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Can we divide $\mathbb{R}^2$ into two connected parts such that each part is not simply-connected?

My attempt

Put $A= \{ (0,0) \} $ and $B$ is the punctured plane.

Since that $S^1$ is a deformation retract of the punctured plane, $B$ is not simply-connected. Thus we can find a division of $\mathbb{R}^2$ such that one part is simply-connected but the other is not.

But how to deal with the problem above which requires that each part is not simply-connected?

It seems to be related to contractible and holes. But I don't know how to convert these ideas into precise mathematical language.

Any hints? Thanks in advance!

Added:

As pointed out in the comment, the counterexample exists.

Now I want to ask another question

Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?

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    $\begingroup$ I suppose something like the deleted comb space $([0, 1] \times \{0\}) \cup (\{1/n\}\times [0, 1])\cup\{(0, 1)\}$ and its complement work. The deleted comb space is connected but not path-connected, hence not simply connected. $\endgroup$
    – Dan Rust
    Commented Apr 30, 2019 at 12:54
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    $\begingroup$ @DanRust Thanks for your comment! How about each part is path-connected but not simply connected? $\endgroup$
    – Chiquita
    Commented Apr 30, 2019 at 12:57
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    $\begingroup$ I suspect that in case both are path-connected (or more generally if there is a non nullhomotopic path) this cannot be done by the Jordan curve theorem. But I can't figure out details yet. $\endgroup$
    – freakish
    Commented Apr 30, 2019 at 15:57
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    $\begingroup$ @DanRust: In case others were baffled by your example like I was: $[0,1]$ denotes an interval in $\Bbb R$, so I assumed that $(0,1)$ denotes an interval too. But here it means the point $(0,1)\in\Bbb R^2$. $\endgroup$
    – TonyK
    Commented Aug 4, 2019 at 11:21
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    $\begingroup$ @MoisheKohan Sorry, that answer was beyond my comprehension at that time. I accept it now. $\endgroup$
    – Chiquita
    Commented Apr 15, 2021 at 12:57

6 Answers 6

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If there is an example then it is nasty. With some mild hypotheses it's not possible. For notation consider $S^2 = \mathbb{R}^2 \cup \{\infty\}$.

Suppose that $\mathbb{R}^2 = X \cup Y$ and that $X$ is path connected but not simply connected. Suppose in addition that $X$ is a compact CW-complex. Then it deformation retracts onto a graph, so its fundamental group is free (and nontrivial by hypothesis), thus its first cohomology group is nontrivial. Then by Alexander duality, $\tilde{H}_0(S^2 \setminus X) \cong \tilde{H}^1(X) \neq 0$.

Note that $S^2 \setminus X = Y \cup \{\infty\}$. Since $X$ is compact, it follows that $Y$ is a neighborhood of $\infty$. Removing a point from an open subset of $S^2$ does not change its connectivity. It follows that $\tilde{H}_0(Y) = \tilde{H}_0(S^2 \setminus X) \neq 0$ and therefore $Y$ is not path-connected.

There is probably a way around the hypothesis that $X$ is compact (if $X$ is noncompact then add $\infty$ to it and then $S^2 \setminus X = Y$). However I'm not sure how to get rid of the assumption that $X$ is nice; it has to be at least locally contractible to apply Alexander duality. Perhaps one can work something out with Čech cohomology, but then I don't know how to see that $\check{H}^1(X) \neq 0$.

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  • $\begingroup$ I wouldn't call the assumption "$X$ is a compact CW-complex" mild. :D $\endgroup$
    – freakish
    Commented May 2, 2019 at 10:37
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Both $\{ (x,\sin (1/x)) \mid x \neq 0 \} \cup \{(0,0)\}$
and it's complement are connected.
Since they are not path connected,
they are not simply connected.

It is an example of two connected sets that pass through each other.

Are there two path connected, not simply connected sets whose union is the plane?

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    $\begingroup$ You certainly mean $X = \{ (x,\sin(1/x)) \mid x \in (0,1] \} \cup \{ ( 0,0) \}$ and its compelement $C$. $X$ is not path conncted, but $C$ is. Nevertless it is a counterexample. $\endgroup$
    – Paul Frost
    Commented Apr 30, 2019 at 14:10
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    $\begingroup$ This answer is not satisfactory (for me) because all paths in $X$ and in $C$ are contractible. It abuses the fact that simply connected has to be path connected. I fully understand why OP edited the question. $\endgroup$
    – freakish
    Commented Apr 30, 2019 at 16:02
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The answer to the modified question:

Can we divide $\mathbb R^2$ into two path connected parts such that each part is not simply connected?

is no, this is not possible. In fact, if $A$ and $B$ are disjoint sets with $A\cup B=\mathbb R^2$ then, if $A,B$ are path connected but $A$ is not simply connected, then $B$ must be bounded. So, if $A$ and $B$ are both path connected but not simply connected, then they must both be bounded, giving a contradiction.

I will make use of the following lemma, from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups.

Lemma 13: Let any set $A\subseteq\mathbb R^2$, and map $\alpha\colon S^1\to A$ be given. Let $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\alpha)$. If $\alpha\colon S^1\to A$ is null-homotopic, then so is $\alpha\colon S^1\to A\setminus U$.

So, suppose $A$ is connected but not simply connected. Then there is a curve $\alpha\colon S^1\to A$ which is not null-homotopic. The complement $\mathbb R^2\setminus {\rm Im}(\alpha)=\bigcup_iU_i$ decomposes as the union of its connected components $U_i$. As $B\subseteq\bigcup_iU_i$ is connected, it will be contained within one of the $U_i$. We show that $U_i$ is bounded: As $\mathbb R^2\setminus U_i\subseteq A$, $\alpha$ is not null-homotopic in $\mathbb R^2\setminus U_i$. If $U_i$ were unbounded, then the quoted result says that $\alpha$ is not null-homotopic in $\mathbb R^2$, a contradiction.

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  • $\begingroup$ I don't understand the last sentence: can you explain how you get from "I can make the points in $F$ as far from the origin as I want" to "$\gamma$ can be deformed to a point in $\mathbb R^2\setminus F$" ? Because $\gamma$ moves too when you deform $U_i$, doesn't it ? $\endgroup$ Commented Aug 4, 2019 at 11:13
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    $\begingroup$ Let $U$ be a connected open set and $P\in U$. For some $r > 0$, the open ball of radius $r$ about $P$ will lie in $U$. Then, by only deforming the interior of this open ball (which lies in $U$, you can move $P$ to any point in this open ball. By covering $U$ by such open balls, you can move $P$ to any other point by deforming $U$ in such a way that only a compact subset of $U$ is moved. This is easily extended to finite sets. Such a deformation will not affect any points outside of $U$ (i.e., $\gamma$ will not be affected). $\endgroup$ Commented Aug 4, 2019 at 11:18
  • $\begingroup$ I tried to clarify the argument a bit. $\endgroup$ Commented Aug 4, 2019 at 11:24
  • $\begingroup$ Thank you, the comment was clear enough ! $\endgroup$ Commented Aug 4, 2019 at 11:25
  • $\begingroup$ I updated my answer to use a different result from the paper, which is more direct, and the argument about deforming $U$ is no longer required. $\endgroup$ Commented Aug 4, 2019 at 12:48
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So here's my partial answer. Assume additionally that

In each space there is a Jordan curve that is not null homotopic.

Let's assume $\lambda$ is such curve in $X$. With that we apply the Jordan curve theorem. Say $\mathcal{I}$ is the interior of $\lambda$ and $\mathcal{E}$ the exterior. If $\mathcal{I}\cap Y\neq\emptyset$ and $\mathcal{E}\cap Y\neq\emptyset$ then $Y$ cannot be path connected because any path (between exterior and interior) would have to cross $\lambda$.

On the other hand if $\mathcal{I}\cap Y=\emptyset$ then $\mathcal{I}\subseteq X$ and so $\lambda$ is null homotopic in $X$ because by the Jordan–Schoenflies theorem the interior of $\lambda$ is homeomorphic to a ball.

So the only possibility is that $\mathcal{E}\cap Y=\emptyset$, i.e. $\mathcal{E}\subseteq X$.

The same reasoning can be applied to $Y$ in order to get that the exterior of some Jordan curve has to be fully contained in $Y$. This contradicts them being disjoint.

So I couldn't find an example of a path-connected, not simply-connected subset of $\mathbb{R}^2$ which doesn't have a not null homotopic Jordan curve. But I also couldn't prove that this is not possible. Any help appreciated. Note that such subsets do exist in $\mathbb{R}^3$ (see here).

Side note: If we don't require that both have to be path-connected, but at least one, then there's a neat counterexample of $X=\mathbb{Q}^2$ and $Y=\mathbb{R}^2\backslash\mathbb{Q}^2$. Note that $Y$ is path-connected but not simply connected.

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It will be very difficult to find an example (and I do not believe it is possible). Here is a partial result.

Let $X \subset \mathbb R ^2$ be compact and path connected. If its complement $\mathbb R ^2 \setminus X$ is connected (which means that $X$ does not separate the plane) , then $X$ is simply connected.

This relies on some deep theorems, and I can only give an outline. The paper

Hagopian, Charles. "The fixed-point property for simply connected plane continua." Transactions of the American Mathematical Society 348.11 (1996): 4525-4548.

contains the following results:

Theorem 1.2. Every arcwise-connected nonseparating plane continuum has the fixed-point property.

Theorem 9.4. Suppose $M$ is an arcwise-connected plane continuum. For $M$ to have the fixed-point property it is necessary and sufficient that the fundamental group of $M$ be trivial.

The above claim is an immediate consequence.

I could imagine that there is a more direct proof (which does not focus on the fixed point property), but I don't have any idea how to do that.

Remark: One can moreover show that if $X$ is a compact connected subset of the plane such that $\mathbb R ^2 \setminus X$ is connected (which is the same as path connected because it is an open set), then $\mathbb R ^2 \setminus X$is homeomorphic to $\mathbb R ^2 \setminus \{ 0 \}$. See my answer to Homotopy type of the complement of a subspace.

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I believe the answer to the question "Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?" is false.

Let the two (disjoint) parts be A and B. I claim that if A is not simply connected then B is not path connected:

If A is not simply connected then there exists a simple closed loop L lying entirely in A s.t L cannot be contracted to a point. This means that the region R surrounded by L must contain a "hole", more precisely there exists a point $p\in R$ which does not lie in A and hence must lie in B. This means that B cannot be path connected since any path that starts at p must cross L (by the Jordan Curve Theorem) and hence can not lie entirely in B.

UPDATE: The answer to this question "In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point?" that I posed on MathOverflow indicates that it is probably true that in a subset of $\mathbb{R}^2$ that is not simply connected there must exist a simple loop that contracts to a point. Therefore this makes the above proof probably true. Note that this is a special property of $\mathbb{R}^2$ since, as freakish points out, there exist spaces in $\mathbb{R}^3$ that are not simply connected but where all simple closed curves contract to a point.

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  • $\begingroup$ How to you know the loop is homeomorphic to $S^1$? $\endgroup$ Commented Aug 3, 2019 at 7:53
  • $\begingroup$ @JasonDeVito That's a great question thank you and definitely paths or loop in general spaces are not necessarily simple as they can have self intersections. I think however that my argument is still OK in the context of a space like $\mathbb{R}^2$ which is Hausdorff since in a Hausdorff space arc-connected is equivalent to path-connected. Therefore in the definition of simply connected we can assume that loops are simple i.e. they are injections from the unit interval and hence homeomorphic to $S^1$. I have updated my answer to include this point. $\endgroup$
    – Ivan
    Commented Aug 3, 2019 at 9:03
  • $\begingroup$ Well, for example, in $\mathbb{R}^2$ with a point removed, I don't think a curve which goes twice around the point is homotopic to Jordan curve. What happens if, e.g., $\pi_1(A)\cong\mathbb{Q}$ so every nontrivial loop is twice another nontrivial loop? $\endgroup$ Commented Aug 4, 2019 at 16:44

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