# I found a proof of the Jordan curve theorem for $C^1$ curves. Can anyone check correctness or generalize method to full JCT?

Bear with me, this is a little long. Let $\gamma: S^1 \to \Bbb{R}^2$ be a continuously differentiable simple closed curve. We will often refer to $S^1$ as $[0, 1]$ with the endpoints identified. Let $G \subset \Bbb{R}^2$ be the image of $\gamma$, and $G^c$ be its complement in $\Bbb{R}^2$. We will show that $G^c$ is disconnected.

Let $A$ be the set of points defined by $A = \{x \in G^c: [\gamma]$ is the identity of $\pi_1(\Bbb{R}^2-\{x\}) \}$, where $\pi_1$ is the first homotopy group and $[\gamma]$ is the homotopy class of $\gamma$. Let $B = G^c - A$ be the set of points around which $\gamma$ has nonzero winding number.

Lemma: $A$ is open.

Proof: Let $x \in A$, and let $h: [0, 1]^2 \to \Bbb{R}^2 - \{x\}$ be a homotopy from $\gamma$ to a constant map. By compactness of $[0,1]^2$, there exists $\delta > 0$ such that all $(s, t) \in [0, 1]^2$, $|x - h(s, t)| > \delta$, since $x$ is not in the image of $h$. It follows that for all $y$ with $|x-y| < \delta$, $h$ maps $[0,1]^2$ into $\Bbb{R}^2 - \{y\}$, and thus $\gamma$ is homotopic to zero in $\Bbb{R}^2 - \{y\}$. Thus $y \in A$ as claimed.

Lemma: $B$ is open.

Proof: It's essentially the same. Let $x \in B$ and let $h$ be a homotopy from $\gamma$ to a circular loop $f(\theta) = e^{in\theta} - x$, $\theta \in [0, 1]$ $0 \neq n \in \Bbb{Z}$, such that $h$ does not pass through $x$. By the same compactness argument, the image of $h$ has minimum distance $\delta > 0$ from $x$. But the loop $f$ has nonzero winding number around points within unit distance of $x$, so if $|x - y| < \min (1, \delta)$, the homotopy $h$ shows that $\gamma$ also has nonzero winding number around $y$. Thus $y \in B$, and $B$ is open.

To show $G$ is disconnected we will have to show both $A$ and $B$ are nonempty. The former claim is easy to prove: let $R = \max_{z \in G} |z|$. If $|x| > R$, the homotopy $h(s, t) = (1-t)\gamma(s)$ takes $\gamma$ to a constant map without passing through $x$, so $x \in A$.

The proof that $B$ is nonempty is more involved. Let $p = \gamma(t_0)$ be a point which attains $|p| = R$, i. e. $p$ is the farthest point $\gamma$ travels from the origin. The maximum is attained by compactness of $G$.

Let $K$ be the ray from $p$ to the origin. The tangent line to $\gamma$ at $t_0$ is perpendicular to $K$, or else the mean value theorem would imply that $p$ was not the point of maximum norm on $\gamma$. Since $\gamma$ locally "looks like" its tangent line around $t_0$, there is a neighborhood $U$ of $t_0$ such that $\gamma(U) \cap K = p$ (you can make this rigorous using the fact that $\gamma$ is $C^1$). Note that $S^1 - U$ is compact, and $\gamma|_{S^1 - U}$ does not go through $p$, by injectivity of $\gamma$. Therefore there exists $\delta > 0$ such that $\min_{t \in S^1 - U} |p - \gamma(t)| = \delta$. It follows that if $q$ is a point on $K$ with $|p-q| < \delta$, then the line segment from $p$ to $q$ only intersects $G$ at $p$.

We claim that every ray $L$ starting at $q$ intersects $G$ at some point. The proof considers three cases.

Case 1: $L$ goes through $p$. Then it trivially intersects $\gamma$ at $p$.

Case 2: $L$ goes through the origin. Then, since $q$ lies on $K$, which also goes through the origin, $L$ intersects $G$ somewhere if and only if $K$ intersects $G$ at some point other than $p$. Assume that this is not the case. Note that $G$ is contained in the disc $D = \{z \in \Bbb{R}^2||z| \le R\}$. Clearly $D-K$ has two connected components. Because $G$ is homeomorphic to a circle and only intersects $K$ in one point, $G-K = G - \{p\}$ is connected, which implies $G-K$ is contained entirely in one of the components of $D-K$. But this is false, because the tangent to $\gamma(t)$ at $t=t_0$ is perpendicular to $K$ at $p$. This implies that $\gamma$ crosses from one side of $K$ to the other, and hence intersects both components of $D-K$, a contradiction. We conclude that $L$ actually does intersect $G$.

Case 3: $L$ does not intersect $p$ or the origin. Assume that $L$ does not intersect $G$. Let $M$ be the ray from $q$ to $p$. Then $X = \Bbb{R}^2 - (L \cup M)$ is clearly disconnected, and $G - (L \cup M)$ is not: $p$ is the point of maximum norm in $G$, so $M$ still only intersects $G$ at $\{p\}$. But once again, $\gamma$ is perpendicular to $M$ at $p$ and hence crosses from one connected component of $X$ to another. This is a contradiction.

We conclude every ray starting at $q$ intersects $G$. Now let $u = \min_{t \in S^1} |q - \gamma(t)|$, and let $v = \max_{t \in S^1} |q - \gamma(t)|$. Define $\Gamma$ to be the region $\bigcup_{1 \le r \le v/u} \{q + r \gamma(t)| t \in S^1\}$. Then, because every ray from $q$ intersects $G$, $\Gamma$ contains the circle $C$ of radius $v$ centered at $q$. Note that $\Gamma$ obviously deformation retracts onto $G$ via radial shrinking centered at $q$. $\Gamma$ also deformation retracts onto $C$, again by radially contracting it towards $C$. Therefore, the inclusion maps $G \subset \Gamma$, $C \subset \Gamma$ induce isomorphisms of the homotopy groups $\pi_1(G) \cong \pi_1(\Gamma) \cong \pi_1(C)$.

Now, $0 \neq [\gamma] \in \pi_1(G)$, as $\gamma$ is a homeomorphism from $S^1$ to $G$. Therefore, the image of $[\gamma]$ under the isomorphism $\pi_1(G) \to \pi_1(C)$ is nonzero. In other words, the homotopy induced by the deformation retract takes $\gamma$ to a nontrivial circular loop around $q$. The homotopy never passes through $q$, so $\gamma$ is not nulhomotopic in $\Bbb{R}^2 - \{q\}$. It follows that $q \in B$. Thus, $B$ is nonempty and $\Bbb{R}^2 - G$ is disconnected as claimed.

Any thoughts, suggestions or critiques are appreciated. I'm still trying to think of a way to dispense with the $C^1$ assumption and extend the proof to the full JCT. Thanks!

• I don't know enough math to understand your proof, but I'm really curious as to whether it's right! Since you haven't received a response here so far, I think you should cross-post this on Mathoverflow. – Train Heartnet Oct 16 '17 at 3:46