Find an arc connecting two Jordan curves on the plane

Question

Let $S_1$ and $S_2$ be disjoint Jordan curves on the plane $\mathbb R^2$ (i.e. homeomorphic images of a circle). I want to prove that for any $x_1\in S_1$ and $x_2\in S_2$ there exists an arc joining $x_1$ and $x_2$ that lies, except for its endpoints, entirely outside $S_1$ and $S_2$.

This seems trivial, but...

Remarks.

1. By Schoenflies theorem there exists a homeomorphism of the plane that sends $S_1$ to the real circle, so we can assume that $S_1$ is actually a circle.

2. Let $K_i$ be the closed region bounded by $S_i$. If $K_1\cap K_2=\emptyset$, then $\mathbb R^2\setminus (K_1\cup K_2)$ is connected (and hence path-connected) by the fact discussed for example here. Similarly, if $K_1\subset K_2$, then $\operatorname{int}K_2\setminus K_1$ is connected. However, the closure of connected open set need not be path-connected, see for example here.

Answer 1

Let us first introduce some notation. For each $r$ with $0 < r < 1$ we define annuli and circles

$$ \begin{array}{lcl} A(r) & = & \{ a \in \mathbb R^2 \mid 1- r \le \lVert a \rVert \le 1 + r \} \\ A^+(r) & = & \{ a \in \mathbb R^2 \mid 1 \le \lVert a \rVert \le 1 + r \} \\ A^-(r) & = & \{ a \in \mathbb R^2 \mid 1 - r \le \lVert a \rVert \le 1 \} \\ B^+(r) & = & \{ a \in \mathbb R^2 \mid \lVert a \rVert = 1 + r \} \\ B^-(r) & = & \{ a \in \mathbb R^2 \mid \lVert a \rVert = 1 - r \} \\ \end{array} $$

The Jordan-Schoenflies theorem allows to find homeomorphisms $h_i : \mathbb R^2 \to \mathbb R^2$ such that $h_i(S_i) = S^1$ = unit circle with center $0$.

Take disjoint open neighborhoods $W_i \subset \mathbb R^2$ of the $S_i$. Then the $h_i(W_i)$ are open neighborhoods of $S^1$ in $\mathbb R^2$ and there exists $r$ with $0 < r < 1$ such that the annulus $A(r)$ is contained in both $h_i(W_i)$. Let

$$A_i = h_i^{-1}(A(r)) .$$ These are disjoint compact subsets of the plane. There exist $a_i \in A_i$ such that $\lVert a_1 - a_2 \rVert$ is minimal. Due to minimality these $a_i$ must be contained in the boundary $B_i$ of $A_i$. Clearly $B_i = B^+_i \cup B^-_i$ with $B^\pm_i = h_i^{-1}(B^\pm(r))$. Thus $a_i \in B^{\sigma(i)}_i$ with a unique sign $\sigma(i)$.

The line segment $[a_1,a_2] = \{ ta_1 + (1-t)a_2 \mid t \in [0,1]\}$ connecting $a_1$ and $a_2$ has the property $[a_1,a_2] \cap A_i = \{a_i\}$.

Now consider $x_i \in S_i$. Obviously there exists an arc $J'_i \subset A^{\sigma(i)}(r)$ such that $J_i \cap S^1 = \{h_i(x_i)\}$ and $J'_i \cap B_i^{\sigma(i)}(r) = \{h_i(a_i)\}$. Then $J_i = h_i^{-1}(J'_i)$ is an arc in $A^{\sigma(i)}_i =h_i^{-1}(A^{\sigma(i)}(r)) \subset A_i$ such that $J_i \cap S_i = \{x_i\}$ and $J_i \cap B^{\sigma(i)}_i = \{a_i\}$.

Concatenating the arcs $J_1, [a_1,a_2]$ and $J_2$ gives an arc $J$ connecting $x_1$ with $x_2$ such that $J \cap S_i = \{x_i\}$.

Drawing a picture is helpful, but it is easy to check everything formally.