How do I show that this function in the p-adic field is continuous

Question

Given

$f:\mathbb{Z}_p \to \mathbb{Q}_p $ such that

$$ f(x)= \begin{cases} 0, & \text{if } x=0 \\ 1/ |x|_p & \text{if } x\neq0\\ \end{cases} $$

How do I show that f is continuous and locally constant on $\mathbb{Z}_p$?

I'm planning to show that f is locally constant first so that it would just imply that it is also continuous. But I'm having a hard time choosing what particular radius I should put so that the function is constant on that open ball.

Any help would be great! :) I really need to understand this for my potential undergraduate thesis topic. :)

Answer 1

I would restate your definition of $f$ slightly, because most properly, $|x|_p$ is not a $p$-adic number but a real number. I would rather define your function as $f(x)=p^{v_p(x)}$, where $v_p:\Bbb Q_p\to\Bbb Z\cup\{+\infty\}$, in other words, $v_p(x)=-\log_p(|x|_p)$, so that $v_p(p)=1$, $v_p(xy)=v_p(x)+v_p(y)$, and $v_p(n)=0$ if $n$ is an integer prime to $p$.

To prove that $f$ was locally constant, I would take $0\ne z\in\Bbb Z_p$, say $v(z)=r\ge0$. Then on the set $\{x\in\Bbb Z_p:v_p(x-z)>r\}$, the function is constant. In the language of absolute values, if $|z|=p^{-r}$, then the function is constant on the “ball” centered at $z$ of radius $p^{-r-1}$. Of course you use the nonarchimedean property to show this.

For continuity at $0$, the only point where things are questionable: take a sequence of points $\{z_m\}$ going to zero, then you see at a glance that $v(z_m)\to\infty$, so that $p^{v(z_m)}\to0$, and all’s right with the world.

Answer 2

Hint for locally constant at $x\neq 0$: What is $\vert x + y\vert_p$, if $\vert y\vert_p$ is less than $\vert x\vert_p$ (i.e., in a small ball around $x$)?

From that, you indeed get continuity at every $x\neq 0$, but still need to show continuity at $0$. Well, what is the $p$-adic limit $\lim_{x \to 0} f(x)$? Concretely, what is

$\lvert 1/\vert x\vert_p \rvert_p$

if $\vert x\vert_p \le \epsilon$?