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the space the Real line with bounded metric (i.e. d/(1+d), d: euclidean)

we know that totally boundedness means that there exists a finite epsilon-net. we first approached to question by directly try to find finitely many points s.t. their metric balls cover the space. then, tried , by contradiction, but failure.

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  • $\begingroup$ Hint: is it complete? $\endgroup$ – Daniel Fischer Nov 24 '13 at 21:38
  • $\begingroup$ yes. since R with |.| is complete and |.| > bounded metric. but could not get the hint. $\endgroup$ – 104078 Nov 24 '13 at 21:42
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    $\begingroup$ Doesn't follow that easily. The bounded metric could have more Cauchy sequences than the Euclidean. That aside, a complete metric space is totally bounded if and only if it is - what? $\endgroup$ – Daniel Fischer Nov 24 '13 at 21:44
  • $\begingroup$ i am not sure. but a good candidate will be "bounded" $\endgroup$ – 104078 Nov 24 '13 at 21:45
  • $\begingroup$ No, something stronger. $\endgroup$ – Daniel Fischer Nov 24 '13 at 21:46
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HINT: Let $d_b$ be the bounded metric, so that $$d_b(x,y)=\frac{|x-y|}{1+|x-y|}$$ for all $x,y\in\Bbb R$. Suppose that $m,n\in\Bbb Z$ and $m\ne n$; then $d_b(m,n)\ge\ldots\;$?

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