In this section, we briefly lay down some foundations of different branches of analysis, including functional analysis, nonarchimedean analysis, and complex analysis. The references are [Pin], [Sten], [Gilt], [Die], [Rub], [Bos], etc.
First, we introduce the notion of distance space, which generalizes Euclidean spaces.
Definition 2.1. A distance space or a metric space is a pair $(X,d)$ consisting of a set $X$ and a function $d:X\times X\rightarrow\mathbb{R}_{\geq0}$ that satisfies the following axioms
We shall call $d$ a distance or a metric on $X$.
Equipping $X$ with a distance $d$ enables us to define a topology on $X$. One just need to let open balls be the sets of the following form
$$ \mathbb{B}(x_{0},r)=\{x\in X\mid d(x_{0},x)<r\} $$
for $x_{0}\in X$ and $r>0$. Then we obtain a topology $\mathscr{O}_{d}$, which is called the induced topology on $X$. A topological space $(X,\mathscr{O})$ is metrizable if $\mathscr{O}=\mathscr{O}_{d}$ for some metric $d$ on $X$.
In a distance space, one has the notions of Cauchy sequence and completeness.
Definition 2.2. Let $(X,d)$ be a distance space.
Now, we make the definition of normed linear spaces, which can be viewed as generalizations of Euclidean spaces. The notion of norm generalizes the notion of length in vector spaces. Moreover, every norm induces a distance in some way such that every normed space is a distance space.
Definition 2.3. A normed linear space (also simply called a normed space) is an $\mathbb{F}$-linear space $X$ endowed with a function $\|\cdot\|:X\rightarrow\mathbb{R}_{\geq0}$ such that for all $\alpha\in\mathbb{F}$ and $x,y\in X$
Remark 2.4. Note that $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. We will call $\|\cdot\|$ a norm on $X$. And we call the value of $\|x\|$ for any $x\in X$ the length of $x$.
Example 2.5. The ordinary absolute values on $\mathbb{R}$ or $\mathbb{C}$ are norms on $\mathbb{R}$ or $\mathbb{C}$.
Clearly, $(\mathbb{R}^{n},d)$ is a distance space and $(\mathbb{R}^{n},\|\cdot\|)$ is a normed linear space. And every norm $\|\cdot\|$ on a linear space $X$ induces a distance $d(x,y)=\|x-y\|$ on $X$. Using a norm on $X$, we can also define an \textbf{induced topology} on $X$. Just define open balls to be the sets of the form
$$\mathbb{B}(x_{0},r)=\{x\in X\mid \|x-x_{0}\|<r\}$$
for some $x_{0}\in X$ and $r>0$.
One also has the notions of Cauchy sequence and completeness in normed spaces.
Definition 2.6. Let $(X,\|\cdot\|)$ be a normed space. A sequence $(a_{n})_{n\in\mathbb{N}}$ in $X$ is a Cauchy sequence if for every $\varepsilon>0$ $$\|a_{n}-a_{m}\|<\varepsilon$$ when $n,m$ sufficiently large. We say that the sequence $(a_{n})_{n\in\mathbb{N}}$ converges to some $a\in X$ if for every $\varepsilon>0$ $$\|a_{n}-a\|<\varepsilon$$ for $n$ sufficiently large. Then $(X,\|\cdot\|)$ is complete if every Cauchy sequence in $X$ is convergent. A complete normed space is called a Banach space.
The ordinary absolute values on the fields $\mathbb{R}$ or $\mathbb{C}$ can be easily generalized to absolute values on arbitrary fields. In fact, viewing a field as vector space over itself, absolute values on a field are special cases of norms on fields.
Definition 2.7. Let $K$ be a field. An absolute value on $K$ is a map $\left|\cdot\right|:K\rightarrow\mathbb{R}_{\geq0}$ such that for all $x,y\in K$
Every absolute value on $K$ induces a distance by defining $d(x,y)=\left|x-y\right|$. And we say that $K$ is complete if it is complete with respect to the induced distance.
Lemma 2.8. Let $\left|\cdot\right|$ be an absolute value on a field $K$. Then we have
Proof. The proof is trivial and will be left for the reader.
However, when the absolute value satisfies an inequality that is stronger than the triangular inequality, i.e. the ultrametric inequality, we get into the so-called nonarchimedean analysis.
First, we state the Archimedean property of $\mathbb{R}$ to motivate the definition of archimedean absolute values.
Lemma 1 (Archimedean property). For every $x>0,y\geq0$ there exists $n\in\mathbb{N}$ such that $nx\geq y$. Equivalently, for every $x\geq0$ there exists $n\in\mathbb{N}$ such that $n\geq x$.
So to define archimedean absolute values, we just need to pass the above condition to the ranges of the absolute values.
Definition 2.9. An absolute value $\left|\cdot\right|$ on a field $K$ is archimedean if for every $x\in K$, there exists $n\in\mathbb{N}$ such that $\left|n\right|\geq\left|x\right|$.
Remark 2.10. Note that $\left|n\right|$ is short for $\left|n\cdot1\right|$.
Example 2.11. The ordinary absolute values on $\mathbb{Q},\mathbb{R},\mathbb{C}$ are clearly archimedean absolute values.
Proposition 2.12. Let $\left|\cdot\right|$ be an absolute value on a field $K$. Then the following conditions are equivalent:
Proof. $(1)\Rightarrow(2)$: Since $\left|\cdot\right|$ is nonarchimedean, there exists $x\in K$ such that $\left|n\right|<\left|x\right|$ for all $n\in\mathbb{N}$. Assume that there exists some integer $m>1$ such that $\left|m\right|>1$. Then we have $$ \lim_{k\to\infty}\left|m\right|^{k}=\lim_{k\to\infty}\left|m^{k}\right|=\infty, $$ which contradicts the nonarchimedean condition. So $\left|n\right|\leq1$ for all $n\in\mathbb{Z},n>1$.
$(2)\Rightarrow(3)$: Since we have $\left|-n\right|=\left|n\right|$ for $n>1$ and $\left|1\right|=1$, $\left|n\right|\leq1$ for all $n\in\mathbb{Z}$.
$(3)\Rightarrow(4)$: For $n>0$, we have $ \begin{align*} \left|x+y\right|^{n}&=\left|\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^{i}\right| \\ &\leq \sum_{i=0}^{n}\left|\binom{n}{i}\right|\left|x^{n-i}\right|\left|y^{i}\right| \\ &\leq \sum_{i=0}^{n}\left|x^{n-i}\right|\left|y^{i}\right| \\ &\leq (n+1)\max\{\left|x\right|^{n},\left|y\right|^{n}\}. \end{align*}$
So we have $\left|x+y\right|\leq(n+1)^{\frac{1}{n}}\max\{\left|x\right|,\left|y\right|\}$. By the following limit $$ \lim_{n\to\infty}(n+1)^{\frac{1}{n}}=1, $$ we conclude that $\left|x+y\right|\leq\max\{\left|x\right|,\left|y\right|\}$.
$(4)\Rightarrow(1)$: Since for $n>0$ $$ \left|n\right|=\left|\underbrace{1+1+...+1}_{n\textrm{ times}}\right|\leq\left|1\right|=1, $$ (1) follows trivially.
We can say that an absolute value on a field $K$ is nonarchimedean if it satisfies one of the equivalent conditions in Proposition 2.12. However, we just utilize one of them.
Definition 2.13. Let $K$ be a field. A nonarchimedean absolute value on $K$ is a map $\left|\cdot\right|:K\rightarrow\mathbb{R}_{\geq0}$ such that for all $x,y\in K$
The pair $(K,\left|\cdot\right|)$ consisting of a field $K$ and a nonarchimedean absolute value $\left|\cdot\right|$ on $K$ will be called a valued field.
Lemma 2.14. For $a,b\in K$, then $a\neq b$ implies $$ \left|a+b\right|=\max\{\left|a\right|,\left|b\right|\}. $$
Proof. Without loss of generality, we assume that $\left|b\right|>\left|a\right|$. Then $$ \left|a+b-a\right|\leq\max\{\left|a+b\right|,\left|a\right|\}. $$
Assume that $\left|a+b\right|<\left|a\right|$, then we have $ \begin{align*} \left|b\right|&\leq\max\{\left|a+b\right|,\left|a\right|\} \\ &\leq\left|a\right|, \end{align*}$ which leads to a contradiction! So we have $\left|a+b\right|\geq\left|a\right|$. Then $\left|a+b\right|\leq b$ implies that $\left|a+b\right|= b$.
Definition 2.15. For a centre $a\in K$ and a radius $r\in \mathbb{R}_{> 0}$, the disk around $a$ without boundary is the set $$ D^{-}(a,r) = \{ x \in K\mid d(x,a)<r \}. $$
Similarly, the disk around $a$ with boundary is the set $$ D^{+}(a,r) = \{ x \in K\mid d(x,a)\leq r\}. $$
And the boundary itself is the set $$ \partial D(a,r) = \{ x \in K\mid d(x,a)= r\}. $$
Corollary 2.16. Let $K$ be a valued field. Every triangle in $K$ is isosceles. And each point of a disk in $K$ serves as the center. In particular, if the intersection of two disks is non-empty, then the two disks are concentric.
Proof. Following the ultrametric inequality, we obtain the following inequality of distances among $x,y,z\in K$: $$ d(y,z) \leq \max\{d(x,y),d(x,z)\}. $$ Then by Lemma 2.14, we can easily deduce the desirable conclusions.
Lemma 2.17. Let $K$ be a valued field. A power series $\sum_{v=0}^{\infty}a_{v}$ with $a_{v}\in K$ is a Cauchy sequence if and only if $\lim\limits_{v\to\infty}a_{v}=0$. Thus, if $K$ is complete, then the power series $\sum_{v=0}^{\infty}a_{v}$ is convergent if and only if $\lim\limits_{v\to\infty}a_{v}=0$.
Proof. Assume that $\sum_{v=0}^{\infty}a_{v}$ is a Cauchy sequence. Then for any $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that $\left|\sum_{v=i}^{j}a_{v}\right|<\varepsilon$ for all $j\geq i\geq N$. So we have $$ \sum_{v=i}^{j}\left|a_{v}\right|\leq\left|\sum_{v=i}^{j}a_{v}\right|<\varepsilon, $$ which implies that $\lim\limits_{v\to\infty}a_{v}=0$.
Conversely, choose any $\varepsilon>0$. We just need to prove that there exists $N\in\mathbb{N}$ such that $\left|\sum_{v=i}^{j}a_{v}\right|<\varepsilon$ for all $j\geq i\geq N$. Since $\lim\limits_{v\to\infty}a_{v}=0$, there exists $N\in\mathbb{N}$ such that $\left|a_{v}\right|<\varepsilon$ for all $v\geq N$. Then for any integers $j\geq i\geq N$, iterated application of ultrametric inequality yields: $$ \left|\sum_{v=i}^{j}a_{v}\right|\leq\max_{i\leq v\leq j}\left|a_{v}\right|<\varepsilon. $$
In this subsection, we collect some fundamental materials of complex analysis.
Lemma 2.18. Let $f:\Omega\rightarrow\mathbb{C}$ be a complex function on a set $\Omega\subset\mathbb{C}$. Let $\xi\in\Omega$ be a point. Then $f$ is continuous at $\xi$ if and only if for every sequence $\{z_{1},...,z_{n}\}$ in $\Omega$ with $\lim_{n\to\infty} z_{n}=\xi$, we have $\lim_{n\to\infty} f(z_{n})=f(\xi)$.