Let $X=\{x_n\}_{n\in \mathbb{N}}$ where $x_n=\sum_{i=1}^{n}(\frac{1}{i})$ given the metric $d$ inherited from $\mathbb{R}$. Also let $\mathcal{U}(d)$ be uniformity generated by metric $d$. We know that every entourage $D\in\mathcal{U}(d)$ is a neighborhood of $\Delta_X$.
In a paper, Authors claimed that since $X$ has discrete topology, there is a neighborhood $U$ of $\Delta_X$ such that $U_x=\{x \}$ for all $x\in X$. In the following, we give a proof: Since $x_{n-1}<x_n<x_{n+1}$, hence we can choice $\delta_n>0$ such that $B(x_n, \delta_n)=\{x_n\}$. Take $U= \bigcup_{n=1}^{\infty}B(x_n, \delta_n)\times B(x_n, \delta_n)$. Then $U$ is an open set and $\Delta_X\subseteq U$.
I do not know if my proof is true and I do not know if there is an entourage $D\in\mathcal{U}(d)$ with $D=U$.
Please help me to know it.
Indeed $X$ as a topological space is discrete: it's an infinite sequence that increases to $\infty$ and all such sequences (considered as sets) are discrete as a subspace of $\mathbb{R}$. Hence so is its square. This means that $D \subseteq X \times X$ is open and so its own open neighbourhood.
But any basic entourage under $d$, so any set $U(\varepsilon)=\{(x,x') \in X \times X: d(x,x') < \varepsilon\}$ can never just contain $D$, as there are always terms $x_n$ in $X$ closer together than $\varepsilon$, for any $\varepsilon>0$. So maximal uniformity for $X$ (the discrete one, of all subsets of $X^2$ containing $D$) is different from the induced by $d$, although the topologies are the same, which is probably the point of this exercise.