In Van der Put's Rigid analytic geometry and its applications, he's trying to define holomorphic functions on an affinoid subset of $\mathbb{P}^1/K$ as follows, where $(K, |\cdot|)$ is an algebraically closed, complete non-Archimedean field.
I'm wondering why such an $f$ has bounded absolute value on $F$. My argument is like:
Suppose $F\subsetneq \mathbb{P}^1$, then it is contained in a closed disk, say of the form $B = \{a\in K: |a|\leq r\}$ with $r\in |K^*|$. Write $f = g/h$ as a quotient of polynomials in the lowest form. Then $g$ clearly has bounded absolute value on $B$ and hence on $F$. We can use a similar argument as in real analysis to show that $h$ is bounded too; $h$ is clearly continuous on $F$, hence attains a non-zero minimum on $F$ if provided that $F$ is compact. But I feel like this argument only holds if every closed disks are compact, which I'm not sure if it is true in general, say if $K$ is not locally compact.
I'm wondering if there are other arguments without using the compactness of closed disks. Also, is $\mathbb{P}^1$ compact w.r.t. the ordinary topology induced from $(K,|\cdot|)$?