It is known that if $\mathcal{U}$ be a uniformity on $X$, then every entourage $N\in\mathcal{U}$ is an open set of diagonal $\Delta_X$, but the converse of it, is not true.

For instance, Consider $\mathbb{R}$ with usual metric $d$. For every $\epsilon>0$, let $$U^d_\epsilon:=\left\{(x, y)\in \mathbb{R}^2 : d(x, y)<\epsilon\right\}$$ Then the collection $$ \mathcal{U}_d=\left\{E\subseteq \mathbb{R}^2 : U_\epsilon^d\subseteq E, \text{ for some } \epsilon>0\right\}$$ is a uniformity on $\mathbb{R}$.

In this example, every element of $\mathcal{U}_d$ is a neighborhood of $\Delta_\mathbb{R}$ in $\mathbb{R}^2$ but $\left\{(x, y) : |x-y|<\frac{1}{1+|y|}\right\}$ is a neighborhood of $\Delta_\mathbb{R}$ but not a member of $\mathcal{U}_d$.

In my research, $(X, d)$ is a locally compact, $N=\{(x, y): d(x, y)<\delta \}$ and $\mathcal{U}$ is an uniformity of $X$. Can I say that there is $D\in\mathcal{U}$ with $D\subseteq N$?


No, an entourage need not be an open neighbourhood of $\Delta_X$, but it is a neighbourhood of $\Delta_X$. E.g. $\{(x,y): d(x,y) \le \varepsilon\}$ is not open in $X \times X$ in general but it is a neighbourhood of the diagonal as it contains $\{(x,y): d(x,y) < \varepsilon\}$ as a subset, which happens to be open. But an (even open) neighbourhood of $\Delta_X$ need not be an entourage, as your example indeed shows.

As to your final question: the answer is no, I think. Use another equivalent metric on $\mathbb{R}$ and its induced uniformity, to see this: if this property were true it would hold for $\mathbb{R}$ (locally compact) and then it would imply that all metrics on the reals would be uniformly equivalent, which they're not.

  • $\begingroup$ Thanks. Can we say that if metric space $(X, d)$ and uniform space $(X, \mathcal{U})$ induce the same topology, then $(X, \mathcal{U})$ is metrizable? $\endgroup$ – user479859 Feb 2 at 6:59
  • 1
    $\begingroup$ @user479859 No. Consider $\mathbb{R}$ in the fine uniformity, which is not metrisable. $\endgroup$ – Henno Brandsma Feb 2 at 7:13
  • $\begingroup$ I need to use your idea. In my research, I have a collection $\mathcal{A}= \{E[x]: x\in X\}$ for some an entourage $E\in\mathcal{U}$ and $(X, \mathcal{U})$ is compact uniform space. Can I say that $\mathcal{A}$ is an open cover of $X$. Thanks for your helps. $\endgroup$ – user479859 Feb 2 at 9:08
  • 1
    $\begingroup$ @user479859 No this need not be an open cover. $E[x]$ is not open, but it is a neighbourhood of $x$. So it is a cover by neighbourhoods so the interiors form an open cover and have a finite subcover. The corresponding $E[x]$, being larger, then also form a subcover. $\endgroup$ – Henno Brandsma Feb 2 at 9:12

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.