Lemma 5.1. Let $X={\rm{Sp}}\ A$ be an affinoid $K$-space and $f_{0}, ..., f_{r}\in A$. Then $f_{0}, ..., f_{r}$ have no common zeros on $X$ if and only if $f_{0}, ..., f_{r}$ generate $A$ as the unit ideal.
Proof. If $f_{0}, ..., f_{r}$ generate $A$ as the unit ideal, assume that $f_{0}, ..., f_{r}$ have some common zero $x_{0}$ on $X$. Then for any $f\in A\backslash x_{0}$,
$$f=\sum_{i=0}^{r}a_{i}f_{i},\ a_{i}\in A,$$
but then $f(x_{0})=0$ implies that $f\in x_{0}$, which is clearly a contradiction! So $f_{0}, ..., f_{r}$ have no common zeros on $X$.
Conversely, if $f_{0}, ..., f_{r}$ have no common zeros on $X$, assume that $f_{0}, ..., f_{r}$ generate an ideal $I\varsubsetneq A$. Since there exists some maximal ideal $x\in X$ such that $I\subseteq x$, we have
$$f_{0}(x)=\cdot\cdot\cdot=f_{r}(x)=0,$$
which shows that $f_{0}, ..., f_{r}$ have a common zero. So by contradiction, we conclude that $f_{0}, ..., f_{r}$ generate the unit ideal in $A$.
$\Box$
Definition 5.2. Let $X=\textrm{Sp }A$ be an affinoid $K$-space.
(1) A subset of $X$ of type $$X(f_{1}, ..., f_{r}):=\{x\in X\mid \left|f_{i}(x)\right|\leq1\}$$ for $f_{1},...,f_{r}\in A,1\leq i\leq r$ is called a Weierstrass domain in $X$.
(2) A subset of $X$ of type $$ X(f_{1},...,f_{r},g_{1}^{-1},...,g_{s}^{-1}):=\{x\in X\mid\left|f_{i}(x)\right|\leq1, \left|g_{j}(x)\right|\geq1 \} $$ for $f_{1},...,f_{r},g_{1},...,g_{s}\in A,1\leq i\leq r,1\leq j\leq s$ is called a Laurent domain in $X$.
(3) A subset of $X$ of type $$ X(\frac{f_{1}}{f_{0}},...,\frac{f_{r}}{f_{0}}):=\{x\in X\mid\left|f_{i}(x)\right|\leq\left|f_{0}(x)\right|\} $$ for $f_{0},...,f_{r}\in A,1\leq i\leq r$ without common zeros is called a rational domain in $X$.
Remark 5.3. The domains in Definition 5.2 are special cases of affinoid subdomains.