# Complex Analysis/Identity theorem, Liouville-type theorems, Riemann's theorem

After now having established the main tools of complex analysis, we may deduce the first corollaries from them, which are theorems about general holomorphic functions; for, it is the case that holomorphic functions follow a wide range of rules just by their holomorphy, and this and the following chapter cover some of them.

## Connectedness and the identity theorem

For the famous identity theorem, we first need to frame the intuitively plausible notion of a connected set in mathematically precise terms.
Definition 7.1:
A topological space $X$ is called connected if and only if the only nonempty set which is simultaneously open and closed is $X$.
It seems somewhat implausible that this is actually the "right" definition of connectedness. The following theorem shows that in fact, if the topological space $X$ equals $\mathbb {R} ^{n}$ (and thus in particular if $X=\mathbb {R} ^{2}=\mathbb {C}$) and the subset is open, then there is a very intuitive characterisation of connectedness.
Theorem 7.2:
Let $X=\mathbb {R} ^{n}$ with the standard Euclidean topology. Then any nonempty open subset $O\subseteq X$ is connected (w.r.t. to the subspace topology induced by the standard Euclidean topology) if and only if for all $x,y\in O$ there exists a continuous function $\gamma :[0,1]\to O$ such that $\gamma (0)=x$ and $\gamma (1)=y$ (such sets are also called path-connected, and the continuous map $\gamma$ is called a path).
Proof:
Assume first that $O$ is path-connected and not connected. We first prove the following claim, which is actually an important alternative characterisation of connectedness:
Claim: A topological space $X$ is connected if and only if there do not exist open $U,V\subset X$ such that $U\neq \emptyset$, $V\neq \emptyset$, $U\cap V=\emptyset$ and $U\cup V=X$ (that is, we may not split up $X$ into two nonempty open disjoint subsets).
Proof of claim: Suppose first that $X$ is connected, and assume for a contradiction that we can write it as $X=V\cup U$ with $U,V$ nonempty, open, disjoint. Then $V=X\setminus U$ is open, closed, nonempty and not the whole space, contradiction. Let now $X$ not be connected. We pick a nontrivial open and closed set $S\subseteq X$, and $X=S\cup (X\setminus S)$ contradicts the condition of the claim.$\Box$
Further, we have the following:
Claim: If $X$ is connected and $f:X\to Y$ is a continuous function between topological spaces, then $f(X)$ is connected.
Proof of claim: Assume otherwise. Then there exists $S\subset f(X)$ open, closed, nontrivial. By the continuity of $f$, $f^{-1}(S)$ is open, closed, nontrivial, contradiction.$\Box$
Claim: The set $[0,1]$ is connected.
Proof:
Let now $O$ be connected, and let $x$ be an arbitrary point within $O$ ($O$ was assumed to be nonempty). We define
    $S:=\{y\in O|{\text{there exists a path from }}x{\text{ to }}y\}$.
This set is certainly open, because if there exists a path from $x$ to $y$ and $B_{\epsilon }(y)$ is any ball around $y$ contained in $O$, then in fact $B_{\epsilon }(y)\subseteq S$ since we may connect $z\in B_{\epsilon }(y)$ to $y$ using a straight line and $y$ to $x$ using the path we have. But it is also closed, since the complement is open for the same reason: Pick $w\in O\setminus S$ and $\eta >0$ such that $B_{\eta }(w)\subseteq O$; if any point in $B_{\eta }(w)$ is in $S$, then, since $B_{\eta }(w)$ is convex, the previous argument applies and $w\in S$, contradiction. Since $O$ is connected and $S$ nonempty ($x\in S$), $S=O$ and thus, by connecting any two points $a,b$ by first connecting $a$ and $x$ and then $x$ and $b$, $O$ is path-connected.
Now we can state and prove the identity theorem.
Theorem 7.3:
An accumulation point of a sequence $(z_{n})_{n\in \mathbb {N} }$ of complex numbers is a point $c\in \mathbb {C}$ such that for all $\epsilon >0$, $B_{\epsilon }(c)$ contains infinitely many distinct points
Let $O\subseteq \mathbb {C}$ be a connected set, and let $f,g:O\to \mathbb {C}$ be holomorphic. If there exists a sequence $(z_{n})_{n\in \mathbb {N} }$ that has an accumulation point such that $\forall n\in \mathbb {N} :f(z_{n})=g(z_{n})$, then $f\equiv g$.
Proof:

## Riemann's theorem

Theorem 7.6:
Let $f:O\setminus \{z_{0}\}\to \mathbb {C}$ be a holomorphic function, where $z_{0}$ is an interior point of $O$ and assume that $f$ is bounded in a neighbourhood of $z_{0}$. Then there exists a holomorphic function ${\tilde {f}}:O\to \mathbb {C}$ which extends $f$, that is,
    ${\tilde {f}}\upharpoonright _{O\setminus \{z_{0}\}}=f$.
Thus, ${\tilde {f}}$ is an extension of $f$ to $z_{0}$.
Proof:
Consider the function
    $g(z):=(z-z_{0})^{2}f(z)$.
We claim that $g$ is holomorphic in $O$. Indeed, for $z\neq z_{0}$ it is differentiable by the product rule, and for $z_{0}$ we have
    $\lim _{h\to 0}{\frac {(z_{0}+h-z_{0})^{2}f(z_{0}+h)-(z_{0}-z_{0})^{2}f(z)}{h}}=\lim _{h\to 0}hf(z_{0}+h)=0$, i.e. $g'(z_{0})=0$.
Hence, we may write $g$ as a Taylor series
    $g(z)=\sum _{n=0}^{\infty }c_{n}(z-z_{0})^{n}$
where $c_{0},c_{1}=0$ due to $g(z_{0})=0$ and $g'(z_{0})=0$. Thus,
    $g(z)=(z-z_{0})^{2}\sum _{n=2}^{\infty }c_{n}(z-z_{0})^{n-2}$,
and hence for $z\neq z_{0}$, $z$ in the domain of convergence of the power series, we have
    $f(z)=\sum _{n=2}^{\infty }c_{n}(z-z_{0})^{n-2}$,
where the latter expression makes sense also at $z_{0}$ and is complex differentiable there. This means that
    ${\tilde {f}}(z)={\begin{cases}f(z)&z\neq z_{0}\\c_{2}&z=z_{0}\end{cases}}$
is the desired continuation of $f$ to $z_{0}$.$\Box$

## Exercises

* Prove that if $f$ is an entire function such that $\forall z\in \mathbb {C} :|f(z)|<{\frac {1}{|\operatorname {Im} (z)|}}$, then $f$ is constant.
* Prove that every uncountable subset of $\mathbb {R} ^{2}$ as an accumulation point. Conclude that if for uncountably many $z\in \mathbb {C}$ we have $f(z)=g(z)$, then $f\equiv g$.