Openness theorem/theorem of territorial loyalty

Statement

Let $U\subseteq ℂ$ be a domain, and let $f:U\to ℂ$ 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 $f:U\to ℂ$ 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 ${\overline{D}}_r(z_0)\subseteq V$.

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 $D_r(z_0)$:

$\epsilon :=\underset{z\in \partial D_r(z_0)}{\inf }|f(z)-w_0|>0$

Here, $\epsilon >0$, because $f$ is continuous and attains a minimum on the compact set $\partial D_r(z_0)$. Since ${\overline{D}}_r(z_0)\subseteq V$, no $w_0$-preimages can lie on the boundary.

Proof 8: Openness - Maximum Principle

We show that $D_{\frac{\epsilon }{3}}(w_0)\subseteq f(U)$. Let $|w-w_0|<\frac{\epsilon }{3}$. We prove by contradiction that this arbitrary $w\in D_{\frac{\epsilon }{3}}(w_0)$ is in the image of $f$.

Proof 9: Openness - Maximum Principle

Assume $f(z)\ne w$ for all $z\in {\overline{D}}_r(z_0)$. 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 $\frac{ϵ}{3}$-neighborhood $D_{\frac{ϵ}{3}}(w_0)\subseteq f(U)$, we obtain $f(U)=\bigcup _{w_0\in f(U)}{D}_{\frac{ϵ}{3}}(w_0)$ as an Norms, metrics, topology, and thus $f(U)$ is open.

See also

• Norms, metrics, topology

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