Wikipedia article on totally bounded spaces states "... the completion of a totally bounded space might not be compact in the absence of choice." Where is the axiom of choice used, and do you need it for metric spaces or only for general uniform spaces?
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asked Jan 16, 2018 at 17:06
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The issue here is that a metric space might not have non-trivial (read: not eventually constant) Cauchy sequences. For example, if the underlying space is a Dedekind finite set.
Indeed it is consistent that there is a dense subset of $[0,1]$ which is Dedekind finite. As a space with the inherited metric it is complete already and totally bounded, but it is not compact as it is not closed on $[0,1]$.
answered Jan 16, 2018 at 17:28
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$\begingroup$ Presumably this is only an issue with Cauchy sequences, and if you use more general nets or filters then the problem goes away? If so, I think it's misleading to present it as "without AC the completion of a totally bounded space might not be compact" -- I would instead say that "without AC it does not suffice to use sequences in defining the completion of a metric space, but one has to use nets or filters instead just as for a general uniform space". $\endgroup$ Commented Jan 16, 2018 at 17:40
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1$\begingroup$ @Mike: At least in the case I gave, nets are enough since the space is linearly ordered, every Dedekind cut corresponds to an obvious net. There is a lot of results about the equivalence between nets and filters, so I imagine the same can be said about filters. I will give it more thought. But in fairness, the completion of a metric space is defined as the Cauchy limits. It just shows there are two ways to complete. Just like there are two ways to define Noetherian without choice. $\endgroup$– Asaf Karagila ♦Commented Jan 16, 2018 at 17:52
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$\begingroup$ @AsafKaragila: Every metric space that is complete and totally bounded is compact. Does your example mean that this statement uses AC? $\endgroup$ Commented Jan 16, 2018 at 18:06
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2$\begingroup$ @C.Eratosthene: Yes. There has been a lot of research into these sort of statements, and their equivalences (or lack thereof) to certain choice principles. Countable choice seems to be a central one in the case of metric and pseudometric spaces (and I'd imagine that generally in first-countable spaces too). $\endgroup$– Asaf Karagila ♦Commented Jan 16, 2018 at 18:06
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1$\begingroup$ @C.Eratosthene: The thing is that it is debatable whether or not the axiom of choice should be taken for granted (or at least countable choice), and only later on its uses should be pointed out, or if it should be explicitly debated from the get go. $\endgroup$– Asaf Karagila ♦Commented Jan 16, 2018 at 19:05