Let be a set. A filter on is a set of subsets of (ie. ) which satisfies the following conditions:
- (closure under supersets)
- (closure under finite intersections)
Example (neighbourhood filter):
Let be a topological space and let be a point. The set of all neighbourhoods of is a filter.
Note that the set of open neighbourhoods of a point does not in general form a filter.
Definition (filter convergence):
Let be a topological space, , and let be a filter on . We say that converges to if and only if .
Definition (filter base):
Let be a filter on a set . A filter base of is a set such that for all , there exists such that .
Note in particular that the bases are required to be contained within the filter.
Definition (filter subbase):
Let be a filter on a set . A set is said to be a subbasis of if and only if for every , there exist such that .
Note in particular that the subbases are required to be contained within the filter. Note also that we use the terms base and basis interchangeably; both versions are in use. Note finally that a filter base of a filter is a filter subbase of .
Proposition (characterisation of filter subbases):
Let be a set and let . is a filter subbase of some filter if and only if for any finite set of elements of , we have that . In this case, is uniquely determined by .
Proof: Suppose first that whenever are elements of , then . Then define
We claim that is a filter. Indeed, the empty set can't be in , since every intersection of elements of is, by assumption, nonempty. Further, whenever and , then . Finally, suppose that , and choose so that and so that , then
so that .
Now suppose that is the subbasis of some filter , and let . Since and is a filter, , so that .
Let now be an arbitrary filter that has as a subbasis. Then by the definition of a subbasis. Now since , we also have , so that .
Proposition (characterisation of filter bases):
Let be a set and . is the basis of some filter if and only if for each , there exists so that . In this case, is uniquely determined by .
Proof: Suppose first that is the filter basis of some filter . Since is a filter and , we find that , so that there exists such that since is a filter basis of . Suppose now that has the property that intersections of finitely many elements of contain an element of . Define
Then is readily seen to be a filter since can't contain the empty set and by some routine arguments, and is a filter base of it, since it is certainly contained in .
Now note that is also a subbasis of , so that is uniquely determined by , using the characterisation of filter subbases.
Proposition (filter basis from filter subbasis):
Let be a filter which is generated by a subbasis . Then a filter basis for is given by
Proof: We have , so that , where is the filter generated by . But since is closed under finite intersections, , and hence . We conclude .
Definition (convergence of filter base):
Let be the base of a filter on the topological space . We say that converges to a point if and only if .
Analogous to real analysis, we can rephrase continuity in terms of filter convergence. In general, filters are supposed to play the role for topological spaces that sequences play for finite-dimensional real normed spaces; we will see many theorems that are analogous to those on , with sequences replaced by filters.
Proposition (characterisation of continuity by filter convergence):
Let be topological spaces and a function. is continuous if and only if for each filter on that converges to a point , the filter base converges to .
Proof: Note that continuity is equivalent to continuity at each point. Further, being continuous at a point means that for each open neighbourhood of we find an open neighbourhood of so that , which in turn implies that is a filter base for a filter that contains . On the other hand, if has the property that every filter of convergent to has the property that converges to , note that converges to , so that for every neighbourhood of we may choose so that , and then, since , we may choose open, and still have , so that is continuous at .
Proposition (intersection of filters is a filter):
Whenever is a family of filters on a set , the intersection
is a filter.
Proof: That and intersections of finite subsets are contained in is a special case of intersection preserving closure properties. Similarly, we can also regard taking supersets as an operation, with input a set of a single set and output a family of sets encompassing all supersets, so that we also obtain closure under taking supersets of , so that a filter becomes a set of sets of sets.
Proposition (family of filters has a greatest lower bound):
Let be a family of filters on a set . Then the intersection of the filters constitutes a least upper bound of the .
Proof: This follows from the fact that the greatest lower bound structure of certain algebraic structures is their intersection, regarding filters as algebraic structures as above.
Proposition (extension of filters by a set):
Let be a filter on a set and let . can be extended to a filter that contains if and only if has nonempty intersection with al elements of .
Proof: Suppose first that intersects nontrivially with all elements of . Then the union of and forms a filter subbasis, so that the filter generated by and is a filter that contains and . Conversely, whenever is a filter that contains and , then for any element we have and so that .
Proposition (existence of least upper bound of two filters):
Let and be two filters on a set . There will exist a filter that contains both and (and hence a minimal such filter) if and only if whenever and , then .
Proof: Necessity is clear as in the extension of filters by a set. Sufficiency follows since whenever the condition is satisfied, forms the subbasis of a filter by the characterisation of filter subbases.
Proposition (filter is ultrafilter iff it contains any set or its complement):
Let be a set and let be a filter on . is an ultrafilter if and only if for every set , contains either or .
Proof: Suppose that is an ultrafilter, and let . Then if neither nor , and if we assume that neither nor intersect all elements of , we find elements so that and , and then , contradicting being a filter. But then either or has nonempty intersection with all elements of , so that we may extend by one of the two sets, in contradiction to being maximal.
Suppose now that contains either or for all . Suppose that was not maximal. Then we find by which we may extend , to obtain a larger filter (after adding suitable supersets). But since , we have and hence and yet , so that is not a filter.
Proposition (characterisation of compactness by filter convergence):
Let be a topological space. is compact if and only if every filter on can be refined to a filter that converges to some point in .
(On the condition of the ultrafilter lemma.)
Proof: Let be a compact topological space, and let be any ultrafilter whose underlying set is . If does not converge to any point, then for each , the set of open neighbourhoods of that are not in is nonempty. Instead of choosing one open neighbourhood from each (which would require the axiom of choice), we define
This is an open cover of , and since is compact, we may choose a finite subcover, that is, some points and open neighbourhoods () such that
Now was supposed to be an ultrafilter, and all ultrafilters have the property that for each subset of the set on which they are defined, they contain either that subset or its complement. Therefore, contains the complement of each () and hence the intersection of all these complements, which is empty by de Morgan's rule.
Conversely, suppose that every filter on may be refined to a filter that converges to a point of , and let be an open cover of . Suppose that doesn't have any finite subcover. Then by de Morgan, finite intersections of sets of the form are never empty, so that these sets form a subbasis of a filter , which by assumption may be refined to contain for some . But then no can contain , for otherwise both and , whereas .
An ultrafilter on a set is a maximal element of the set of all filters on ordered by inclusion.
Theorem (ultrafilter lemma):
Let be a set and let be a filter on . Then there exists a filter which contains and is an ultrafilter.
(On the condition of the axiom of choice.)
Proof: Observe that the set of filters that contain has the property that every ascending chain has an upper bound; indeed, the union of that chain is one, since it is still a filter and contains . Hence, Zorn's lemma yields a maximal element among those filters that contain , and this filter must also be maximal, since any larger filter would also contain .
Note: It has been proven that one can't prove the ultrafilter lemma from Zermelo–Fraenkel axioms alone, but some form of the axiom of choice is needed. Still, the ultrafilter lemma does not imply the axiom of choice in ZF, that is, it is strictly weaker than the axiom of choice.
Proposition (filter limits in R1 spaces are topologically indistinguishable):
Let be an R1 topological space, let be a filter in and let finally so that converges to both and . Then and are topologically indistinguishable.
Proof: If were topologically distinguishable, the R1 property allows to pick open so that and and . Since converges to and , and in contradiction to .
Proposition (filter limits in Hausdorff spaces are unique):
Let be a Hausdorff space, and let be a filter on . Then may converge at most to one point.
Proof: If and , then are topologically distinguishable since is Hausdorff and in particular T0. Hence, as Hausdorff spaces are also R1, we apply the fact that filter limits in R1 spaces are topologically indistinguishable.
Proposition (characterisation of closed sets by filter convergence):
Let be a topological space and . Then is closed if and only if for every filter generated by subsets of that converges to some , we have .
Proof: Suppose first that is closed. Then let be a filter as in the theorem statement, and one of its limits. Suppose that . By the convergence, , being a neighbourhood of . But then is not generated by subsets of , because these are closed under finite intersection, and contains none of them.
Suppose now that has the given property. Recall that a set is closed if and only if it contains its boundary. Hence, let , and consider the filter
- generated by ,
where is the neighbourhood system of ; this is a filter because , so that . converges to , because every neighbourhood of is a superset of some set of , so that . Hence, .
Let be a set, a filter on , and . is said to cluster at iff for all .
In this way, we may reformulate the criterion for the existence of a filter containing another filter and a given set: If is a filter and a set, then there exists a filter with and iff clusters at .
Proposition (characterisation of closedness by clustering):
Let be a topological space and . Then is closed if and only if whenever so that clusters at , then .
Proof: Recall that a set is closed if and only if it contains its boundary. The proposition is a mere reformulation of that statement, because the union of and are the so that clusters at .
Definition (principal ultrafilter):
Let be a set. A principal ultrafilter is a filter on such that there exists an element so that .
Note that principal ultrafilters are ultrafilters, because either or for all .
Definition (non-principal ultrafilter):
Let be a set. A non-principal ultrafilter on is an ultrafilter on that is not a principal ultrafilter.
Proposition (non-principal ultrafilters exist):
Let be an infinite set. Then there exists a non-principal ultrafilter on .
(On the condition of the ultrafilter lemma.)
Proof: Consider the filter on that is given by the cofinite sets. Upon extending it to an ultrafilter by the ultrafilter lemma, we obtain an ultrafilter that contains all cofinite sets. In particular, it cannot contain a point set , since otherwise and .
- Let be a topological space which is not compact. Prove that the set of all complements of compact subsets of is a filter. Prove that if is compact, then is not a filter.
- Suppose that is an ultrafilter on a set , and let be finitely many subsets of . Prove that if does not contain any of the , it does not contain .
- Let be sets and a function. Prove that is injective iff for all filters in , is a filter.