nLab
uniformly regular space

Uniformly regular spaces

Idea

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.)

Definition

Let XX be a uniform space. When the uniformity of XX is defined in terms of entourages, then XX is uniformly regular if it satisfies the following property:

  • For every entourage UU, there exists an entourage VV such that ¬VU=X×X\neg{V} \cup U = X \times X, where ¬V\neg{V} is the complement of VV. That is,
    U,V,x,y,x Uyx Vy. \forall U,\; \exists V,\; \forall x, y,\; x \approx_U y \;\vee\; x \mathbin{≉}_V y .

With excluded middle, we can take VV to be UU 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 XX is defined in terms of uniform covers, then the equivalent condition is:

  • If CC is a uniform cover, there exists a uniform cover CC' such that, for any two points x,yx, y, either x,yAx, y \in A for some ACA \in C or x,yBx, y \in B for no BCB \in C'.

With excluded middle, we can take CC' to be CC.

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.

Examples

  • 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 U d,ε={(x,y)d(x,y)<ε}U_{d,\varepsilon} = \{ (x,y) \mid d(x,y)\lt\varepsilon \}, in which case we can take V={(x,y)d(x,y)<ε/2}V = \{ (x,y) \mid d(x,y)\lt\varepsilon/2 \} and use the fact that every located real number is either <ε\lt \varepsilon or >ε/2\gt \varepsilon/2.

  • 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 UU is any neighborhood of the identity ee in a topological group XX, by regularity we have a neighborhood VV of ee and an open set GG with VG=V\cap G = \emptyset and GU=XG\cup U = X, and in particular ¬VU=X\neg V\cup U = X, whence the entourages defined by UU and VV 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.

Compact spaces

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.

Theorem

Every compact regular space admits a unique compatible uniformity that is uniformly regular.

Proof

Let XX be compact regular. Since compactness and regularity are properties of the open-set lattice, XX 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 X×XX\times X coincide. Furthermore, X×XX\times X is also compact regular.

Define the entourages to be all the neighborhoods of the diagonal in X×XX\times X; this obviously satisfies all the axioms of a uniformity except possibly the square-root (and uniform regularity).

Let UU be a neighborhood of the diagonal in X×XX\times X. Since X×XX\times X is regular, and using the definition of the product topology, for each xXx\in X there is a square neighborhood V x×V xV_x\times V_x and an open set G xX×XG_x\subseteq X\times X such that (V x×V x)G x=(V_x\times V_x)\cap G_x = \emptyset and G xU=X×XG_x \cup U = X\times X. The open sets V xV_x cover the compact space XX, hence finitely many of them suffice. Let V=(V x 1×V x 1)(V x n×V x n)V = (V_{x_1}\times V_{x_1}) \cup \cdots \cup (V_{x_n}\times V_{x_n}), and G=G x 1G x nG = G_{x_1} \cap \cdots \cap G_{x_n}. Then VV is a neighborhood of the diagonal, and GG an open set such that GU=X×XG\cup U = X\times X and VG=V \cap G = \emptyset. In particular, ¬VUGG=X×X\neg V \cup U \subseteq G\cup G = X\times X, so this uniformity (if it exists) is uniformly regular.

Now, GG is the union of rectangular open neighborhoods A×BA\times B. Note that since GG is disjoint from the diagonal, any A×BGA\times B \subseteq G satisfies AB=A\cap B = \emptyset. Moreover, since XX is regular, each AA and BB is the union of open sets that are well-inside it; thus GG is the union of rectangular open neighborhoods C×DC\times D such that there are opens A,BA,B with CAC\triangleleft A and DBD\triangleleft B and AB=A\cap B = \emptyset. Since GU=X×XG\cup U = X\times X and X×XX\times X is compact, we can cover it by UU together with finitely many such rectangles C×DC\times D, say X×X=U(C 1×D 1)(C m×D m)X\times X = U \cup (C_1\times D_1) \cup\cdots \cup (C_m \times D_m).

For each of these rectangles C i×D iC_i\times D_i, let E iE_i be an open set such that E iA iB i=XE_i \cup A_i \cup B_i = X and E i(C iD i)=E_i \cap (C_i \cup D_i) = \emptyset, and let W i=(A i×A i)(B i×B i)(E i×E I)W_i = (A_i\times A_i) \cup (B_i \times B_i) \cup (E_i \times E_I). Then W iW_i is a neighborhood of the diagonal; let W=W 1W mW = W_1 \cap \cdots \cap W_m.

We claim WWUW\circ W \subseteq U. Let (x,z)WW(x,z) \in W\circ W, so that there is a yy with (x,y)W(x,y)\in W and (y,z)W(y,z) \in W. Now we have either (x,z)U(x,z)\in U (which is what we want) or (x,z)C i×D i(x,z) \in C_i\times D_i for some ii, so it suffices to rule out the latter possibility. Since (x,y)W(x,y)\in W, we have (x,y)W i(x,y) \in W_i, hence (x,y)A i×A i(x,y)\in A_i\times A_i or (x,y)B i×B i(x,y)\in B_i\times B_i or (x,y)E i×E i(x,y)\in E_i\times E_i. But xC ix\in C_i, which is disjoint from B iB_i and E iE_i, so it must be that (x,y)A i×A i(x,y)\in A_i\times A_i and hence yA iy\in A_i. Similarly, we conclude that (y,z)B i×B i(y,z)\in B_i\times B_i so that yB iy\in B_i. But A iB i=A_i\cap B_i = \emptyset, a contradiction.

It remains to show that the uniform topology on XX is the original one. Clearly if UU is an entourage then each U[x]U[x] is a neighborhood of xx. Conversely, if VV is a neighborhood of xx, then by regularity we have a neighborhood WW of xx and an open GG with WG=W\cap G = \emptyset and GV=XG\cup V = X. Then Z=(V×V)(G×G)Z = (V\times V) \cup (G\times G) is a neighborhood of the diagonal, i.e. an entourage, such that Z[x]=VZ[x] =V.

Finally, we show this is the unique uniformity compatible with a compact regular topology. Suppose XX is any compact regular uniform space, and UU any neighborhood of the diagonal; we want to show UU is an entourage. For any xXx\in X, there is an entourage VV such that V x[x]×V x[x]UV_x[x] \times V_x[x] \subseteq U. Let W xW_x be an entourage such that W xW xV xW_x \circ W_x \subseteq V_x. Since XX is compact, it is covered by finitely many of the neighborhoods W x[x]W_x[x], say X=W x 1[x 1]W x k[x k]X = W_{x_1}[x_1] \cup \cdots\cup W_{x_k}[x_k]. Let W=W 1W kW = W_1 \cap \cdots \cap W_k; we claim WUW\subseteq U. For let (y,z)W(y,z)\in W; then we have yW x k[x k]y\in W_{x_k}[x_k] for some kk, whence also zV x k[x k]z\in V_{x_k}[x_k], and thus (y,z)V x k[x k]×V x k[x k]U(y,z) \in V_{x_k}[x_k]\times V_{x_k}[x_k] \subseteq U.

Properties

Regularity

Every uniformly regular uniform space is a regular topological space in its uniform topology. For if U[a]U[a] is a uniform neighborhood of aa, let VV be an entourage such that ¬VU=X×X\neg V\cup U = X\times X, and WW an entourage such that WWVW\circ W \subseteq V. Then W[a]W[a] is a neighborhood of aa, and the interior of W[¬V[a]]W[\neg V[a]] is an open set GG, such that GW[a]=G\cap W[a] = \emptyset and GU[a]=XG\cup U[a] = X.

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).

Apartness

Recall that every uniform space XX has a inequality relation (an irreflexive symmetric relation) where “x#yx # y” means that there exists an entourage UU such that x Uyx \mathbin{&#8777;}_U y. If XX is uniformly regular, then this is an apartness relation, i.e. it is also a comparison. For if x Uzx \mathbin{&#8777;}_U z, let VVUV \circ V \subseteq U and ¬WV=X×X\neg{W} \cup V = X \times X. Then for any yy, we cannot have both x Vyx \approx_V y and y Vzy\approx_V z, so we must have either x Wyx \mathbin{&#8777;}_W y or y Wzy \mathbin{&#8777;}_W z, whence either x#yx # y or y#zy # z. (In fact, this is a special case of the fact that any regular space admits an apartness relation, discussed at regular space).

Locatedness

Uniform regularity has some nice consequences for the existence of (almost/uniformly) located subsets: see there for details.

Uniform apartness spaces

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 ε\varepsilon apart” rather than “at least ε\varepsilon close”.

Definition

A uniform apartness space is a set XX equipped with a collection of binary relations AX×XA\subseteq X\times X called anti-entourages such that

  1. (x,x)A(x,x)\notin A for xXx\in X and anti-entourages AA, i.e. each anti-entourage is irreflexive.
  2. For every anti-entourage AA, there exists an anti-entourage BB such that whenever (x,z)A(x,z)\in A and (y,z)B(y,z)\notin B we have (x,y)B(x,y)\in B.
  3. For every anti-entourage AA, there exists an anti-entourage BB such that AB opA\subseteq B^{op} (hence, by axiom (6), A opA^{op} is an anti-entourage).
  4. There exists an anti-entourage (hence, by axiom (6), the empty set is an anti-entourage).
  5. If AA and BB are anti-entourages, so is some superset of ABA\cup B (hence, by axiom (6), ABA\cup B is itself).
  6. If AA is an anti-entourage and BAB\subseteq A, then BB is an anti-entourage.

If XX 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. AA is an anti-entourage if A¬U=(XU)A \subseteq \neg U = (X\setminus U) for some entourage UU. Most of the axioms are obvious. For axiom (2), we use transitivity of a uniformity: if VVUV\circ V \subseteq U, then if (x,z)U(x,z)\notin U and ¬((y,z)V)\neg((y,z)\notin V), then if (x,y)V(x,y)\in V we can derive a contradiction by assuming (y,z)V(y,z)\in V (double negations can be removed constructively when the goal is a contradiction) and obtaining (x,y)U(x,y)\in U from VV=UV\circ V = U; thus (x,y)V(x,y)\notin V. Note also that axiom (5) needs only the direction of de Morgan’s law that is true constructively, ¬U¬V¬(UV)\neg U \cup \neg V \subseteq \neg (U\cap V).)

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 AA there is an anti-entourage BB such that ¬AB=X×X\neg A\cup B = X\times X: if A¬UA\subseteq \neg U, let B=¬VB = \neg V where U¬V=X×XU\cup \neg V = X\times X.

(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 AA, there exists an anti-entourage BB such that whenever (x,z)A(x,z)\in A, for any yy we have either (x,y)B(x,y)\in B or (y,z)B(y,z)\in B. To see this, first let CC be for AA as in the standard axiom (2), then let BB be for CC as in uniform regularity. Now given (x,z)A(x,z)\in A, we have either (y,z)B(y,z)\in B or (y,z)C(y,z)\notin C; but the latter implies (x,y)C(x,y)\in C and hence (x,y)B(x,y)\in B.)

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 UU in a uniformly regular uniform space contains the double-complement ¬¬V\neg\neg V of some entourage VV, and dually for any anti-entourage AA in a uniformly regular uniform apartness space there is an anti-entourage BB containing ¬¬A\neg\neg A. But for the first it suffices to have U¬V=X×XU\cup \neg V = X\times X, and for the second it suffices to have ¬AB=X×X\neg A \cup B = X\times X. Thus, there is a bijection between uniformly regular uniformities and uniformly regular uniform apartnesses on any set XX.

We define a function f:XYf:X\to Y between uniform apartness spaces to be uniformly continuous if for every anti-entourage BB in YY, there is an anti-entourage AA in XX such that if (f(x),f(y))B(f(x),f(y))\in B then (x,y)A(x,y)\in A. Since a conditional implies its contrapositive, the constructions above become functors UnifAUnifUnif \to AUnif and AUnifUnifAUnif \to Unif. Moreover, we have:

Theorem

The above functors form an idempotent adjunction UnifAUnifUnif \rightleftarrows AUnif over SetSet, 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 SetSet.

Proof

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.

category: topology

Last revised on April 5, 2017 at 23:52:05. See the history of this page for a list of all contributions to it.