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Dieudonné completion $\mu X$ of a space $X$ is the completion of $X$ with respect to the maximal uniformity $U_X$ on $X$ compatible with the topology of $X$.

I failed to find references to explain this definition. How can we define such a completion for topological space $X$? $X$ must be a completely regular space?

TXC
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    Yes, $X$ must be completely regular: every uniformizable space is completely regular, so if there is a uniformity compatible with the topology of $X$, then $X$ is completely regular. – Brian M. Scott May 04 '13 at 11:56
  • Which part of the definition is causing trouble: getting the maximal uniformity, or getting the completion with respect to a particular uniformity? – Brian M. Scott May 04 '13 at 11:59
  • By theorem28 p196 Kellye's general topology, Each uniform space has a completion and by your note Space $X$ must be completely regular therefore $X$ has a completion but why completion of $X$ with respect to the maximal uniformity exists? This definition is too complicated for me, especially given that this definition is equivalent to other references. for example topologically completeness and topologically completion. – TXC May 04 '13 at 12:08
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    If $\langle X,\mathscr{U}\rangle$ is any uniform space, it has a uniform completion; that completion is constructed from the uniformity $\mathscr{U}$. If you start with a completely regular topological space $\langle X,\tau\rangle$, there are many uniformities on $X$ that generate $\tau$; one of them is the maximal uniformity, and that’s the one used here. (I can’t say anything about Kelley’s specific construction, since I don’t have the book.) – Brian M. Scott May 04 '13 at 12:45
  • Thank you. I understand exactly what you mean (First choose maximal uniformity on space $X$ then construct a completion on this uniformity). Dieudonné topological completion of space $X$ is explain in this paper. these two definitions are equivalent? – TXC May 04 '13 at 13:34
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    Yes, they are equivalent. – Brian M. Scott May 04 '13 at 13:49
  • @BrianM.Scott: Can i ask you to express your opinion about this question, Is Brian Rushton's answer true? – TXC Jun 01 '13 at 07:19
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    Sorry to be so late getting to this. It’s partly wrong, even with the added comments, and partly too vague to be very helpful. Making $X$ $T_1$ and perfectly normal is sufficient. – Brian M. Scott Jun 02 '13 at 04:50

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