# In $\mathbb Q_p$, proving every open ball is the disjoint union of more than one open ball

I'm reading the Foundations chapter of Gouvea's p-adic Numbers: An Introduction, and I'm trying to solve the following problem he poses to the reader:

Take the $p-$adic absolute value on $\mathbb Q$. Show that with respect to this absolute value, every open ball is the disjoint union of [more than one open ball].

In the hints, he suggests that the reader use "scaling and translating" of results from earlier problems about $\mathbb Q_p$: First, the problem of showing that the closed ball $\overline B(0,1)$ is the disjoint union of open balls $$\overline B(0,1) = B(0,1)\cup B(1,1)\cup B(2,1)\cup \cdots\cup B(p-1,1),$$ and second, the problem of showing that the open ball of radius $1$ about $0$ is equivalent to the closed ball of radius $1/p$ about $0$.

The first problem can be solved by supposing $x\in \overline B(0,1)$, noting that in lowest terms $x = a/b$ with $p\nmid b$, and then showing that the terms $$a, a-b, a-2b, \ldots, a-(p-1)b$$ are all unequal modulo $p$. The second problem can be solved simply by noting that the $p-$adic absolute value can only take values of the form $p^n$, $n\in\mathbb Z$, and thus there are no elements of $\mathbb Q_p$ with absolute values strictly between $1/p$ and $1$.

What I'm having trouble doing is "scaling and translating" the above two solutions to the general cases required to solve the present problem. My work so far for each part is as follows:

For the first part: Let $x\in \overline B(x_0, \varepsilon)$. Then $|x-x_0| \leq \varepsilon$. Write $x-x_0$ in lowest terms as $a/b$ and suppose that the $p-$adic valuation $v_p(a/b)\geq 0$, so that $p\nmid b$. Now consider the $p$ terms $$a, a-b, a-2b, \ldots, a-(p-1)b.$$ If any pair of these were congruent modulo $p$, then we would have their difference being divisible by $p$, so $p|(kb)$ where $0 < k < p$, and then $p|b$, a contradiction. Thus, $p$ divides precisely one of these terms, $a - jb$. Then $v_p(\frac{a-jb}{b}) > 0$, so $|\frac{a-jb}{b}| < 1$, that is, $|\frac{a}{b} - j| < 1$, so $x-x_0 = a/b \in B(j,1)$. However, I think I'm supposed to end up with $x$ being in some sort of open ball, rather than $x-x_0$ being in the open ball, so I'm stuck. I am intending to look at the case of $v_p(a/b) < 0$ separately, once I get past this point.

For the second part, I know that given $B(x_0,\varepsilon)$, if I let $n$ be the largest integer strictly less than $\log_p \varepsilon$, then $\overline B(x_0, p^n) = B(x_0,\varepsilon)$, and then since every closed ball is the disjoint union of open balls from the preceding part, I am done.

• I'm a bit confused - is the problem asking you to show that every open set is a disjoint union of open balls? (Otherwise it seems trivially true.) – Noah Schweber May 23 '15 at 20:55
• Sorry, more than one. I'll fix it. – justin May 23 '15 at 20:57

Every ball (open or closed) of $\Bbb Q_p$ looks like $a+p^n\Bbb Z_p$ for some $a\in\Bbb Q_p$ and $n\in\Bbb Z$. This is the image of $\Bbb Z_p$ under the map $x\mapsto a+p^nx$, and affine transformations send balls to balls...

Note $\displaystyle\Bbb Z_p=\bigsqcup_{a=0}^{p-1} (a+p\Bbb Z_p)=\bigsqcup_{a=0}^{p-1}\bigsqcup_{b=0}^{p-1}(a+bp+p^2\Bbb Z_p)=\bigsqcup_{a=0}^{p-1}\bigsqcup_{b=0}^{p-1}\bigsqcup_{c=0}^{p-1}(a+bp+cp^2+p^3\Bbb Z_p)=\cdots$

To see why, think about what the expansions of $p$-adic numbers look like.

I highly recommend familiarizing yourself with the tree picture of $p$-adics. For instance,

$\hskip 1in$

(extended infinitely downward) would represent $\Bbb Z_2$. Every $2$-adic number is represented by an infinite path down the tree - every choice of going left signifies a digit of $0$ and every choice of going right signifies a digit of $1$ in its $2$-adic expansion. If we wanted to depict $\Bbb Z_p$, every node would have a total of $p$ children instead of just the $p=2$ here. And if we want to depict $\Bbb Q_p$, then the tree must also extend upward infinitely as well, where going backwards represents the coefficients of the negative powers $p^{-1}$, $p^{-2}$, $p^{-3}$, $\cdots$ in a $p$-adic expansion.

The valuation of a $p$-adic number can be determined by looking at where the first non-$0$ digit is, i.e. the first not-full-left choice going down. And every ball is just the set of all paths through some fixed node. Of course, such a ball is the union of the $p$ balls below it!