Let $d(x,y)= \frac {|x-y|} {1+|x-y|}$. Is the space $(\mathbb{R},d)$ totally bounded? Compact?

Question

Let $d(x,y)= \frac {|x-y|} {1+|x-y|}$. Is the space $(\mathbb{R},d)$ totally bounded and compact?

Clearly it is bounded since being metric ensures that distance between any point is less than 1. But how is total boundedness showed? After proving total boundedness I need to show the space is complete to prove it is compact but I am not sure how to show all Cauchy sequences converges to a point.

Answer 1

No, it is not totally bounded. It is complete, though (for a real sequence, being convergent and being a Cauchy sequence means the same thing for this distance and for the usual one), and this can be used to prove that it is not totally bounded. In fact, if it was both complete and totally bounded, then it would be compact. But it is not; for intance,$$\left\{(-n,n)\,|\,n\in\mathbb{N}\right\}$$is an open cover without a finite sub-cover.

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You can also prove directly that it is not totally bounded. For this distance, if $x\in\mathbb R$, then $B\left(x,\frac12\right)=(x-1,x+1)$. And you cannot express $\mathbb R$ as the union of finitely many intervals of this type.

Answer 2

Every (R,d)-sphere around x contains a standard-R sphere around x and vice versa, so (R,d) is homeomorphic with standard-R