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 $X$ be a uniform space. When the uniformity of $X$ is defined in terms of entourages, then $X$ is uniformly regular if it satisfies the following property:
With excluded middle, we can take $V$ to be $U$ 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 $X$ is defined in terms of uniform covers, then the equivalent condition is:
With excluded middle, we can take $C'$ to be $C$.
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 $U_{d,\varepsilon} = \{ (x,y) \mid d(x,y)\lt\varepsilon \}$, in which case we can take $V = \{ (x,y) \mid d(x,y)\lt\varepsilon/2 \}$ and use the fact that every located real number is either $\lt \varepsilon$ or $\gt \varepsilon/2$.
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 $U$ is any neighborhood of the identity $e$ in a topological group $X$, by regularity we have a neighborhood $V$ of $e$ and an open set $G$ with $V\cap G = \emptyset$ and $G\cup U = X$, and in particular $\neg V\cup U = X$, whence the entourages defined by $U$ and $V$ 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 $X$ be compact regular. Since compactness and regularity are properties of the open-set lattice, $X$ 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 $X\times X$ coincide. Furthermore, $X\times X$ is also compact regular.
Define the entourages to be all the neighborhoods of the diagonal in $X\times X$; this obviously satisfies all the axioms of a uniformity except possibly the square-root (and uniform regularity).
Let $U$ be a neighborhood of the diagonal in $X\times X$. Since $X\times X$ is regular, and using the definition of the product topology, for each $x\in X$ there is a square neighborhood $V_x\times V_x$ and an open set $G_x\subseteq X\times X$ such that $(V_x\times V_x)\cap G_x = \emptyset$ and $G_x \cup U = X\times X$. The open sets $V_x$ cover the compact space $X$, hence finitely many of them suffice. Let $V = (V_{x_1}\times V_{x_1}) \cup \cdots \cup (V_{x_n}\times V_{x_n})$, and $G = G_{x_1} \cap \cdots \cap G_{x_n}$. Then $V$ is a neighborhood of the diagonal, and $G$ an open set such that $G\cup U = X\times X$ and $V \cap G = \emptyset$. In particular, $\neg V \cup U \subseteq G\cup G = X\times X$, so this uniformity (if it exists) is uniformly regular.
Now, $G$ is the union of rectangular open neighborhoods $A\times B$. Note that since $G$ is disjoint from the diagonal, any $A\times B \subseteq G$ satisfies $A\cap B = \emptyset$. Moreover, since $X$ is regular, each $A$ and $B$ is the union of open sets that are well-inside it; thus $G$ is the union of rectangular open neighborhoods $C\times D$ such that there are opens $A,B$ with $C\triangleleft A$ and $D\triangleleft B$ and $A\cap B = \emptyset$. Since $G\cup U = X\times X$ and $X\times X$ is compact, we can cover it by $U$ together with finitely many such rectangles $C\times D$, say $X\times X = U \cup (C_1\times D_1) \cup\cdots \cup (C_m \times D_m)$.
For each of these rectangles $C_i\times D_i$, let $E_i$ be an open set such that $E_i \cup A_i \cup B_i = X$ and $E_i \cap (C_i \cup D_i) = \emptyset$, and let $W_i = (A_i\times A_i) \cup (B_i \times B_i) \cup (E_i \times E_I)$. Then $W_i$ is a neighborhood of the diagonal; let $W = W_1 \cap \cdots \cap W_m$.
We claim $W\circ W \subseteq U$. Let $(x,z) \in W\circ W$, so that there is a $y$ with $(x,y)\in W$ and $(y,z) \in W$. Now we have either $(x,z)\in U$ (which is what we want) or $(x,z) \in C_i\times D_i$ for some $i$, so it suffices to rule out the latter possibility. Since $(x,y)\in W$, we have $(x,y) \in W_i$, hence $(x,y)\in A_i\times A_i$ or $(x,y)\in B_i\times B_i$ or $(x,y)\in E_i\times E_i$. But $x\in C_i$, which is disjoint from $B_i$ and $E_i$, so it must be that $(x,y)\in A_i\times A_i$ and hence $y\in A_i$. Similarly, we conclude that $(y,z)\in B_i\times B_i$ so that $y\in B_i$. But $A_i\cap B_i = \emptyset$, a contradiction.
It remains to show that the uniform topology on $X$ is the original one. Clearly if $U$ is an entourage then each $U[x]$ is a neighborhood of $x$. Conversely, if $V$ is a neighborhood of $x$, then by regularity we have a neighborhood $W$ of $x$ and an open $G$ with $W\cap G = \emptyset$ and $G\cup V = X$. Then $Z = (V\times V) \cup (G\times G)$ is a neighborhood of the diagonal, i.e. an entourage, such that $Z[x] =V$.
Finally, we show this is the unique uniformity compatible with a compact regular topology. Suppose $X$ is any compact regular uniform space, and $U$ any neighborhood of the diagonal; we want to show $U$ is an entourage. For any $x\in X$, there is an entourage $V$ such that $V_x[x] \times V_x[x] \subseteq U$. Let $W_x$ be an entourage such that $W_x \circ W_x \subseteq V_x$. Since $X$ is compact, it is covered by finitely many of the neighborhoods $W_x[x]$, say $X = W_{x_1}[x_1] \cup \cdots\cup W_{x_k}[x_k]$. Let $W = W_1 \cap \cdots \cap W_k$; we claim $W\subseteq U$. For let $(y,z)\in W$; then we have $y\in W_{x_k}[x_k]$ for some $k$, whence also $z\in V_{x_k}[x_k]$, and thus $(y,z) \in V_{x_k}[x_k]\times V_{x_k}[x_k] \subseteq U$.
Every uniformly regular uniform space is a regular topological space in its uniform topology. For if $U[a]$ is a uniform neighborhood of $a$, let $V$ be an entourage such that $\neg V\cup U = X\times X$, and $W$ an entourage such that $W\circ W \subseteq V$. Then $W[a]$ is a neighborhood of $a$, and the interior of $W[\neg V[a]]$ is an open set $G$, such that $G\cap W[a] = \emptyset$ and $G\cup U[a] = X$.
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).
Recall that every uniform space $X$ has a inequality relation (an irreflexive symmetric relation) where “$x # y$” means that there exists an entourage $U$ such that $x \mathbin{≉}_U y$. If $X$ is uniformly regular, then this is an apartness relation, i.e. it is also a comparison. For if $x \mathbin{≉}_U z$, let $V \circ V \subseteq U$ and $\neg{W} \cup V = X \times X$. Then for any $y$, we cannot have both $x \approx_V y$ and $y\approx_V z$, so we must have either $x \mathbin{≉}_W y$ or $y \mathbin{≉}_W z$, whence either $x # y$ or $y # z$. (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 $\varepsilon$ apart” rather than “at least $\varepsilon$ close”.
A uniform apartness space is a set $X$ equipped with a collection of binary relations $A\subseteq X\times X$ called anti-entourages such that
If $X$ 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. $A$ is an anti-entourage if $A \subseteq \neg U = (X\setminus U)$ for some entourage $U$. Most of the axioms are obvious. For axiom (2), we use transitivity of a uniformity: if $V\circ V \subseteq U$, then if $(x,z)\notin U$ and $\neg((y,z)\notin V)$, then if $(x,y)\in V$ we can derive a contradiction by assuming $(y,z)\in V$ (double negations can be removed constructively when the goal is a contradiction) and obtaining $(x,y)\in U$ from $V\circ V = U$; thus $(x,y)\notin V$. Note also that axiom (5) needs only the direction of de Morgan’s law that is true constructively, $\neg U \cup \neg V \subseteq \neg (U\cap V)$.)
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 $A$ there is an anti-entourage $B$ such that $\neg A\cup B = X\times X$: if $A\subseteq \neg U$, let $B = \neg V$ where $U\cup \neg V = X\times X$.
(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 $A$, there exists an anti-entourage $B$ such that whenever $(x,z)\in A$, for any $y$ we have either $(x,y)\in B$ or $(y,z)\in B$. To see this, first let $C$ be for $A$ as in the standard axiom (2), then let $B$ be for $C$ as in uniform regularity. Now given $(x,z)\in A$, we have either $(y,z)\in B$ or $(y,z)\notin C$; but the latter implies $(x,y)\in C$ and hence $(x,y)\in B$.)
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 $U$ in a uniformly regular uniform space contains the double-complement $\neg\neg V$ of some entourage $V$, and dually for any anti-entourage $A$ in a uniformly regular uniform apartness space there is an anti-entourage $B$ containing $\neg\neg A$. But for the first it suffices to have $U\cup \neg V = X\times X$, and for the second it suffices to have $\neg A \cup B = X\times X$. Thus, there is a bijection between uniformly regular uniformities and uniformly regular uniform apartnesses on any set $X$.
We define a function $f:X\to Y$ between uniform apartness spaces to be uniformly continuous if for every anti-entourage $B$ in $Y$, there is an anti-entourage $A$ in $X$ such that if $(f(x),f(y))\in B$ then $(x,y)\in A$. Since a conditional implies its contrapositive, the constructions above become functors $Unif \to AUnif$ and $AUnif \to Unif$. Moreover, we have:
The above functors form an idempotent adjunction $Unif \rightleftarrows AUnif$ over $Set$, 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 $Set$.
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 April 5, 2017 at 23:52:05. See the history of this page for a list of all contributions to it.