Let $\mathcal{G} $ be the collection of nonempty, open, connected subsets of $\mathbb{C} $. For each $\mathcal{S}\subset\mathcal{G} $, we may ask whether $\mathcal{S} $ is 'closed' under the following two 'operations'.
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If $D_1, D_2\in\mathcal{S} $ and $D_1\cap D_2 $ is nonempty and connected, then $D_1\cup D_2\in\mathcal{S} $.
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If $D_1, D_2, D_3,\ldots\in\mathcal{S} $ and $D_n\subset D_{n+1} $ for all $n=1,2, 3,\ldots $, then $\bigcup_{n=1}^\infty D_n\in\mathcal{S} $.
In page 87 of Complex Analysis by Freitag and Busam, the authors remark that the following can be proven in a non-trvial way:
If $\mathcal{S}\subset\mathcal{G} $ contains all (nonempty open) disks and is closed under the above two operations, then $\mathcal{S} $ contains all simply connected domains in $\mathbb{C} $
So, how do we prove this statement? Can we generalize this to higher dimensions?
Best Answer
Let $(z_0,\ldots,z_{n - 1}) \in \mathbb{C}^n$ and $\gamma : \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{C}$ is the only affine by parts function that is affine on each $\left[\frac{k}{n},\frac{k + 1}{n}\right]$ such that for all $k$, $\gamma\!\left(\frac{k}{n}\right) = z_{k\%n}$. If $\gamma$ is injective, it is a Jordan curve and we denote by $[z_0,\ldots,z_{n - 1}]$ its interior which is a polygon with $n$ vertices which are the $z_k$. By Jordan's theorem, such polygons are biholomorphic to the open unit disk $\mathbb{D} = B(0,1)$. Let us introduce the following propositions,
$P_1$ : $\mathcal{S}$ contains all triangles and if $P = [z_0,\ldots,z_{n - 1}]$ is a polygon that belongs to $\mathcal{S}$ and $z_n \in \mathbb{C}$ is such that $Q = [z_0,\ldots,z_n]$ is still a polygon (which is the union of $P$, the triangle $[z_{n - 1},z_n,z_0]$ and the open edge $]z_{n - 1},z_0[$), then $Q \in \mathcal{S}$.
$P_2$ : $\mathcal{S}$ contains all polygons.
$P_3$ : $\mathcal{S}$ contains all open subsets $\Omega$ such that there exists a diffeomorphism $f : \Omega \rightarrow \mathbb{D}$ that extends to $f : \overline{\Omega} \rightarrow \overline{\mathbb{D}}$, which is still a diffeomorphism.
$P_4$ : $\mathcal{S}$ contains all non-empty open simply connected domains.
Let us show that for all $0 \leqslant i \leqslant 3$, $P_i \Rightarrow P_{i + 1}$, hence it is sufficient to prove $P_1$ (which I have been unable to prove but seems more reachable than the initial statement) to get the wanted result.
First of all, if $P_3$ is true, let $\Omega$ be non-empty open and simply connected. By Riemman uniformisation theorem, it is diffeomorphic to the disk $\mathbb{D}$. Let $f : \Omega \rightarrow \mathbb{D}$ a diffeomorphism (that we can choose to be a biholomorphism if $\Omega \neq \mathbb{C}$). For all $n$, $\Omega_n = B(0,1 - 2^{-n})$ and $f_n = \frac{1}{1 - 2^{-n}}f_{|\Omega_n}$, which is a diffeomorphism $\Omega_n \rightarrow \mathbb{D}$ that extends to the border and the extention is a diffeomorphism. By $P_3$, $\Omega_n \in \mathcal{S}$ and for all $n$, $\Omega_n \subset \Omega_{n + 1}$. Therefore, $\mathcal{S}$ contains $\Omega = \bigcup_{n \geqslant 0} \Omega_n$, which proves $P_4$.
Then, assume $P_1$ is true. Let us show by induction on the number $n$ of edges that $\mathcal{S}$ contains all polygons. If is true for $n = 3$ by hypothesis and $\mathcal{S}$ contains all the $n$-gons, let $Q = [z_0,\ldots,z_n]$ be an $n + 1$-gon. Up to permuting the vertices, we can assume that the open segment $]z_{n - 1},z_0[$ is included in $Q$ (it is true for at least one vertex because each polygon admits a triangulation). Let $P = [z_0,\ldots,z_{n - 1}]$ which belongs to $\mathcal{S}$ by induction. By $P_1$, $Q$ also belongs to $\mathcal{S}$, which proves $P_2$.
Now, let us do the hard part : $P_2 \Rightarrow P_3$. Assume $P_2$ and let $\Omega$ be a domain of $\mathbb{C}$ and $f : \overline{\Omega} \rightarrow \overline{\mathbb{D}}$ given by the hypothesis of $P_3$. Let for all $m \geqslant 0$, $\alpha : t \mapsto f^{-1}((1 - 2^{-m})e^{2i\pi t})$, which is a loop in $\Omega$ close to the edge. Let $n \geqslant 0$ and for all $k \in \mathbb{Z}$, $z_k = f^{-1}((1 - 2^{-m})e^{2i\pi k/n}) \in \mathrm{Im}(\alpha)$.
$\mathrm{Im}(\alpha)$ is compact and included in the open set $\Omega$ so there exists an $\varepsilon > 0$ such that for all $x \in \mathrm{Im}(\alpha)$, $|x - y| \leqslant \varepsilon \Rightarrow y \in \Omega$. And by Heine theorem, $f$ and $f^{-1}$ are uniformly continuous. Therefore, there exists $\delta_1 > 0$ such that for all $(x,y) \in \overline{\Omega}^2$, $|x - y| \leqslant \delta_1 \Rightarrow |f(x) - f(y)| \leqslant 2^{-m - 2}$ and there exists $\delta_2 > 0$ such that for all $(x,y) \in \overline{\mathbb{D}}^2$, $|x - y| \leqslant \delta_2 \Rightarrow |f^{-1}(x) - f^{-1}(y)| \leqslant \min\{\delta_1,\varepsilon\}$.
Choose $n$ such that $|e^{2i\pi/n} - 1| \leqslant \delta_2$. It implies that for all $k$, $|f(z_k) - f(z_{k + 1})| = |e^{2i\pi/n} - 1| \leqslant \delta_2$ so $|z_k - z_{k + 1}| \leqslant \min\{\delta_1,\varepsilon\}$. Therefore, for any point $z$ on the edge $[z_k,z_{k + 1}]$, $|z_k - p| \leqslant |z_k - z_{k + 1}| \leqslant \min\{\delta_1,\varepsilon\}$ thus $p \in \Omega$ and $|f(z_k) - f(p)| \leqslant 2^{-m - 2}$. In particular, $|f(p)| \leqslant |f(z_k)| + 2^{-m - 2} = 1 - 2^{-m} + 2^{-m - 2}$ and $|f(p)| \geqslant |f(z_k)| - 2^{-m - 2} = 1 - 2^{-m} - 2^{-m - 2}$.
As $\gamma$ is smooth ($f$ is a diffeomorphism), up to increasing $n$, we may assume that $[z_0,\ldots,z_{n - 1}]$ defines a polygon $P_m$, that belongs in $\mathcal{S}$ by proposition $P_2$. If $p$ is a point in the edge of $P_m$ and $P_{m + 1}$, we have $|f(p)| \leqslant 1 - 2^{-m} - 2^{-m - 2}$ because $p$ is in the edge of $P_m$ and $|f(p)| \geqslant 1 - 2^{-(m + 1)} + 2^{-(m + 1) - 2} = 1 - 2^{-m - 1} + 2^{-m + 1}$. It implies $1 - 2^{-m - 1} + 2^{-m + 1} \leqslant 1 - 2^{-m} - 2^{-m - 2}$ i.e. $(8 - 2)2^{-m - 2} \leqslant (4 + 1)2^{-m - 2}$, which is a contradiction. It shows that for all $m$, $P_m \subset P_{m + 1}$ hence $\mathcal{S}$ contains $\Omega = \bigcup_{m \geqslant 0} P_m$, which proves $P_3$.
Now the difficulty is to prove $P_1$, for this, I guess we must add progressively disks to the polygon $P$ in order to fill the triangle $[z_{n - 1},z_n,z_0]$ progressively and use the stability of $\mathcal{S}$ under increasing countable union to add the whole triangle, but I don't see how to show it formally.
Edit : I found a way to prove $P_1$ even if some parts are not $100\%$ rigourous. Let us introduce,
$P_0$ : $\mathcal{S}$ contains all triangles.
We have $P_0 \Rightarrow P_1$. Indeed, if $P$ and $Q$ are polygons like in the statement of $P_1$, let $x_0$ be a point of $]z_{n - 1},z_0[$ and $r_0$ be the largest $r$ such that $B(x_0,r) \subset Q$. Then, let $x_1$ be the only point of $]x_0,z_0[ \cap \partial B(x_0,r_0)$ and $r_1$ the largest $r$ such that $B(x_1,r_1) \subset Q$.
Like this, we build a sequence of elements on $]x_0,z_0[$ that converge toward some $x_\infty \in ]x_0,z_0]$. If $x_\infty \neq z_0$, then it belongs to $Q$ thus it exists some $\varepsilon > 0$ such that $B(x_\infty,\varepsilon) \subset Q$. For $n$ large enough, $x_n \in B\!\left(x_\infty,\frac{\varepsilon}{2}\right)$ so $B\!\left(x_n,\frac{\varepsilon}{2}\right) \subset B(x_\infty,\varepsilon) \subset Q$ hence $r_n \geqslant \frac{\varepsilon}{2}$ which implies that $x_{n + 1}$ is strictly closer from $z_0$ than $x_\infty$, which is a contradiction. Therefore, $x_\infty = z_0$.
Let for all $n$, $D_n = B(x_n,r_n)$ and $U_n = \bigcup_{k = 0}^n D_k$. For all $n$, $U_n \cap D_{n + 1}$ is non-empty and connected so by induction, each $U_n$ belongs de $\mathcal{S}$. Thus $U = \bigcup_{n \geqslant 0} U_n = \bigcup_{n \geqslant 0} D_n \in \mathcal{S}$. Similarly, we can build a countable union of open disks on the other side of $x_0$ (toward $z_{n - 1}$ this time) and build an open set $V \in \mathcal{S}$. $U \cap V = B(x_0,r_0)$ is non-empty and connected thus $W = U \cap V \in \mathcal{S}$.
The fact that $x_n \rightarrow z_0$ (and the same argument in the other side) imply that $W$ contains the open edge $]z_{n - 1},z_0[$. Let $T = [z_{n - 1},z_n,z_0]$ that belongs to $\mathcal{S}$ by hypothesis. $W \cap T$, which is the "half" of $W$ is non-empty and connected. Same thing for $P \cap W$, which is the other "half" of $W$. Therefore, $P \cup W$ and $Q = P \cup W \cup T$ are in $\mathcal{S}$. It proves $P_1$.
Now, to prove $P_0$, take a triangle $T_{0,0} = [z_0,z_1,z_2]$ and consider $D_{0,0}$ the disk inside its incircle. $\partial D_{0,0} \cap \partial T_{0,0}$ are three points, one on each edge of the triangle. Let us call them $z_{0,1} \in ]z_0,z_1[$, $z_{1,2} \in ]z_1,z_2[$ and $z_{2,0} \in ]z_2,z_0[$. Let $T_{1,0} = [z_0,z_{0,1},z_{2,0}]$, $T_{1,1} = [z_1,z_{1,2},z_{0,1}]$ and $T_{1,2} = [z_2,z_{2,0},z_{1,2}]$.
Let for all $0 \leqslant i \leqslant 2$, $D_{1,i}$ be the disk inside the incircle of $T_{1,i}$. Clearly, the $D_{0,0} \cap D_{1,i}$ are non-empty, connected and pairwise disjoint thus $U_1 = D_{0,0} \cup D_{1,0} \cup D_{1,1} \cup D_{1,2}$ is in $\mathcal{S}$.
Repeating this operation, we build $U_n \in \mathcal{S}$ that is the union of $\sum_{k = 0}^n 3^k = \frac{3^{n + 1} - 1}{2}$ open disks (it is probably pretty long and hard to check it rigourously by I am pretty sure it works) and $T = \bigcup_{n \geqslant 0} U_n$ (once more, it is absolutely not trivial but intuitively true), thus $T \in \mathcal{S}$, which proves $P_0$.