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The long line $L$ is uniformizable; in fact, as $L$ is a locally compact Hausdorff space we can explicitly write down a uniformity for it: If $\hat{L}$ is the one-point compactification of $L$, then $\hat{L}$ is compact Hausdorff, and so it has a natural uniformity that we can restrict to $L$.

But this is not a complete uniformity; its completion is $\hat{L}$. Is there a complete uniformity on $L$?

One well-known result is that any paracompact Hausdorff space is completely uniformizable. In particular, any manifold is completely uniformizable. Does the same apply to spaces like $L$, or are they "too noncompact" to have a complete uniformity?

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There is only one compatible uniformity with the topology on $L$. This follows from this paper or the well-known result (reference anyone?) that locally compact Tychonoff spaces have a unique uniformity iff the one-point compactification is the same as the Cech-Stone compactification. So $L$ has a unique uniformity that is totally bounded and not complete.

Added: I misinterpreted $L$ as being the long ray $R$ (which is the space $\omega_1 \times [0,1)$, in the lexicographic order and order topology) and then the result holds (the unicity). The long line is two copies of the long ray, one ordered reversely, the other as above, glued together at their endpoints $(0,0)$.

$L$ has two compactifications: the ordered compactification, which adds two points (one at $+\omega_1$ the other $-\omega_1$, as it were), and the one-point compactification of $L$. The former also equals the Cech-Stone compactification $\beta(L)$. This is by the well-known fact that every real-valued function on the long ray is eventually constant. The constants can differ between the left and the right side, and so the classification in the previous paper (the last theorem) shows that $L$ has at least two compatible uniformities (e.g. there are continuous functions $f$ to the reals such that neither $\{x: f(x) \le 0 \}$ and $\{x: f(x) \ge 1\}$ are compact). (The Cech-Stone compactification for $L^n$ is $\beta(L)^n$ because of Glicksberg's theorem (as $L$ is pseudo-compact and locally compact)).

So this still leaves open the question whether $L$ has a compatible complete uniformity (my guess is still no). Does it have exactly two (my guess is yes)?

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  • $\begingroup$ This is a very nice answer, though it seems to apply to the long ray rather than the long line. I assume that there is a simple argument to extend it. $\endgroup$ – Slade Nov 17 '15 at 17:47
  • $\begingroup$ Is there an easy way to see that the one-point compactification of the long ray is universal, or does it satisfy a simple sufficient condition for this type of behavior? $\endgroup$ – Slade Nov 17 '15 at 17:48
  • $\begingroup$ The long line extends to two sides? Then it has two compactifications (a two-point one which is the Cech-Stone as well, and a one-point one which is a long circle, I suppose, so the argument from the paper does not apply. So it must also have more than one uniformity, probably two? $\endgroup$ – Henno Brandsma Nov 17 '15 at 18:40
  • $\begingroup$ I found this paper with Google, and it appears to show that the universal compactification of the long line has two extra points. Intuitively, I can see how this should imply that there are exactly two uniformities on $L$, since a uniform completion should give an embedding into a compact Hausdorff space. But I do not really grasp what geometric properties of $L$ cause its compactifications to be so tame. $\endgroup$ – Slade Nov 17 '15 at 19:07
  • $\begingroup$ @Slade The continuous functions being eventually constant plays a large part in the tameness, I think. $\endgroup$ – Henno Brandsma Nov 17 '15 at 19:35

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