \operatorname {ker} P\in P,}$
• principal at a point if ${\displaystyle \operatorname {ker} P\in P}$  and ${\displaystyle \operatorname {ker} P}$  is a singleton set; in this case, if ${\displaystyle \operatorname {ker} P=\{x\}}$  then ${\displaystyle P}$  is said to be principal at ${\displaystyle X.}$

If a family of sets ${\displaystyle P}$  is fixed then ${\displaystyle P}$  is ultra if and only if some element of ${\displaystyle P}$  is a singleton set, in which case ${\displaystyle P}$  will necessarily be a prefilter. Every principal prefilter is fixed, so a principal prefilter ${\displaystyle P}$  is ultra if and only if ${\displaystyle \operatorname {ker} P}$  is a singleton set. A singleton set is ultra if and only if its sole element is also a singleton set.

Every filter on ${\displaystyle X}$  that is principal at a single point is an ultrafilter, and if in addition ${\displaystyle X}$  is finite, then there are no ultrafilters on ${\displaystyle X}$  other than these.[10] If there exists a free ultrafilter (or even filter subbase) on a set ${\displaystyle X}$  then ${\displaystyle X}$  must be infinite.

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.

Proposition — If ${\displaystyle U}$  is an ultrafilter on ${\displaystyle X}$  then the following are equivalent:

1. ${\displaystyle U}$  is fixed, or equivalently, not free.
2. ${\displaystyle U}$  is principal.
3. Some element of ${\displaystyle U}$  is a finite set.
4. Some element of ${\displaystyle U}$  is a singleton set.
5. ${\displaystyle U}$  is principal at some point of ${\displaystyle X,}$  which means ${\displaystyle \operatorname {ker} U=\{x\}\in U}$  for some ${\displaystyle x\in X.}$
6. ${\displaystyle U}$  does not contain the Fréchet filter on ${\displaystyle X}$  as a subset.
7. ${\displaystyle U}$  is sequential.[9]

## Examples, properties, and sufficient conditions

If ${\displaystyle U}$  and ${\displaystyle S}$  are families of sets such that ${\displaystyle U}$  is ultra, ${\displaystyle \varnothing \not \in S,}$  and ${\displaystyle U\leq S,}$  then ${\displaystyle S}$  is necessarily ultra. A filter subbase ${\displaystyle U}$  that is not a prefilter cannot be ultra; but it is nevertheless still possible for the prefilter and filter generated by ${\displaystyle U}$  to be ultra.

Suppose ${\displaystyle U\subseteq \wp (X)}$  is ultra and ${\displaystyle Y}$  is a set. The trace ${\displaystyle U\cap Y:=\{B\cap Y:B\in U\}}$  is ultra if and only if it does not contain the empty set. Furthermore, at least one of the sets ${\displaystyle [U\cap Y]\setminus \{\varnothing \}}$  and ${\displaystyle [U\cap (X\setminus Y)]\setminus \{\varnothing \}}$  will be ultra (this result extends to any finite partition of ${\displaystyle X}$ ). If ${\displaystyle F_{1},\ldots ,F_{n}}$  are filters on ${\displaystyle X,}$  ${\displaystyle U}$  is an ultrafilter on ${\displaystyle X,}$  and ${\displaystyle F_{1}\cap \cdots \cap F_{n}\leq U,}$  then there is some ${\displaystyle F_{i}}$  that satisfies ${\displaystyle F_{i}\leq U.}$ [11] This result is not necessarily true for an infinite family of filters.[11]

The image under a map ${\displaystyle f:X\to Y}$  of an ultra set ${\displaystyle U\subseteq \wp (X)}$  is again ultra and if ${\displaystyle U}$  is an ultra prefilter then so is ${\displaystyle f(U).}$  The property of being ultra is preserved under bijections. However, the preimage of an ultrafilter is not necessarily ultra, not even if the map is surjective. For example, if ${\displaystyle X}$  has more than one point and if the range of ${\displaystyle f:X\to Y}$  consists of a single point ${\displaystyle \{y\}}$  then ${\displaystyle \{y\}}$  is an ultra prefilter on ${\displaystyle Y}$  but its preimage is not ultra. Alternatively, if ${\displaystyle U}$  is a principal filter generated by a point in ${\displaystyle Y\setminus f(X)}$ then the preimage of ${\displaystyle U}$  contains the empty set and so is not ultra.

The elementary filter induced by an infinite sequence, all of whose points are distinct, is not an ultrafilter.[11] If ${\displaystyle n=2,}$  then ${\displaystyle U_{n}}$  denotes the set consisting all subsets of ${\displaystyle X}$  having cardinality ${\displaystyle n,}$  and if ${\displaystyle X}$ contains at least ${\displaystyle 2n-1}$  (${\displaystyle =3}$ ) distinct points, then ${\displaystyle U_{n}}$  is ultra but it is not contained in any prefilter. This example generalizes to any integer ${\displaystyle n>1}$  and also to ${\displaystyle n=1}$  if ${\displaystyle X}$  contains more than one element. Ultra sets that are not also prefilters are rarely used.

For every ${\displaystyle S\subseteq X\times X}$  and every ${\displaystyle a\in X,}$  let ${\displaystyle S{\big \vert }_{\{a\}\times X}:=\left\{y\in X~:~(a,y)\in S\right\}.}$  If ${\displaystyle {\mathcal {U}}}$  is an ultrafilter on ${\displaystyle X}$  then the set of all ${\displaystyle S\subseteq X\times X}$  such that ${\displaystyle \left\{a\in X~:~S{\big \vert }_{\{a\}\times X}\in {\mathcal {U}}\right\}\in {\mathcal {U}}}$  is an ultrafilter on ${\displaystyle X\times X.}$ [12]

The functor associating to any set ${\displaystyle X}$  the set of ${\displaystyle U(X)}$  of all ultrafilters on ${\displaystyle X}$  forms a monad called the ultrafilter monad. The unit map

${\displaystyle X\to U(X)}$

sends any element ${\displaystyle x\in X}$  to the principal ultrafilter given by ${\displaystyle x.}$

This monad admits a conceptual explanation as the codensity monad of the inclusion of the category of finite sets into the category of all sets.[13]

## The ultrafilter lemma

The ultrafilter lemma was first proved by Alfred Tarski in 1930.[12]

The ultrafilter lemma/principle/theorem[5] — Every proper filter on a set ${\displaystyle X}$  is contained in some ultrafilter on ${\displaystyle X.}$

The ultrafilter lemma is equivalent to each of the following statements:

1. For every prefilter on a set ${\displaystyle X,}$  there exists a maximal prefilter on ${\displaystyle X}$  subordinate to it.[2]
2. Every proper filter subbase on a set ${\displaystyle X}$  is contained in some ultrafilter on ${\displaystyle X.}$

A consequence of the ultrafilter lemma is that every filter is equal to the intersection of all ultrafilters containing it.[14][note 2]

The following results can be proven using the ultrafilter lemma. A free ultrafilter exists on a set ${\displaystyle X}$  if and only if ${\displaystyle X}$  is infinite. Every proper filter is equal to the intersection of all ultrafilters containing it.[5] Since there are filters that are not ultra, this shows that the intersection of a family of ultrafilters need not be ultra. A family of sets ${\displaystyle \mathbb {F} \neq \varnothing }$  can be extended to a free ultrafilter if and only if the intersection of any finite family of elements of ${\displaystyle \mathbb {F} }$  is infinite.

### Relationships to other statements under ZF

Throughout this section, ZF refers to Zermelo–Fraenkel set theory and ZFC refers to ZF with the Axiom of Choice (AC). The ultrafilter lemma is independent of ZF. That is, there exist models in which the axioms of ZF hold but the ultrafilter lemma does not. There also exist models of ZF in which every ultrafilter is necessarily principal.

Every filter that contains a singleton set is necessarily an ultrafilter and given ${\displaystyle x\in X,}$  the definition of the discrete ultrafilter ${\displaystyle \{S\subseteq X~:~x\in S\}}$  does not require more than ZF. If ${\displaystyle X}$  is finite then every ultrafilter is a discrete filter at a point; consequently, free ultrafilters can only exist on infinite sets. In particular, if ${\displaystyle X}$  is finite then the ultrafilter lemma can be proven from the axioms ZF. The existence of free ultrafilter on infinite sets can be proven if the axiom of choice is assumed. More generally, the ultrafilter lemma can be proven by using the axiom of choice, which in brief states that any Cartesian product of non-empty sets is non-empty. Under ZF, the axiom of choice is, in particular, equivalent to (a) Zorn's lemma, (b) Tychonoff's theorem, (c) the weak form of the vector basis theorem (which states that every vector space has a basis), (d) the strong form of the vector basis theorem, and other statements. However, the ultrafilter lemma is strictly weaker than the axiom of choice. While free ultrafilters can be proven to exist, it is not possible to construct an explicit example of a free ultrafilter; that is, free ultrafilters are intangible.[15]Alfred Tarski proved that under ZFC, the cardinality of the set of all free ultrafilters on an infinite set ${\displaystyle X}$  is equal to the cardinality of ${\displaystyle \wp (\wp (X)),}$  where ${\displaystyle \wp (X)}$  denotes the power set of ${\displaystyle X.}$ [16]

Under ZF, the axiom of choice can be used to prove both the ultrafilter lemma and the Krein–Milman theorem; conversely, under ZF, the ultrafilter lemma together with the Krein–Milman theorem can prove the axiom of choice.[17]

#### Statements that cannot be deduced

The ultrafilter lemma is a relatively weak axiom. For example, each of the statements in the following list can not be deduced from ZF together with only the ultrafilter lemma:

1. A countable union of countable sets is a countable set.
2. The axiom of countable choice (ACC).
3. The axiom of dependent choice (ADC).

#### Equivalent statements

Under ZF, the ultrafilter lemma is equivalent to each of the following statements:[18]

1. The Boolean prime ideal theorem (BPIT).
• This equivalence is provable in ZF set theory without the Axiom of Choice (AC).
2. Stone's representation theorem for Boolean algebras.
3. Any product of Boolean spaces is a Boolean space.[19]
4. Boolean Prime Ideal Existence Theorem: Every nondegenerate Boolean algebra has a prime ideal.[20]
5. Tychonoff's theorem for Hausdorff spaces: Any product of compact Hausdorff spaces is compact.[19]
6. If ${\displaystyle \{0,1\}}$  is endowed with the discrete topology then for any set ${\displaystyle I,}$  the product space ${\displaystyle \{0,1\}^{I}}$  is compact.[19]
7. Each of the following versions of the Banach-Alaoglu theorem is equivalent to the ultrafilter lemma:
1. Any equicontinuous set of scalar-valued maps on a topological vector space (TVS) is relatively compact in the weak-* topology (that is, it is contained in some weak-* compact set).[21]
2. The polar of any neighborhood of the origin in a TVS ${\displaystyle X}$  is a weak-* compact subset of its continuous dual space.[21]
3. The closed unit ball in the continuous dual space of any normed space is weak-* compact.[21]
• If the normed space is separable then the ultrafilter lemma is sufficient but not necessary to prove this statement.
8. A topological space ${\displaystyle X}$  is compact if every ultrafilter on ${\displaystyle X}$  converges to some limit.[22]
9. A topological space ${\displaystyle X}$  is compact if and only if every ultrafilter on ${\displaystyle X}$  converges to some limit.[22]
• The addition of the words "and only if" is the only difference between this statement and the one immediately above it.
10. The Ultranet lemma: Every net has a universal subnet.[23]
• By definition, a net in ${\displaystyle X}$  is called an ultranet or an universal net if for every subset ${\displaystyle S\subseteq X,}$  the net is eventually in ${\displaystyle S}$  or in ${\displaystyle X\setminus S.}$
11. A topological space ${\displaystyle X}$  is compact if and only if every ultranet on ${\displaystyle X}$  converges to some limit.[22]
• If the words "and only if" are removed then the resulting statement remains equivalent to the ultrafilter lemma.[22]
12. A convergence space ${\displaystyle X}$  is compact if every ultrafilter on ${\displaystyle X}$  converges.[22]
13. A uniform space is compact if it is complete and totally bounded.[22]
14. The Stone–Čech compactification Theorem.[19]
15. Each of the following versions of the compactness theorem is equivalent to the ultrafilter lemma:
1. If ${\displaystyle \Sigma }$  is a set of first-order sentences such that every finite subset of ${\displaystyle \Sigma }$  has a model, then ${\displaystyle \Sigma }$  has a model.[24]
2. If ${\displaystyle \Sigma }$  is a set of zero-order sentences such that every finite subset of ${\displaystyle \Sigma }$  has a model, then ${\displaystyle \Sigma }$  has a model.[24]
16. The completeness theorem: If ${\displaystyle \Sigma }$  is a set of zero-order sentences that is syntactically consistent, then it has a model (that is, it is semantically consistent).

#### Weaker statements

Any statement that can be deduced from the ultrafilter lemma (together with ZF) is said to be weaker than the ultrafilter lemma. A weaker statement is said to be strictly weaker if under ZF, it is not equivalent to the ultrafilter lemma. Under ZF, the ultrafilter lemma implies each of the following statements:

1. The Axiom of Choice for Finite sets (ACF): Given ${\displaystyle I\neq \varnothing }$  and a family ${\displaystyle \left(X_{i}\right)_{i\in I}}$  of non-empty finite sets, their product ${\displaystyle \prod _{i\in I}X_{i}}$  is not empty.[23]
2. A countable union of finite sets is a countable set.
• However, ZF with the ultrafilter lemma is too weak to prove that a countable union of countable sets is a countable set.
3. The Hahn–Banach theorem.[23]
• In ZF, the Hahn–Banach theorem is strictly weaker than the ultrafilter lemma.
5. Every set can be linearly ordered.
6. Every field has a unique algebraic closure.
7. The Alexander subbase theorem.[23]
8. Non-trivial ultraproducts exist.
9. The weak ultrafilter theorem: A free ultrafilter exists on ${\displaystyle \mathbb {N} .}$
• Under ZF, the weak ultrafilter theorem does not imply the ultrafilter lemma; that is, it is strictly weaker than the ultrafilter lemma.
10. There exists a free ultrafilter on every infinite set;
• This statement is actually strictly weaker than the ultrafilter lemma.
• ZF alone does not even imply that there exists a non-principal ultrafilter on some set.

## Completeness

The completeness of an ultrafilter ${\displaystyle U}$  on a powerset is the smallest cardinal κ such that there are κ elements of ${\displaystyle U}$  whose intersection is not in ${\displaystyle U.}$  The definition of an ultrafilter implies that the completeness of any powerset ultrafilter is at least ${\displaystyle \aleph _{0}}$ . An ultrafilter whose completeness is greater than ${\displaystyle \aleph _{0}}$ —that is, the intersection of any countable collection of elements of ${\displaystyle U}$  is still in ${\displaystyle U}$ —is called countably complete or σ-complete.

The completeness of a countably complete nonprincipal ultrafilter on a powerset is always a measurable cardinal.[citation needed]

## .mw-parser-output .vanchor>:target~.vanchor-text{background-color:#b1d2ff}Ordering on ultrafilters

The Rudin–Keisler ordering (named after Mary Ellen Rudin and Howard Jerome Keisler) is a preorder on the class of powerset ultrafilters defined as follows: if ${\displaystyle U}$  is an ultrafilter on ${\displaystyle \wp (X),}$  and ${\displaystyle V}$  an ultrafilter on ${\displaystyle \wp (Y),}$  then ${\displaystyle V\leq {}_{RK}U}$  if there exists a function ${\displaystyle f:X\to Y}$  such that

${\displaystyle C\in V}$  if and only if ${\displaystyle f^{-1}[C]\in U}$

for every subset ${\displaystyle C\subseteq Y.}$

Ultrafilters ${\displaystyle U}$  and ${\displaystyle V}$  are called Rudin–Keisler equivalent, denoted URK V, if there exist sets ${\displaystyle A\in U}$  and ${\displaystyle B\in V}$  and a bijection ${\displaystyle f:A\to B}$  that satisfies the condition above. (If ${\displaystyle X}$  and ${\displaystyle Y}$  have the same cardinality, the definition can be simplified by fixing ${\displaystyle A=X,}$  ${\displaystyle B=Y.}$ )

It is known that ≡RK is the kernel of ≤RK, i.e., that URK V if and only if ${\displaystyle U\leq {}_{RK}V}$  and ${\displaystyle V\leq {}_{RK}U.}$ [27]

## Ultrafilters on ${\displaystyle \wp (\omega )}$

There are several special properties that an ultrafilter on ${\displaystyle \wp (\omega ),}$  where ${\displaystyle \omega }$  extends the natural numbers, may possess, which prove useful in various areas of set theory and topology.

• A non-principal ultrafilter ${\displaystyle U}$  is called a P-point (or weakly selective) if for every partition ${\displaystyle \left\{C_{n}:n<\omega \right\}}$  of ${\displaystyle \omega }$  such that for all ${\displaystyle n<\omega ,}$  ${\displaystyle C_{n}\not \in U,}$  there exists some ${\displaystyle A\in U}$  such that for all ${\displaystyle A\cap C_{n}}$  is a finite set for each ${\displaystyle n.}$
• A non-principal ultrafilter ${\displaystyle U}$  is called Ramsey (or selective) if for every partition ${\displaystyle \left\{C_{n}:n<\omega \right\}}$  of ${\displaystyle \omega }$  such that for all ${\displaystyle n<\omega ,}$  ${\displaystyle C_{n}\not \in U,}$  there exists some ${\displaystyle A\in U}$  such that ${\displaystyle A\cap C_{n}}$  is a singleton set for each ${\displaystyle n.}$

It is a trivial observation that all Ramsey ultrafilters are P-points. Walter Rudin proved that the continuum hypothesis implies the existence of Ramsey ultrafilters.[28] In fact, many hypotheses imply the existence of Ramsey ultrafilters, including Martin's axiom. Saharon Shelah later showed that it is consistent that there are no P-point ultrafilters.[29] Therefore, the existence of these types of ultrafilters is independent of ZFC.

P-points are called as such because they are topological P-points in the usual topology of the space βω \ ω of non-principal ultrafilters. The name Ramsey comes from Ramsey's theorem. To see why, one can prove that an ultrafilter is Ramsey if and only if for every 2-coloring of ${\displaystyle [\omega ]^{2}}$  there exists an element of the ultrafilter that has a homogeneous color.

An ultrafilter on ${\displaystyle \wp (\omega )}$  is Ramsey if and only if it is minimal in the Rudin–Keisler ordering of non-principal powerset ultrafilters.[citation needed]

## Notes

1. ^ a b Properties 1 and 3 imply that ${\displaystyle A}$  and ${\displaystyle X\setminus A}$  cannot both be elements of ${\displaystyle U.}$
2. ^ Let ${\displaystyle {\mathcal {F}}}$  be a filter on ${\displaystyle X}$  that is not an ultrafilter. If ${\displaystyle S\subseteq X}$  is such that ${\displaystyle S\not \in {\mathcal {F}}}$  then ${\displaystyle \{X\setminus S\}\cup {\mathcal {F}}}$  has the finite intersection property (because if ${\displaystyle F\in {\mathcal {F}}}$  then ${\displaystyle F\cap (X\setminus S)=\varnothing }$  if and only if ${\displaystyle F\subseteq S}$ ) so that by the ultrafilter lemma, there exists some ultrafilter ${\displaystyle {\mathcal {U}}_{S}}$  on ${\displaystyle X}$  such that ${\displaystyle \{X\setminus S\}\cup {\mathcal {F}}\subseteq {\mathcal {U}}_{S}}$  (so in particular ${\displaystyle S\not \in {\mathcal {U}}_{S}}$ ). It follows that ${\displaystyle {\mathcal {F}}=\bigcap _{S\subseteq X,S\not \in {\mathcal {F}}}{\mathcal {U}}_{S}.}$
Proofs
1. ^ Suppose ${\displaystyle {\mathcal {B}}}$  is filter subbase that is ultra. Let ${\displaystyle C,D\in {\mathcal {B}}}$  and define ${\displaystyle S=C\cap D.}$  Because ${\displaystyle {\mathcal {B}}}$  is ultra, there exists some ${\displaystyle B\in {\mathcal {B}}}$  such that ${\displaystyle B\cap S}$  equals ${\displaystyle B}$  or ${\displaystyle \varnothing .}$  The finite intersection property implies that ${\displaystyle B\cap S\neq \varnothing }$  so necessarily ${\displaystyle B\cap S=B,}$  which is equivalent to ${\displaystyle B\subseteq C\cap D.}$

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