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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?

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  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Oct 12 '18 at 21:54
  • $\begingroup$ @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? $\endgroup$ – Keshav Srinivasan Oct 12 '18 at 22:38
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Oct 13 '18 at 4:46
  • $\begingroup$ @HennoBrandsma I posted a question on this, if you want to elaborate: math.stackexchange.com/q/2953476/71829 $\endgroup$ – Keshav Srinivasan Oct 13 '18 at 5:02
  • $\begingroup$ I expanded the above comment for this question. $\endgroup$ – Henno Brandsma Oct 13 '18 at 5:53
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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.

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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)

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