In classical mathematics, every uniform space is a regular space, and indeed a completely regular space. In constructive mathematics, however, this fails to be true. The notion of uniformly regular uniform space is an intermediate notion in between regularity and complete regularity for uniform spaces, which is frequently useful in the constructive theory of uniform spaces; classically every uniform space is uniformly regular. (Indeed, some authors include uniform regularity in the constructive definition of a uniform space.)
Let be a uniform space. When the uniformity of is defined in terms of entourages, then is uniformly regular if it satisfies the following property:
With excluded middle, we can take to be itself, so every uniform structure is uniformly regular in classical mathematics. Note that any uniformity induced by a gauge is uniformly regular (as long as the gauging distances take values in located real numbers as is usually understood).
When the uniformity of is defined in terms of uniform covers, then the equivalent condition is:
With excluded middle, we can take to be .
Note that uniform regularity is not literally a “uniformization” of regularity but rather of local decomposability; so it could also be called uniform local decomposability. In the case of uniform spaces, it implies regularity (see below), so we generally call it uniform regularity; however, for quasi-uniform spaces (and uniform convergence spaces) it would be more proper to say ‘uniformly locally decomposable’ instead.
Every metric space, and indeed any gauge space, is uniformly regular, as long as the metric(s) take values in located real numbers. It suffices to consider a basic entourage , in which case we can take and use the fact that every located real number is either or .
A discrete uniform space is uniformly regular if and only if the underlying set has decidable equality. (Indeed, a discrete topological space is regular if and only if it has decidable equality.) Thus, the statement “every uniform space is (uniformly) regular” is equivalent to excluded middle.
In particular, not every topological group is uniformly regular: any group without decidable equality can be equipped with the discrete topology. However, the classical examples of Lie groups and topological vector spaces (TVSs) are uniformly regular.
However, if a topological group is a regular space, then it is uniformly regular. For if is any neighborhood of the identity in a topological group , by regularity we have a neighborhood of and an open set with and , and in particular , whence the entourages defined by and satisfy the desired condition.
Every completely regular uniform space should be uniformly regular, although the correct constructive definition of “completely regular” is not entirely clear.
The examples above show that uniform regularity coincides with regularity for discrete uniform spaces and topological groups. Another class of spaces for which this holds are compact ones.
Every compact regular space admits a unique compatible uniformity that is uniformly regular.
Let be compact regular. Since compactness and regularity are properties of the open-set lattice, is still compact regular when regarded as a locale, and hence also Hausdorff as a locale. Moreover, it is locally compact, and hence the locale and spatial products coincide. Furthermore, is also compact regular.
Define the entourages to be all the neighborhoods of the diagonal in ; this obviously satisfies all the axioms of a uniformity except possibly the square-root (and uniform regularity).
Let be a neighborhood of the diagonal in . Since is regular, and using the definition of the product topology, for each there is a square neighborhood and an open set such that and . The open sets cover the compact space , hence finitely many of them suffice. Let , and . Then is a neighborhood of the diagonal, and an open set such that and . In particular, , so this uniformity (if it exists) is uniformly regular.
Now, is the union of rectangular open neighborhoods . Note that since is disjoint from the diagonal, any satisfies . Moreover, since is regular, each and is the union of open sets that are well-inside it; thus is the union of rectangular open neighborhoods such that there are opens with and and . Since and is compact, we can cover it by together with finitely many such rectangles , say .
For each of these rectangles , let be an open set such that and , and let . Then is a neighborhood of the diagonal; let .
We claim . Let , so that there is a with and . Now we have either (which is what we want) or for some , so it suffices to rule out the latter possibility. Since , we have , hence or or . But , which is disjoint from and , so it must be that and hence . Similarly, we conclude that so that . But , a contradiction.
It remains to show that the uniform topology on is the original one. Clearly if is an entourage then each is a neighborhood of . Conversely, if is a neighborhood of , then by regularity we have a neighborhood of and an open with and . Then is a neighborhood of the diagonal, i.e. an entourage, such that .
Finally, we show this is the unique uniformity compatible with a compact regular topology. Suppose is any compact regular uniform space, and any neighborhood of the diagonal; we want to show is an entourage. For any , there is an entourage such that . Let be an entourage such that . Since is compact, it is covered by finitely many of the neighborhoods , say . Let ; we claim . For let ; then we have for some , whence also , and thus .
Every uniformly regular uniform space is a regular topological space in its uniform topology. For if is a uniform neighborhood of , let be an entourage such that , and an entourage such that . Then is a neighborhood of , and the interior of is an open set , such that and .
It appears to be unknown whether there exist regular uniform spaces that are not uniformly regular. Regularity implies uniform regularity for at least three large classes of uniform spaces: discrete ones, compact ones, and topological groups (plus, trivially, metric and gauge spaces, which are automatically uniformly regular).
Recall that every uniform space has a inequality relation (an irreflexive symmetric relation) where “” means that there exists an entourage such that . If is uniformly regular, then this is an apartness relation, i.e. it is also a comparison. For if , let and . Then for any , we cannot have both and , so we must have either or , whence either or . (In fact, this is a special case of the fact that any regular space admits an apartness relation, discussed at regular space).
Uniform regularity has some nice consequences for the existence of (almost/uniformly) located subsets: see there for details.
A uniformly regular uniform space can be equivalently described in terms of the complements of entourages, i.e. by giving notions of when two points are “at least apart” rather than “at least close”.
A uniform apartness space is a set equipped with a collection of binary relations called anti-entourages such that
If is a uniform space, we can make it a uniform apartness space by declaring the anti-entourages to be the sets contained in the complement of some entourage, i.e. is an anti-entourage if for some entourage . Most of the axioms are obvious. For axiom (2), we use transitivity of a uniformity: if , then if and , then if we can derive a contradiction by assuming (double negations can be removed constructively when the goal is a contradiction) and obtaining from ; thus . Note also that axiom (5) needs only the direction of de Morgan’s law that is true constructively, .)
Thus, any uniform space gives rise to a uniform apartness space. If our original uniform space was uniformly regular, then this uniform apartness space is again “uniformly regular” in the sense that for any anti-entourage there is an anti-entourage such that : if , let where .
(We note in passing that a uniformly regular uniform apartness space also satisfies a stronger version of axiom (2), namely that for every anti-entourage , there exists an anti-entourage such that whenever , for any we have either or . To see this, first let be for as in the standard axiom (2), then let be for as in uniform regularity. Now given , we have either or ; but the latter implies and hence .)
In the converse direction, any uniform apartness space gives rise to a uniform space whose entourages are the supersets of the complements of the anti-entourages, and uniform regularity is again preserved. These constructions are not inverse in general, but they are in the uniformly regular case. For this it suffices to show that any entourage in a uniformly regular uniform space contains the double-complement of some entourage , and dually for any anti-entourage in a uniformly regular uniform apartness space there is an anti-entourage containing . But for the first it suffices to have , and for the second it suffices to have . Thus, there is a bijection between uniformly regular uniformities and uniformly regular uniform apartnesses on any set .
We define a function between uniform apartness spaces to be uniformly continuous if for every anti-entourage in , there is an anti-entourage in such that if then . Since a conditional implies its contrapositive, the constructions above become functors and . Moreover, we have:
The above functors form an idempotent adjunction over , whose fixed points contain the uniformly regular spaces on each side. Thus, in particular, the categories of uniformly regular uniform spaces and uniformly regular uniform apartness spaces are isomorphic over .
The unit and counit are the identity functions, which are uniformly continuous since a statement implies its double negation, and obviously satisfy the triangle identities. The adjunction is idempotent since a negated statement is equivalent to its double negation, and we have seen above that uniformly regular spaces are among the fixed points.
Last revised on December 5, 2022 at 04:18:34. See the history of this page for a list of all contributions to it.