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I know that a sequence in $\mathbb{R}^d$ is Cauchy iff it is convergent, and also convergent sequences are Cauchy in a general metric space. How does the converse fail in a general metric space? In other words, what part of the proof that Cauchy sequences in $\mathbb{R}^d$ are convergent cannot be generalized to an arbitrary metric space?

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The sequence $$\left\{\frac1n\right\}_{n=1}^\infty$$ is Cauchy, but not convergent, in $(0,\infty)$ equipped with the standard metric.

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  • $\begingroup$ Thanks, this a great example but not exactly what I was asking for. I am hoping to understand why the proof fails in the general case. $\endgroup$ – b_pcakes Nov 7 '16 at 8:37
  • $\begingroup$ @b_pcakes Well, did you try to apply the proof to my example? $\endgroup$ – 5xum Nov 7 '16 at 8:38
  • $\begingroup$ Ah, B.W. only holds in $\mathbb{R}^d$ $\endgroup$ – b_pcakes Nov 7 '16 at 8:42

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