View source for Openness theorem/theorem of territorial loyalty

==Statement==
Let <math>U \subseteq\mathbb C</math> be a domain, and let <math>f \colon U \to \mathbb C</math> be a [[w:en:Holomorphicfunction|holomorphic]], non-constant function. Then, <math>f(U)</math> is a domain.

==Proof==
According to the theorem of domain preservation, one must show that <math>f(U)</math> is a domain, i.e., the set <math>f(U)</math>
*is connected, and
*is open.

The proof is divided into these two parts.
=== Proof 1: Connectedness ===
We show that if <math>f</math> is continuous and <math>U</math> is connected, then <math>f(U)</math> is also connected.

=== Proof 2: Connectedness ===
Let <math>w_1, w_2 \in f(U)</math> be arbitrarily chosen. Then, there exist <math>z_1,z_2 \in U</math> such that <math>f(z_1)=w_1</math> and <math>f(z_2)=w_2</math>. Since <math>U</math> is connected, there exists a path <math>\gamma:[a,b]\to U</math> such that <math>\gamma(a)= z_1</math> and <math>\gamma(b)= z_2</math>.

=== Proof 3: Connectedness ===
Because <math>f</math> is continuous and <math>\gamma:[a,b]\to U</math> is a continuous path in <math>U</math>, the composition <math>\gamma_f := f \circ \gamma</math> is a continuous path in <math>f(U)</math>, for which:
: <math>\gamma_f(a) = f(\gamma (a)) = f(z_1)=w_1</math> and <math>\gamma_f(b) = f(\gamma (b)) = f(z_2)=w_2</math>.

=== Proof 4: Openness ===
It remains to show that <math>f(U)</math> is open. Let <math>w_0 \in f(U)</math> and <math>z_0 \in U</math> such that <math>f(z_0) = w_0</math>.
Now, consider the set of <math>w_0</math>-preimages:
: <math> S(f,w_0) := { z \in U \ | \ f(z) = w_0 } </math>

=== Proof 5: Openness - Identity Theorem ===
According to the [[Complex Analysis/Identity Theorem|Identity Theorem]], the set <math>S(f,w_0) := { z \in U \ | \ f(z) = w_0 } </math> cannot have accumulation points in <math>f(U)</math>. If <math>S(f,w_0) \subseteq f(U)</math> had accumulation points in <math>f(U)</math>, the holomorphic function <math>f \colon U \to \mathbb C</math> would be constant with <math>f(z) = w_0</math> for all <math>z \in U</math>.

=== Proof 6: Openness - Neighborhoods ===
If the set <math>S(f,w_0)</math> of <math>w_0</math>-preimages of <math>f</math> has no accumulation points, one can choose a neighborhood <math>V \subseteq U</math> of <math>z_0</math> where <math>z_0</math> is the only <math>w_0</math>-preimage. Let <math>r > 0</math> be such that <math>\bar D_r(z_0) \subseteq V</math>.

=== Proof 7: Openness ===
We then define the smallest lower bound for the distance of <math>f(z)</math> to <math>w_0</math>, where <math>z</math> lies on the boundary of the disk <math>D_r(z_0)</math>:
: <math> \varepsilon := \inf_{z \in \partial D_r(z_0)} |f(z) - w_0| > 0 </math>
Here, <math>\varepsilon > 0</math>, because <math>f</math> is continuous and attains a minimum on the compact set <math>\partial D_r(z_0)</math>. Since <math>\bar D_r(z_0) \subseteq V</math>, no <math>w_0</math>-preimages can lie on the boundary.

=== Proof 8: Openness - Maximum Principle ===
We show that <math>D_{\frac{\varepsilon}{3}}(w_0) \subseteq f(U)</math>. Let <math>|w - w_0| < \frac{\varepsilon}{3}</math>. We prove by contradiction that this arbitrary <math>w \in D_{\frac{\varepsilon}{3}}(w_0)</math> is in the image of <math>f</math>.

=== Proof 9: Openness - Maximum Principle ===
Assume <math>f(z) \neq w</math> for all <math>z \in \bar D_r(z_0)</math>. Then, <math>|g(z)|</math> with <math>g(z):=f(z) - w</math> attains a nonzero minimum on <math>\overline{D_r(z_0)}</math>.
Since <math>f</math> is not constant, this minimum must lie on <math>\partial D_r(z_0)</math> (otherwise <math>h(z):=\frac{1}{f(z)-w}</math> would be constant by the [[Maximum Principle|Maximum Principle]]. If <math>h</math> were constant, <math>f</math> would also have to be constant—a contradiction to the assumption).

=== Proof 9: Openness ===
Since <math>w_0 \in f(U)</math> was chosen arbitrarily, and for every <math>w_0 \in f(U)</math>, there exists a <math>\frac{\epsilon}{3}</math>-neighborhood <math>D_{\frac{\epsilon}{3}}(w_0) \subseteq f(U)</math>, we obtain <math>f(U) = \bigcup_{w_0 \in f(U)} D_{\frac{\epsilon}{3}}(w_0)</math> as an [[Norms, metrics, topology|Norms, metrics, topology]], and thus <math>f(U)</math> is open.

==See also==
*[[Norms, metrics, topology|Norms, metrics, topology]]
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