# Jordan curve theorem: Can every point on the curve be reached from each region?

The Jordan-Schönflies theorem states that any simple closed curve $C$ divides the plane into two connected regions $U$ and $V$, and that any point $x \in C$ is a boundary point of both $U$ and $V$. A stronger statement would be that for any point $x$ in $C$, the set $U \cup V \cup \{x\}$ is pathwise connected. Equivalently, any point in $U$ can be connected to $x$ by a continuous path lying entirely inside $U$ (except for its ending point, of course) and likewise for $V$.

I haven't run across any proof of this stronger statement. Has it been proved? If not, is there a counterexample?

• The follows from Caratheodory's theorem: en.wikipedia.org/wiki/… – Autumn Kent May 9 '13 at 20:08
• A better question is if the same is true for, say, codimension 1 topological spheres in $R^n$, since these can be wild (unlike for $n=2$). – Misha May 9 '13 at 21:53
• Actually, what you state is not the full "Jordan Schonflies theorem". The full statement of the Schonflies theorem is that there is a homeomorphism of the plane taking $C$ to $S^1$, from which your property follows easily. – Lee Mosher May 9 '13 at 22:48

The answers by Kent and Eremenko are, of course, absolutely correct. Here is the more general statement: Let $M$ be a closed connected topological $n-1$-dimensional manifold embedded in ${\mathbb R}^n$ (note that $M$ can be wild for $n\ge 3$). Let $U$ be a component of ${\mathbb R}^n -M$. Then for every point $x\in M$ there exists a continuous path $p: [0,1)\to U$, so that

$$\lim_{t\to 1} p(t)=x$$

In particular, $({\mathbb R}^n -M) \cup \{x\}$ is path-connected. The proof is based on the following observation:

Observation. There exists a function $\phi(r), r\in [0, \infty)$, satisfying $\phi(r)\ge r$ and
$$\lim_{r\to 0+} \phi(r)=0$$ so that: For every $r>0$ the map $$\tilde{H}_0(U \cap B(x, r))\to \tilde{H}_0(U \cap B(x, \phi(r)))$$ (induced by inclusion) is zero. In other words, any two points in $U \cap B(x, r)$ can be connected by a path in $U \cap B(x, \phi(r))$.

The above observation follows from the Poincare/Alexander duality in ${\mathbb R}^n$ (in conjunction with the fact that $M$ is a manifold).

Given the above observation, take a sequence $x_k\in U\cap B(x, \frac{1}{k})$ and construct the path $p$ by concatenating paths $p_i$ in $$B(x, \phi(\frac{1}{i}))\cap U$$ connecting points $x_i, x_{i+1}$, $i\in {\mathbb N}$.

Yes. One way to prove this (and the Jordan theorem too) is to use Complex Variables:-) A good reference is Milnor, MR2193309 Dynamics in one complex variable. Third edition. Annals of Mathematics Studies, 160. Princeton University Press, Princeton, NJ, 2006.

A chapter in this book contans the best exposition of these questions that I know.

Let y be a point in, say, the exterior E of C, and x be a point in C. Then x belongs to the closure of E, hence, there is a sequence of points {x_n} in E which converges to x. As E is connected, there is a chain y --- x_0 --- x_1 --- x_2 --- of continuous curves lying entirely within E.

To see that their union connects y to x, we have to use a trick.

Namely, after a line s_0 connecting y to x_0 within E is defined, let r_0 be its distance to C, S_0 be the open (r_0/2) nbhd of s_0, and X_0 be the open (r_0/2) nbhd of x_0 - now we claim that the line s_1 connecting x_0 to x_1 be defined so that it does not contain points in S_0 except of those in X_0 - which is easily possible by path-connectedness.

A suitable iteration of this trick yields the result.

• You did not use the fact that $C$ is a curve in your argument; for general compact subsets the statement is false (e.g., "topologist's sine curve"). – Misha May 11 '13 at 23:28