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I'm looking at some old question papers and here's a problem:

If $n$ and $m$ are integers and $d(n, m) = 2^{-r}$ if $|n - m| = 2^{r}t$ with $t$ odd if $n \neq m$ and $d_2(n, n) = 0$. Prove that the metric space $(\mathbb Z, d)$ is not complete. (Hint: Consider the sequence $a_n = \frac{4^{n} - 1}{3}$.)

The given hint doesn't really make sense to me, as every element of the form $a_n = \frac{4^n - 1}{3}$ can be shown to be odd. Then the distance between any two elements of $(a_n)$ would be $2^{-0} = 1$ according to the given metric $d(\cdot, \cdot)$. So this is not even a Cauchy sequence in $(\mathbb Z, d)$.

Edit: I realized that this is incorrect, after seeing the comments. $(a_n)$ is indeed a Cauchy sequence -- one should consider the difference between the elements of $(a_n)$.

Is there any other sensible Cauchy sequence that I can use to prove that this metric on $\mathbb Z$ is not complete?

S.D.
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  • Yes, but what is $d(a_n,a_m)?$ You need a Cauchy sequence which doesn't converge. – Thomas Andrews Oct 12 '22 at 20:46
  • In fact, you are describing the well-known dyadic or (2-adic) metric. See here for p-adic numbers in general. – Jean Marie Oct 12 '22 at 20:49
  • @ThomasAndrews Sorry, you're right. I should consider the distance between the elements which is of the form $4^mt$ where $t$ is an odd integer. Indeed, this distance gets smaller as we move along the tail of the sequence. What do you suggest to check that the sequence doesn't converge to an integer? – S.D. Oct 12 '22 at 20:53
  • Show that if $a_n$ converges to an integer $a,$ than $3a_n$ converges to $3a,$ and then show that $4^n-1$ converges to $-1.$ – Thomas Andrews Oct 12 '22 at 20:56
  • How big is the integer to which this sequence converges, if there is such a thing? – Igor Rivin Oct 12 '22 at 20:56
  • @IgorRivin That doesn't work, at least not obviously. $2^n$ converges to $0,$ for example. "Big" is not much of a useful concept in this metric. – Thomas Andrews Oct 12 '22 at 20:57
  • @ThomasAndrews But it is a useful concept in the Archimedean metric. We have limit $l$ such that $|l - (4^n-1)/3)|$ goes to zero, so the difference is divisible by a large power of two, which means that the difference is at least as large as that power of 2 (that is exactly the point of subtracting the $1$ and dividing by $3.$ – Igor Rivin Oct 12 '22 at 21:01
  • Thanks for all the comments. There is an explanation for a more general version of this question here. The context is indeed that of $p$-adic metrics on $\mathbb Z$ not being complete. – S.D. Oct 12 '22 at 21:06
  • @IgorRivin We are explicitly in a non-archimedian metric. – Thomas Andrews Oct 12 '22 at 21:09
  • @ThomasAndrews You are not understanding what I said. – Igor Rivin Oct 12 '22 at 23:15

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