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Let $(X,\mathcal U)$ be a Hausdorff uniform space. Can $(X,\mathcal U)$ have a non-hausdorff completion?

  • Maybe Arthur Fischer's answer here can be adapted: http://math.stackexchange.com/questions/320629/is-the-closure-of-a-hausdorff-space-hausdorff – Seirios Mar 05 '13 at 23:13
  • have you used the characterization of Hausdorff uniform spaces?; $(X,\mathcal U)$ is Hausdorff iff $\bigcap \mathcal U=\Delta$. – Camilo Arosemena Serrato Mar 05 '13 at 23:22

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Let $d$ be the Euclidean metric on $\Bbb R$. Let $\mathcal D_d$ be the uniformity made by $d$ on the interval $X= (0,1]$. Let $$Y=[0,1]\cup \{10\}$$ and define $$\rho:Y^2\to[0,\infty)$$ $$(\forall x,y \in [0,1])\left(\rho (x,y)=d(x,y),\quad \rho (10,y)=d(0,y),\quad \rho (x,10)=d(x,0) \right)$$ $$\rho(10,10)=0$$ I think, I'm not sure, $\rho$ is a pseudometric on $Y$ and is an extension for $d$.

So the metric space $(X,d)$ is a subspace of the pseudometric space $(Y,\rho)$. I think $(X,\mathcal D_d)$ is a subsapce of $(Y,\mathcal D_\rho)$ too. The Sequence $(\frac{1}{n})$ tends to $0$ and $10$. So $X$ is dense in $Y$.

Let $(a_n)$ be a Cauchy sequence in $Y$. Then $a_n$ is convergent in $([0,1],d)$ and so in $(Y,\rho)$ . Therefore $(Y,\rho)$ is complete.

So $(Y,\rho)$ is a completion of $(X,d)$ and clearly it is not Hausdorff.

  • Is it true that a pseudometric space is complete if its Cauchy sequences converge? Is it sufficient? Whatever, we can say that $(Y,\rho)$ is complete thanks to compactness. – Seirios Apr 07 '13 at 09:52
  • yes in fact if any Cauchy sequence in a pseudometric space is convergent, then any Cauchy net is convergent too. –  Apr 07 '13 at 10:10
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    In fact, more generally, your argument can be used to show that every noncomplete metric space has a nonHausdorff completion. +1. – Seirios Apr 07 '13 at 12:09
  • for me there's a question: Is a minimal completion of a Hausdorff space, Hausdorff? –  Apr 08 '13 at 16:31
  • How do you define the minimality? – Seirios Apr 08 '13 at 20:30
  • $(Y,\mathcal D)$ is a minimal completion iff it is not a proper (uniform) subspace of any other completion. –  Apr 08 '13 at 20:35
  • But if $Y \subset Z$ are two completions of $X$, then $Z= \overline{X} \subset Y \subset Z$ because $Y$ is complete and so closed in $Z$, therefore $Z=Y$. Did I misunderstood something? – Seirios Apr 08 '13 at 21:13
  • closure in which space?! in the above example $[0,1]$ seems to be a minimal completion. –  Apr 08 '13 at 21:24
  • In fact I made a mistake: $Y$ is not closed in $Z$ if it is not Hausdorff. A possible solution may be: Let $Y$ be a completion of $X$. Define $\sim$ on $Y$ by $x \sim y$ iff $y \in V$ for all neighborhood of $x$. Then pick out one element modulo $\sim$ to get $Z \subset Y$. Because $X$ is Hausdorff, $X \subset Z$. Moreover, $Z$ seems to be complete (and it is Hausdorff by construction). – Seirios Apr 09 '13 at 07:02