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It is known that for any open bounded connected subset $U$ of the plane, it is simply connected if and only if its complement is connected, which can be proved thanks to a little bit of complex analysis.

I was wondering to what extent I can loosen the hypothesis on $U$ to be open. I struggle to find a counterexample, because all I can think about are closed curves which satisfy this property up to considering an open tubular neighbourhood. Could I have some hints?

I add another question since the first was answered so quickly. Is the converse also invalid, that is to say: can I find a bounded connected (or even path-connected) subset of the plane which is not simply connected but whose complement is connected?

Loulou
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  • Without open, you need at least need path-connectedness instead of connectedness. – Sassatelli Giulio Mar 12 '25 at 12:08
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    The usual example of a set which is simply connected but has disconnected complement is the Warsaw circle. I expect that this has come up before on this site but don't have time to look ATM! – Izaak van Dongen Mar 12 '25 at 12:08
  • @IzaakvanDongen I'm not convinced, because to my comprehension, the Warsaw circle circle is not simply connected since it is not connected. Even if connectedness is not required in you definition of simple connectedness, still I was looking to loosen only the hypothesis on openness. – Loulou Mar 12 '25 at 12:19
  • @SassatelliGiulio Do you mean it is sufficient? – Loulou Mar 12 '25 at 12:20
  • @Loulou the Warsaw circle is path connected, and in fact simply connected. Unlike classical circle. And its complement is connected. – freakish Mar 12 '25 at 12:21
  • Sorry, in my previous comment I wrote connected and I meant path-connected each time. There are two variants of the Warsaw circle - I just learned it checking on nLab, one is path connected and the other is not (in France, that is the latter we call cercle polonais). On the other hand, you too do not agree on whether its complement is connected or not... – Loulou Mar 12 '25 at 12:28
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    Yes sorry, I meant this one, which is path-connected and simply connected. It's fairly clear that the complement is disconnected (eg because it's open and not path-connected), I think freakish made a typo in their last comment. – Izaak van Dongen Mar 12 '25 at 12:42
  • Yes, I meant "disconnected". My mistake. Otherwise it wouldn't be a counterexample. – freakish Mar 12 '25 at 12:52
  • Ok, thank you all! I'll look into it. By the way, it implies that my intuition about tubular neighbourhood of curves is wrong. – Loulou Mar 12 '25 at 13:06
  • The "tubular neighbourhood" term applies to (sub)manifolds only, no? The Warsaw circle is not a manifold. Either way, the Warsaw circle is simply connected, but its small enough (e.g. not containing say $0$) neighbourhoods are not. – freakish Mar 12 '25 at 14:01
  • For your second question, do you assume that your subset is path-connected? Otherwise you could use the topologist’s sin curve (that is essentially the Warsaw circle minus a point). – Christophe Boilley Mar 12 '25 at 19:57
  • @ChristopheBoilley You're right! Although as you spotted it is the path-connected case that keeps me awake. I'll precise it in the question. – Loulou Mar 12 '25 at 20:01

2 Answers2

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To add to @PaulFrost answer: there is no subset $X\subseteq \mathbb{R}^2$ which is bounded, path connected, not simply connected and its complement is connected.

Indeed, if $X$ is path connected, but not simply connected, then we can find a non null-homotopic Jordan curve in $X$ (see this: https://mathoverflow.net/questions/338073/in-a-subset-of-mathbbr2-which-is-not-simply-connected-does-there-exist-a-s and note that it is important that we deal with $\mathbb{R}^2$, there are counterexamples in $\mathbb{R}^3$). Then the Jordan curve divides entire $\mathbb{R}^2$ into two open subsets: the unbounded exterior and the bounded interior. The interior together with the curve is homeomorphic to a closed disc, and the homeomorphism maps the curve onto the circle (this is Jordan-Schoenflies theorem).

Now the exterior has to intersect the complement $X^c$, otherwise $X$ wouldn't be bounded. On the other hand if $X^c$ doesn't intersect the interior, then this means that entire interior is a subset of $X$. But then the boundary, i.e. the Jordan curve (which topologically is just a circle) can be contracted to a point inside $X$, contrary to being non null-homotopic.

And therefore $X^c$ intersects both the interior and the exterior, which are open. And it is separated by the curve itself. Meaning it is disconnected.

freakish
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For compact subsets it is not true. As an example take the Warsaw circle. For a picture see here.

This space is simply connected, but its complement has two components.

Another example is the closed topologist's sine curve.

This space is connected, but not path-connected, and its complement is connected.

Paul Frost
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  • Thanks. I wonder now if you can have a path-connected but not simply connected subset whose complement is connected. – Loulou Mar 13 '25 at 08:58
  • @Loulou I don't think this is possible. Take a path that is not null-homotopic. Such path can be reparametrized to be an embedding, because the initial path is not constant. This is true for all Hausdorff spaces I think. And then if it is an embedding then by Jordan curve theorem it separates the complement into at least two components. – freakish Mar 13 '25 at 12:13
  • I understand what you are saying. I'm not aware of anything concerning your key argument though, I must lack knowledge about that. Is it because of something like: multiple points of the path must be isolated, or not at all? – Loulou Mar 13 '25 at 13:55
  • @Loulou if a loop touches two distinct points $a\neq b$, then it can be decomposed as glueing of two paths: $a\to b$ and $b\to a$. Then it is a property of Hausdorff spaces that path connectedness is equivalent to arc connectedness. Meaning I can choose both those paths to be injective. And then their composition (seen as function on $S^1$) is injective as well. Which in the case of $\mathbb{R}^2$ can be restated as: it is a Jordan curve. – freakish Mar 13 '25 at 15:56
  • @freakish It is not that easy. If you start with an essential loop in $X$, you want to "shorten" it to an injective loop. But it is conceivable that you cut off the part which made the loop essential and end with a Jordan curve whose interior is contained in $X$. – Paul Frost Mar 13 '25 at 16:18
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    @PaulFrost yes, you're right. I've been careless. Regardless, a non-simply connected subset of $\mathbb{R}^2$ always has an injective essential loop: https://mathoverflow.net/questions/338073/in-a-subset-of-mathbbr2-which-is-not-simply-connected-does-there-exist-a-s As an interesting side note: this is not true for subsets of $\mathbb{R}^3$. – freakish Mar 13 '25 at 16:38