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I recently heard a lecture where the following fact was mentioned without further comment:

The space $\omega_1$ of all countable ordinals is not completely uniformizable. (I.e., there is no complete uniformity on $\omega_1$ which induces the usual (order) topology on $\omega_1$.)

This seemed interesting, and I thought (naively and incorrectly, as it turns out) that it wouldn't be too difficult to prove. Honestly, I haven't got anywhere with it. Virtually all of my knowledge of uniform spaces comes from I.M. James' "Topologies and Uniformities".

I'm hoping that someone here can at least outline a proof of this fact. Thanks in advance!

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  • $\begingroup$ Kelley, General Topology, 1955, Exercise 6.E, p. 204. There is an outline of the proof. The citation is Dieudonné, C.R.A.S. Paris 209 (1939) 145--147. $\endgroup$ – GEdgar Dec 4 '19 at 16:44
  • $\begingroup$ If $X$ admits such a uniformity and $X$ is countably compact then $X$ is compact. (quoting Engelking, chapter 7, exercise on Dieudonné complete spaces, as these are called). $\omega_1$ is not compact. $\endgroup$ – Henno Brandsma Dec 4 '19 at 17:54
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The filter generated by the final segments $[\alpha,\omega_1)$ is Cauchy and it does not converge. To see the former you need to show that every (uniform) neighbourhood of the diagonal contains a `corner' $[\alpha,\omega_1)^2$. You can do this using Fodor's Pressing Down Lemma, or by arguing by contradiction: if $U$ is a set that contains the diagonal but does not contain a corner then for every $\alpha$ there is a point $(\beta,\gamma)$ with $\beta,\gamma>\alpha$ that is not in $U$. Then you get sequences $\langle\beta_n:n\in\omega\rangle$ and $\langle\gamma_n:n\in\omega\rangle$ such that $(\beta_n,\gamma_n)$ is not in $U$ and $\max\{\beta_n,\gamma_n\}<\min\{\beta_{n+1},\gamma_{n+1}\}$ for all $n$. Let $\alpha$ be the supremum of the $\beta_n$ and $\gamma_n$. Then $(\alpha,\alpha)$ is not in the interior of $U$.

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