# How kinky can a Jordan curve get?

Let s be a simple arc (with distinct end-points) in the Euclidean plane E. If there exists a straight line k in E such that every straight line parallel to k intersects s in at most one point, we will call s a Cartesian arc. This is to bring out the fact that an x-y Cartesian co-ordinate system can be set up in E-with the y-axis parallel to k-in terms of which s becomes the graph of a function of the form y=f(x).

QUESTION: If J is a Jordan curve in E and p is any point of J, is it true that a Cartesian c always exists exists such that (1) c is a sub-arc of J and (2) p is a point of c that is distinct from each of its end-points.

If the answer is yes, then, from the fact that J is compact and homeomorphic to a circle, it should follow that every Jordan curve in E is the union of a finite set of Cartesian arcs.

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Isn't the Koch snowflake a counterexample? –  Darsh Ranjan May 8 '10 at 20:13
If it were so, the Jordan curve theorem would be a lot easier to prove. Also, what you call a Cartesian arc has zero area. So the Osgood curve is a counterexample too. See mathoverflow.net/questions/4722/boundary-of-planar-region/… and the reference therein. –  Harald Hanche-Olsen May 8 '10 at 20:46