# Openness theorem/theorem of territorial loyalty
## Statement

Let $U\subseteq \mathbb {C}$ be a domain, and let $f\colon U\to \mathbb {C}$ be a holomorphic, non-constant function. Then, $f(U)$ is a domain.

## Proof

According to the theorem of domain preservation, one must show that $f(U)$ is a domain, i.e., the set $f(U)$
* is connected, and
* is open.
The proof is divided into these two parts.

### Proof 1: Connectedness

We show that if $f$ is continuous and $U$ is connected, then $f(U)$ is also connected.

### Proof 2: Connectedness

Let $w_{1},w_{2}\in f(U)$ be arbitrarily chosen. Then, there exist $z_{1},z_{2}\in U$ such that $f(z_{1})=w_{1}$ and $f(z_{2})=w_{2}$. Since $U$ is connected, there exists a path $\gamma :[a,b]\to U$ such that $\gamma (a)=z_{1}$ and $\gamma (b)=z_{2}$.

### Proof 3: Connectedness

Because $f$ is continuous and $\gamma :[a,b]\to U$ is a continuous path in $U$, the composition $\gamma _{f}:=f\circ \gamma$ is a continuous path in $f(U)$, for which:
    $\gamma _{f}(a)=f(\gamma (a))=f(z_{1})=w_{1}$ and $\gamma _{f}(b)=f(\gamma (b))=f(z_{2})=w_{2}$.

### Proof 4: Openness

It remains to show that $f(U)$ is open. Let $w_{0}\in f(U)$ and $z_{0}\in U$ such that $f(z_{0})=w_{0}$. Now, consider the set of $w_{0}$-preimages:
    $S(f,w_{0}):={z\in U\ |\ f(z)=w_{0}}$

### Proof 5: Openness - Identity Theorem

According to the Identity Theorem, the set $S(f,w_{0}):={z\in U\ |\ f(z)=w_{0}}$ cannot have accumulation points in $f(U)$. If $S(f,w_{0})\subseteq f(U)$ had accumulation points in $f(U)$, the holomorphic function $f\colon U\to \mathbb {C}$ would be constant with $f(z)=w_{0}$ for all $z\in U$.

### Proof 6: Openness - Neighborhoods

If the set $S(f,w_{0})$ of $w_{0}$-preimages of $f$ has no accumulation points, one can choose a neighborhood $V\subseteq U$ of $z_{0}$ where $z_{0}$ is the only $w_{0}$-preimage. Let $r>0$ be such that ${\bar {D}}_{r}(z_{0})\subseteq V$.

### Proof 7: Openness

We then define the smallest lower bound for the distance of $f(z)$ to $w_{0}$, where $z$ lies on the boundary of the disk $D_{r}(z_{0})$:
    $\varepsilon :=\inf _{z\in \partial D_{r}(z_{0})}|f(z)-w_{0}|>0$
Here, $\varepsilon >0$, because $f$ is continuous and attains a minimum on the compact set $\partial D_{r}(z_{0})$. Since ${\bar {D}}_{r}(z_{0})\subseteq V$, no $w_{0}$-preimages can lie on the boundary.

### Proof 8: Openness - Maximum Principle

We show that $D_{\frac {\varepsilon }{3}}(w_{0})\subseteq f(U)$. Let $|w-w_{0}|<{\frac {\varepsilon }{3}}$. We prove by contradiction that this arbitrary $w\in D_{\frac {\varepsilon }{3}}(w_{0})$ is in the image of $f$.

### Proof 9: Openness - Maximum Principle

Assume $f(z)\neq w$ for all $z\in {\bar {D}}_{r}(z_{0})$. Then, $|g(z)|$ with $g(z):=f(z)-w$ attains a nonzero minimum on ${\overline {D_{r}(z_{0})}}$. Since $f$ is not constant, this minimum must lie on $\partial D_{r}(z_{0})$ (otherwise $h(z):={\frac {1}{f(z)-w}}$ would be constant by the Maximum Principle. If $h$ were constant, $f$ would also have to be constant—a contradiction to the assumption).

### Proof 9: Openness

Since $w_{0}\in f(U)$ was chosen arbitrarily, and for every $w_{0}\in f(U)$, there exists a ${\frac {\epsilon }{3}}$-neighborhood $D_{\frac {\epsilon }{3}}(w_{0})\subseteq f(U)$, we obtain $f(U)=\bigcup _{w_{0}\in f(U)}D_{\frac {\epsilon }{3}}(w_{0})$ as an Norms, metrics, topology, and thus $f(U)$ is open.