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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 $D\subseteq N$?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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