Closed subset of $\mathbb R^2$ disconnecting $\mathbb R^2$ must contain a curve?

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Edit: Per Alex S's answer, my first statement was false, so I've moved it to the bottom.

Here is the weaker statement that I am trying to prove (this was included in my original question): Let $C \subset \mathbb R^2$ be a closed subset such that $\mathbb R^2-C$ is disconnected. I suspect that $C$ should contain a connected component with more than one element. Edited addendum: As an intermediate between this question and my first one below, one could also ask whether $C$ has a path-connected component with more than one element.

Context: This the approach I wanted to take in answering this question.

Original question: Let $C \subset \mathbb R^2$ be a closed subset such that $\mathbb R^2-C$ is disconnected. I strongly suspect that $C$ should contain a curve $\gamma: \mathbb R \to C$ which also disconnects the plane, that is, $\mathbb R^2-\gamma(\mathbb R)$ is disconnected, but I don't know how to prove it. It seems to me like one should be able to try and "travel along a boundary" of $C$, but these kinds of things are always difficult to make precise in topology.

Note that $C$ is precisely the complement of two disjoint nonempty open subsets of $\mathbb R^2$; this perspective may be easier to approach conceptually. Also, if $C$ has nonempty interior, then this property is automatically satisfied - simply take a circle in the interior. As that example illustrates, I'm not requiring that the connected components of $\mathbb R^2 - \gamma(\mathbb R)$ "respect" any disconnecting of $\mathbb R^2-C$.

Do you assume that the image of $\gamma$ is homeomorphic to $S^1$? — Moishe Kohan, Apr 14 '16 at 17:58
I see: What do you mean by a "curve" then? A continuous map of the line? An injective continuous map of the line? .... — Moishe Kohan, Apr 14 '16 at 18:07
@studiosus If I had meant for the image to be homeomorphic to $S^1$, I wouldn't have made the domain of $\gamma$ be all of $\mathbb R$. Continuous map is enough for me; I don't need injectivity. — Dustan Levenstein, Apr 14 '16 at 18:18

Plutoro · Answer 1 · 2016-04-14T18:13:19.473

4

Your primary claim is false, even you require that $C$ be compact. Consider the Warsaw curve:

It is the closure of the graph of $\sin(1/x)$ for $x\in[\pi,\infty)$, together with a simple arc joining $(1,0)$ and $(0,0)$. This separates the plane, but it does not contain a loop that also separates the plane.

Of course, you are looking for the image of a map $\gamma:\mathbb R\to C$ to separate the plane, not just a loop. For that we have to change the example. Start with the closure of the graph of $\sin(1/x)$ for $x\in[\pi,\infty)$, and union that with the closure of the graph of $\sin(1/(x-1))$ for $x\in[\pi,\infty)$, and finally union with a simple arc from $(2,0)$ to $(0,0)$. We can avoid one of the problem points by having $\gamma(t)$ limit to, but never reach that point at $t\to\pm\infty$. But we cannot avoid both points.

At this time, I do not know about your weaker claim.

edited Apr 14 '16 at 18:13

answered Apr 14 '16 at 18:01

Plutoro

23,378

Thanks for the counterexample! I figured there was a good chance I was stretching with that part. – Dustan Levenstein Apr 14 '16 at 18:04
Actually, with OP's definition, Warsaw circle is a curve! (It is the image of the real line under a continuous map.) – Moishe Kohan Apr 14 '16 at 18:05
Wait. How is it a curve? I don't think it's continuous at (0,0). – Seven Apr 14 '16 at 18:08
What if you change $C$ at $(1,0)$ to another $ sin(\frac{1}{x})$ like curve? – DLN Apr 14 '16 at 18:09
@Seven $C$ is not a curve. It is a closed subset of the plane. – Plutoro Apr 14 '16 at 18:10
@DLN I think this works. I will have to make a picture. – Plutoro Apr 14 '16 at 18:10
I think the problem is caused by the lack of clarity of what OP means by a "curve". – Moishe Kohan Apr 14 '16 at 18:11
I thought he meant a continuous function from $\mathbb{R} $. – DLN Apr 14 '16 at 18:11
1

no need to delete your answer. To make sure it's not a curve, all you need to do is add another topologist's sine curve in there, so that there's more than one path component. – Dustan Levenstein Apr 14 '16 at 18:19
@DustanLevinstein That's what I did. See my recent edit. – Plutoro Apr 14 '16 at 18:20
Just saw that. :) – Dustan Levenstein Apr 14 '16 at 18:20

score 1 · Accepted Answer · edited Apr 13 '17 at 12:58

1

R. H. Bing constructed a counter example to your intermediate question, see Eric Wosey's answer here: It is a totally path disconnected compact subset of $R^2$ which separates $R^2$ (a pseudocircle). However, a totally disconnected closed subset of $R^2$ cannot separate, see George Lowther's answer to the linked question.

edited Apr 13 '17 at 12:58

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answered Apr 16 '16 at 16:10

Moishe Kohan

111,854

Closed subset of $\mathbb R^2$ disconnecting $\mathbb R^2$ must contain a curve?

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