Let $(X,\mathcal{U})$ be a uniform space with Hausdorff completion (X',\mathcal{U}') (made by the minimal Cauchy filters). Since $X$ is uniform, $\mathcal{U}$ is generated by pseudometrics $(d_i)_{i \in I}$, which are uniformly continuous for $\mathcal{U}$. Consequently, there are for every $i \in I$ extensions d_i': X'\times X'\rightarrow \mathbb{R}_{\geq 0}, which can be proven to be pseudometrics using the fact that the image of $X$ in X' is dense. Is it true that U' is generated by the (d_i')_{i \in I} and if so, how do you prove it ? Necessarily, the uniformity generated by (d_i')_{i \in I} is part of \mathcal{U}', but how do you prove the converse ?
Any help will be appreciated.