Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.