TITLE: non-hausdorff completion of a uniform space.
QUESTION [7 upvotes]: Let $(X,\mathcal U)$ be a Hausdorff uniform space. Can $(X,\mathcal U)$ have a non-hausdorff completion?

REPLY [4 votes]: 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.