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Is $\mathbb{R}^2\setminus f([0,1])$ connected if $f:[0,1]\to\mathbb{R}^2$ is an embedding?

It seems that this is clearly true but I am having a hard time proving it. The only things that I know is that $f([0,1])$ is compact and simply connected.

Since $H=\mathbb{R}^2\setminus f([0,1])$ is open in $\mathbb{R}^2$, we can prove that $H$ is locally path-connected and therefore its connected components and path-connected components are the same, or more precisely $A\subset H$ is a connected component if and only if is a path-connected component.

Since $f([0,1])$ is bounded, it is contained in a disk $D$. Hence if $H$ has a separation $\{U,V\}$ (both non-empty, open and disjoint with union $H$), then $U\cap (\mathbb{R^2}\setminus D)$ and $V\cap (\mathbb{R^2}\setminus D)$ are open, disjoint with union $\mathbb{R^2}\setminus D$. Thus we can assume without lose of generality that $U\subset D$. In fact, either $U\cap (\mathbb{R^2}\setminus D)$ or $V\cap (\mathbb{R^2}\setminus D)$) is empty because $\mathbb{R^2}\setminus D$ is connected.

Finally we have that $\{U,V,f([0,1])\}$ is a partition of $\mathbb{R}^2$ where $U$ and $V$ are open and $f([0,1])=\partial U \cup \partial V$ and $U$ is bounded.

It sounds like I have to use the Jordan Curve Theorem somehow.

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I think you are right when you say you can use the Jordan Curve Theorem. First, we want to create a close curve, so we take a tubular neighborhood of the image of $f$ and we draw a parallel copy of $f$ in this tubular neighborhood. You can then join the end points of both segments by a straight line in your tubular neighborhood.

Then, by the Jordan Curve Theorem, this curve separates you plane in two connected region $\mathcal{U},$ $\mathcal{V}$. Moreover, you can prove that by construction, they will both be locally path connected and hence path connected. Therefore, you only need to prove that you can join an element from $\mathcal{U}$ to an element of $\mathcal{V}$ by a path that do not cross $f([0,1]).$ In order to do that, take a small neighborhood of the parallel copy of $f$ that you drew and take you line to be there. You then prove that $\mathbb{R}^2 \backslash f([0,1])$ is path connected, and hence connected

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  • $\begingroup$ Great idea! But I am not really sure if I can draw a parallel copy of $f$ without intersecting $f$ which is needed for the strategy. I think we can use a translation of the plane, say $T:\mathbb{R}^2\to\mathbb{R}^2$, $T(u)=u+v$, for a vector $v$ of small norm such that $T\circ f([0,1])$ doesn't intersect $f([0,1])$, but I have my doubts about this plan because $f([0,1])$ can be an ugly spiral. $\endgroup$ – Chilote Mar 27 '15 at 17:35
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    $\begingroup$ So the reason why you can do it is because you are actually embedding a compact space in $\mathbb{R}^2$, you can apply the tubular neighborhood theorem to find an epsilon neighborhood of your that will be embedded in $\mathbb{R}^2$. $\endgroup$ – Maxime Scott Mar 28 '15 at 3:51
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    $\begingroup$ @djvyu72 I don't believe you can apply the tubular neighborhood to an embedding that is not differentiable. Take for instance $t\mapsto (t, \sqrt{|t-1/2|})$. More pathologically, one might have something like the Julia set of $z\mapsto z^2 + c$, some $c$ so that it's connected, with an open interval removed, c.f., upload.wikimedia.org/wikipedia/commons/b/b1/… $\endgroup$ – Neal Mar 29 '15 at 0:09
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You can use the result to prove the Jordan curve theorem (for example, see Fulton's book on Algebraic Topology).

Suppose, for the sake of a contradiction, that $I$ is homeomorphic to $[0,1]$ and $U\subseteq \Bbb R^2-I$ has two connected components. Divide $[0,1]$ into two halves intersecting at a point, and let $I_1$ and $I_2$ be the corresponding arcs. Then I claim that if $x,y$ are in different connected components of $U$, they must be in either different connected components of $U_1=\Bbb R^2-I_1$ or $U_2=\Bbb R_2-I_2$. We know there exists a locally constant function $f:U\to \Bbb R$ such that $f(x)\neq f(y)$. I claim that $f$ is of the form $f_1-f_2$ where $F_i:U_i\to\Bbb R$ is continuous locally constant and $f_i=F_i\mid U_1\cap U_2$. Indeed, we can construct a partition of unity $\psi_1,\psi_2$ subordinate to the open cover $U_1,U_2$ of $U_1\cup U_2$ and consider $F_1=f\psi_1$, $F_2=-f\psi_2$.

Knowing this, the claim follows, for if $x,y$ were in the same connected components of both $I_i$, we could write $f=f_1-f_2$ and we would have $f_i(x)=f_i(y)$ contradicting our choice of $f$.

We can continue this, to obtain a sequence of intervals $I_1,I_2,I_3,\ldots$ that halve in length at every step for which $x,y$ are in distinct connected components of each $\Bbb R^2-I_i$. Now $\bigcap I_i=\{p\}$ is a point and $\Bbb R^2-p$ is path connected, so we can take a path $x\to y$, and we may take a ball $B$ such that $p\in B$ doesn't intersect this path. But $\Bbb R^2-B$ is also path connected, and we can take $J$ big enough so that $I_J\subset B$, meaning that $\Bbb R^2-B\subseteq \Bbb R^2-I_J$. This contradicts the fact that $x,y$ were in different connected components, and finishes the proof.

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  • $\begingroup$ Nice idea but I am a little confused with your second claim. What is defined first? $f_1,f_2$ or $F_1,F_2$? Also, could you please fix the typos with the $\in$ symbol and the subindex of $U_2$? $\endgroup$ – Chilote Mar 29 '15 at 2:48
  • $\begingroup$ @Chilote $F_i$ is defined first, $f_i$ are restrictions of said functions. $\endgroup$ – Pedro Tamaroff Mar 29 '15 at 2:56
  • $\begingroup$ Why $F_i$ is locally constant if $\psi_i$ is not locally constant necessarily? $\endgroup$ – Chilote Mar 29 '15 at 6:01
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I was reading a little and I found out that this problem is far from being trivial, and actually can be used as the core of the proof of the Jordan curve theorem. I am going to post some links that I found in the web with five different proofs of the Jordan arc theorem (which states that $\mathbb{R}^2\setminus f([0,1])$ is connected if $f$ is and embedding) and the Jordan Curve Theorem, respectively. One of them using the Janizewski's theorem, other using the Brouwer's fixed-point theorem in which shows that if $\mathbb{R}^2\setminus f([0,1])$ is not connected then we can build a retraction from the disk $D^2$ to the circumference $S^1$, which is impossible. http://www.helsinki.fi/~luisto/JordanCurveTheorem.pdf

Also I found an interesting proof using gratings: http://pages.vassar.edu/mccleary/files/2011/04/FinalChapter9.pdf

This is another approach using standard paths and standard curves: http://www.maths.ed.ac.uk/~aar/jordan/dosttind.pdf

And here using graphs.

But I am still interested in a proof using the Jordan Curve Theorem. @djvyu72 gave an idea of how we can apply it but still is incomplete.

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