Unfortunately, I have some difficulties understanding the following
proof from Profinite Groups (https://link.springer.com/book/10.1007/978-3-662-04097-3)
on page 123.
Let $G$ be a profinite group and $x \in G$. Since
profinitely completed integers $\widehat{\mathbb{Z}}=
\prod_{p \in \mathbb{P}} \mathbb{Z}_p$ is
a free profinite group on $\{1\}$, there is a unique epimorphism
$$ \varphi: \widehat{\mathbb{Z}} \to \overline{\langle x \rangle} $$
such that $\varphi(1) = x$. Given $\lambda \in \widehat{\mathbb{Z}}$,
define $x^{\lambda} := \varphi(\lambda)$.
In Lemma 4.1.1 (a) is stated that
if $n_1 n_2, ... \in \mathbb{Z}$ is a sequence of integers converging
to $\lambda \in \widehat{\mathbb{Z}}$, then
$$ \lim_{i \to \infty} x^{n_i} =x^{\lambda} $$
The book says that this clear since $\widehat{\mathbb{Z}}$ and $\overline{\langle x \rangle}$ are metric spaces.
I not see how the property of beeing metric spaces can effectively applied here to obtain the claimed statement about convergence? Isn't this a question of completedness with respect of underlying topology and not if the two spaces carry a metric or not? Therefore I not understand this argument. Could somebody clarify that point?