totally disconnected compact Hausdorff uniform space Let $(X, \mathcal{U})$ be a compact and Hausdorff  uniform space. For $D\in\mathcal{U}$, the sequence $\{x_i\}_{i\in\mathbb{Z}}$
is called a $D$- chain if $(x_i, x_{i+1})\in D$ for all $i\in\mathbb{Z}$. 
Let $X$ be totally disconnected and $E\in\mathcal{U}$ be given. 
Is it true that 
There is $D\in\mathcal{U}$ such that for every $D$- chain $\{x_i\}_{i\in\mathbb{Z}}$, we have $(x_i, x_j)\in E$ for all $i, j\in\mathbb{Z}$?
Please help me to understand it.
 A: Yes, this is true.
The slightly tricky part is showing that a compact Hausdorff totally disconnected space is totally separated. See here: Any two points in a Stone space can be disconnected by clopen sets. This means for any distinct $x,y$ there is a clopen $U$ containing $x$ but not $y.$ Since the space is compact, $U$ is "uniformly clopen" meaning there is $D\in\mathcal U$ such that there is no $D$-chain from $x$ to $y.$
For each $D\in\mathcal U$ let $C_D\subseteq X^2$ be the set of pairs connected by a $D$-chain. Each $C_D$ is clopen, and the family $\{C_D\}$ is closed under intersection: $C_{D\cap D'}=C_D\cap C_{D'}.$ We're given a neighborhood $E$ of the diagonal $\Delta=\{(x,x)\mid x\in X\}$ and we want to show that some $C_D\setminus E$ is empty.
 We can assume $E$ is open which makes these sets $C_D\setminus E$ closed.
By the previous paragraph $\bigcap_{D\in\mathcal U} C_D=\Delta,$ so $\bigcap_{D\in\mathcal U} C_D\setminus E$ is empty. By compactness some $C_D\setminus E$ is empty.
