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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.

Ridvan
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    Please ask one question per post, or, if you see a relationship between the questions that makes it helpful to consider them together, please point it out. – joriki Dec 10 '12 at 02:55
  • For the metric, take the usual metric and the "british rail metric" given by $d(x,x) = 0$ and $d(x,y) = |x| + |y|$ for all $x \not= y$ – Tom Oldfield Dec 10 '12 at 03:43

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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 $h:\Bbb N\to X$.

  • 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.

Brian M. Scott
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  • Why does $\pi$ being irrational mean that the numbers are dense in $[0,1]$? Liouville's constant $l$ is irrational, and yet $nl - \lfloor nl \rfloor$ is always $0$. Also, why do we use $A$ countable to get a contradiction? What does it contradict? Isn't it enough to say that any proper dense subset is not closed? – Tom Oldfield Dec 10 '12 at 03:42
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    @Tom: $x-\lfloor x\rfloor=0$ iff $x\in\Bbb Z$, so your statement about Liouville’s constant is clearly false. The result that ${nx\bmod 1:n\in\Bbb Z}$ is dense in $[0,1]$ iff $x$ is irrational is well-known and has been proved here. The countability of $A$ is the easiest way to demonstrate that it’s a proper dense subset. – Brian M. Scott Dec 10 '12 at 03:48
  • Sorry, I meant to say that it is always less than $0.2$ rather than is always $0$, I was thinking about $10^n$, not $n$, sorry... I think I see what you're saying about the contradiction, but surely it's clear that it's a proper subset since $0$ isn't in $A$? It just doesn't feel like a proof by contradiction, but I guess that's just opinion. – Tom Oldfield Dec 10 '12 at 03:59
  • @Tom: $0$ is in $A$: $\Bbb Z\subseteq A$. – Brian M. Scott Dec 10 '12 at 04:03
  • Okay, $\frac{1}{2}$ then, or any non-integer rational. – Tom Oldfield Dec 10 '12 at 04:06
  • @Tom: Yes, one could do it that way; I saw the cardinality argument instantly and didn’t bother looking for another. – Brian M. Scott Dec 10 '12 at 04:09