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  • $(X,\mathcal{D})$ is a uniform space.
  • for each nonempty subset $A\subseteq X$, if the subspace $(A,\mathcal{D}_A)$ is complete then $A$ is closed.

Is $(X,\mathcal{D})$ Hausdorff?

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I am not sure, but for me the second condition seems to be always satisfied, no? – Seirios Feb 5 '13 at 18:12
@Seirios: I have a proof for 2 which uses Hausdorffness. I wonder if it is needed really. – user59671 Feb 5 '13 at 18:18
As you noticed, my proof used Hausdorffness; I removed it. In fact, there exist (nonHausdorff) uniform spaces in which the property fails. – Seirios Feb 5 '13 at 23:07
but it contained useful info. – user59671 Feb 5 '13 at 23:12
up vote 3 down vote accepted

First, notice that $X$ is $T_0$: the uniform structure induced on a singleton make it a complete uniform space, so any singleton is closed. Then, Hausdorffness follows from the results:

Theorem 1: A topological space is uniformizable iff it is $T_{3 \frac{1}{2}}$.

Theorem 2: $T_0+ T_{3 \frac{1}{2}}$ is equivalent to $T_2+T_{3 \frac{1}{2}}$.

I can add some indications if needed. For example, the second theorem results from a construction which showed that any $T_0+T_{3 \frac{1}{2}}$ topological space is homeomorphic to a subspace of some cube $[0,1]^I$.

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I think again you use Hausdorffness.$\mathfrak{B}$ is Cauchy on $A$ so it is convergent in $A$ but why does it converge to $x$? it may converge to some other point in $A$ if the space is not Hausdorff. – user59671 Feb 5 '13 at 19:45
@CutieKrait is right. Let $X=\{0,1\}$ with the trivial uniformity that induces the indiscrete topology, and let $A=\{0\}$. – Brian M. Scott Feb 5 '13 at 22:57
I completely edited my answer. – Seirios Feb 5 '13 at 23:27
great. any $T_0$ uniform space is $T_2$ too. – user59671 Feb 5 '13 at 23:46
+1. Just a comment: I much prefer the notation that makes the $T_\alpha$ properties a hierarchy, so that $T_{3 \frac{1}{2}}$ is completely regular + $T_0$. I would say that $X$ is uniformizable iff it’s completely regular, and Hausdorff and uniformizable iff it’s Tikhonov = $T_{3 \frac{1}{2}}$. – Brian M. Scott Feb 5 '13 at 23:50

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