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Given a sequence $(x_n)$ in a metric space $M$, call it uniformly separated if all pairwise distances $d(x_n,x_m)$ between distinct terms are uniformly bounded away from zero.

Suppose that a given sequence $(x_n)$ has the property that it has no Cauchy subsequence. Must it have a uniformly separated subsequence?

A couple of my students asserted this in homework without watertight proof, and proof or disproof is eluding me. Clearly any counterexample cannot be in ${\mathbb R}^n$. Enjoy!

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up vote 4 down vote accepted

A proof can be inferred from the answers to an earlier question.

Which metric spaces are totally bounded?

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Indeed, thanks! And distinctly cute, since rearrangements apparently are involved. – Bob Pego Nov 12 '10 at 5:38

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