I am currently learning the basics of $p$-adic analysis, and its similarities (or dissimilarities) to $\Bbb{R}$. The Archimedean topology on $\Bbb{Z}$ is discrete, so there's nothing interesting in the continuity of a function $f:\Bbb{Z}\to \Bbb{Z}$. However, one may ask the following question:
Let $f:\Bbb{Z}\to\Bbb{Z}$ be a continuous function w.r.t. the $p$-adic metric. Must there exists a continuous function $g:\Bbb{Q}_p\to\Bbb{Q}_p$ s.t. $g\restriction_{\Bbb{Z}} = f$?
Since $\Bbb{N}$ is dense in $\Bbb{Z}_p$, we can extend $f$ (uniquely) into $\Bbb{Z}_p$. But I see no reason for there to be an extension to $\Bbb{Q}_p$. Is there indeed a counterexample?
Any function on $\mathbb Z_p$ can be easily extended to $\mathbb Q_p$, because $\mathbb Z_p$ is a clopen subset of $\mathbb Q_p$. In particular $f(x)\equiv 0$ for all $x\not\in\mathbb Z_p$ can get the job done.
However, it's not the case that any continuous function on $\mathbb Z$ can be extended to $\mathbb Z_p$, just like not every continuous function on $(0, 1)$ can be extended to $[0, 1]$.
For a concrete counter-example, consider the function $$f(x):=\nu_p(x-\frac{1}{1+p})$$
which is locally constant on $\mathbb Z$ hence continuous. However since $x_n=\sum_{i=0}^n (-p)^n$ converges to $\frac{1}{1+p}$, and $f(x_n)=n+1$ is not Cauchy, $f(x)$ cannot be continuously extended to be defined at $x=\frac{1}{1+p}$.
However, Lipschitz functions on $\mathbb Z$ can be easily extended to $\mathbb Z_p$.