Conditions for validity of interior test In the case of a polygon (bounded by finitely many straight lines) in the plane, points can be tested for being inside the region bounded by the polygon by casting a ray from the point in any direction (not going through a vertex of the polygon) and inspecting the parity of the number of intersections with the polygon. In this question this algorithm was used to characterize the interior of a region bounded by a smooth curve. So my follow up question is:

Which conditions do we have to impose on a Jordan curve, such that the test described above works?

It is clear that any line through the region is supposed to intersect the curve only finitely many times, since otherwise parity is not defined. I just noted that the counterexamples I gave in the original questions (topologists sine circle / circle with antenna) are not given by Jordan curves, so I now tend to believe that a continuous embedding of the circle is enough. I would like to know this for certain though.
Thank you very much for your time!
 A: You have not stated the test carefully enough. As stated, that test can be invalid even for polygons or smooth curves. For example, your Jordan curve could be a round circle, $c$ could be a point outside the circle, and the ray you pick could just happen to be a ray through $c$ that is tangent to the circle, intersecting that circle in exactly one point. Something similar could happen to a polygonal closed curve.
If you reformulate the test, it will work. For a smooth curve, the test works as long as the ray is transverse to the curve. For a polygonal curve, the test works as long as the ray passes through no vertex of the curve, which implies that the ray crosses the curve transversely. On top of that, for each $p$, a generic ray based at $p$ will satisfy this transversality condition. That makes the test very useful: as said the test is invalid for some special rays, but for generic rays it is valid.
There is indeed a general version of the test. On a Jordan curve $c$, every point $x \in c$ has a neighborhood $U$ and a homeomorphism $\phi : U \to (-1,+1) \times (-1,+1)$ such that $\phi(U \cap c) = (-1,+1) \times \{0\}$ and $\phi(x)=(0,0)$. Let's say that the ray $R$ crosses $c$ transversely at $x$ if $U$ and $\phi$ exist so that $\phi(R \cap c) - (0,0)$ has two components, one in $(-1,+1) \times (-1,0)$ and the otehr in $(-1,+1) \times (0,+1)$. Then the general test says: if a $R$ is a ray based at $p$, and if $R$ crosses $c$ transversely at each point of $R \cap c$, then $p$ is inside $c$ if and only if the number of crossings equals zero. In fact you could even apply this to a very general concept of "ray", namely any continuous, proper injection $R : [0,\infty) \to \mathbb R^2$ such that $R(0)=p$.
The difficulty, though, is that it that for general Jordan curves, there is no statement saying that "generic" rays cross the curve transversely. That means transverse rays to which the test for which the test is valid are hard to find.
