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.
Best Answer
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 $U_n[0]\times U_n[0]\not\subset U_1$.