How to prove that in p-adic rationals every sphere is open?

Question

I‘m a little bit confused with this proof question, because it does intuitively not make any sense:

Let r = p^m for m in $\mathbb{Z}$ and p prime, let q be a rational number and |.|_p be the p-adic norm on $\mathbb{Q}$.

Let S(q,r) denote the sphere around q with radius r, that is S = {t in $\mathbb{Q}$ | |q-t| _p = r}. Show that S(q,r) is open with respect to |.| _p

Hint: Choose 0< $\varepsilon$ < r. Show that q₁ $\in$ S(q,r) => B(q₁, $\varepsilon$) $\subset$ S(q,r).

Answer 1

Hints
Of course it "does not intuitively make sense" if your intuition is only Euclidean space.

Your hint says that if $|q-q_1|_p=r$ and $|x-q_1|_p=\varepsilon$, then $|x-q|_p = r$. Try proving this using the definition of $|\cdot|_p$.

That is:
If $p^m$ is the highest power of $p$ that divides $q-q_1$ and $p^n$ (with $n>m$) is the highest power of $p$ that divides $x-q_1$, then $p^m$ is the highest power of $p$ that divides $x-q$.

Answer 2

We can take any point in the sphere $q_1 \in S(q,r)$ and show that the open ball $B(q_1,r):=\{t \in \mathbb{Q}_p : |t-q_1|_p<r\}$ is contained in it, $B(q_1,r) \subseteq S(q,r)$. That will allow us to say, $$S(q,r) = \bigcup_{q_1 \in S(q,r)}B(q_1,r)$$ Which means $S(q,r)$ is open because it's an arbitrary union of open sets.

To show $B(q_1,r) \subseteq S(q,r)$, let's pick an arbitrary $t \in B(q_1,r)$ and show that $t \in S(q,r)$. We know $|q-q_1|_p=r$ and $|q_1-t|_p<r$. This means by the ultrametric inequality,

$$|q-t|_p = |q-q_1+q_1-t| \le \max(|q-q_1|_p, |q_1-t|_p)$$

But the ultrametric inequality has the strong property, that if $|x|_p \ne |y|_p$ then $|x+y|_p = \max(|x|_p, |y|_p)$. In our case $|q_1-t|_p<|q-q_1|_p$ so we actually have equality,

$$|q-t|_p = \max(|q-q_1|_p, |q_1-t|_p) = r$$

So $t\in S(q,r)$, and that concludes the proof.

---

If you're unfamiliar with the ultrametric inequality and its strong property, you should know that it's not too difficult to prove the strong property of being equal to the maximum follows from the regular ultrametric inequality. I only bring this up because it's true for all ultrametric spaces, and personally it helps clear my mind which might be clouded by trying to prove this strong property at the level of playing with prime factorizations of rational numbers or something cumbersome like that.