# What is an example of a compact non-uniformizable space?

Every compact Hausdorff space is uniformizable. But I don’t think every compact space is uniformizable. So my question is, what is an example of a compact non-uniformizable space?

• As an aside: every compact Hausdorff space is even uniquely uniformisable: there is exactly one uniformity that will induce its topology. This also holds for the non-compact space $\omega_1$ in the order topology. – Henno Brandsma Oct 12 '18 at 21:54
• @HennoBrandsma What other examples are there of noncompact spaces which are uniquely uniformizable? Is there a general result about under what conditions a space is uniquely uniformizable? – Keshav Srinivasan Oct 12 '18 at 22:38
• They are exactly the spaces that are so-called "almost compact": there is a unique compactification of $X$ (up to equivalence). Exercises 8.5.11 and 12 in Engelking "General Topology" (2nd ed.) have references and a bit more. It's not a well-studied class.See also Gilman and Jerrison exercises 6J and 15R for more equivalent formulations. – Henno Brandsma Oct 13 '18 at 4:46
• @HennoBrandsma I posted a question on this, if you want to elaborate: math.stackexchange.com/q/2953476/71829 – Keshav Srinivasan Oct 13 '18 at 5:02
• I expanded the above comment for this question. – Henno Brandsma Oct 13 '18 at 5:53

## 2 Answers

Recall that the topology on every uniform space is completely regular, and hence regular.

Let $$X$$ be Sierpinski space, i.e. $$X=\{0,1\}$$ with the topology $$\{\emptyset,X,\{0\}\}$$. Then $$X$$ is trivially compact, but is not regular because $$\{1\}$$ is closed and doesn't contain $$0$$, but every open set containing $$\{1\}$$ also contains $$0$$. Thus, $$X$$ is not uniformizable.

• Is there a less trivial example? – Keshav Srinivasan Oct 12 '18 at 13:51
• – Robert Thingum Oct 12 '18 at 13:53
• OK thanks for your answer! – Keshav Srinivasan Oct 12 '18 at 13:59

A topology from a uniform space is $$R_0$$ (also called symmetric). This means that topologically distinct points can be separated by disjoint open sets, where points $$x \neq y$$ are topologically indistinguishable iff $$\overline{\{x\}} = \overline{\{y\}}$$ (or equivalently if for every open set $$O$$ we have $$x \in O$$ iff $$y \in O$$). So e.g. a $$T_1$$ space that is $$R_0$$ is also Hausdorff (because then the closures of singletons are singletons, so all points are distinguishable and must be "separatable"). The same holds for $$T_0$$ spaces (because $$T_0$$ can be seen as equivalent to "all distinct points of $$X$$ are topologically distinguishable"). This means that the following are standard examples of compact non-$$R_0$$ spaces: The cofinite topology on any infinite set, the excluded point topology on any set, as they are $$T_0$$ and not $$T_2$$.

That a topology from a uniformity is $$R_0$$ is a standard fact (and follows from the fact that every entourage $$D$$ contains a symmetric entourage $$E$$, i.e. one where $$E = E^{-1}$$).

Any indiscrete space is uniformisable (trivially) and is an example of a compact uniformisable space that is not Hausdorff. But if $$X$$ is uniformisable then $$T_0$$, $$T_1$$, $$T_2$$, $$T_3$$ and Tychonoff are all equivalent properties (if you have one, you have them all), so in a way, indiscrete is the best you can do as far as non-Hausdorff examples go (finite sums of them, too, if you like)