# About compactness and completeness of set $X = \mathbb{R}$ with the metric $d(x, y) = \frac{|x-y|}{1+|x-y|}$

What can we say about the compactness and completeness of the set $X = \mathbb{R}$ with the metric $$d(x, y) = \frac{|x-y|}{1+|x-y|}\;?$$

I tried by showing that $d(x, y)<1$ and thus it is bounded. But how to show rest of the things? Can anybody help me? Thanks

• Is this homework? May 13, 2012 at 17:34
• No that was asked in a entrance exam? And i am preparing for that. May 13, 2012 at 17:35
• @ srijan , ( not related to question) which entrance exam are u preparing for :) ? May 13, 2012 at 17:47
• The topology induced by this metric is the usual one. May 13, 2012 at 17:48
• Closed and bounded under the usual metric. Every set is bounded under the metric you gave. May 13, 2012 at 17:58

$\langle\Bbb R,d\rangle$ is Cauchy, but it’s not compact. It’s easy to see that it’s not compact: $\langle n:n\in\Bbb N\rangle$ is a sequence with no convergent subsequence.

To see that it’s Cauchy, suppose that $\langle x_n:n\in\Bbb N\rangle$ is $d$-Cauchy, so that for every $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge n_\epsilon$. Then for $m,n\ge n_\epsilon$ we have $$\frac{|x_m-x_n|}{1+|x_m-x_n|}<\epsilon\;,$$ so $|x_m-x_n|<\epsilon(1+|x_m-x_n|)$, and $$|x_m-x_n|<\frac{\epsilon}{1-\epsilon}\;.\tag{1}$$ We’re interested only in small $\epsilon$, so there’s no harm in assuming that $\epsilon<\frac12$, in which case $(1)$ implies that $$|x_m-x_n|<2\epsilon\;.$$ It follows that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy with respect to the usual metric on $\Bbb R$ and therefore converges to some $x\in\Bbb R$ in the usual metric. From here it’s easy to show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in the metric $d$ as well since the two are approximatly equal for small distances.

• @ BrianThank you very much sir for helping me every time. May 13, 2012 at 18:16

Denote $$d_1(x,y)=|x-y|\qquad d(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ You should check that $(\mathbb{R},d)$ and $(\mathbb{R},d_1)$ are metric spaces. Moreover, from the point of view of theory of metric spaces, they are essentially the same! Indeed, consider maps $$f:(\mathbb{R},d)\to(\mathbb{R},d_1):x\mapsto x\qquad g:(\mathbb{R},d_1)\to(\mathbb{R},d):x\mapsto x$$ You should check that they are continuous. Then we easily see that $f(g(x))=x$ and $g(f(x))=x$, so this maps are inverse to each other. Hence $f$ and $g$ are isomorphisms between metric spaces $(\mathbb{R},d)$ and $(\mathbb{R},d_1)$.

For isomorphic spaces notion of compactness coinside (but not completeness!). Recall what you know about compactness/non-compactnes of $\mathbb{R}$ with standard metric.

• Rest of the things are cleared by your answer. Thank you very much. May 13, 2012 at 18:47
• Certainly $\mathbb{R}$ is not compact with standard metric. May 13, 2012 at 18:55
• Yes, you are right May 13, 2012 at 18:56

I'm somewhat rusty on this stuff, so check all the details, but I think this should do the completeness: Consider a sequence $\{x_n\}$ that is Cauchy in the $d$ metric. If you solve for $|x_n-x_m|$ in the definition of $d(x_n,x_m)$, you get $|x_n-x_m| = \frac{d(x_n,x_m)}{1-d(x_n,x_m)}$. This shows that $\{x_n\}$ is a Cauchy sequence in $\mathbb{R}$ with the Euclidean metric. Since $\mathbb{R}$ is complete with the Euclidean metric, there's an $x$ in $\mathbb{R}$ that the sequence converges to in the Euclidean sense. But it shouldn't be too hard to show that $\{x_n\}$ converges to $x$ in $d$ sense as well. That shows that $\mathbb{R}$ is complete with $d$ as the metric.

• that is also useful dear May 13, 2012 at 18:17