# Is there an entorage $N\subseteq \{(x, y): d(x, y)<\delta\}$ in an uniformity of locally compact metric space $(X, d)$?

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.
• 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? – user479859 Feb 2 at 6:59
• @user479859 No. Consider $\mathbb{R}$ in the fine uniformity, which is not metrisable. – Henno Brandsma Feb 2 at 7:13
• 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. – user479859 Feb 2 at 9:08
• @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. – Henno Brandsma Feb 2 at 9:12