1
$\begingroup$

It is well known that every topological space is quasi-uniformizable, and every quasi-uniform space $(X, \mathcal U)$ induces a natural topology on $X$. Moreover, if $(X, \mathcal U)$ is a quasi-uniform space, then there is a family of quasi-pseudometrics $\{p_{\alpha}\}_{\alpha \in A}$ which generates the quasi-uniformity on $X$, in the sense that the sets $$V_{\alpha, \varepsilon} := \{(x,y) \in X^2 | p_{\alpha}(x,y) < \varepsilon\}$$ form a base for the quasi-uniformity $\mathcal U$ (one can also replace $<$ by $\leq$ in the definition of the $V_{\alpha, \varepsilon}$). Moreover, these quasi-pseudometrics are quasi-uniformly continuous, in the sense that to every $\varepsilon>0$, there is $U \in \mathcal U$ such that if $((x,y), (x', y') ) \in U \times U$, then $|p_{\alpha}(x,y) - p_{\alpha}(x', y') | < \varepsilon$. Further, if $\mathcal U$ is a uniformity, it is possible to take the $p_{\alpha}$ to be pseudometrics.

As per usual, when speaking of the topology on a (quasi)uniform space, the canonical topology generated by the (quasi)uniformity will always be meant.

Every uniform space is completely regular (hence regular); this is not true for quasi-uniform spaces by the preliminary remarks above. One way to show uniform spaces are regular would be as follows:

Let $(X, \mathcal U)$ be a uniform space. Take a family of uniformly continuous (in the sense described above) pseudometrics $\{p_{\alpha}\}_{\alpha}$ generating the uniformity. The sets $$F_{\alpha, \varepsilon}: = \{(x,y) \in X^2 | p_{\alpha}(x,y) \leq \varepsilon\}$$ form a base for the uniformity. These sets are closed in the product topology of $X \times X$, because they are the preimages of $[0, \varepsilon]$ under the maps $X \times X \rightarrow [0, \infty): (x,y) \mapsto p_{\alpha}(x,y)$, which are uniformly continuous (hence continuous). So for every $x \in X$, $\{F_{\alpha, \varepsilon}[x] \}_{\alpha, \varepsilon}$ forms a local base at $x$ consisting of closed sets. Regularity follows.

My question: where does this argument fail in the case of quasi-uniform spaces?

I suspect the argument fails because the following may not hold in quasi-uniform spaces (I do not have a counterexample, though): if $V$ is an entourage which is closed in the product topology $X \times X$, then $V[x] $ is a closed subset of $X$ for every $x \in X$. In uniform spaces, one way to show this is to use the fact that if $E \subseteq X \times X$, then $$\overline E = \bigcap_{U~ \text{symmetric entourage}} U \circ E \circ U.$$ Since $U^{-1}$ need not be an entourage whenever $U$ is an entourage in a quasi-uniformity, I do have my suspicions about $V[x]$ being closed whenever $V \in \mathcal U$ is closed in $X \times X$. Besides this point, I do not know where else the argument goes wrong, since any quasi-uniformly continuous map between quasi-uniform spaces is continuous in the induced topologies.

Thanks in advance for reading my question. If I need to clarify anything or add more details, do let me know.

$\endgroup$
2
$\begingroup$

if $V$ is an entourage which is closed in the product topology $X \times X$, then $V[x] $ is a closed subset of $X$ for every $x \in X$.

This is not the problem because $V[x]$ is closed as a homeomorphic copy of a closed subset $(\{x\}\times X)\cap V$ of $\{x\}\times X$. The problem is that a quasi-pseudometric generating a quasi-uniformity on $X$ is not necessarily continuous (because of assymetry of quasiuniforities). For instance, let $X$ be $\Bbb R^2$ endowed with a quasiuniformity with a base $\{U_n:n\in\Bbb N\}$, where $U_n=\big\{((x_1, y_1), (x_2, y_2))\in\Bbb R^2\times \Bbb R^2: (x_1,y_1)= (x_2,y_2) \vee \big(x_1<x_2<x_1+\tfrac 1n \wedge$ $ y_1<y_2<y_1+\tfrac 1n\big)\big\},$ for each $n\in\Bbb N$. Then $\mathcal U$ generates a non-regular topology on $X$. We cannot use a continuous function $f:X\times X\to\Bbb R_+$ such that $f(0,0)=0$ to define a base of $\mathcal U$ because for any $\varepsilon>0$ there exists $n\in\Bbb N$ such that $f(z,z’)<\varepsilon$ for any $z,z’\in U_n[0]$, but $U_n[0]\times U_n[0]\not\subset U_1$.

$\endgroup$
0
$\begingroup$

This is really meant as a (perhaps rambling) comment to Alex Ravsky's answer as well as some observations, but it is too long alas.

I think I found the source of my confusion: first, if $A$ and $B$ are quasi-uniform spaces and $f:A \to B$ is quasi-uniformly continuous, it's continuous in the induced topologies. Any quasi-uniform space $X$ is generated by a family of quasi-pseudometrics, but these quasi-pseudometrics are not necessarily quasi-uniformly continuous (or even continuous) as maps $X \times X \to [0, \infty)$, as Ravsky's example shows (here, $X \times X$ is given the initial quasi-uniformity with respect to the projection maps, which coincides with the product topology).

This contrasts with the uniform case, since the pseudometrics generating the uniformity can always be taken to be continuous (even uniformly continuous) with respect to the product $X \times X$. In the book Asymmetric Functional Analysis by Cobzas (this is where I first read about all this), a different definition of product quasi-uniformity is used: $X \times X$ is given quasi-uniformity generated by the base $\{U \times U | U \in \mathcal U\}$ (where $\mathcal U$ is a quasi-uniformity on $X$); see the proof of Prop. 1.1.50. Under this definition, any quasi-uniformity arises from a collection of quasi-pseudometrics which are quasi-uniformly continuous with respect to this quasi-uniformity generated by the base $\{U \times U | U \in \mathcal U\}$. However, this quasi-uniformity does not necessarily coincide with the product topology (this is a consequence of Ravsky's example), and as far as I know, it doesn't seem standard.

This also clears another point which was confusing me. Consider the following "proof" that quasi-uniform spaces are completely regular:

Let $(X, \mathcal U)$ be a quasi-uniform space. Suppose $x \in X$ and $F \subseteq X$ is closed. Take an entourage $V$ such that $V[x] \cap F = \emptyset$ (this is possible because $X \setminus F$ is open). By the pseudo-metrization lemma (see the theorem mentioned here Question on the Proof to the Pseudometrization Lemma for (Quasi) Uniform Spaces), we can take a quasi-pseudometric $d$ such that $\{(a,b) | d(a,b) < 1\} \subseteq V$. Then, $f: X \rightarrow [0,1], y \mapsto 1- \min\{1, d(x, y)\}$ is the desired map.

The error, though, is that $d$ may not be continuous, and so $f$ may not be continuous.

_

Now suppose a family of quasi-pseudometrics $\{d_{\alpha}\}$ generates a quasi-uniformity $\mathcal U$ on $X$. (That is, $V_{\alpha, \varepsilon} := \{(x,y) | d_{\alpha}(x,y) < \varepsilon\}$ is a base for $\mathcal U$). In the uniform case, we can immediately deduce that the $d_{\alpha}$ are uniformly continuous on $X \times X$ due to the following lemma (from Kelley):

In a uniform space, a pseudometric $d$ on $X$ is uniformly continuous iff $V_{d, \varepsilon} \in \mathcal U$ for every $\varepsilon>0$.

As it happens, symmetry of $d$ is used in the proof of the converse.

In short, the asymmetry of quasi-uniformities prevents us from repeating the argument from the uniform case verbatim, and it is not generally true that there is always a family of quasi-uniformly continuous quasi-pseudometrics generating a given quasi-uniformity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.