# How to show the space is totally bounded?

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.

• Hint: is it complete? – Daniel Fischer Nov 24 '13 at 21:38
• yes. since R with |.| is complete and |.| > bounded metric. but could not get the hint. – 104078 Nov 24 '13 at 21:42
• 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? – Daniel Fischer Nov 24 '13 at 21:44
• i am not sure. but a good candidate will be "bounded" – 104078 Nov 24 '13 at 21:45
• No, something stronger. – Daniel Fischer Nov 24 '13 at 21:46

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 $d_b(m,n)\ge\ldots\;$?