# Quasi-Uniform Spaces I: Pervin quasi-uniformities, Pervin spaces

A uniform space is a natural generalization of the notion of a metric space, on which completeness still makes sense . The non-Hausdorff variant of this is called a quasi-uniform space, and the purpose of this post is to introduce some of the features of those spaces, with a particular stress on a now very classical construction due to William Pervin . It is rather puzzling that I managed to avoid the subject of quasi-uniform spaces in something like the 7 years that this blog existed!

## The definition of quasi-uniform spaces

Let X be a set. A binary relation R on X is a subset of X × X. Binary relations compose: R o S is the set of those pairs (x,z) such that (x,y) ∈ R and (y,z) ∈ S for some y in X. Note that R o RR if and only if R is transitive. A relation R is reflexive if and only if it contains the diagonal Δ ≝ {(x,x) | xX}.

A quasi-uniformity U on X is a filter of reflexive binary relations on X that satisfies the following relaxed transitivity condition: for every relation S in U, there is a relation R in U such that R o RS. Explicitly, it is a family of binary relations on X such that:

1. (reflexivity) for every R in U, for every x in X, (x,x) is in R;
2. (filter 1) X × X is in U;
3. (filter 2) for every R in U, every binary relation S on X such that RS is in U;
4. (filter 3) for all R, S in U, RS is in U;
5. (relaxed transitivity) for every S in U, there is a relation R in U such that, for all points x, y, z in X such that (x,y) ∈ R and (y,z) ∈ R, we have (x,z) ∈ S.

A quasi-uniform space is a set with a quasi-uniformity U. The relations R in U are called entourages, a French word that means surroundings. (And also environment, neighborhood, relatives, acquaintances. As most French words, the word also exists in English.)

The primary example is given by hemi-metric spaces. Given any hemi-metric d on X, one defines a quasi-uniformity Ud as follows. A basic entourage is a relation of the form [<r] ≝ {(x,y) | d(x,y) < r}, where r varies over the positive real numbers (namely, r>0). Then Ud is the family of relations that contains some basic entourage [<r].

It is interesting to see why relaxed transitivity holds. Let S be in Ud, so S contains some basic entourage [<s], with s>0. Define rs/2 and R as [<r]. For all points x, y, z in X such that (x,y) ∈ R and (y,z) ∈ R, we have d(x,y)<r and d(y,z)<r, so d(x,z)≤d(x,y)+d(y,z)<s, showing that (x,z) is in S. In other words, relaxed transitivity stems from the triangular inequality.

By the way, the way we have built Ud is a practical way of defining a quasi-uniformity. A base (of entourages) is any non-empty family B of binary relations on X satisfying:

1. (reflexivity) for every R in B, for every x in X, (x,x) is in R;
2. (filter 3) for all R, S in B, there is a relation T in B such that TRS;
3. (relaxed transitivity) for every S in B, there is a relation R in B such that R o RS.

Then the set of binary relations that contain some element of B forms a quasi-uniformity. This is the quasi-uniformity generated by B. This is what we just did, with B consisting of the relations [<r], r>0.

Before we go on, I should mention the original notion of uniformity. Given any binary relation R on X, let Rop be its converse: (x,y) is in Rop if and only if (y,x) is in R. A quasi-uniformity U is a uniformity if and only if it satisfies the following extra condition:

1. (symmetry) for every R in U, Rop is in U.

The prime example of a uniformity is Ud provided that d is a pseudo-metric now.

## Quasi-uniform spaces and topological spaces

There is a traditional way to extract a topology from a quasi-uniformity U.

Given any point x in X, and any binary relation R on X, let R[x] denote the set {yX | (x,y) ∈ R}; namely, the image of x by R. We say that a U-neighborhood of x is a set of the form R[x] for some entourage RU. Next, we call U-open any subset of X that is a U-neighborhood of each of its points. This forms a topology, which I will call the induced topology, or the topology induced by U. One also says that the quasi-uniformity U is compatible with a topology if and only if that topology is the induced topology.

For example, if d is a hemi-metric, a Ud-neighborhood of x is simply a subset of X that contains the open ball Bdx,<r for some r>0. The topology induced by Ud is therefore the open ball topology of d.

By the way, for every xX, the Uneighborhoods R[x] (RU) are neighborhoods of x in the topology induced by U. Indeed, consider the set U of all the points y ∈ X such that there is an SU such that S[y] is included in R[x]. This is open in the induced topology underlying U, by definition. In fact, it is easy to check that U is the interior of R[x] in that topology. Additionally, x is in U, because one can take R for S in that case. Hence R[x] indeed contains an open set U that contains x.

A natural question arises: which topologies arise as topologies induced by quasi-uniformity? The question was solved by Császár : all topologies whatsoever! The simplest known argument is due to Pervin , and is as follows. There are in general several quasi-uniformities that induce the same topology, and we will see an example below; another way of stating the following theorem is to say that there is at least one.

Theorem (Császár , Pervin ). Let X be a topological space. For every open subset U of X, let RU be the binary relation {(x,y) | xUyU}. The finite intersections of relations RU, UOX, form a base B of entourages of a quasi-uniformity on X, whose induced topology is exactly the topology of X. This quasi-uniformity is the Pervin quasi-uniformity of the topology of X.

Proof. We check the axioms for a base of entourages. First, (reflexivity) and (filter 3) are clear. In order to verify (relaxed transitivity), we verify the much stronger statement that RU o RURU for every open subset U of X. That inclusion just means that if xUyU and if yUzU, then xUzU. The relation RU o RURU will play an important rôle later in this post.

We now consider an arbitrary element SRU1 ∩ … ∩ RUn in B. We look for an element R of B such that R o RS, and we simply define R as S. Every pair (x,z) in R o R is such that there is a point y such that (x,y) and (y,z) are in R. Hence (x,y) and (y,z) are in RUi for every i, 1≤in, which shows that (x,z) is in RUi o RUiRUi for every i, 1≤in, hence that (x,z) is in R=S.

Let U be the quasi-uniformity generated by B.

Given any point x of X, and any basic entourage SRU1 ∩ … ∩ RUn, the image S[x] is equal to the intersection of the opens sets Ui that contain x. It follows that every U-open set is a neighborhood of each of its points, hence is open in the original topology on X. Conversely, let U be any open set in the original topology on X. For every x in U, RU[x]=U is a U-neighborhood of x included in U, so U is U-open. ☐

You may have noticed the similarity with Wilson’s theorem (Theorem 6.3.13 in the book), which says that every countably-based topological space is hemi-metrizable. You may have noticed that the proof strategy is very similar, too!

Indeed, each open subset U of X gives rise to a binary relation RU, but that relation has many properties. First and foremost, it is a preordering. Indeed, RU o RURU, as we have shown in the proof above, and that means that RU is transitive; and every entourage is reflexive by definition. Being a preordering, it induces a hemi-metric dU, defined by dU(x,y) ≝ 0 if (x,y) ∈ RU, and 1 (or ∞) otherwise. The open ball topology of that hemi-metric contains only U, the empty set, and the whole space.

Now remember that the topology defined by a countable family of hemi-metrics is hemi-metrizable (Lemma 6.3.11). What we have just said shows the following.

Proposition. Every topology is defined by a family of hemi-metrics.

Proof. Just take the hemi-metrics dU, where U ranges over the open subsets. ☐

As an aside, the topologies defined by a family of pseudo-metrics (symmetric hemi-metrics) are exactly the completely regular topologies, and are also exactly the topologies induces by uniformities. Let me allow not to say why this month, and to postpone it to a later post.

## Morphisms

At this point, one may be tempted to think that there is no point in inventing a new notion (quasi-uniform spaces), since quasi-uniform spaces and topological spaces seem to be one and the same thing.

The difference is in morphisms. While the morphisms in the category Top of topological spaces are the continuous maps, the morphisms from X to Y in the category QUnif of quasi-uniform spaces are the uniformly continuous maps, namely the maps f : XY such that (f×f)–1 (S) is an entourage of X for every entourage S of Y.

Every uniformly continuous map f is continuous for the underlying induced topologies. Indeed, let V be any open subset of Y. For every xf–1(V), we need to show that there is a neighborhood of X whose image by f is included in V. We use the fact that f(x) is in V, and the definition of the induced topology on Y, to obtain an entourage S in Y such that S[f(x)] ⊆ V. Let R be the entourage (f×f)–1 (S). Then R[x] is a neighborhood of x, by definition of the induced topology on X, and every element of its image by f is of the form f(y) with (x,y) ∈ R; so (f(x),f(y)) is in S, by definition of R, showing that f(y) is in S[f(x)], hence in V.

For quasi-metric spaces X and Y, with respective quasi-metrics dX and dY, a uniformly continuous map f : XY between the corresponding quasi-uniform spaces is a map such that for every s>0, (f×f)–1 ([<s]) contains a basic entourage [<r] for some r>0. In other words: for every s>0, there is an r>0 such that for all points x, y such that dX(x,y)<r, dY(f(x),f(y))<s. This is the usual definition of uniform continuity in (quasi-)metric spaces.

It is well-known that there are continuous maps between metric spaces that are not uniformly continuous, for example the map x ↦ 1/x on the positive real numbers. Hence that also holds in the realm of (quasi-)uniform spaces.

Rather remarkably, uniform continuity and continuity coincide on Pervin quasi-uniformities arising from topologies. Indeed, let X and Y be two topological spaces, and consider them with their Pervin quasi-uniformities. Let f : XY be a continuous map. For every basic entourage SRV1 ∩ … ∩ RVn in Y, (f×f)–1 (S) is the set of pairs (x,y) of points of X such that f(x) ∈ V1 implies f(y) ∈ V1 and … and f(x) ∈ Vn implies f(y) ∈ Vn, in other words (f×f)–1 (S) is the basic entourage RU1 ∩ … ∩ RUn where U1f–1(V1), …, Unf–1(Vn).

What all that means, categorically, is:

• There is a (forgetful) functor U : QUnifTop that maps every quasi-uniform space to the underlying topological space (with the induced topology), and every uniformly continuous map f to itself, seen as a continuous map.
• There is a functor Perv : TopQUnif that maps every topological space to itself, seen as a quasi-uniform space with the Pervin quasi-uniformity of its topology. It maps every continuous map f to itself, since f is uniformly continuous as well, as we have just seen.
• The composite functor U o Perv is the identity functor on Top: indeed, that is what the Császár-Pervin theorem above states.

The composite functor Perv o U does not seem to have any notable feature, and in particular there does not seem to be any adjunction involving Perv and U. Notably, the Pervin quasi-uniformity of a topology O is in general neither the smallest nor the largest quasi-uniformity that induces O.

Here is a counterexample, again due to Pervin . Consider the quasi-metric dR on the real numbers: dR(a,b)≝max(ab,0). That induces a quasi-uniformity, whose basic entourages are [<r] = {(a,b) ∈ R × R | a<b+r}, r>0. The topology induced by this quasi-uniformity is the open ball topology of dR, which is the Scott topology on R. The Pervin quasi-uniformity of this Scott topology has basic entourages of the form RU1 ∩ … ∩ RUn where each Ui is of the form ]ai, ∞[.

• No such entourage contains any quasi-metric entourage [<r]: any pair (a,b) where b is strictly larger than every ai, and ab+r is in RU1 ∩ … ∩ RUn but not in [<r].
• In the other direction, we claim that no quasi-metric entourage [<r] contains any Pervin entourage RU1 ∩ … ∩ RUn (where Ui = ]ai, ∞[). if n=0, this is clear, since the empty intersection is the whole of R × R. Otherwise, let ε>0 be chosen such that ε<r. Then the point (a1+ε,a1) is in [<r], but not in RU1 ∩ … ∩ RUn, in fact not even in RU1.

It follows that the two (Pervin and quasi-metric) quasi-uniformities on R are incomparable.

One may, at least, attempt to identify what kinds of quasi-uniform spaces can be obtained by applying Perv to arbitrary topological spaces. I do not know the answer to that question. However, if we are prepared to replace topological spaces by sets equipped with a lattice of subsets (not necessarily closed under arbitrary unions), Perv generalizes naturally, and the answer to the corresponding question is known, as I will now argue. The corresponding quasi-uniform spaces are called Pervin spaces.

## Pervin spaces

As far as I know, Pervin spaces were introduced by Mai Gehrke, Serge Grigorieff and Jean-Éric Pin, see . They form a generalization of the Pervin quasi-uniformities introduced in the Császár-Pervin theorem: we now define the Pervin quasi-uniformity of any family of subsets of a set X whatsoever, not necessarily a topology. The formula is the same: for any subset B of X, RB is the set of pairs (x,y) such that xB implies yB. Accordingly, the Császár-Pervin theorem has the following generalization.

Theorem. Let F be a family of subsets of a set X. The finite intersections of relations RB, BF, form a base B of entourages of a quasi-uniformity U on X, whose induced topology is exactly the topology O generated by F on X. This quasi-uniformity is the Pervin quasi-uniformity of the family F.

Proof. The proof that U is a quasi-uniformity is as in the Császár-Pervin theorem. Notably, we need to observe that RB o RBRB, namely that RB is a transitive binary relation on X for every B in F. This implies that every basic entourage RB1 ∩ … ∩ RBn is transitive as well.

Given any point x of X, and any basic entourage SRB1 ∩ … ∩ RBn, the image S[x] is equal to the intersection of the sets Bi that contain x. It follows that every U-open set is a neighborhood of each of its points (relative to the topology O), hence is open in O. Conversely, let U be any open set in O. For every x in U, there is a finite intersection B1 ∩ … ∩ Bn of elements of F that contains x and is included in U. Let SRB1 ∩ … ∩ RBn. Then S[x] = B1 ∩ … ∩ Bn is a U-neighborhood of x included in U, so U is U-open. ☐

In the proof, we have noticed that the Pervin quasi-uniformity U of F has a base of transitive entourages (namely, those of the form RB1 ∩ … ∩ RBn). The quasi-uniformities with that property are called… transitive.

Here is another property of U. Given any entourage R in U, R contains some basic entourage RB1 ∩ … ∩ RBn. For every subset I of {1, …, n}, let CI be the set obtained as the intersection of the sets Bi with iI, and of the complements of the sets Bi with iI. The family of all the sets CI, where I ranges over the subsets of {1, …, n}, forms a finite cover of the whole space X. Moreover, CI × CI is included in RB1 ∩ … ∩ RBn, hence in R. Indeed, for every pair (x,y) in CI × CI, for every index i, if iI then (x,y) is in RBi because y is in Bi, and if iI then (x,y) is in RBi because x is not in Bi.

A quasi-uniformity U on a set X is totally bounded if and only if for every entourage RU, there is a finite cover of X such that, for every element C of that cover, C × C is included in R. We have just argued that the Pervin quasi-uniformity of a family F of subsets is always transitive and totally bounded.

It is instructive to see what this notion of ‘totally bounded’ means when restricted to a hemi-metric space X, d. For every subset C of X, for every r>0, C × C is included in [<r] if and only if, for all x, y in C, d(x,y)<r. Every such subset C is either empty, or included in the open ball with center x and radius r under the symmetrized pseudo-metric dsym, where x is any given point of C. Hence, if X, d is totally bounded qua quasi-uniform space, then for every r>0, there is a finite cover of X by (sets included in) open balls of radius r under dsym. In other words, X, d is totally bounded qua hemi-metric space (compare with Definition 6.7.4 in the book).

Conversely, if X, d is totally bounded qua hemi-metric space, then for every r>0, X has a cover by open balls B of radius r/2 under dsym, and each of these open balls B satisfies B × B ⊆ [<r]. Therefore X, d is totally bounded qua quasi-uniform space. In summary, ‘totally bounded’ really means the same thing as the usual notion on hemi-metric spaces.

Let us return to Pervin quasi-uniformities. The following is due to Mai Gehrke, Serge Grigorieff and Jean-Éric Pin (April 2012; no, I will not give any more precise reference, because the document I have says that it is not finished and therefore must remain confidential—it is fairly easy to find it on the Web nonetheless). The proof I will give is different, though.

Proposition. A quasi-uniformity U on a set X is the Pervin quasi-uniformity of a family of subsets of X if and only if it is transitive and totally bounded. In that case:

1. U is the Pervin quasi-uniformity of the family F of subsets of X of the form R[E] ≝ {yX | ∃xE, (x,y) ∈ R}, where R ranges over the transitive entourages in U and E ranges over the subsets of X;
2. F is a subbase of the topology O induced by U;
3. U is also the Pervin quasi-uniformity of the lattice L of finite unions of finite intersections of elements of F, and L is a base of O.

Proof. We have already proved the ‘only if’ direction. Let us assume that U is transitive and totally bounded.

We first note that every element of F is open (in O). Indeed, let us pick any such element. It is of the form R[E], where R is a transitive entourage, and E is a subset of X. For every point y of R[E], every point zR[y] is such that (y,z) ∈ R, and since (x,y) ∈ R for some x in E and R is transitive, (x,z) is also in R, showing that z is in R[E]. This shows that R[y] is a neighborhood of y included in R[E]. Since y is arbitrary in R[E], R[E] is open in O.

We claim that, given any element BR[E] of F, RB is in U. (Remember that R is assumed to be transitive, and in U, here.) Every pair (y,z) in R is in RB: if y is in B=R[E], then (x,y) is in R for some x in E, hence (x,z) is also in R, since R is transitive; so z is in B. This shows that R is included in RB. By (filter 2), RB is in U.

A similar argument shows that RA is in U for every A in L. Let us write A as the finite union (over i=1…m) of the finite intersection (over j=1…ni) of elements BijRij[Eij] of F. Let R be the intersection of all the relations Rij, when i and j vary. We claim that R is included in RA. As above, this will entail that RA is in U. Given any pair (y,z) in R such that y is in A, there is an index i such that for every j (1≤jni), y is in Bij=Rij[Eij]. Hence for every j, there is a point xj in Eij such that (xj,y) is in Rij. Since (y,z) is in R, hence in Rij, and since Rij is transitive, (xj,z) is also in Rij for every j. This shows that z is in the intersection of the sets Bij=Rij[Eij], 1≤jni, hence in A.

In the converse direction, we claim that every entourage R (in U) contains some basic Pervin entourage RB1 ∩ … ∩ RBn, where each Bi is in F. Since U is transitive, and up to the replacement of R by a smaller entourage, we may assume that R is transitive. We now use total boundedness. Let {C1, …, Cn} be some finite cover of X such that each product Ci × Ci is included in R. For each i, R[Ci] is in F, by definition. We let BiR[Ci]. For every pair (x,y) in RB1 ∩ … ∩ RBn, we claim that (x,y) is in R. Since {C1, …, Cn} is a cover, there is an index i such that x is in Ci, hence in the larger set Bi = R[Ci]. (Note that Ci is included in R[Ci] because R is reflexive.) Now (x,y) is in RBi, so y is also in Bi. It follows that there is a point z in Ci such that (z,y) is in R. Therefore (x,z) is in Ci × Ci, hence in R. It follows that (x,y) is also in R, since R is transitive. This finishes to show that RB1 ∩ … ∩ RBn is included in R.

All this establishes item 1, and also item 3. Item 2 is a consequence of the previous Theorem. ☐

Warning. The previous proposition does not show that U is the Pervin topology of the topology O, only of its subbase F, or of its base L. The latter is closed under finite unions, but not under arbitrary unions in general.

There is much more that one can say about quasi-uniform spaces… enough to fill a book, quite certainly. One should certainly read Hans-Peter Künzi’s introduction to the subject , which is already quite encyclopedic! I was sad to hear that he had recently passed away (see this page). I had met him once or twice; the last time in Leicester, UK, in 2016, and I certainly enjoyed the lively breakfast we had there one morning with his many students and postdocs.

1. Nicolas Bourbaki. Topologie générale (éléments de mathématique), chapitre 2. Springer, 2007. (Many other, previous editions, too.)
2. Ákos Császár. Fondements de la topologie générale. Gauthier-Villars, Paris, 1960.
3. William Joseph Pervin. Quasi-Uniformisation of Topological Spaces. Mathematische Annalen 147:316–317, 1962.
4. Jean-Éric Pin. Pervin Spaces. Slides of a talk given in Coimbra, Portugal, September 2016.
5. Hans-Peter Albert Künzi. An Introduction to Quasi-Uniform Spaces. In Beyond Topology, pages 239-304, volume 486 of the Contemporary Mathematics series, American Mathematical Society, 2009, edited by Frédéric Mynard and Elliott Pearl.

Jean Goubault-Larrecq (October 19th, 2020)