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Can we prove that the following statement is equivalent to the axiom of countable choice (CC)?

If every sequence in a metric space $X$ has a Cauchy subsequence, then $X$ is totally bounded.

Note: CC is known to be equivalent to the above condition with "metric space" replaced by "pseudometric space".

Nate Eldredge
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Sushil
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  • Can you please give the definition of CC, what is it? – ThePortakal Aug 20 '15 at 08:12
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    @ThePortakal: Axiom of countable choice. – Asaf Karagila Aug 20 '15 at 08:12
  • @AsafKaragila thanks – ThePortakal Aug 20 '15 at 08:13
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    Where did you come across this question, and where did you come across the solution you posted in your answer? You say it's from an answer I wrote, but it's not in there. – Asaf Karagila Aug 20 '15 at 15:44
  • same paper you mentioned. There it was for pseudometric space. I thought this must be true for metric space but as in your comments. I think it may not be. But in y question I'll add words, is it true in metric spaces. @AsafKaragila – Sushil Aug 20 '15 at 15:57
  • I should have been careful while asking the question. @AsafKaragila – Sushil Aug 20 '15 at 16:02
  • I took the liberty of rephrasing your question; I had a little trouble understanding the original wording. Feel free to make further edits if you prefer something different. – Nate Eldredge Aug 20 '15 at 17:21
  • @AsafKaragila although you deleted your answer. But can we prove countable choice for finite set exist using trick in your answer under given assumption – Sushil Aug 21 '15 at 03:20

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I found a proof in case we work with pseudometric space instead of metric space. If CC fails then PCC fails. Equivalent of the countable axiom of choice?

So there exist sequence of non empty sets such that each sequence meets only finitely many of Xn. Now we define sequence X = Xn*{n) with metric d[(x,m), (y,n)] = 0 if n=m and 1 otherwise.

(It works in pseudometric space, can we use same set with slight change in metric for metric space case)

Community
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Sushil
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    I don't see any obvious way to adapt this proof to metric spaces. Intuitively, I would like to equip each $X_n$ with a totally bounded metric $d_n$, and then consider the metric $d(x,y) = d_n(x,y)$ if $x,y \in X_n$ and $d(x,y) = 1$ otherwise. But at least in ZFC, sets of cardinality greater than $\mathfrak{c}$ don't have totally bounded metrics (if it did, its completion would be compact, and compact metric spaces are separable and so have cardinality at most $\mathfrak{c}$). – Nate Eldredge Aug 20 '15 at 17:25
  • @NateEldredge sorry didn'y get what you want to say. Totally bounded implies sequential bounded. This is true in ZF only. So we should rather try to find a sequential bounded set which is not totally bounded(if possible) – Sushil Aug 20 '15 at 20:03