A Jordan arc passing through prescribed points

Question

Let $0=t_0<t_1<\dots<t_{n-1}<t_n=1$ and $x_j\in \mathbb R$, $x_j\not= x_k$ for $j\not=k$, and distinct from $0,1$, $(j,k=1,\dots,n-1)$. Is it known under which conditions on the $x_j$ (in terms of permutations of the tuple $(x_1,\dots,x_{n-1})$ for instance, there is a Jordan arc $\phi:[0,1]\to \mathbb C$ with end points $x_0:=0$ and $x_n:=1$ such that $\phi(t_j)= x_{j}$? I am looking for a reference. Thanks.

Answer 1

I don't have a reference but here is a sketch of a proof that shows this is always possible. By induction on the number of points, we claim that there exists a piecewise linear curve $\phi$ passing through $x_0,x_1,\ldots, x_n=1$ in order such that its union with another piecewise curve connecting 1 to 0 is a Jordan curve. For $n=1$ this is easy to see. Suppose the claim is true for $n$ and consider the Jordan curve $\gamma$ that is claimed to exist for $x_0,\ldots, x_{n-1},1$ by the induction hypothesis. There are three cases.

• $x_{n}$ lies on $\gamma$. In this case, the line segment in $\gamma$ containing $x_{n}$ can be replaced by a union of two line segments such that $x_{n}$ lies inside the new Jordan curve $\gamma'$. Then wlog we can assume we are in the next case.

• $x_{n}$ is inside $\gamma$. The interior of $\gamma$ is path connected and so $x_{n}$ can be connected by a piecewise linear path $\gamma_1$ to $x_{n-1}$ on $\gamma$ such that $\gamma_1$ intersects $\gamma$ only at $x_{n-1}$. The interior of $\gamma$ with $\gamma_1$ removed is still connected and so $x_{n}$ can be connected to $1$ by a piecewise linear path $\gamma_2$ such that $\gamma_2$ intersects $\gamma$ only at 1 and intersects $\gamma_1$ only at $x_{n}$. Now, in $\gamma$ replace the part that connects $x_{n-1}$ to 1 by the union of $\gamma_1$ and $\gamma_2$. The new curve passes through $x_0,\ldots, x_{n},1$.

• $x_n$ is outside $\gamma$. A similar proof as in the previous case works.

Note that we can always reparametrize the curve to satisfy the conditions $\gamma(t_i)=x_i$. Finally, one can smoothen the path and obtain a smooth curve.