Proving a corollary to the Jordan Curve Theorem

Question

A Jordan curve is a continuous closed curve in $\Bbb R^2$ which is simple, i.e. has no self-intersections. The Jordan curve theorem states that the complement of any Jordan curve has two connected components, an interior and an exterior.

I would like to prove the following statement, which I think is a corollary to the Jordan Curve Theorem:

Let X, Y, Z, and W be four consecutive points (in order) of a Jordan curve C. Then if C1 is a continuous curve from X to Z and C2 is a continuous curve from Y to W, and C1 and C2 are either both entirely in the interior of C, or both entirely in the exterior of C, then C1 and C2 must intersect.

This statement is intuitively obvious. Suppose C1 and C2 are both on the exterior of C, for instance. Then we can form a Jordan curve D consisting of C1 and the segment of C between X and Z. And intuitively, it seems like any curve starting from Y which is entirely on the exterior of C must pass through the interior of D. And it also seems intuitively clear that W cannot be on the interior D. Therefore, C2 must intersect D, and since it can't intersect C, it must intersect C1.

But how would I prove those two statements I said were intuitively obvious?

Any help would be greatly appreciated.

Thank You in Advance.

P.S. The reason I'm asking this is that this statement is used in this webpage's solution of the famous three utilities problem.

Answer 1

If the conclusion does not hold, place a fifth point in the component that does not contain the curves C1 and C2. Connect this fifth point to the four points X, Y, Z, W by connected arcs that do not intersect each other or meet the Jordan curve or cross it. The result is a complete graph $K_5$ on five vertices in the plane with no self intersections. This is impossible.

There was no real trouble for the MO question because the Jordan curve was required to be an ordinary square, https://mathoverflow.net/questions/35514/pair-of-curves-joining-opposite-corners-of-a-square-must-intersect-proof/35517#35517

In comparison, think of this Jordan curve as, say, the Koch snowflake, which is far from the worst possible. There is now a lemma that needs proving, that there really is a position to place the fifth point and four continuous arcs that reach W,X,Y,Z, without touching the Jordan curve. It is true, but subtle. Probably best to take the Jordan curve as piecewise real analytic, you get local convexity that way for each component.

Answer 2

Note that this is essentially the Lovers & Haters problem in disguise. Ref e.g. Lovers and haters via the Jordan curve theorem . While not strictly necessary, the problem dramatically simplifies by using the Schönflies theorem (an involved extension of JCT).

Case 1: $C_1,C_2$ are both interior
The Schönflies theorem tells us there is a homeomorphism $f$ that conformally maps the interior of the Jordan Curve , $\gamma$, to the unit disc, and maps the image of the Jordan curve to the unit circle, i.e. there is a homeomorphism from the Jordan Curve's image $[\gamma]\cup$ its interior, to the closed unit disc. The 'alternating' structure of $W, X,Y,Z$ is preserved under homeomorphism since $[\gamma]-\big\{W,Y\big\}$ has two components, with $X$ in one and $Z$ in the other and this is preserved under homeomoprhism.

Thus the problem reduces to labeling $W',X',Y',Z'$ along the unit circle and constructing not necessarily simple closed curves from $W'\to Y'$ and $Z'\to X'$ inside the unit disc (except for end points). Must these curve intersect? The answer is yes via a winding number argument that is essentially identical to that in the Lovers and Haters problem. Details: take the curve $Z'\to X'$ and draw another curve $X'\to Z$ outside the unit disc (except for endpoints) such that it hits the ray from $0$ through $W'$ but not the ray from $0$ through $Y'$. Call the combined close curve $\sigma$. Then $\big\vert n\big(\sigma, W'\big)\big\vert =1$ by construction but $ n\big(\sigma, Y'\big) =0$ since $\sigma$ does not intersect said ray running through $Y'$. If $W'\to Y'$ did not intersect $[\sigma]$ then $n\big(\sigma,Y'\big)=n\big(\sigma,W'\big)$ by path connectivity; hence the curve does intersect $[\sigma]$ and the intersection cannot happen at endpoints hence it occurs inside the unit disc i.e. the curve from $W'\to Y'$ must intersect the curve from $Z'\to X'$.

Case 2: $C_1,C_2$ are both exterior
via translation we can assume WLOG that $0$ is in the bounded component of $\mathbb C-[\gamma]$ (i.e. if not, then select some $p$ in the bounded component of $\mathbb C- [\gamma]$ and apply the map $z\mapsto z-p$).

Now working on the sphere, i.e. $\cong \mathbb C_\infty$ the one point compactification, consider the map $z\mapsto z^{-1}$. This sends Jordan curve $\gamma$ to another Jordan curve $\gamma^*$ and $\gamma^*$'s interior, exterior is the exterior, interior of $\gamma$ (since $0\mapsto \infty$ under this Möbius transform). Now Case 1 tells us the the interior curves for $\gamma^*$ from $X^*$ to $Z^*$ and $W^*$ to $Y^*$ must intersect and this is preserved under homeomorphism hence the exterior curves for curves for $\gamma$ from $X$ to $Z$ and $W$ to $Y$ must intersect as well.