Metric spaces Equivalent I have two question could not figure out about their solutions.
First, give an example of two metrics on Z that are not equivalent?
Secondly,  A= {a+πm a,m in Z} Is A complete with respect to usual metric on R.
thank you for everyone share their opinion. 
 A: First, you can give $\Bbb Z$ the usual metric that it inherits from $\Bbb R$: the distance between $m,n\in\Bbb Z$ is simply $|m-n|$.
The simplest way to find another metric for $\Bbb Z$ is start with a metric space $\langle X,d\rangle$ such that $X$ is countably infinite. Since $X$ is countably infinite, there is a bijection $h:\Bbb N\to X$.


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*Prove that the function $$\rho:\Bbb N\times\Bbb N\to\Bbb R:\langle m,n\rangle\mapsto d\big(h(m),h(n)\big)$$ is a metric on $\Bbb N$. Basically you’re just using $h$ to copy $d$ over to $\Bbb N$; $\langle\Bbb N,\rho\rangle$ is isometric to $\langle X,d\rangle$ $-$ the ‘same’ metric space under a different name.

*Find a countably infinite metric space $\langle X,d\rangle$ that is not discrete, and show that the corresponding metric $\rho$ on $\Bbb N$ is not equivalent to the usual one. (There are at least two very natural ones that are subspaces of $\Bbb R$.)

For the second question, note that a complete subset of $\Bbb R$ must be closed in $\Bbb R$. (Why?) Thus, if $A$ is complete, it must be closed, and therefore $A\cap[0,1]$ must be closed. $A\cap[0,1]$ contains every number of the form $n\pi-\lfloor n\pi\rfloor$ (why?), and since $\pi$ is irrational, these numbers are dense in $[0,1]$. Now use the fact that $A$ is countable to get a contradiction.
