# When is a uniform space complete

From Wikipedia:

a uniform space is called complete if every Cauchy filter converges.

1. I was wondering if the following three are equivalent in a uniform space:

Or, which one implies which but doesn't imply which? For example, are the first two equivalent, while the third is implied by but does not implie any of the first two?

2. How about in a metric space?

Thanks and regards!

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Example: Let $X$ be $\omega_1$ with the order topology. $X$ is Tikhonov, so it has a compatible uniformity $\mathscr{U}$. $X$ is countably compact, so $\langle X,\mathscr{U}\rangle$ is totally bounded, but $X$ is not compact, so $\langle X,\mathscr{U}\rangle$ is not complete: a uniform space is compact iff it is complete and totally bounded. Thus, $\langle X,\mathscr{U}\rangle$ must have Cauchy filters/nets that do not converge. However, every sequence in $X$ is contained in a compact subspace of $X$, so every Cauchy sequence does converge.
"In a metric space all three properties are equivalent" is equivalent to Countable Choice $\hspace{1 in}$ (en.wikipedia.org/wiki/Axiom_of_countable_choice). $\;$ –  Ricky Demer Feb 2 '12 at 3:57
@Tim: I don’t offhand see how to do it for metric spaces, but for pseudometric spaces it’s not too hard. If CC fails, one can construct a pseudometric space $X=\bigsqcup_nX_n$ that has no Cauchy sequences (so vacuously all Cauchy sequences converge!) but is such that $\mathscr{F}=\{\bigcup_{k\ge n}X_k:n\in\omega\}$ is a Cauchy filter; clearly $\mathscr{F}$ does not converge. –  Brian M. Scott Feb 2 '12 at 22:06