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Hi I am working on the following problem Let $\mathbb{Z}$ be the set of integers and $p$ be a fixed prime number. Represent any $x\in\mathbb{Z}$ as follows $$x=p^k\cdot y\qquad \text{where }y\not\equiv 0\,\,\,(\text{mod }p)$$ Put $$\|x\|_{(p)}:=\begin{cases}p^{-k},\,\,\,\,\text{if }x\neq 0\\0,\,\,\,\,\,\,\,\,\,\,\text{if }x=0\end{cases}$$ and define the function $\rho_{(p)}(x,y):=||x-y||_{(p)}$

(i) Show that $\rho_{(p)}$ is a metric on $\mathbb{Z}$

(ii) Is $\mathbb{Z}$ bounded in this metric?

(iii) Is $\mathbb{Z}$ totally bounded in this metric?

(iv) Is $\mathbb{Z}$ complete in this metric?

(v) Is $\mathbb{Z}$ compact in this metric?

I know how to do (i), but I don't know how to do the rest. Any help would be appreciated.

Jyrki Lahtonen
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mint
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    ii) is there a maximum or supreme value of ||x||? That should be a straightforward question to answer. – fleablood Sep 06 '16 at 05:04
  • @fleablood has pointed you in the right direction for (ii); note that (v) will follow immediately once you answer (iii) and (iv), since it’s a standard theorem that a metric space is compact if and only if it complete and totally bounded. – Brian M. Scott Sep 06 '16 at 05:06
  • Just go through the text book definitions of all those term and see if they hold or not. You may need to also review the definition of open to see which sets are open and closed. – fleablood Sep 06 '16 at 05:07
  • @fleablood Thank you very much for the hints. I tried to find out what could be the value for k, is it integer or natural number? – mint Sep 06 '16 at 06:19

1 Answers1

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HINTS:

  • Is there a maximum possible value of $\|x\|_{(p)}$?
  • For any given positive integer $k$ let $R_k=\{0,1,2,\ldots,p^k-1\}$; given $x\in\Bbb Z$, can you find $y\in R_k$ such that $\rho_{(p)}(x,y)\le p^{-k}$?
  • Corrected and extended: Let $x_n=\sum_{k=0}^np^k$ for $n\in\Bbb N$. Show that if $k,\ell\ge n$, then $\rho_{(p)}(x_k,x_\ell)<p^{-n}$, so the sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ is Cauchy. Let $x\in\Bbb Z$ be arbitrary, and fix $n\in\Bbb N$ such that $|x|<p^n$. Show that if $k>n$, then $\rho_{(p)}(x_k,x)\ge p^{-(n+1)}$, so $\sigma$ does not converge to $x$. Conclude that the space is not complete.
  • A metric space is compact if and only if it is complete and totally bounded.
Brian M. Scott
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  • Thank you very much for the hints. I tried to find out what could be the value for k, is it integer or natural number? – mint Sep 06 '16 at 06:18
  • @mint: You’re welcome. In the second point $k$ should have been a positive integer; I’ll correct that as soon as I post this comment. If $p=3$, say, then $R_1={0,1,2}$, and $R_2={0,1,2,3,4,5,6,7,8}$. – Brian M. Scott Sep 06 '16 at 06:25
  • If k is a positive integer then the maximum value of $||x||_{(p)}$ will be $\frac{1}{p}$, Is it right? – mint Sep 06 '16 at 06:44
  • @mint: Yes, but I was talking about my second bullet point there. For part (ii) of the question $k$ can be any non-negative integer, so it can also be $0$; for instance, $|5|{(3)}=3^{-0}=1$, since $5=3^0\cdot5$. Thus, the largest possible value is actually $1$, not $\frac1p$. But it’s still true that $|x|{(p)}$ is bounded. – Brian M. Scott Sep 06 '16 at 06:48
  • Sorry that's my bad. At first I got 1, but as you mentioned k can be only positive then I changed the answer to $\frac{1}{p}$. Also the smallest value will be 0, right? So it is bounded below as well as bounded above. is it right? – mint Sep 06 '16 at 06:52
  • @mint: Yes that’s correct. But the important thing is that since $|x|{(p)}\le 1$ for all $x$, we must have $d(x,y)=|x-y|{(p)}\le 1$ for any points $x,y\in\Bbb Z$. That is, there is a finite upper bound on the distances between points of the space, and that’s what we mean when we say that a metric is bounded. – Brian M. Scott Sep 06 '16 at 06:55
  • Sorry to disturb you, I couldn't solve the totally bounded and compact problem. Could I have any more help? – mint Sep 06 '16 at 08:11
  • @mint: The sets $R_k$ show that the metric is totally bounded: for any $\epsilon>0$ there is a $k\in\Bbb Z^+$ such that $\frac1k<\epsilon$, and then for each $x\in\Bbb Z$ there is a $y\in R_k$ such that $\rho_{(p)}(x,y)<\epsilon$. (Or did you mean that you don’t see how to show that there is such a $y$?) As for compactness, if you’ve shown that the metric is not complete, you’re done: every compact metric space is complete. – Brian M. Scott Sep 06 '16 at 16:42
  • Are you saying that this metric is not complete? I can't see that. – mint Sep 06 '16 at 19:27
  • @mint: I am indeed saying that it’s not complete, but my hint had a fatal typo that made it nearly useless. Give me a few minutes to correct and expand it. – Brian M. Scott Sep 06 '16 at 23:43