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On $\mathbb{Z}$, we define the $p$-adic metric $d_p$ (for $p$ prime) as follows, for $m,n \in \mathbb Z$:

  • If $m=n$ then $d_p(m,n) =0$
  • If $m \neq n$ then $d_p(m,n) = \tfrac{1}{r+1}$ where $p^r \mid (m-n)$ but $p^{r+1} \not \mid(m-n)$

It's fairly straightforward to show that $d_p$ is a metric space (and infact a ultrametric space, I believe), however I'm having a bit of trouble showing that the space is not complete via the example

$$ a_n = 1 + p + p^2 + \cdots + p^n = \frac{1}{p-1}(p^{n+1} - 1) $$

While I've managed to show that $a_n$ is Cauchy, I'm finding it difficulty to try and prove that the sequence has no limit in $\mathbb{Z}$. Although the closed form given above makes it seem very intuitive that $a_n \to \tfrac{1}{1-p}$, and hence the limit is not in $\mathbb Z$, I can't see any way of doing this with the way $d_p$ is defined (as much as I would love to use the fact that $d_p(p^n,0) \to 0$, I can't see a valid way of taking the $\tfrac{1}{p-1}$ factor out). Is there any way I would be able to do this via the method above, or would I have to try some other method to show non-convergence?

Andrew D
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  • Out of curiosity, which textbook defines $d_p=1/(r+1)$? – Chris Culter Sep 18 '13 at 16:58
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    @ChrisCulter It's not a textbook; our lecturer defined the metric in one of the exercises in the following document: https://www.dpmms.cam.ac.uk/~twk/Top.pdf (to save you and others having to find it themselves, the exercise is on page 49). I can't say I know why it's defined in such a way, to be honest (especially having briefly read up on the generalisation of the p-adic metric to $\mathbb Q$). – Andrew D Sep 18 '13 at 17:29
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    So, you already know the extension of this metric space on $\Bbb Z$ to $\Bbb Q$. Doesn't this sequence $a_n$ tend to $\displaystyle\frac1{1-p}$ in this metric on $\Bbb Q$? Can it have another limitpoint? – Berci Sep 18 '13 at 17:33
  • @Berci While I know that there is an extension of this metric space to $\mathbb Q$, I can't think of a way of extending the definition of the metric in order to do so (in some respects, I think its definition is given such to try and avoid people from making the jump to $\mathbb Q$). As for the limit-points, having had a quick look at the definition that uniqueness of limit points is true for Hausdorff and hence metric spaces, so while what you say is certainly the preferred method of attack, it's just how to do the first of those steps which bothers me. – Andrew D Sep 18 '13 at 17:44

1 Answers1

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Note that the statement is only true for $2\not=p$ since if $2=p$ then $a_n \rightarrow -1$.

Suppose $a_n \rightarrow a\in \mathbb{Z}$. Then by the work in your question, the sequence $(p-1)a_n \rightarrow -1$. But by properties of metric spaces, $(p-1)a_n\rightarrow(p-1)a$. Thus $p=2$ and $a=-1$, so this is the only possible case.

Alternatively: Suppose $a_n \rightarrow a\in\mathbb{Z}$. Write $a=\pm(\sum_{i=0}^Nb_ip^i)$ where $N$ is the largest integer such that $p^N\leq |a|$ and $0\leq b_i\leq p-1$ (which is possible to do by the Euclidean algorithm, we are really writing $a$ in base $p$ if you prefer to think of it like that.)

Now, what does using this representation of $a$ tell us about $d(a,a_n)$? (In particular, for $n > N$?)

Edit: The more I think about it, the more difficult it becomes to complete the proof simply with the alternative method. If anyone can do it in a line or two please leave a comment, otherwise I think that the first way is a lot neater.

Tom Oldfield
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  • I agree with the first method being a lot more neater - I think while the alternative method is perhaps more instructive as to why the completeness fails, the first method is just generally a lot slicker. Anyways, thanks a lot! – Andrew D Sep 18 '13 at 21:37