Equivalence of definitions of simple connectedness of a region $\Omega$ in $\overline{\mathbb{C}}$ I've started reading up on complex analysis recently, and many books give their own definition of simple connectedness, all of which I assume to be equivalent, so I'm trying to prove their equivalence. Here are the definitions I've encountered (I'm excluding all the definitions which I've proven to be equivalent to one of the following three):


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*A region (i.e. open and connected subset of $\overline{\mathbb{C}}$) $\Omega$ is simply connected in $\overline{\mathbb{C}}$ if $\partial\Omega$ is connected in $\overline{\mathbb{C}}$.

*A region $\Omega$ is simply connected in $\overline{\mathbb{C}}$ if $\Omega^\complement$ is connected in $\overline{\mathbb{C}}$.

*(A very unusual definition in my opinion, but interesting nonetheless) A region $\Omega$ in $\overline{\mathbb{C}}$ is simply connected if for any closed, simple (simple meaning no self-intersections) polygonal line $\Lambda \subset \Omega$, $\Lambda^{\circ}\subset \Omega$ holds, where ${}^{\circ}$ denotes the interior of a set.


I've managed to figure out $1. \implies 3$, $1. \implies 2.$ and partially $3. \implies 2$, where I've managed to figure out that if I can construct a polygon $\Lambda \subset \Omega$ around a compact connected component $C$ of $\Omega^{\complement}$ (which exists because either $\Omega$ or one component of $\Omega^{\complement}$ will contain $\infty$, and any closed set in $\overline{\mathbb{C}}$ is compact if we look at $\overline{\mathbb{C}}$ as a sphere; I'm trying $\neg 2. \implies \neg 3$, so there's more than one connected component), I'll have a point in $C \subset \Lambda^{\circ}$ outside of $\Omega$, which is exactly what was needed.
So my problems here are: 


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*The proof $2. \implies 1$, which I don't know how to approach because the only thing that occurs to me is to try $\neg 1. \implies \neg 2,$ which gets me nowhere because if I assume that the boundary is disconnected, i.e. $\partial \Omega = A \cup B$ where $A$, $B$ are disjoint, non-empty and open, I have no idea how to construct a set whose boundary is $A$ or $B$, so I can't turn these "lines" into "areas" (I don't mean this in a literal sense; I'm just saying I don't know how to use the disconnectedness of the boundary to prove the disconnectedness of the complement).

*The construction of a polygon $\Lambda$. Now, I've been reading something in Ahlfors, where in the proof of Theorem 14, pages 139-140, the author constructs such a polygon (for a different reason altogether, though) under the assumption that if $\Omega^{\complement} = A \cup B$ where $A$, $B$ are disjoint, non-empty and clopen in $\Omega^{\complement}$, then $d(A, B)>0$, which I'm not entirely clear on. There are examples of disjoint, closed sets with distance $0$, such as $\mathbb{N}$ and $\{n+\frac{1}{n} | n \in \mathbb{N}\}$, so just saying that they're closed and disjoint isn't enough. I'm looking for a clarification of $d(A, B) > 0$, which seems intuitive, but I'm mentally blocked on how to prove it.

 A: We can show $1 \implies 2$ and $2 \implies 1$ with essentially the same construction, both by proving the contrapositive.
For the case $\lnot 2 \implies \lnot 1$ we first note that if $\Omega^c = A \cup B$ with $A,B$ being disjoint nonempty closed sets, then each of the two intersects $\partial \Omega$. For let $p \in \partial A$. Since $B$ is closed and $p \notin B$, there is an $\varepsilon > 0$ such that $D_{\varepsilon}(p) \cap B = \varnothing$, hence $D_{\varepsilon}(p) \subset A \cup \Omega$. Since $p \in \partial A$, we have $D_{\rho}(p) \not\subset A$ for every $\rho > 0$, and that implies $D_{\rho}(p) \cap \Omega \supset D_{\min \{ \rho, \varepsilon\}}(p) \cap \Omega \neq \varnothing$. That means $p \in \overline{\Omega}$, and since $p \notin \Omega$ we finally conclude $p \in \partial \Omega$.
Now we make our construction. Suppose that we have a decomposition of $\partial \Omega$ or $\Omega^c$ respectively into two nonempty disjoint closed sets $A$ and $B$. Choose the labelling so that $\infty \notin A$, i.e. $A$ is a compact subset of the plane. Then
$$\delta := \operatorname{dist}(A,B) = \min \: \{ \operatorname{dist}(z,B) : z \in A\} > 0.$$
Choose a $k \in \mathbb{N}$ such that $\eta := 2^{-k} < \delta/2$. Consider the lattice $\Lambda = \eta\cdot \mathbb{Z}[i]$. Each closed square determined by the lattice can meet at most one of $A$ and $B$, since the diameter of such a square is $\sqrt{2}\,\eta < \delta$. Let $\mathcal{Q}$ be the (finite) set of closed squares of $\Lambda$ intersecting $A$. For each $Q \in \mathcal{Q}$, let $\partial Q$ the positively oriented boundary of $Q$, consisting of four axis-parallel line segments. If such a line segment $S$ intersects $A$, then the square $Q' \neq Q$ having $S$ in its boundary also belongs to $\mathcal{Q}$, and the orientation of $S$ in $\partial Q'$ is opposite to its orientation in $\partial Q$. Thus,
$$\Gamma = \sum_{Q \in \mathcal{Q}} \partial Q$$
is a polygonal cycle consisting entirely of line segments in $\mathbb{C} \setminus (A \cup B)$, and $n(\Gamma, p) = 1$ for every $p\in A$, while $n(\Gamma,q) = 0$ for every $q\in B$ (see below). Let
$$U = \{ z \in \mathbb{C}\setminus \Gamma : n(\Gamma,z) = 1\}\qquad \text{and}\qquad V = \{ z \in \mathbb{C}\setminus \Gamma : n(\Gamma,z) = 0\}.$$
Then $U$ and $V$ are disjoint open (in $\mathbb{C}$) sets with $\mathbb{C} \setminus \Gamma = U \cup V$, and we have $A \subset U$ and $B \subset V$.
Now $\lnot 2 \implies \lnot 1$ follows by noting that $U \cap \partial \Omega = A \cap \partial \Omega \neq \varnothing$ and $V \cap \partial \Omega = B \cap \partial \Omega \neq \varnothing$ and
$$\partial \Omega = (A \cap \partial \Omega) \cup (B\cap \partial \Omega),$$
since in this case $\Gamma$ is a cycle in $\Omega$.
For $\lnot 1 \implies \lnot 2$, we don't know a priori that $\Gamma$ is a cycle in $\Omega$. But since $\Gamma$ doesn't intersect $\partial \Omega = A \cup B$, each loop in $\Gamma$ is either a loop in $\Omega$, or it is a loop in $W = \operatorname{int}(\Omega^c)$, for any path intersecting both $\Omega$ and $W$ must intersect $\partial \Omega$. Since $n(\Gamma,a) = 1$ for all $a\in A$, there is at least one loop $\Gamma_1$ in $\Gamma$ with $n(\Gamma_1,a_1) = 1$ for some $a_1\in A$. Define $U_1, V_1$ analogous to $U,V$ above using $\Gamma_1$ instead of $\Gamma$, and let $A_1 = \partial \Omega \cap U_1$, $B_1 = \partial \Omega \cap V_1$. By the choice of $\Gamma_1$, we have $\varnothing \neq A_1 = A \cap U_1$. Also, $\varnothing \neq B \subset B_1$. Hence $\Omega \cap U_1 \neq \varnothing$ and $\Omega \cap V_1 \neq \varnothing$. Since $\Omega$ is connected, it follows that $\Gamma_1$ intersects $\Omega$ (otherwise $U_1,V_1$ would be a decomposition of $\Omega$ into two disjoint nonempty open sets), and therefore $\Gamma_1$ is a loop in $\Omega$. But then $\Omega^c \cap U_1$ and $\Omega^c \cap V_1$ give a decomposition of $\Omega^c$ into two relatively open nonempty disjoint sets, i.e. $\Omega^c$ is not connected.

For $Q \in \mathcal{Q}$, we have $n(\partial Q,z) = 1$ if $z$ lies in the interior of $Q$, and $n(\partial Q, z) = 0$ if $z \notin Q$. So
$$n(\Gamma,z) = \sum_{Q \in \mathcal{Q}} n(\partial Q, z)$$
is $1$ for $z \in \bigcup_{Q\in \mathcal{Q}} \operatorname{int} Q$, and it is $0$ for $z \in \mathbb{C}\setminus \bigcup_{Q \in \mathcal{Q}} Q\supset B$. Since $A \subset \bigcup \mathcal{Q}$, this shows that $n(\Gamma,a) = 1$ for all $a\in A$ which don't lie on the boundary of some $Q\in \mathcal{Q}$. If $a\in A$ is a vertex of some  $Q$, then all four squares having $a$ as a vertex belong to $\mathcal{Q}$. Calling these squares $Q_1,Q_2,Q_3,Q_4$, we have $n(\partial Q_1 + \partial Q_2 + \partial Q_3 + \partial Q_4,a) = 1$, and $n(\partial Q,a) = 0$ for all other $Q\in \mathcal{Q}$, so overall $n(\Gamma,a) = 1$. If $a\in A$ lies on a boundary segment $S$ of $Q_1 \in \mathcal{Q}$ but is not a vertex, then the square $Q_2\neq Q_1$ also having $S$ as a boundary segment also belongs to $\mathcal{Q}$, and $n(\partial Q_1 + \partial Q_2,a) = 1$ while $n(\partial Q,a) = 0$ for all other squares, hence $n(\Gamma,a) = 1$ also in this case.
