Is the set of integers with respect to the p-adic metric compact?

Given the integers and a prime $p$. I thought I had successfully shown that $\mathbb{Z}$ was compact with respect to the metric $|\cdot |_p$, by showing that the open ball centered at zero contained all integers with more than a certain number of factors of $p$, and then showing that the remaining integers took on a finite number of possible p-adic absolute values and thus fell into a finite number of balls.

Now if the integers are compact with respect to $|\cdot |_p$, then that means they are complete with respect to $|\cdot |_p$.

But then I read that the p-adic integers $\mathbb{Z}_p$ are defined to be the completion of the integers with respect to $|\cdot |_p$, and include in their completion all the rational numbers with p-adic absolute value less than or equal to one. So this means that the integers with respect to the p-adic metric are not complete, and thus not compact, and hence there must be something wrong with my proof, correct?

Edit: Ok upon typing this up I realized that my proof is most likely wrong as there's no reason to conclude that two elements with the same absolute value are necessarily in the same ball.

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$\mathbb{Z}$ is not closed $p$-adically. For example, the geometric series $$1+p+p^2+p^3+\cdots=\frac1{1-p}$$ of integers that convegers w.r.t the $p$-adic metric has a sum that is not an integer. The space is Hausdorff, so a compact set should be closed. – Jyrki Lahtonen Mar 1 '13 at 8:30

You don't prove compactness "by showing that the open ball centered at zero contained all integers with ..., and then showing that the remaining integers ... fell into a finite number of balls", i.e. by showing that there is a finite number of open balls covering the space.

Actually you can show more : $\Bbb Z$ with the $p$-adic metric is a bounded metric space : every integer is at a distance less than $1$ from $0$.

Instead, to prove compactness you have to show that for any covering of $\Bbb Z$ by open balls, you can select a finite number of those open balls and still cover $\Bbb Z$. For example, pick the covering of $\Bbb Z$ by placing an open ball on $n$ with radius $p^{-|n|}$. For most $p$ ($p \ge 5$) , you can't extract a finite cover of $\Bbb Z$ from this cover.

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@Electromagnetism : Pick the space of functions $\Bbb N \to [0;1]$ with $d(f,g) = \sup |f(n)-g(n)|$. It is bounded (everyone is at distance less than $1$ from the $0$ function) but there are sequences of functions that don't have any Cauchy sequence. The space is complete, but not locally compact. – mercio Mar 1 '13 at 17:30

Jyrki showed you a specific example of a cauchy sequence not converging in $\mathbb{Z}$, but there is a wholly more dramatic answer. Let's suppose that $\mathbb{Z}$ was complete. Then, every infinite series $\sum_{n=0}^{\infty}a_np^n$ with $a_n\in\{0,1,\ldots,p-1\}$ would converge.(because each such series has partial sums that are Cauchy). Moreover, two such infinite series are equal if and only if their coefficients (of $p^n$) are equal (just check their valuations). Thus, we'd have an injection $(\mathbb{Z}/p\mathbb{Z})^\mathbb{N}\to \mathbb{Z}$ which is problematic due to carinality issues.

This also shows that $\mathbb{Q}$ is not complete

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