$|f(\mathbb{Z}_p)|$ has at most $0$ as a limit point in the non-zero reals

Question

I am currently reading through Robert's A Course in p-adic Analysis textbook and got stuck on one of the steps on his proof of the Mahler Theorem (p173). I can embed images, so here is an excerpt of the proof:

I understand that continuous functions send compact sets to compact sets (so $f(\mathbb{Z}_p)$ is compact), but how does one show that $|f(\mathbb{Z}_p)|$ (p-adic absolute value) has at most $0$ as a limit point in the non-zero reals?

I think I start with supposing that there is some $r > 0$ that is a limit point, so there is sequence of points $x_1, x_2, \dots \in \mathbb{Z}_p$ so that $|f(x_n)| \to p^{-r}$, but I'm not sure where to go from there.

Answer 1

The key concept is this: the non-zero accumulation points are deceptive, in the sense that if $x_n \to x$ in $\Bbb C_p$, then either $|x_n|$ is eventually constant, or $|x_n| \to 0$; in other words, if $|x_n|$ tends to say $1$, and $|x_n|$ is strictly increasing, then there is no chance that $x_n$ actually converges.

With this in mind, it is now a trivial application of the "Bolzano–Weierstrass theorem":

Assume $y \in \Bbb R$ is a non-zero accumulation point, so there are $x_n \in \Bbb Z_p$ such that $|f(x_n)| \to y$ and $|f(x_n)|$ are all distinct. By "Bolzano–Weierstrass theorem", $x_n$ has a convergent subsequence, so we might WLOG assume that $x_n \to x$; since $f(x_n) \to f(x)$ and $|f(x_n)|$ are all distinct, we must have $|f(x_n)| = y = 0$, contradiction.

Answer 2

Hint 1: First, let $S \subseteq \mathbb C_p$ be any subset and assume some $p^r$ is an accumulation point of $\lvert S \rvert_p \subseteq p^{\mathbb Q} \cup \{0\}$. That means there is a sequence of mutually distinct rational numbers $r_n$ such that $\lim_{n \to \infty} r_n= r$ and corresponding elements $s_n \in S$ with $\lvert s_n \rvert_p =p^{r_n}$.

Hint 2: Show that a sequence $s_n$ as constructed above does not have any convergent subsequence. (Use that in $\mathbb C_p$, $\lim_{n \to \infty} c_n = c \neq 0$ implies that $\lvert c_{n} \rvert_p = \lvert c \rvert_p$ for high enough $n$. Compare "Added" in https://math.stackexchange.com/a/3280427/96384)

So, can such an $S$ be compact?

Apply to $S := f(\mathbb Z_p)$.