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 $\omega\_1$ ?
> 
> 
> 



Including a Jordan arc into a Jordan loop (Can the Magi go home by another way?)
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)](http://www.math.auckland.ac.nz/class750/) -- 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.


