弦圈

In this section, we would like to review some interesting results about the Euclidean space $\mathbb{R}^{n}$. By a Euclidean space $\mathbb{R}^{n}$, we may mean the real inner product space $\mathbb{R}^{n}$ or the vector space $\mathbb{R}^{n}$ depending on the context. Let $U\subset\mathbb{R}^{n}$ be a subset. The Euclidean norm is given by$$ \|x\|=\sqrt{x_{1}^{2}+...+x_{n}^{2}} $$for any $x=(x_{1},...,x_{n})\in\mathbb{R}^{n}$ and the Euclidean distance is defined by$$d(x,y)=\sqrt{(x_{1}-y_{1})^{2}+...+(x_{n}-y_{n})^{2}}$$for any $x=(x_{1},...,x_{n}),y=(y_{1},...,y_{n})\in\mathbb{R}^{n}$.Definition 1.1. We say that $U$ is open if for every $x\in U$ there exists $\mathbb{B}(x,r)\subset U$, where $\mathbb{B}(x,r):=\{y\in\mathbb{R}^{n}\mid\|y-x\|<r\}$ for some $r>0$. And we say that $U$ is closed if $U^{c}$ is open.An (open) neighborhood of a point $x\in\mathbb{R}^{n}$ is an open set containing $x$. A point $p\in\mathbb{R}^{n}$ is a limit point of $U$, if for every neighborhood $V$ of $p$, we have $(V-p)\cap U\neq\varnothing$. The set of all limit points of $U$ is called the derived set of $U$, which is denoted by $U'$.The closure of $U$ denoted by $\overline{U}$ is the union of $U$ and all of its limits points, i.e. $\overline{U}=U\cup U'$.Remark 1.2. By convention, the empty set is an open subset. And we would only consider open neighborhood since it will make no difference.Proposition 1.3. Let $U\subset\mathbb{R}^{n}$ be a subset. Then the following are equivalent:$U$ is a closed subset;$\overline{U}=U$.Proof. First we assume that $U$ is a closed subset, i.e. $U^{c}$ is open. Then for every $x\in U^{c}$, there exists $\mathbb{B}(x,r)\subset U^{c}$. So there is no limit point of $U$ in $U^{c}$. Consequently, we have $\overline{U}=U$ as desired.Conversely, assume that $\overline{U}=U$, then we consider the complement $U^{c}$ of $U$. Clearly, for any $x\in U^{c}$, there exists $\mathbb{B}(x,r)\subset U^{c}$. So $U^{c}$ is open, which means $U$ is closed.By Proposition 1.3, we can easily deduce the following proposition. Proposition 1.4. Let $U$ be a subset of $\mathbb{R}^{n}$, the closure $\overline{U}$ of $U$ is the smallest closed set containing $U$, or equivalently, the intersection of all closed sets containing $U$.By Definition 1.1, we can easily deduce the following proposition, which indicates the topological properties of $\mathbb{R}^{n}$.Proposition 1.5. The following properties of $\mathbb{R}^{n}$ are satisfied:$\mathbb{R}^{n},\varnothing$ are open sets;Any finite intersection of open sets is open;The union of an arbitrary family of open sets is open.So we have the dual proposition about closed sets:Proposition 1.6. The following properties of $\mathbb{R}^{n}$ are satisfied:$\mathbb{R}^{n},\varnothing$ are closed sets;Any finite union of closed sets is closed;Arbitrary intersections of closed sets are closed.Proposition 1.5 leads to the notion of a topology:Definition 1.7. Let $X$ be a set and $\tau$ be a set of subsets of $X$. Then $\tau$ is called a topology on $X$, whose elements are called open sets, if it is subject to the following axioms:$X,\varnothing$ are open sets;Any finite intersection of open sets is open;Arbitrary unions of open sets are open.A topological space is a pair $(X,\tau)$ consisting of a set $X$ and a topology $\tau$ on $X$. By abuse of notation, we will often simply denote by $X$ when no confusion will arise.So $\mathbb{R}^{n}$ is a topological space with the standard topology $\mathscr{O}$ defined in Proposition 1.5. And we will always hold the assumption that $\mathbb{R}^{n}$ has the standard topology $\mathscr{O}$.Next, we discuss more about the topological properties of $\mathbb{R}^{n}$. It is observed that $\mathbb{R}^{n}$ cannot be expressible as a union of two non-empty disjoint open sets, i.e. the only subsets of $\mathbb{R}^{n}$ that are both open and closed are $\mathbb{R}^{n}$ and $\varnothing$. In fact, if we assume that $\mathbb{R}^{n}=U_{1}\cup U_{2}$ for two non-empty disjoint open sets $U_{1},U_{2}$, we readily see that $U_{1}\cup U_{2}$ cannot be equal to $\mathbb{R}^{n}$, which is a contradiction. However, disconnected regions in $\mathbb{R}^{n}$ can be expressible as a union of two non-empty disjoint open sets. We generalize this property to arbitrary topological spaces.Definition 1.8. A topological space $X$ is connected if $X\neq U_{1}\cup U_{2}$ where $U_{1},U_{2}$ are non-empty disjoint open sets.So $\mathbb{R}^{n}$ is a connected topological space! In fact, $\mathbb{R}^{n}$ satisfies a stronger condition.Definition 1.9. A topological space $X$ is path-connected if any pair of points in $X$ can be joined by a curve.Remark 1.10. To help understand, we use a more intuitive word "curve" instead of a topological terminology "path".Clearly, $\mathbb{R}^{n}$ is a path-connected space! Next, take a subset $U$ of $\mathbb{R}^{n}$. It is natural to ask whether properties of $\mathbb{R}^{n}$ in Proposition 1.5 can be passed down to $U$. Next, we construct the subspace topology so that every subset can be equipped with a natural topology.Definition 1.11. Let $(X,\mathscr{O})$ be a topological space and let $S\subset X$ be a subset. We define the induced or subspace topology on $S$ as follows:$$\mathscr{O}_{S}:=\{S\cap U\mid U\in\mathscr{O}\}.$$The topological space $(S,\mathscr{O}_{S})$ is called a subspace of $(X,\mathscr{O})$.Remark 1.12. In the sequel, without explicitly mentioned, every subset of a topological space naturally carries the subspace topology. Note that for any closed $U\subset X$, $U\cap S$ is a closed set in the induced topology.Then recall that in the real line case $\mathbb{R}^{1}$, every open covering consisting of open intervals of a closed interval has a finite subcovering. In fact, the $\mathbb{R}^{1}$ case can be generalized to $\mathbb{R}^{n}$ by the Heine-Borel theorem in the following.We first introduce the concept of compactness in $\mathbb{R}^{n}$.Definition 1.13. A subset $U\subset\mathbb{R}^{n}$ is bounded if there exists $\mathbb{B}(0,r)$ such that $\mathbb{B}(0,r)\supset U$. And we say that $U$ is compact if it is closed and bounded. Let $(U_{i})_{i\in I}$ be a family of subsets $U_{i}\subset\mathbb{R}^{n}$. Then we say that $(U_{i})_{i\in I}$ is a covering of $U$ if $U\subset\bigcup_{i\in I}U_{i}$. The covering is open if each $U_{i}$ is open. A subcovering of $(U_{i})_{i\in I}$ is a subfamily $(U_{j})_{j\in J}$ of $(U_{i})_{i\in I}$ with $J\subset I$ such that $U\subset\bigcup_{j\in J}U_{j}$.Theorem 1.14 (The Heine-Borel theorem). A subset $U\subset\mathbb{R}^{n}$ is compact if and only if every open covering of $U$ has a finite subcovering.This leads to the following concept of compactness of topological spaces.Definition 1.15. Let $X$ be a topological space.Let $(U_{i})_{i\in I}$ be a family of subsets $U_{i}\subset X$. Then we say that $(U_{i})_{i\in I}$ is a covering of $U$ if $U\subset\bigcup_{i\in I}U_{i}$. The covering is open if each $U_{i}$ is open in $X$. A subcovering of $(U_{i})_{i\in I}$ is a subfamily $(U_{j})_{j\in J}$ of $(U_{i})_{i\in I}$ with $J\subset I$ such that $U\subset\bigcup_{j\in J}U_{j}$. The subcovering is finite if the index set $J$ is finite.The topological space $X$ is compact if every open covering of $X$ has a finite subcovering. If $S\subset X$ is a subset, then $S$ is compact if every open covering of $S$ has a finite subcovering, or equivalently, $S$ is a compact as a subspace of $X$.And we say that $X$ is locally compact if every point in $X$ has a compact neighborhood.Remark 1.16. In the sequel, by compactness, we will always mean compactness of topological spaces. Note that $S$ is a compact as a subspace of $X$ if every open covering of $S$ whose members are open in the induced topology, has a finite subcovering.It is easy to see that $\mathbb{R}^{n}$ is locally compact.For any pair of points $x\neq y$ in $\mathbb{R}^{n}$, we can always find $\mathbb{B}(x,r_{1})\cap\mathbb{B}(y,r_{2})=\varnothing$ for some $r_{1},r_{2}>0$, which implies separated property.Definition 1.17. A topological space $X$ is Hausdorff if for every pair of points $x\neq y$ in $X$, there exist open sets $x\in U,y\in V$ such that $U\cap V=\varnothing$.Remark 1.18. For any pair of points in a Hausdorff space, we can always find two disjoint neighborhoods. Clearly, $\mathbb{R}^{n}$ is a Hausdorff space.It is observed that there is some kind of metric structure on $\mathbb{R}^{n}$, i.e. the Euclidean norm and the Euclidean distance on $\mathbb{R}^{n}$. We will generalize this property in the following section.

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$d(x,y)=0\Leftrightarrow x=y$ for all $x,y\in X$;$d(x,y)=d(y,x)$ for all $x,y\in X$;$d(x,y)\leq d(x,z)+d(y,z)$ for all $x,y,z\in X$.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.A sequence $(a_{n})_{n\in\mathbb{N}}$ in $X$ is a Cauchy sequence if for every $\varepsilon>0$ $$d(a_{n},a_{m})<\varepsilon$$ 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 $\varepsilon>0$ $$d(a_{n},a)<\varepsilon$$ for $n$ sufficiently large.$(X,d)$ is said to be complete if every Cauchy sequence in $X$ is convergent.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$$\|x\|=0\Leftrightarrow x=0$;$\|\alpha x\|=\left|\alpha\right|\|x\|$;$\|x+y\|\leq \|x\|+\|y\|$.Remark 2.4. Note that $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. We will call $\|\cdot\|$ a norm on $X$. And we call the value of $\|x\|$ for any $x\in X$ the length of $x$.Example 2.5. The ordinary absolute values on $\mathbb{R}$ or $\mathbb{C}$ are norms on $\mathbb{R}$ or $\mathbb{C}$.Clearly, $(\mathbb{R}^{n},d)$ is a distance space and $(\mathbb{R}^{n},\|\cdot\|)$ is a normed linear space. And every norm $\|\cdot\|$ on a linear space $X$ induces a distance $d(x,y)=\|x-y\|$ on $X$. Using a norm on $X$, we can also define an \textbf{induced topology} on $X$. Just define open balls to be the sets of the form$$\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$$ for $n$ sufficiently large. Then $(X,\|\cdot\|)$ is complete if every Cauchy sequence in $X$ is convergent. A complete normed space is called a Banach space.The ordinary absolute values on the fields $\mathbb{R}$ or $\mathbb{C}$ can be easily generalized to absolute values on arbitrary fields. In fact, viewing a field as vector space over itself, absolute values on a field are special cases of norms on fields.Definition 2.7. Let $K$ be a field. An absolute value on $K$ is a map $\left|\cdot\right|:K\rightarrow\mathbb{R}_{\geq0}$ such that for all $x,y\in K$$\left|x\right|=0\Leftrightarrow x=0$,$\left|xy\right|=\left|x\right|\left|y\right|$,$\left|x+y\right|\leq\left|x\right|+\left|y\right|$.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$\left|1\right|=1$,$\left|x^{-1}\right|=\left|x\right|^{-1}$ for all $x\in K$,$\left|-x\right|=\left|x\right|$ for all $x\in K$.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.§2.1. Nonarchimedean analysisFirst, 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:$\left|\cdot\right|$ is nonarchimedean;$\left|n\right|\leq1$ for all $n\in\mathbb{Z},n>1$;$\left|n\right|\leq1$ for all $n\in\mathbb{Z}$;$\left|x+y\right|\leq\max\{\left|x\right|,\left|y\right|\}$.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$$\left|x\right|=0\Leftrightarrow x=0$,$\left|xy\right|=\left|x\right|\left|y\right|$,$\left|x+y\right|\leq\max\{\left|x\right|,\left|y\right|\}$.The pair $(K,\left|\cdot\right|)$ consisting of a field $K$ and a nonarchimedean absolute value $\left|\cdot\right|$ on $K$ will be called a valued field.Lemma 2.14. For $a,b\in K$, then $a\neq b$ implies $$ \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. $$§2.2. Complex analysisIn 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 $\lim_{n\to\infty} f(z_{n})=f(\xi)$.

In this section, we will review more general topology.First, we take real functions $f:\mathbb{R}\supset D\rightarrow\mathbb{R}$ as an example to motivate the definition of continuous functions between topological spaces.Definition. We say that $f$ is continuous at a point $x_{0}\in D$ if for every $\varepsilon>0$, there exists $\delta>0$ such that for all $x\in D$, $\left|x-x_{0}\right|<\delta$ implies $\left|f(x)-f(x_{0})\right|<\varepsilon$.Lemma. Let $x\in D$. Then $f$ is continuous at $x$ if and only if for every neighborhood $V\ni f(x)$ there exists a neighborhood $U\ni x$ such that $f(U)\subset V$.Proof. Assume that $f$ is continuous at $x$. By definition, for every neighborhood $\mathbb{B}(f(x),\varepsilon)$ there exists a neighborhood $\mathbb{B}(x,\delta)$ such that $f(\mathbb{B}(x,\delta))\subset\mathbb{B}(f(x),\varepsilon)$.Conversely, for every $\left|f(x)-f(x_{0})\right|<\varepsilon$, where $f(x)\in V$ and $\varepsilon>0$, there exists $\delta>0$ with $\left|x-x_{0}\right|<\delta,x\in U$. $\Box$So we can make use of neighborhoods to define continuity of a real function. We generalize this to topological spaces. Let $X,Y$ be two topological spaces.Definition 3.1. We say that $f:X\rightarrow Y$ is continuous at $x\in X$ if for every open set $V\subset Y$ containing $f(x)$, there exists an open set $U\subset X$ containing $x$ such that $f(U)\subset V$. And we say that $f$ is continuous on $X$ or simply continuous if it is continuous at every point in $X$.Proposition 3.2. A function $f:X\rightarrow Y$ is continuous if and only if for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.Proof. Assume that $f$ is continuous. Then for every neighborhood $V\subset Y$ of some point $y\in Y$, there exists a neighborhood $U\subset X$ of $x$ such that $f(x)=y,f(U)\subset V$. Then we have $x\in U\subset f^{-1}(f(U))\subset f^{-1}(V)$, which implies that $f^{-1}(V)$ is open in $X$.Conversely, assume that for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$. Let $x\in f^{-1}(V)$ be some point, then $f(x)\in V$. So we have $f(f^{-1}(V))=V\subset V$, which shows that $f$ is continuous.$\Box$Definition 3.3. By Proposition 3.2, we can equivalently define continuous functions in terms of open sets: We say that a function $f:X\rightarrow Y$ is continuous if for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.Next, we introduce a map that preserves topological property.Definition 3.4. A function $f:X\rightarrow Y$ is called a homeomorphism if $f$ is bijective and $f,f^{-1}$ are both continuous. We say that $X$ is homeomorphic to $Y$, which is denoted by $X\cong Y$, if there exists a homeomorphism $f:X\xrightarrow{\sim}Y$.Remark 3.5. Abstractly, two homeomorphic topological spaces have no difference.Proposition 3.6. Every compact subset of a Hausdorff space $X$ is closed.Proposition 3.7. Every singleton of a point in a Hausdorff space $X$ is closed.Definition 3.8. Let $X$ be a topological space. A subset $U\subset X$ is dense in $X$ if $\overline{U}=X$.

In this section, we briefly introduce some basic concepts in group theory.Recall that a binary operation $*$ on a set $G$ is a map $*:G\times G\rightarrow G$ that assigns to each $(x,y)\in G\times G$ an element $x*y:=*(x,y)\in G$. A binary operation in additive notation is called addition, and a binary operation in multiplicative notation is called multiplication.Definition 4.1. A group is an ordered pair $(G,*)$ consisting of a set $G$ and a binary operation $*:G\times G\rightarrow G$ such that the following axioms are satisfied:Associativity: $(x*y)*z=x*(y*z)$ for all $x,y,z\in G$;Existence of identity: There exists an element $e\in G$ such that $x*e=e*x=x$ for all $x\in G$;Existence of inverses: For each $x\in G$, there exists $y\in G$ such that $x*y=y*x=1$. Then $y$ will be denoted by $x^{-1}$.A group whose binary operation is addition will be called an additive group. And a group whose binary operation is multiplication will be called a multiplicative group. The set $G$ is called the underlying set of the group $(G,*)$. Moreover, when no confusion arises, we will simply denote the group by $G$.Example 4.2. $(\mathbb{Z},+),(\mathbb{R},+),(\mathbb{Q},+),(\mathbb{C},+)$ are all additive groups. And $(\mathbb{Z}\backslash\{0\},\cdot)$, $(\mathbb{R}\backslash\{0\},\cdot)$, $(\mathbb{Q}\backslash\{0\},\cdot)$, $(\mathbb{C}\backslash\{0\},\cdot)$, $(\mathbb{N}_{>0},\cdot)$, $(\mathbb{R}_{>0},\cdot)$, $(\mathbb{Q}_{>0},\cdot)$, $(\mathbb{Z}_{>0},\cdot)$, $(\mathbb{C}_{\left|\ \right|>0},\cdot)$ are all multiplicative groups. A trivial group is a group consisting of only identity.Definition 4.3. A subgroup of a group $G$ is a subset $H\subset G$ such that $H$ forms a group under the induced operation from $G$. The resulting group is also called a subgroup of $G$.Remark 4.4. In the sequel, we will always use the latter subgroup.In the sequel, without explicitly mentioned, most groups are multiplicative groups.Definition 4.5. A homomorphism of groups $f:G\rightarrow H$ is a function $G\rightarrow H$ such that $f(xy)=f(x)f(y)$ for all $x,y\in G$.Definition 4.6. A group $G$ is abelian or commutative if $xy=yx$ for all $x,y\in G$.Definition 4.7. Let $f:G\rightarrow H$ be a homomorphism of groups. The kernel of $f$ is $$ \textrm{Ker }f:=\{x\in G\mid f(x)=1\}. $$The image or range of $f$ is $$ \textrm{Im }f:=\{f(x)\in H\mid x\in G\}. $$Moreover, the cokernel of $f$ is $$ \textrm{Coker }f:=H/\textrm{Im }f, $$ and the coimage of $f$ is $$ \textrm{Coim }f:=G/\textrm{Ker }f. $$Proposition 4.8. Let $f:G\rightarrow H$ be a homomorphism of groups. Then ${\rm{Ker}}f=0$ if and only if $f$ is injective and ${\rm{Coker}}f=0$ if and only if $f$ is surjective.Definition 4.9. A homomorphism of groups is a monomorphism/an epimorphism/an isomorphism if it is injective/surjective/bijective. Two groups $A,B$ are isomorphic, denoted by $A\cong B$, if there is an isomorphism between them.Proposition 4.10. If $f$ is an isomorphism of groups, then its bijective inverse $f^{-1}$ is also an isomorphism of groups.Definition 4.11. Let $G$ be a group and let $N\subset G$ be a subgroup. Then $N$ is normal if $xN=Nx$ for all $x\in G$. Moreover, the sets of the form $xN$ are called left-cosets of $N$ and the sets of the form $Nx$ are called right-cosets of $N$.Definition/Proposition 4.12. Let $G$ be a group and let $N$ be a normal subgroup of $G$. The set of all cosets of $N$ forms a group under the operation $xNyN=xyN$ for all $x,y\in G$. Such a group is called the quotient group of $G$ by $N$, and is denoted by $G/N$.If $N=G$, then $G/G\cong1$. And if $N=1$, then $G/1\cong G$.Example 4.13. The additive group of integers modulo $n$ for some $n\in\mathbb{N}$, denoted by $\mathbb{Z}_{n}$, is the quotient group $\mathbb{Z}/n\mathbb{Z}$ of $\mathbb{Z}$.

Definition 5.1. A ring a triple $(R,+,\cdot)$ consisting of a set $R$, an addition $+$, and a multiplication $\cdot$, such that the following axioms are verified:(1) $(R,+)$ is an abelian additive group;(2) Associativity: $(xy)z=x(yz)$ for all $x,y,z\in R$;(3) Distributivity: $x(y+z)=xy+xz$ and $(x+y)z=xz+yz$ for all $x,y,z\in R$;We will simply denote by $R$ when there is no confusion. We say that $R$ is commutative if $xy=yx$ for all $x,y\in R$. And we say that $R$ is a ring with identity if there exists a multiplicative identity in $(R,\cdot)$, which is denoted by $1$. And we denote the additive identity in $(R,+)$ by $0$ and call it the zero element of $R$. The ring consisting of only 0 is called the null ring, which is also denoted by $0$.Example 5.2. $(\mathbb{Z},+,\cdot)$, $(\mathbb{Q},+,\cdot)$, $(\mathbb{R},+,\cdot)$, $(\mathbb{C},+,\cdot)$, $(\mathbb{Z}/n\mathbb{Z},+,\cdot)$ are all rings with identity.Definition 5.3. A homomorphism of rings $f:R\rightarrow S$ is a function $R\rightarrow S$ such that $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for all $x,y\in R$. Moreover, if $R,S$ are rings with identity, then the homomorphism $f$ also preserves identity: $f(1)=1$. A homomorphism of rings is an isomorphism if it is bijective.Definition 5.4. Let $R$ be a ring. A subring of $R$ is a subset $S\subset R$ such that $S$ forms a ring under the induced operations from $R$. An ideal of $R$ is a subset $I\subset R$ such that $(I,+)$ is an additive subgroup and $xy\in I,yx\in I$ for all $x\in R,y\in I$.Definition 5.5. An ideal $I$ of a ring $R$ is a maximal ideal if for an ideal $J\subset R$, $I\subset J$ implies $I=J$.Definition/Proposition 5.6. Let $R$ be a ring and let $I$ be an ideal of $R$. The set of all cosets of $(I,+)$ forms a ring under the operations $(x+I)+(y+I)=(x+y)+I$ and $(x+I)(y+I)=xy+I$ for all $x,y\in R$. Such a ring is called the quotient ring of $R$ by $I$, and is denoted by $R/I$.Example 5.7. The ring of integers modulo $n$ for some $n\in\mathbb{N}$, denoted by $\mathbb{Z}_{n}$, is the quotient ring $\mathbb{Z}/n\mathbb{Z}$ of $\mathbb{Z}$.Definition 5.8. Let $R$ be a ring with identity. If there exists a least positive integer $n$ such that $n1=0$, then we say that $R$ has characteristic $n$. If there for all integers $n>0$, $n1\neq0$, then we say that $R$ has characteristic 0.Definition 5.9. An integral domain or simply a domain is a commutative ring $R$ with identity such that $xy\neq0$ for all $x\neq0,y\neq0$ in $R$.Definition 5.10. An ideal $\mathfrak{p}$ of a ring $R$ is a prime ideal if $xy\in \mathfrak{p}$ implies $x\in\mathfrak{p}$ or $y\in\mathfrak{p}$ for all $x,y\in R$.Proposition 5.11. Let $R$ be a commutative ring with identity and let $I$ be an ideal of $R$. Then $R/I$ is a domain if and only if $I$ is prime. And $R/I$ is a field if and only if $I$ is a maximal ideal. Consequently, every maximal ideal of $R$ is prime.Definition 5.12. An ideal is principle if it is generated by a single element.Definition 5.13. A principle ideal domain or PID for short is a domain whose ideals are all principle ideals.Example 5.14. The ring $\mathbb{Z}$ and the polynomial ring $K[X]$ for a field $K$ are PIDs.

Definition 6.1. A field is a triple $(K,+,\cdot)$ consisting of a set $K$ such that $(K,+)$ is an additive abelian group and $(K\backslash\{0\},\cdot)$ is an abelian multiplicative group.Example 6.2. $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_{p}$ are all fields. In particular, they are fields of characteristic 0.Definition 6.3. A homomorphism of fields $f:K\rightarrow L$ is a function $K\rightarrow L$ such that $f(x+y)=f(x)+f(y)$, $f(xy)=f(x)f(y)$, and $f(1)=1$ for all $x,y\in K$. A homomorphism of fields is an isomorphism if it is bijective.Proposition 6.4. Every homomorphism of fields is injective.Proof. Since there is no non-zero proper ideal of a field.Definition 6.5. Let $K$ be a field. A field extension of $K$ is a field $E$ such that $K$ is a subfield of $E$. We denote this by $K\subset E$.Example 6.6. $\mathbb{Q}\subseteq\mathbb{R}$, $\mathbb{R}\subseteq\mathbb{C}$, $\mathbb{Q}\subseteq\mathbb{Q}_{p}$ are field extensions.Definition 6.7. Let $K\subset E$ and $K\subset L$ be two field extensions. A $K$-homomorphism $f:E\rightarrow L$ is a field homomorphism $E\rightarrow L$ such that $f(x)=x$ for all $x\in K$. A $K$-isomorphism is a $K$-homomorphism that is a field isomorphism.§6.2. Fraction fieldsA subring of a domain is clearly a domain. And we want to show that every domain is a subring of a field. This motivates us to construct the fraction field of a domain by ``adjoining inverses''. Let $R$ be a domain. We define a relation on $R\times R\backslash\{0\}$ by $$ (x,y)\sim(z,h)\textrm{ if and only if }xz=yh. $$It is easy to check that the relation is an equivalence relation. Then we write $\frac{x}{y}$ or $x/y$ for an equivalence class of $(x,y)\in R\times R\backslash\{0\}$ and called it fraction. We denote by $\textrm{Frac}(R)$ the set of all equivalence classes. Next, we define addition and multiplication on $\textrm{Frac}(R)$ by $$ \frac{x}{y}+\frac{z}{h}=\frac{xh+zy}{yh}, \ \ \ \frac{x}{y}\frac{z}{h}=\frac{xz}{yh}, \ \ x,z\in R, y,h\in R\backslash\{0\}. $$Definition/Proposition 6.8. Let $R$ be a domain. Then ${\rm{Frac}}(R)$ with the operations defined above is a field, which is called the fraction field of $R$. And there is a canonical injective homomorphism $$ R\rightarrow {\rm{Frac}}(R),\ \ x\mapsto \frac{x}{1}, $$ which shows that $R$ is a subring of the field ${\rm{Frac}}(R)$.Remark 6.9. If $R$ is not a domain, but a commutative ring with identity, then we can construct the fraction ring in a similar process.Example 6.10. If $R=\mathbb{Z}$, then $\textrm{Frac}(\mathbb{Z})=\mathbb{Q}$. Moreover, if $R=\mathscr{O}_{K}$ is the valuation ring of the field $K$, then $\textrm{Frac}(\mathscr{O}_{K})=K$. In particular, if $R=\mathbb{Z}_{p}$, then $\textrm{Frac}(\mathbb{Z}_{p})=\mathbb{Q}_{p}$.Proposition 6.11. Let $R$ be a domain. The fraction field ${\rm{Frac}}(R)$ corresponds to a field $K$ whose elements are of the form $xy^{-1}$ for $x,y\in R$, i.e. there is an isomorphism $$ {\rm{Frac}}(R)\xrightarrow{\sim} K, \ \ \frac{x}{y}\mapsto xy^{-1}. $$Proof. Let $K$ be a field containing $R$. So it must contain all elements of the form $xy^{-1}$ for $x,y\in R$. Clearly, the map ${\rm{Frac}}(R)\rightarrow K$ is a homomorphism. If every element in $K$ can be written in the form $xy^{-1}$, then we obtain the isomorphism as desired.Let $f:R\rightarrow R'$ be a homomorphism of domains. Then we have a homomorphism of fraction fields $$\textrm{Frac}(f):\textrm{Frac}(R)\rightarrow \textrm{Frac}(R'),\ \ \frac{x}{y}\mapsto\frac{f(x)}{f(y)}.$$Thus we can view Frac(-) as a functor, which assigns to each domain $R$ its fraction field Frac$(R)$ and to each homomorphism of domains $f$ a homomorphism of fraction fields Frac$(f)$, i.e. Frac(-) is a functor from the category of domains to the category of fraction fields $$\textrm{Frac}: Domains\rightarrow Fraction-Fields.$$§6.3. Algebraic extensions and algebraic closuresDefinition 6.12. Let $K\subset E$ be a field extension. An element $\alpha\in E$ is algebraic over $K$ if there exists $f\in K[X]$ such that $f(\alpha)=0$. And $\alpha$ is transcendental over $K$ if $f(\alpha)\neq0$ for all $f\in K[X]$. An algebraic extension of $K$ is a field extension $K\subset L$ whose elements are all algebraic over $K$. A transcendental extension of $K$ is a field extension $K\subset L$ if there exists $\alpha\in L$ that is transcendental over $K$.Definition 6.13. A field $K$ is algebraically closed if every non-constant polynomial over $K$ has a root in $K$.Example 6.14. $\mathbb{C}$ is an algebraically closed field.Definition 6.15. Let $K$ be a field. The algebraic closure of $K$, denoted by $\overline{K}$, is the smallest algebraically closed field that contains $K$.Remark 6.16. The algebraic closure of $K$ is indeed not unique. But since all algebraic closures are $K$-isomorphic, we shall call "the algebraic closure" of $K$ rather than "an algebraic closure".Example 6.17. $\mathbb{C}=\overline{\mathbb{R}}$ is the algebraic closure of $\mathbb{R}$