# The space $\omega_1$ of all countable ordinals is not completely uniformizable

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!

• 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. – GEdgar Dec 4 '19 at 16:44
• 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. – Henno Brandsma Dec 4 '19 at 17:54

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 $$U$$.