# GATE(competitive Exam in India for PhD admission) 2010 Question:-$(\mathbb{R},d)$ be a metric space. Which of the following are correct?

$(\mathbb{R},d)$ be a metric space. Where $d(x,y)=\frac{|x-y|}{1+|x-y|}$. Which of the following are correct? Metric space $(\mathbb{R},d)$ is

(a)bounded, not compact

(b)bounded, not complete

(c)compact, not complete

(d)complete, not bounded

I could prove $(X,d)$ is bounded. Checking the compactness and completenes using their respective definition is time consuming. How to eliminate the wrong answers from the options by applying any theorems of completeness and compactness with lesser time? Please help me.

• Does $d$ determine the usual topology on $\Bbb R$? – Lord Shark the Unknown Aug 14 '17 at 11:32
• Yes, thank you very much. So , (a) must be the correct answer? – Unknown x Aug 14 '17 at 11:50
• – Martin Sleziak Oct 21 '17 at 12:50

Clearly $(\mathbb{R}, d)$ is bounded, since $d(x,y) < 1$ for every $x,y\in\mathbb{R}$.
Let us prove that $(\mathbb{R}, d)$ is complete. Namely, let $(x_n)\subset\mathbb{R}$ be a Cauchy sequence. Given $\epsilon\in (0,1)$, let $\eta := \epsilon / (1-\epsilon)$, so that $\eta/(1+\eta) = \epsilon$. Since $(x_n)$ is a Cauchy sequence, there exists $N\in\mathbb{N}$ such that $$d(x_j, x_k) < \eta, \qquad \forall j,k\geq N,$$ i.e. $$\frac{|x_j - x_k|}{1+|x_j-x_k|} < \frac{\epsilon}{1+\epsilon} \qquad \forall j,k\geq N.$$ Since the function $t\mapsto t/(1+t)$ is strictly increasing in $[0,+\infty)$, the last condition is equivalent to $$|x_j - x_k| < \epsilon \qquad \forall j,k\geq N.$$ In other words, we have proved that $(x_n)$ is a Cauchy sequence in $(\mathbb{R}, |\cdot|)$. Since $(\mathbb{R}, |\cdot|)$ is complete, the sequence $(x_n)$ is convergent in $(\mathbb{R}, |\cdot|)$, i.e. there exists $x\in\mathbb{R}$ such that $$\lim_{n\to +\infty} |x_n - x| = 0.$$ But this implies that $$\lim_{n\to +\infty} \frac{|x_n - x|}{1+|x_n-x|} = 0,$$ hence $(x_n)$ is convergent also in $(\mathbb{R}, d)$.
Finally, the two metrics generate the same topology, so that $(\mathbb{R}, d)$ is not compact.