The Interior of a Jordan Curve versus the Interior of a Polygonal Approximation Let $\gamma:[0,1]\to J$ with $J\subseteq\Bbb C$ is a Jordan curve (so $\gamma(0)=\gamma(1)$). Let $m>0$, $N\in\Bbb N$ and $t_0,t_1,\dots,t_N\in\Bbb R$ be such that $0=t_0<t_1<\dots<t_N=1$.  Suppose there are open discs $D_0,D_1,\dots,D_N$ of radius $m$ such that $\gamma([t_k,t_{k+1}])\subseteq D_k$ for $k=0,1,\dots,N-1$. Let $\pi^*$ be the "polygon" consisting of the line segments from $\gamma(t_k)$ to $\gamma(t_{k+1})$ for $k=0,1,\dots,N-1$.  Assume $\pi$, too, is a Jordan curve (i.e., it does not touch itself except at the beginning and end $\gamma(0)=\gamma(1)$).
By the Jordan Curve Theorem, both $\Bbb C\setminus J$ and $\Bbb C\setminus \pi^*$ have a bounded connected component, $I(J)$ and $I(\pi^*)$, respectively.  
Is it the case that $I(\pi^*)\subseteq\bigcup_{k=0}^{k=N}D_k \cup I(J)$? (Maybe we need to say that "$m$ is small enough." Maybe we need to assume that $\gamma$ is piecewise continuously differentiable.)
This isn't homework for a class! 
 A: It is true. Let $D = \bigcup_{k=0}^N D_k$. We shall show that each $\xi \in I(\pi^*) \setminus D$ is contained in $I(J)$.
We use the concept of the winding number $n(u, x)$ of a closed curve $u : [0,1] \to \mathbb R ^2$ which is defined for all $x \in \mathbb R ^2 \setminus u([0,1])$. Here are two well-known facts:


*

*The winding number is homotopy invariant. That is, if $u_1,u_2$ are closed curves not going through $x$ and being homotopic in $\mathbb R ^2 \setminus \{ x \}$, then $n(u_1,x) = n(u_2,x)$.

*The interior (= bounded component of the complement) of a Jordan curve $u: [0,1] \to \mathbb R ^2$ is the set of $x \in \mathbb R ^2 \setminus u([0,1])$ such that $n(u,x) \ne 0$.
By construction, the closed paths $\gamma$ and $\pi$ are homotopic in $D$ via a homotopy which is stationary on the finite set $\{ 0,t_1,\dots,t_{N-1},1 \}$ (this is true because the $D_k$ are convex). Hence they are homotopic in $\mathbb R ^2 \setminus \{ \xi \}$ and therefore $n(\gamma,\xi) = n(\pi,\xi) \ne 0$ since $\xi \in I(\pi^*)$. Thus $\xi \in I(J)$.
