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$(\mathbb{R},d)$ 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.

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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.

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Actually the metric given in your question is equivalent to the usual metric in real numbers. Real numbers under usual metric is complete but not compact so the same property(property means topological property) carry out by the given metric.

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  • $\begingroup$ Could you prove the equivalence by giving explicit bounds? $\endgroup$ – Miguel Aug 14 '17 at 13:20
  • $\begingroup$ considering an open set in usual metric and showing that this is open in given metric and conversely. $\endgroup$ – GOBINDA GARAI Aug 14 '17 at 13:30

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