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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):

  1. 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 $\overline{\mathbb{C}}$.
  2. A region $\Omega$ is simply connected in $\overline{\mathbb{C}}$ if $\Omega^\complement$ is connected in $\overline{\mathbb{C}}$.
  3. (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:

  • 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.
José Carlos Santos
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Matija Sreckovic
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    It's funny: the definition that you describe as “very unusual” is, for me, the natural one (except that I would use general curves, not just polygons). – José Carlos Santos Aug 15 '17 at 14:30
  • Yes, it's true that it's completely natural if it were curves, what's unusual to me is the polygons, which probably indicate a sort of axiomatic Euclidean element to the idea, where polygons are defined before arbitrary curves. – Matija Sreckovic Aug 15 '17 at 14:33
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    It may be because the Jordan curve theorem for simple polygons is much simpler to prove than the Jordan curve theorem for continuous closed simple curves. – Neal Aug 15 '17 at 14:44
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    Polygonal curves are much easier to handle than general curves (no pathologies can occur, the intersection behaviour of line segments is simple). And since you can approximate every curve with a polygonal curve as closely as you wish, it suffices to consider polygonal curves. And then with some subdividing and a little of moving around the line segments, every closed polygonal curve is homotopic to a chain of simple closed polygonal curves. – Daniel Fischer Aug 15 '17 at 14:44
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    You have $d(A,B) > 0$ in the given situation because $A$ and $B$ are compact, assuming you take the complement of $\Omega$ with respect to the sphere. But one must do that, because otherwise $A\cup B \cup {\infty}$ might be connected. If you exclude that situation, however, then one of $A$ and $B$ is compact, and that suffices for a positive distance. – Daniel Fischer Aug 15 '17 at 14:48
  • Excellent, I completely forgot about compactness! I'm still getting used to "closed implies compact" because of $\overline{\mathbb{C}}$ being homeomorphic to a sphere. Do you have any kind of suggestion on tackling $2. \implies 1$, no matter how small? – Matija Sreckovic Aug 15 '17 at 16:06
  • A region $\Omega$ is simply connected if $\int_a^z f(s)ds$ is always analytic when $f$ is. Equivalently $\int_a^z \frac{ds}{s-b}$ is always analytic when $b \not \in \Omega$. – reuns Aug 16 '17 at 00:47

1 Answers1

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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.

Daniel Fischer
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  • Why is every loop in $\Gamma$ a loop in $\Omega$, which you essentially proved by choosing an arbitrary (if I understood correctly) loop in $\Gamma$, $\Gamma_{1}$ and proved it's in $\Omega$? It's easy to come up with a picture where there's one loop in $\Omega$ and one in $A$, for example. – Matija Sreckovic Aug 21 '17 at 11:08
  • Otherwise, we can take a loop $\Gamma_{1}$ and, at the start, say that it's in $\Omega$, because we can make $\eta$ arbitrarily small and add another "layer" of squares if necessary so that $\Gamma$ definitely has a loop in $\Omega$ (if, for example, $A$ is a polygonal line so that the squares covering it cover exactly $A$ and don't lie in $\Omega$ at all. Is it possible to add a picture in the comments to show you what I'm thinking of?). So then we don't need to show $\Gamma_{1} \subset \Omega$. – Matija Sreckovic Aug 21 '17 at 11:17
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    By the construction of $\Gamma$, all segments that intersect $A$ are cancelled, since they occur exactly twice, with opposite orientations, in $\sum \partial Q$, so we can't have a loop in $A$. It would not matter if some loops of $\Gamma$ lay in $\mathbb{C}\setminus \overline{\Omega}$, a single loop in $\Omega$ winding around some point of $A$ (and none of $B$) gives a separation of $\Omega^c$. But in fact all of $\Gamma$ lies in $\Omega$ since every loop in $\Gamma$ must intersect $\Omega$ and therefore lie completely in $\Omega$ since it doesn't intersect $\partial \Omega$. – Daniel Fischer Aug 21 '17 at 12:15
  • I'm sorry, when I said one loop in $A$, I meant one loop in $\mathbb{C} \setminus \overline{\Omega}$, as you correctly assumed. However, why must every loop in $\Gamma$ intersect $\Omega$? I'm looking at an example I drew up where $\Gamma$ is disconnected, and one part is entirely in the bounded part of $\Omega^{c}$ and another in $\Omega$, sort of "imprisoning" the closed curve $A$ from both sides. If I understood correctly, $\Gamma$ is all the square sides of $\mathcal{Q}$ which don't "cancel out", and in the case $2 \implies 1$, both the "inside" and "outside" parts don't "cancel out". – Matija Sreckovic Aug 21 '17 at 12:58
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    This cannot happen with the construction here because $\Omega$ is connected, and every loop belonging to $\Gamma$ winds around some points of $A$, but none of $B$. So for each loop $\Gamma_1$, some points of $\partial \Omega$ lie in the inside, and some points of $\partial \Omega$ lie in the outside. As there are points of $\Omega$ arbitrarily close to every boundary point, it follows that some points of $\Omega$ lie in the inside of $\Gamma_1$, and some points of $\Omega$ lie in the outside. Thus $\Gamma_1$ must intersect $\Omega$, for otherwise the inside and the outside of $\Gamma_1$ would – Daniel Fischer Aug 21 '17 at 13:12
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    provide a separation of $\Omega$ into two disjoint nonempty open sets, contradicting the connectedness of $\Omega$. If we allowed parts of $\Gamma$ to be farther away from $A$, so that some loops could either not wind around any boundary point, or wind around points in $A$ and points in $B$, then such loops could lie in $\mathbb{C}\setminus \overline{\Omega}$. But $\eta$ is chosen too small for that to happen. – Daniel Fischer Aug 21 '17 at 13:12
  • I see, thank you for your patience; I think I've pinpointed the last part that's unclear to me. Why must every loop in $\Gamma$ wind around some point of $A$, i.e. which part of the definition of $\Gamma$ forbids its loops to be completely inside the bounded part of $\Omega^{c}$? $\eta$ is only chosen so that the squares don't intersect $B$, it doesn't contradict the existence of an "inside" loop (i.e. the one from $\Gamma$ which encloses the smallest possible area, for example). I'm sorry for the extended questioning, I'm just having a little trouble with the visualization. – Matija Sreckovic Aug 21 '17 at 13:31
  • Consider a segment $S$ in $\Gamma$ that is not cancelled. Let $Q_S$ be the (unique, otherwise $S$ would be cancelled) square in $\mathcal{Q}$ having $S$ in its boundary. By definition of $\mathcal{Q}$, we have $Q_S\cap A \neq \varnothing$. Suppose first that there is an $a\in A \cap \operatorname{int} Q_S$. Then $n(\partial Q_S, a) = 1$, and $n(\partial Q,a) = 0$ for all $Q \in \mathcal{Q}\setminus {Q_S}$. So for every $F \subset \mathcal{Q}$, we have $n(\Gamma_F,a) = 1$ if $Q_S \in F$ and $n(\Gamma_F,a) = 0$ if $Q_S\notin F$, where $\Gamma_F = \sum_{Q\in F} \partial Q$. – Daniel Fischer Aug 21 '17 at 14:05
  • Now let $\mathcal{Q}0 = {Q_S}$. Then iteratively let $\mathcal{Q}{n+1}$ be the union of $\mathcal{Q}n$ and all $Q\in \mathcal{Q}$ such that $\partial Q$ has a common segment with some $Q'\in\mathcal{Q}_n$. There is an $n$ such that $\mathcal{Q}_n = \mathcal{Q}{n+1}$ (since $\mathcal{Q}$ is finite, that happens for $n = \operatorname{card} \mathcal{Q}$ or earlier). For such an $n$, $\Gamma_n = \sum_{Q\in \mathcal{Q}_n} \partial Q$ is the loop of $\Gamma$ containing the segment $S$, and by construction $n(\Gamma_n,a) = 1$. – Daniel Fischer Aug 21 '17 at 14:06
  • If $A\cap Q_S$ doesn't contain an interior point of $Q_S$, start the construction with $\mathcal{Q}_0 = {Q_S, Q1}$ or $\mathcal{Q}_0 = {Q_S,Q_1,Q_2,Q_3}$, depending on whether there is an $a\in A$ that lies in the "interior" of a boundary segment of $Q_S$ (then $Q_1$ is the square sharing that segment) or $A\cap Q_S$ consists only of vertices of $Q_S$ (then $Q_1,Q_2,Q_3$ are the other squares having the chosen vertex). – Daniel Fischer Aug 21 '17 at 14:06
  • Let us continue this discussion in chat. – Matija Sreckovic Aug 21 '17 at 14:44