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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.

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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 4 '12 at 11:36
For the first problem: consider $x_n=-n$. For the second problem: you can quite explicitly describe Cauchy sequences. – Norbert Jun 4 '12 at 11:37
@srijan, yes.${}{}{}{}$ – Norbert Jun 4 '12 at 11:45
@srijan: yes, that's correct. – T. Eskin Jun 4 '12 at 13:58
@ThomasE. Thanks again. – srijan Jun 4 '12 at 14:05

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