The title refers, of course, to Matthew (2:12) ''And being warned in a dream not to return to Herod, they departed to their own country by another way''. To be honest, it is not that specific particular case I'm more interested in.

I'd like to have a reference, or a hint here, for a simple proof of the following fact (intuitive, but not easy to proof, as usual in these matters).

Let $\Gamma$ be a simple Jordan arc in $\mathbb{R}^2$ (a homeomorphic image of the interval $[0,1]$). Then, $\Gamma$ can be included in a simple Jordan loop $ \Sigma $ (a homeomorphic image of $\mathbb{S}^1$).

By the (generalized) Jordan's theorem, we know that $\mathbb{R}^2\setminus\Gamma$ is connected; and being open, it is even connected by piece-wise linear paths. The difficulty is that we need a path connecting the end-points of $\Gamma$; in other words, the question is how to show that $(\mathbb{R}^2\setminus\Gamma)\cup \partial\Gamma $ is path-connected (after that, an injective path could always be extracted).

It seems to me everything would follow easily from this lemma:

Assume that $B(0,2)\setminus \Gamma$ has at least two connected components that meet $B(0,1)$. Then, there are three consecutive points of $\Gamma$, resp. $y_1$, $y_2$, and $y_3$ such that $\|y_2\|=1$, and $\|y_1\|=\|y_3\|=2$

I'm also a bit puzzled by the quantitative aspect of this problem:

Assume that $\Gamma$ is parametrized by a homeomorphism $\gamma:[0,1]\to\Gamma$ with a modulus of continuity $\omega(t)$ (say, a continuous concave function vanishing at $t=0$) and let $\omega_1$ be another modulus of continuity such that $\omega_1(t) > \omega(t)$ for $t > 0$. Is there a Jordan loop $\Sigma$ with parametrization $\gamma_1[0,2 > ]\to\Sigma$ that has modulus of continuity $\omega_1$ ?


1 Answer 1


Yes they can! Every Jordan arc can be included in a Jordan loop. I'll give you the simplest proof I could think of, which only uses very basic topology. It also shows that the lemma you propose does indeed hold, although I'll skip the quantitative aspect. [I see that Jim Conant linked to a very similar question in a comment, but the proof I'll give here is very different from the answers given to the other question. It is also, in my opinion, more elementary, and I prove the lemma asked for here.]

The idea is to use the Jordan arc $\Gamma\subseteq\mathbb{R}^2$ to construct an undirected graph $G_\Gamma$ as follows. Write $S^1=\bar B_1(0)\setminus B_1(0)$ for the unit circle. Then, let the nodes $V_\Gamma$ of the graph consist of the connected components of $\mathbb{R}^2\setminus(\Gamma\cup S^1)$, which naturally split into those inside the circle and those outside. Let the edges $E_\Gamma$ of the graph consist of the connected components of $S^1\setminus\Gamma$. Each edge lies in the closure of precisely two elements of $V_\Gamma$ (one inside and one outside the circle), and we consider it to join these two nodes. Then, the main fact used to prove your lemma is the following.

1) $G_\Gamma$ is a tree.

I'll give a proof of this statement below. As it seems like a rather simple fact, there should be a reference somewhere - maybe some knows one?

Using this, we can easily dispatch with your lemma, which I'll state as follows. I'm using $\Gamma={\rm Im}(\gamma)$ for a continuous one to one map $\gamma\colon[0,1]\to\mathbb{R}^2$.

2) If $\epsilon < 1$ and at least two connected components of $B_1(0)\setminus\Gamma$ meet $B_\epsilon(0)$ then there exists $t_1 < t_2 < t_3\in[0,1]$ with $\Vert\gamma(t_1)\Vert,\Vert\gamma(t_3)\Vert\ge1$ and $\Vert\gamma(t_2)\Vert < \epsilon$.

If the curve $\Gamma$ did not intersect with $S^1$ then $G_\Gamma$ would only have one edge and, hence, only two nodes (the inside and outside of the circle), so only one region could intersect with $B_\epsilon(0)$. On the other hand, suppose that $\Gamma\cap S^1$ is nonempty. Then, let $t_1,t_3\in[0,1]$ be respectively the first and last times at which $\gamma(t)\in S^1$. Let $\gamma^\prime=\gamma\vert\_{[t_1,t_3]}$ and $\Gamma^\prime={\rm Im}(\gamma^\prime)$. Note that $\Gamma\cap S^1=\Gamma^\prime\cap S^1$, so $E_\Gamma=E_{\Gamma^\prime}$. Also, we can map $V_\Gamma$ to $V_{\Gamma^\prime}$ by taking each connected component $U\in V_\Gamma$ to the connected connected component of $\mathbb{R}^2\setminus(\Gamma^\prime\cup S^1)$ containing it. This is one-to-one on the components outside the circle (as $\Gamma\setminus B_1(0)=\Gamma^\prime\setminus B_1(0)$. By (1), it must be one-to-one on all of $V_\Gamma$. Otherwise, if $U,V\in V_\Gamma$ mapped to the same element of $V_{\Gamma^\prime}$, then a simple path from $U$ to $V$ in $G_\Gamma$ would map to a closed loop in $G_{\Gamma^\prime}$, which must pass through a node outside of the unit circle (as any path alternates between inside and outside the circle), and only passes through this node once (as $V_\Gamma\to V_{\Gamma^\prime}$ is 1-1 on the regions outside the unit circle). This would contradict the statement that $G_{\Gamma^\prime}$ is a tree unless $U=V$. So, $G_{\Gamma^\prime}$ must also have at least two components intersecting with $B_\epsilon(0)$, so $\Gamma^\prime\cap B_\epsilon(0)\not=\emptyset$ and $t_2\in[t_1,t_3]$ exists as required.

As you suggest, this shows that any point $P\in\mathbb{R}^2\setminus\Gamma$ can be connected to $\gamma(0)$ by a by a curve not intersecting with $\Gamma$. Wlog, suppose that $\gamma(0)=0$ and let $t_n\in(0,1]$ be the first time at which $\gamma$ exits from $B_{1/n}(0)$. Then, choose points $P_n\in\mathbb{R}^2\setminus\Gamma$ with $\Vert P_n\Vert\le\min\_{t\ge t_n}\Vert\gamma(t)\Vert$. We can define $f\colon[0,1]\to\mathbb{R}^2$ by letting $f\vert\_{[1/2,1]}$ be a path in $\mathbb{R}^2\setminus\Gamma$ joining $P_2$ to $P$ and, for each $n\ge 2$, let $f\vert\_{[1/{(n+1)},1/n]}$ be a path in $B_{1/n}(0)\setminus\Gamma$ joining $P_{n+1}$ to $P_n$ (which exists, by the lemma). Then, setting $f(0)=0$, this gives the desired curve.

I'll now give a proof of (1), that $G_\Gamma$ is a tree, based on the following facts.

  1. $\mathbb{R}^2\setminus\Gamma$ is connected.
  2. If $P,Q,R,S$ are points occuring clockwise (in that order) on $S^1$ and $C_1,C_2\subseteq \bar B_1(0)$ are curves joining $P$ to $R$ and $Q$ to $S$ respectively, then $C_1$ and $C_2$ intersect.

Neither of these facts requires anything particularly difficult. Both of them follow from the methods in the notes Brouwer's Fixed Point Theorem and The Jordan Curve Theorem from this link (Section 5) -- see the proof of Lemma 5.6 for the first fact and, reducing to the case $P=(0,1)$, $Q=(1,0)$, $R=(0,-1)$, $S=(-1,0)$, see Lemma 5.5 for the second fact.

The first fact immediately implies that $G_\Gamma$ is connected, so we just need to show that it has no simple closed loops. Suppose, on the contrary, that there was a simple loop with edges $E_1,E_2,\ldots,E_n$ (which are just open segments of $S^1$). Then, as any path alternates between inside and outside the unit circle, $n$ is even. So, the connected components of $S^1\setminus(E_1\cup\cdots\cup E_n)$ consists of $n$ closed connected segments $I_1,\ldots,I_{n/2},J_1,\ldots,J_{n/2}$. I'm labeling these alternately by $I_k$ and $J_k$ as you go round the circle. As each of these closed segments intersects with $\gamma$, there must exist $t_1,t_2\in[0,1]$ with $\gamma(t_1)\in\bigcup_k I_k$ and $\gamma(t_2)\in\bigcup_k J_k$. Wlog, suppose that $t_1 < t_2$. Also, choose $t_1$ maximal in the range $[0,t_2]$ and, then, choose $t_2$ minimal in $[t_1,1]$. Then, $\gamma^\prime=\gamma\vert\_{[t_1,t_2]}$ is a curve joining a point in $I_i$ to $J_j$ (say), and intersects $S^1$ only at its end points. Wlog, we can assume that $\Gamma^\prime={\rm Im}(\gamma^\prime)$ lies in $\bar B_1(0)$ (otherwise, circular inversion can be applied to reduce to this case).

Then, $S^1\setminus(I_i\cup J_j)$ consists of two connected open arcs each of which contains an odd number of the intervals $E_k$. However, the edges of $G_\Gamma$ joining a single node $V\in V_\Gamma$ must all lie in the same arc otherwise, by connectedness of $V$, we would contradict the second fact above. So, each pair of edges $E_k,E_{k+1}$ in the path joining to a node inside the circle lies in the same arc, and hence each arc must contain an even number of the edges $E_k$. This gives the desired contradiction.

  • 2
    $\begingroup$ You can't get much more elementary than Thurston's idea for letting the arc run from $0$ to $\infty$ and taking the preimage under the (branched) double cover of the Riemann sphere $z\mapsto z^2$. I'm still amazed by this. $\endgroup$
    – Jim Conant
    Commented Sep 13, 2011 at 21:39
  • $\begingroup$ Ah, I missed that one. Yes, it is very nice (I think you need to apply the Jordan curve theorem to show that the domain is simply connected, then apply Schoenflies to show that the preimage can be extended into the interior of the circle). $\endgroup$ Commented Sep 13, 2011 at 22:18

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