# A kind of converse to the Jordan curve theorem

The Jordan curve theorem in $\mathbb{R}^2$ says that if $S$ is a closed curve in $\mathbb{R}^2$. Then $S$ splits $\mathbb{R}^2$ into exactly two connected components $A$ and $B$.

I was thinking about a kind of converse to this problem which is as follows.

Let $S$ be a closed and bounded set in $\mathbb{R}^2$ and let $x,y\in\mathbb{R}^2 \setminus S$. Define

$$A:=\{a\in\mathbb{R}^2 \setminus S: a\text{ and } x \text{ are path connected in } \mathbb{R}^2\setminus S\}$$

$$B:=\{b\in\mathbb{R}^2\setminus S: b\text{ and } y\text{ are path connected in }\mathbb{R}^2 \setminus S\}$$

If $A\cap B=\emptyset$ does there exists a connected $T\subset S$ such that $\mathbb{R}^2 \setminus T$ is split into exactly two connected components one which contains $A$ and the other that contains $B$.

Any help or references on this would be much appreciated

Note: Ive made an edit to reflect the comments using the topologists sine curve. I believe that now this will not be a counter-example because the 'quasi-circle" is obviously connected. Also I apologize for not being more specific originally.

• It follows from Alexander duality that the fundamental group of S is non-trivial, at least, and has elements of infinite order. Jan 26, 2016 at 19:13
• Hi thanks for the comment, Is this saying that there is a closed loop in S that is not contractible to either x or y Jan 26, 2016 at 19:46
• @MarianoSuárez-Alvarez But Alexander duality works only if $S$ is locally contractible, no? Jan 26, 2016 at 20:46
• R.L. Moore's student J.R. Kline gave a characterization of such sets $T$ back in 1917 in this paper jstor.org/stable/1988859?seq=1#page_scan_tab_contents. Perhaps this is useful to you. Jan 27, 2016 at 5:14
• Thank you, I will take a look. I feel like if this were not true it would have an off the shelf example Jan 27, 2016 at 16:41

Consider this variant of the extended topologist's sine curve: (picture borrowed from this question) It is the closure of the graph of the function $x\mapsto \sin(\pi/x)$ defined on the half open interval $(0,1]$ with the points $(0,0)$ and $(1,0)$ connected by a simple arc. It is a closed and bounded subset $S$ of $\mathbf R^2$. Its complement has exactly $2$ path-wise connected components, but there is no simple closed curve $T$ contained in $S$.

• So Thank you for the example. I Should have specified more clearly. These "infinite closed curves" actually work fine in my setting, and I didn't know of any way to exclude them. Could you maybe comment on the viability of the following method. Instead of requiring T to be a closed curve. Instead require that T is a connected set that splits $\mathbb{R}^2$ into exactly two regions one that contains A and the other that contains B. The example above is connected (not path connected) and so would work under this new definition right? Jan 26, 2016 at 20:55

Here's a counterexample which I think works.

$S \subset \Bbb R^2$ be the subspace consisting of points $(x, \sin(1/x))$ for $x \in (0, 1]$ and points in $\{0\} \times [-1, 1]$ union an arc joining $(0, 0)$ with $(1, \sin(1))$, namely, the quasi-circle.

$S$ is a closed and bounded subset of $\Bbb R^2$. One can see that $S$ separates $\Bbb R^2$ into two disjoint components by noting that image of any path $\gamma$ not hitting $S$ has a positive distance from $S$, as both are compact subspaces of $\Bbb R^2$. Then one can - with care - replace the $\{0\} \times [-1, 1]$ part of $S$ (because $\gamma$ has distance at least $\delta > 0$ from it) to turn it into a circle and invoke Jordan curve theorem.

However, there is no such $T$: any closed loop in $S$ must miss the origin $(0, 0)$ as $S$ is not path connected at that point. But then obviously such loops do not separate $\Bbb R^2$.

• Thanks to @DanielFischer for the proof of separation of the quasi-circle. Jan 26, 2016 at 20:47
• An exercise in Engelking, General Topology, is to show that $\{(x,\sin 1/x ):x\in (0,1]\}\cup (\{0\}\times [-1,1])$ is connected but not path-connected. from which it follows that the quasi-circle is not a continuous image of [0,1], so it is not a closed loop. Jan 26, 2016 at 21:38
• Yes but if we instead remove the point that T is a closed loop and instead impose that it is a connected set which splits the space into two regions will this be true Jan 26, 2016 at 21:46

Hint: Modify the topologist's sine curve to construct a counter example to your conjecture. The true statement is that $\check{H}^1(R^2 -S)$ is nonzero (by the Alexander duality).

Edit. This is a partial answer to the latest version of the question. (The accepted answer by Lukas does not really settle the problem.)

First, some definitions. A continuum is a metrizable compact connected set. (If you do not know what metrizable means, just think of a subset of $R^n$.) A Lakes of Wada continuum is a continuum $C$ in $R^2$ which has empty interior and whose complement consists of more than 2 components $A_i$ such that the boundary of each $A_i$ is $C$. At the first glance, such thing is impossible; take a look at the Wikipedia picture to get some idea of what it looks like.

Now, suppose that $T\subset C$ is a proper closed subset of a Lakes of Wada continuum $C$. Pick a point $p\in C - T$. Then, since $p$ lies on the boundary of all components of $R^2 - C$ it follows that $R^2 -T$ is connected as all components of $R^2-C$ can be connected to $p$ by paths disjoint from $T$. The conclusion is that such $C$ cannot contain a closed subset whose complement in $R^2$ consists of exactly two components.

This means that the answer to Daniel's question is negative if one also requires $T\subset S$ to be closed. (And the subset constructed in Lukas' answer is closed!) As for noncompact connected subsets, I am not sure, but if I were to bet, I would guess that the answer is again negative.

You can find more and references in

History of continuum theory, by J.Charatonik, in "Handbook of the History of General Topology", Vol. 2.