Is a smooth simple closed curve the union of finitely many arcs?

Question

This is a problem from Pugh's Real Mathematical Analysis [Chapter 5].

If $C$ is a smooth simple closed curve in the plane, show that it is the union of finitely many arcs $C_l$, each of which is the graph of a smooth function $y = h(x)$ or $x = h(y)$, and the arcs $C_l$ meet only at common endpoints.

Attempt : Let $C$ be parametrized by $t$, that is, $(x(t),y(t))$ be its coordinates, where $t$ changes in $[0,1]$. We can start by letting $t_0$ = $0$ and continue traversing the curve (either clockwise or counter-clockwise) until one of the coordinates of the derivate changes sign. Let's call this point $t_1$. With continuing the same algorithm, we have partitioned $C$ into a (not necessarily finite) number of arcs with their domain $[t_i,t_{i+1}]$. But here two things remain to be proved:

First : How to revise the algorithm so that the sign-changing points are finite? Or how to show that they are finite, if it is the case?

Second : It is easy to show that in each arc either $x$ is a function of $y$ or vice versa. The question is why one of them is an smooth function of the other.

Thanks for your suggestions in advance. Note that I'm taking an Advanced Calculus course and I'm not familiar with manifolds or algebraic topology. So keep your answers as simple as possible.

Answer 1

As we discussed in chat, we need a definition of a smooth simple closed curve, for starters. To most of us, it will be a compact $1$-dimensional submanifold (without boundary) of $\Bbb R^2$. You indicated that you want to define it as the image of a parametrization $\phi\colon [0,1]\to\Bbb R^2$ with $\phi(0)=\phi(1)$ and the regularity condition that $\phi'(t)\ne 0$ for all $t$. [The latter condition is required to deduce that the curve has a tangent line at each point.] (Missing, also, is the "gluing" condition that the derivatives at $0$ and $1$ match up correctly.)

The main idea is this: At each point $p$ of the smooth curve $C$ we have a tangent line, and that tangent line projects isomorphically onto (at least one of) the $x$-axis or the $y$-axis. (Only if the line is horizontal or vertical will only one work.) Let $\pi_1\colon\Bbb R^2\to\Bbb R$ be projection onto the $x$-axis and $\pi_2$ projection onto the $y$-axis. Note that if the tangent line at $p$ is non-vertical (resp., non-horizontal), then there is an open arc of $C$ around $p$ on which $\pi_1$ (resp., $\pi_2$) has non-vanishing derivative (i.e., the restriction of $\pi_1$ to the tangent lines of $C$ is an isomorphism at each point). It follows that this arc of $C$ is given as the graph of a smooth function $y=f(x)$.

Let's justify that more carefully. Suitably interpreted, it is just the Inverse Function Theorem. Let $\phi(t_0)=p=(x_0,y_0)$, and suppose that for $t\in (t_0-\epsilon,t_0+\epsilon)$ the vector $\phi'(t)$ is non-vertical. (This is just a consequence of the continuity of $\phi'$.) Rephrasing, $\pi_1(\phi'(t))\ne 0$ for $t$ in this interval. By the chain rule, $(\pi_1\circ\phi)'(t)\ne 0$ for $t$ in the interval, and so the function $\pi_1\circ\phi\colon (t_0-\epsilon,t_0+\epsilon)\to\Bbb R$ has a smooth inverse function [by just single-variable results] $\psi\colon J\to (t_0-\epsilon,t_0+\epsilon)$ for some open interval $J$ containing $x_0$. Now let's consider $\phi\circ\psi\colon J\to\Bbb R^2$. Since $\psi = (\pi_1\circ\phi)^{-1}$, we see that $\phi(\psi(x))=(x,f(x))$ for some (a fortiori) smooth function $f$ defined on $J$.

Now we're done. By compactness of the curve, finitely many such (open) arcs will cover $C$, and so we represent $C$ as the union of finitely many (open) arcs of graphs. You can choose closed arcs by appropriately choosing endpoints within the overlap intervals.