this is my first question here! Hopefully it is appropriate. Let $\mathbb{A}$ be the punctured plane, i.e. the 'standard' annulus. For compact, connected subsets of the plane (planar continua) $X \subset \mathbb{R}^2$, I know it's not necessarily true that $\mathbb{R}^2 \setminus X \simeq \mathbb{A}$ (homeomorphism), for example taking $X$ as the Warsaw Circle.

I was wondering what sort of additional assumptions we can make so that, in fact, $\mathbb{R}^2 \setminus X \simeq \mathbb{A}$.

1) Is it good enough that $\mathbb{R}^2 \setminus X$ be connected? This seems true using some classical separation arguments in the sphere. [EDIT: This is true, as shown in an answer below]

2) Is it good enough that $X$ be unicoherent, i.e. for each pair of compact, connected subsets $A, B \subset X$ with $A \cup B = X$, their intersection is connected? [EDIT: See my comment below. Unicoherence is not sufficient]

3) What if $X$ is hereditarily unicoherent, i.e. all its closed subsets are unicoherent?

It seems to me that "open-unicoherence" where the closedness of the sets $A, B$ is replaced by openness should be enough by Cech Homology, but I was unable to find any results on "open-unicoherence" except in the locally connected setting, which is too permissive for what I'm looking at (dendroids). I suppose a side-question that would be relevant for me is whether hereditary unicoherence implies (hereditary) open-unicoherence.

4) What conditions on a (hereditarily) unicoherent planar continuum are sufficient for it to be (hereditarily) open-unicoherent and vice-versa when not necessarily in the locally connected case (this case has been heavily explored)

So a more specific question would be a reference request for either a proof or counterexample when $X$ is a planar dendroid, i.e. a path-connected and hereditarily unicoherent planar continuum.

EDIT: The question concerning dendroids is known. It is more strongly known that planar tree-like continua do not separate the plane (but I can't find a reference; anyone know of one?), and since dendroids are tree-like we can apply the answer to #1. In Kuratowski Topology II, p. 506 Thm. 4 it is known that if $\mathbb{R}^2 \setminus X$ is connected, then $X$ is unicoherent. The converse is not given, but neither is a counterexample.

So, still stuck on questions 3 and 4, and a reference for the fact that planar tree-like continua don't separate the plane.

Thanks in advance!

  • $\begingroup$ There are unicoherent, one-dimensional planar continua which separate the plane. Take a circle with an arc spiraling closer and closer around it. In the paper "A Survey on Unicoherence" by Garcia-Maynez and Illanes, planar continua which are open-unicoherent but not unicoherent, and unicoherent but not open-unicoherent, are exhibited. Still don't have a counterexample or proof in the hereditarily unicoherent case. $\endgroup$ Jun 25, 2017 at 3:00
  • $\begingroup$ In Q2 you've defined the HEREDITARILY unicoherent continuum instead of simply unicoherent. $\endgroup$
    – Wlod AA
    Jun 25, 2017 at 14:27
  • 1
    $\begingroup$ Oops, you're right! Fixed. $\endgroup$ Jul 1, 2017 at 23:56

4 Answers 4


Moore's theorem says that if $\sim$ is an equivalence relation on $\mathbb{S}^2$ such that any equivalence class is closed connected and has connected complement then the quotient space $\mathbb{S}^2/\sim$ is homeomorphic to $\mathbb{S}^2$.

In particular the answer to your first question is "yes".


After a year the question may be "cooled"; on the other hand good questions never cool off.

Your question concerns the relationship between continua $X \subset \mathbb{R}^2$ amd their complements $CX = \mathbb{R}^2 -X$.

Shape theory provides an answer. We have $CX \approx CY$ iff $X$ and $Y$ have the same shape. There are countably many shapes of continua $X \subset \mathbb{R}^2$: These are represented by $X_0$ = one-point-space, $X_n =$ wedge of $n$ circles and $X_\infty =$ Hawaiin earring. Therefore $CX \approx \mathbb{A}$ iff $X$ has trivial shape (i.e. the shape of $X_0$).

In particular, $CX \approx \mathbb{A}$ iff $CX$ is connected.


Here are some thoughts concerning tree-like continua and dendroids. Let me begin by quoting some facts, although most of this will be known to you.

1) From

Cook, Howard. "Tree-likeness of dendroids and λ-dendroids." Fundamenta Mathematicae 68 (1970): 19-22

we know that hereditarily indecomposable continua are tree-like.

2) $X$ being tree-like means that every open cover of $X$ can be refined by a finite open cover having as nerve a tree. This shows that the Cech expansion of $X$ (which is an inverse system in the homotopy category) has a cofinal subsystem consisting of contractible spaces. Therefore $X$ has trivial shape.

3) We conclude that the complement of a tree-like plane continuum is an annulus.


1) is true by the answer above. Thus in the other cases it's sufficient to show that they don't separate the plane.

2) There is a unicoherent continuum, namely a circle with an arc spiraling closer and closer to it, that separates the plane.

3) There is a counterexample, the pseudo-circle (careful, googling will reveal several things called a 'pseudo-circle'). Here, the pseudo-circle refers to a circle-like continuum written as a union of circular chains $C_i$ such that $C_{i+1}$ is crooked in $C_i$ in the same sense as the pseudo-arc. This continuum is hereditarily indecomposable. Thus the new question would be whether there are decomposable hereditarily unicoherent continua which separate the plane.

4) Seems to be largely open, there is a thesis and a couple of papers by Ganea that I'm going to read which might shed light on this.

A reference for the fact that tree-like continua don't separate the plane is in 'Results and Problems in Fixed Point Theory for Tree-Like Continua', theorem 1.5, by Roman Manka (though not first proven here, it seemed to be 'folklore').

  • $\begingroup$ Re (3), attaching an arc to the pseudo-circle is decomposable, hereditarily unicoherent and separates the plane. A hereditarily decomposable, hereditarily unicoherent continuum is called a $\lambda$-dendroid, and Cook proved that these are tree-like: eudml.org/doc/214232 $\endgroup$ Jun 21, 2018 at 13:02

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