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In complex analysis, there is a classical result concerning simple connectivity.

A set $E \subset \mathbb{C}$ is open, bounded and connected. Then $E$ is simply connected iff $E^c$ is connected.

I know a proof by means of complex analysis. Its sketch is as follows:

"$\Leftarrow$": By the general Cauchy Theorem, for a closed curve $\gamma$ in $E$ and a function $f \in H(E)$, \begin{equation} \int_\gamma f \operatorname{d\!} z=0. \end{equation} Hence $E$ is holomorphically simply connected. Due to the Riemann Mapping Theorem, $E$ is homeomorphic to $\mathbb{D}$, providing that $E$ is simply connected.

"$\Rightarrow$": If $E^c$ is not connected, we can construct a closed curve $\gamma$ in $E$ such that $\operatorname{Ind}_\gamma(a)\ne 0$ for some $a \in E^c$, which contradicts the fact that $E$ is simply connected.

My question is whether there is a topological proof of it. Thank you very much!

Cameron L. Williams
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Andrew
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  • What do you mean by a "topological proof" ? this isn't true in most metric (to make sense of bounded) spaces, so a proof has to use something specific to $\mathbb{C}$ (for instance it is blatantly false in $\mathbb{R}^3$ and higher dimensions, so not only does it have to use something specific to the reals, but it has to use dimension $2$) – Maxime Ramzi Mar 03 '19 at 14:00
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    It is a result just involving the structure of $\mathbb{R}^2$. But the proof by complex analysis adds some additional structures on $\mathbb{R}^2$ (such as making it become a field). I want to find a proof not using complex analysis but using, for example, algebraic topology. – Andrew Mar 03 '19 at 14:19
  • I guess you can use Jordan's theorem, but I don't know its proof so I don't know if this would fit into what you call a topological proof – Maxime Ramzi Mar 03 '19 at 14:34
  • As far as I know, the Jordan Curve Theorem has a proof using algebraic topology. I want to know how to use it to prove this result. – Andrew Mar 03 '19 at 14:48
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    You should note that the complement in question is $\Bbb C\cup{\infty}\setminus E$. (Consider ${0<y<1}$ for example...) – David C. Ullrich Mar 03 '19 at 15:36
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    You should state the definition of "simply connected" that you are using. The reason is that the definitions of simple connectivity given in complex analysis texts tend to be rather nonstandard. See also: https://math.stackexchange.com/questions/330479/complement-is-connected-iff-connected-components-are-simply-connected?rq=1 , https://math.stackexchange.com/questions/2415064/bounded-components-of-complement-of-bounded-planar-domain-are-simply-connected?rq=1 – Moishe Kohan Mar 04 '19 at 00:12
  • Do anyone know that will it be true in higher dimension? – xxxg Feb 21 '24 at 12:59
  • It might not be true. Since in the following link, it says that ``the boundary of a simply-connected domain in the (open) plane $\mathbb{R}^2$ or $C=C^1$ can consist of any number $k$ of connected components, $0 \leq k \leq \infty$.'' https://encyclopediaofmath.org/wiki/Domain – xxxg Feb 21 '24 at 13:23
  • (i) How do you use general Cauchy Theorem (2) What is $Ind_\gamma(z)$. – xxxg Apr 14 '24 at 19:46
  • Do you mean iff compliment is simply connected ? or else take 2 disjoint open balls, compliment is connected but 2 disjoint open balls is not simply connected. –  Oct 03 '24 at 12:16
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    @Balajisb This is why it is assumed that $E$ is connected. – porridgemathematics Oct 03 '24 at 12:46
  • @porridgemathematics Sorry but if you take a unit disk and puncture the center and take this punctured disk as $E$ then $E$ is simply connected and open and bounded but compliment is not connected ? –  Oct 03 '24 at 13:10
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    @Balajisb If you take a unit disk and puncture the center as $E$, i.e. $E = \mathbb{D} \setminus { 0 }$ , then $E$ is not simply connected. – porridgemathematics Oct 03 '24 at 13:24
  • What is the simplest counterexample you can find in dimension $3$? – Loulou Mar 09 '25 at 09:18

2 Answers2

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Boundedness isn't a topological property, but a metric property. The closest corresponding topological property is compactness.

Something like this holds for compact sets: when you consider the sphere $\mathbb{C} \cup \{\infty\}$ as $S^2$, you can apply Alexander duality (here is the statement in Hatcher's Algebraic Topology book):

If $K$ a a compact, locally contractible, nonempty, proper subspace of $S^n$, then $\tilde H_i(S^n \setminus K; \mathbb{Z}) \cong \tilde{H}^{n-i-1}(K; \mathbb{Z})$ for all $i$.

In particular, taking $i=0$ and $n=2$, we get $\tilde H_0(S^2 \setminus K; \mathbb{Z}) \cong \tilde{H}^{1}(K; \mathbb{Z})$.

If $K$ is simply connected, then the right-hand side is trivial, so the left-hand side is trivial, and so $S^2 \setminus K$ has exactly one path-component.

There are probably various ways of generalizing this to certain non-compact but nonetheless bounded subsets of $\mathbb{C}$, such as taking a closure of your open set, just beware that taking closure is not always a homotopy equivalence, so some care is needed.

Edit: Actually, it works out better if we let $K = S^n \setminus U$, where $U$ is your open set. Then $K$ is a closed subset of $S^2$ and is therefore compact, so taking $i=1$, we get $\tilde H_1(U; \mathbb{Z}) \cong \tilde H^{0}(S^n \setminus U; \mathbb{Z})$. There are a few more steps required, like showing local contractibility, showing the fact that $H_1$ measures simple-connectedness for subsets of $\mathbb{C}$, and so on, but all of this should be possible.

Dennis
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Sketching an idea for proof for one direction: Since $E$ is open and connected in $\mathbb{C}$, its path connected. If $E$ is not simply connected, then there exists 2 paths $p_{ab}$ and $q_{ab}$ both of which lies in $E$ which cannot be continiuous transformed to one another. The set/curve $\gamma = [p_{ab} q_{ba}]$ gives $\mathbb{C} \setminus \gamma$ which divides into a some bounded connected sets enclosed by $\gamma$ and an unbounded set outside $\gamma$ by jordan curve theorem. Since there is no continuous transform between $p_{ab}$ and $q_{ab}$ there is a point $x \in E^c$ which is in bounded connected enclosed by $\gamma$. But then there exists a point $y \in E^c$ in unbounded component. Both $x,y$ are separated by $\gamma$ and hence $E^c$ is not connected. So we have sketched a proof for $E$ is not simply connected implies $E^c$ is not connected.

  • Doesn’t Jordan apply only to simple curves? – Lorenzo Pompili Oct 03 '24 at 16:03
  • that is why i mentioned several bounded components rather than single. I just gave a sketch. –  Oct 03 '24 at 23:00
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    I could be wrong, but I still don't think you can use Jordan's theorem in any way, unless you assume something additional on the two paths. Non injective loops can be very ugly. – Lorenzo Pompili Oct 03 '24 at 23:26
  • @LorenzoPompili thanks for the comment. The argument goes like, the curve is bounded set. So only one unbounded component in $\mathbb{C} \setminus \gamma$. Furthet there are finite or infinite number of bounded connected components and since $E$ is not simply connected and because choice of curve there is atleast one bounded component with point from $E^c$ and hence $E^c$ is not connected. –  Oct 04 '24 at 03:14
  • I see what you mean, but for example, the number of bounded components could be zero for non injective loops (see Hilbert's curve for instance). On the same lines, I think it is rather unclear and non obvious how to show rigorously that there is a point in one of the bounded components that belongs to $E^c$. – Lorenzo Pompili Oct 04 '24 at 08:58
  • @LorenzoPompili if you are talking about space filling curve then you can replace the space filling loop by boundary of the space which is filled by the loop which will fail simple connectedness condition for $E$ . Again as a sketch proof. –  Oct 04 '24 at 09:59