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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.

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