A rigid-analytic space is an analogue of a complex analytic space over a non-archimedean field, which first made its appearance in John Tate's thesis \cite{Tate}. In this section, we will introduce the notion of rigid space, using \cite{Tate} as our reference. All rings are commutative with identity and $K$ is the fixed ground field, which is complete with respect to a non-trivial rank-1 valuation denoted by $\left|\cdot\right|$. The valuation ring of $K$ is denoted by $V$ and its maximal ideal is denoted by $\mathfrak{m}$. And $x$ is a fixed non-zero element in $\mathfrak{m}$, i.e. $0<\left|x\right|<1$.
Let $E$ be a vector space over $K$. Then $E$ is also a $V$-module if we restrict the scalar multiplication to $V\times E$. We let $E_{0}$ be a $V$-submodule of $E$ such that $E=KE_{0}$.
Set $E_{n}:=x^{n}E_{0}$ for $n\in \mathbb{Z}$. We obtain a filtration
$$ E\supset \cdot\cdot\cdot\supset E_{-n}\supset E_{-n+1}\supset\cdot\cdot\cdot\supset E_{0}\supset\cdot\cdot\cdot\supset E_{n}\supset E_{n+1}\supset \cdot\cdot\cdot\supset0 $$
of $E$ (note that we have $x^{-\infty}E_{0}=E$ and $x^{\infty}E_{0}=0$). It is easy to see that using such filtration, we can form a fundamental system of neighborhoods of 0 in $E$.
So one can define a topology on $E$ by saying a subset $U\subset E$ is open if for every $y\in U$, $y+x^{n}E_{0}\subset U$ for some $n\in\mathbb{Z}$. Such topology is said to be defined by $E_{0}$.
Lemma 7.1. Let $F$ be another $K$-vector space whose topology is defined by $F_{0}$. A $K$-linear map $\varphi:E\rightarrow F$ is continuous if and only if there exists some integer $n$ such that $\varphi(E_{0})\subset x^{-n}F_{0}$.
Proof. Let $p\in E$ be any point. If $\varphi$ is continuous, then for every neighborhood $\varphi(p)\in V$ in $F$, there exists a neighborhood $p\in U$ in $E$ such that $U\subset V$. This implies that for a neighborhood $F_{0}$ of 0 in $F$, there exists integer $n$ such that $\varphi(x^{n}E_{0})\subset F_{0}$. Hence we have $\varphi(E_{0})\subset x^{-n}F_{0}$.
Conversely, if there is some integer $n$ such that $\varphi(E_{0})\subset x^{-n}F_{0}$, then we have $\varphi(x^{n}E_{0})\subset F_{0}$. Thus, for every $m\in\Z$, we have $\varphi(x^{n+m}E_{0})\subset x^{m}F_{0}$, which shows that for every open $V\subset F$ and $y\in\textrm{Im }\varphi\cap V$, we have $\varphi(y_{0}+x^{n+m}E_{0})\subset y+x^{m}F_{0}\subset V$ where $y=\varphi(y_{0})$.
$\Box$