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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.

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    $\begingroup$ Do not think in terms of algorithms. Instead, think about the implicit function theorem and compactness. $\endgroup$ Commented Jun 4, 2023 at 20:40

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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.

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  • $\begingroup$ Thanks for the detailed answer. I learned a lot from it. By the way, I wonder what happens if we remove the assumption of regularity? Can you give an example of a simple closed smooth curve ( being infinitely differentiable) which is non-regular, but is not the union of finite arcs with the properties that the question says? Does such an example even exist? $\endgroup$ Commented Jun 6, 2023 at 0:06
  • $\begingroup$ Yes. Here’s a clue for you. As we discussed, if the parametrization is allowed to be non-regular, the curve may not be smooth at all. In particular, it can have any number of corners — points where it has no tangent line — even with a smooth parametrization. $\endgroup$ Commented Jun 6, 2023 at 0:22
  • $\begingroup$ Yeah. That's what I was thinking about. To build an example, I aim to imagine a smooth simple closed curve with an infinite number of corners (where we have a zero derivative in any of them) . Since the curve is a bounded subset of $\Bbb R^2$, these points will have a limit point in $\Bbb R^2$. That's the point from where I get lost and stuck. $\endgroup$ Commented Jun 6, 2023 at 0:34
  • $\begingroup$ Even without the limit point, you can’t have a finite number of graph arcs. $\endgroup$ Commented Jun 6, 2023 at 0:53
  • $\begingroup$ Talking about having a finite or infinite number of corners, I was thinking of an example (not closed) in which we have one corner but the result is true: Define your curve $C$ by $C(t) = (t^3,t^2)$ for t in $[-1,1]$. This is infinitely differentiable but has a corner at $t=0$. At this corner the derivative is $(0,0)$ .The curve is non-regular. I observe that by "cutting" the curve at the point $(0,0)$, you can consider the left part and the right part of the curve as a smooth function of $y$, where the derivative of each function (one variable, regarding to $y$) is $0$ at point $x = 0$. $\endgroup$ Commented Jun 6, 2023 at 11:14

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