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I have to check for completeness of following metric spaces

$1)$ : $\mathbb{R}$ with metric defined by $d_1(x,y) =\mid e^x - e^y \mid$ for all $x, y \in \mathbb{R}$.

$2)$: $\mathbb{Q}$ with metric defined by $d_2(x,y) = 1$, for all $x, y\in \mathbb{Q}, ~~ x \neq y$.

I have shown that $(\mathbb{R}, d_1)$ is complete metric space. For that I took Cauchy sequence $(x_n)\subset\mathbb{R}$ that must converge to some point say $x$ with respect to standard metric say $d$ of $\mathbb{R}$ (Since $\mathbb{R}$ is complete then).

The exponential function is continuous so the sequence $e^{(x_n)}$ converges to $e^{x}$ with respect to standard metric of $\mathbb{R}$

from here we can conclude that $(x_n)$ converges to x w.r.t. $d_1$.

Hence $(\mathbb{R}, d_1)$ is complete.

For second I am not able to figure out how to proceed. But intuitively I think it is complete. Because given metric looks like discreet metric.

Please correct me if I am wrong. Is there any alternate way to solve this problem?

Thanks for giving me time.

mich95
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Srijan
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    For $d_{1}$, note that you must start with an arbitrary Cauchy sequence in $(\mathbb{R},d_{1})$ and not in the standard metric. The current argumentation does not show that $d_{1}$ is complete. – T. Eskin Jun 04 '12 at 11:36
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    For the first problem: consider $x_n=-n$. For the second problem: you can quite explicitly describe Cauchy sequences. – Norbert Jun 04 '12 at 11:37
  • @Norbert For second part Cauchy sequences are eventually constant sequences . Am i correct? – Srijan Jun 04 '12 at 11:40
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    @srijan, yes.${}{}{}{}$ – Norbert Jun 04 '12 at 11:45
  • @Norbert Thank you very much. – Srijan Jun 04 '12 at 11:51
  • For the first part note that it is isometric to $\mathbb{R}+$ with the standard metric under the exponential map $\mathop{exp}:\mathbb{R}\to\mathbb{R}+$ and its inverse $\ln:\mathbb{R}_+ \to\mathbb{R}$. – Willie Wong Jun 04 '12 at 13:16
  • @ThomasE. Thanks for your comment.$X_n = -n$ is cauchy sequence with respect to metric $d_1$ but it is not convergent and hence not complete. right? – Srijan Jun 04 '12 at 13:55
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    @srijan: yes, that's correct. – T. Eskin Jun 04 '12 at 13:58
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    @ThomasE. Thanks again. – Srijan Jun 04 '12 at 14:05
  • @WillieWong That mean all the metric $\mathbb{R}$ and $\mathbb{R}_{+}$ will share same metric properties. Am i right? I am sorry I didn't get your comment? Could you please explain little more? – Srijan Jun 04 '12 at 14:09
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    A Cauchy sequence in $\mathbb{R}+$ (with the standard metric) which _doesn't converge will, under the maps indicated, become a Cauchy sequence in $(\mathbb{R},d_1)$ which doesn't converge. It may be easier to visualize what are the required Cauchy sequences on $\mathbb{R}_+$ with the standard metric. – Willie Wong Jun 04 '12 at 14:11
  • @WillieWong Thanks for your comment. Now I got your point. Definiely exp map will determine isometry between $(\mathbb{R}, d_1)$ and $(\mathbb{R}_{+}, d)$. Then we can take cauchy sequence say $(1/n)$ in $\mathbb{R}$ which doesn't converges. And by isometry we may conclude that what you have said above. Am i right? – Srijan Jun 04 '12 at 14:36
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    @srijan: yes. Note that Thomas E's example corresponds to the Cauchy sequence $e^{-n}$ in $\mathbb{R}_+$. With the above you see that you get a wealth of examples. Now that you understood everything, I suggest you write up a consolidated answer and post it below (once the software allows you to). You can even earn the "self-learner" badge! – Willie Wong Jun 04 '12 at 15:14
  • @WillieWong Got much more than I was expected. Thank you very much.:) – Srijan Jun 04 '12 at 15:16
  • Could you please explain me why $x_n=-n$ does not converge in $d_1$ metric? Does not it converges to $0$ in that metric? – Myshkin Jun 06 '12 at 09:28
  • @Mex if it converges to zero then $lim_{n\rightarrow \infty} d_1 (-n, 0)\rightarrow 0$. Is this happening here? – Srijan Jun 06 '12 at 09:42
  • O ha thik kotha srijan bujhlam, $d_1(-n,0)\rightarrow 1$ right? – Myshkin Jun 06 '12 at 09:56

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