On Sat, May 31, 2014 at 8:05 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote: . Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to R contains the origin
I asked earlier whether "if" could be strengthened to "if and only if"; a counter-example is easy to find. However, the following "INTRINSIC" criterion -- which is anyway what I have actually been experimenting with -- seems more robust in this respect, besides (like the rane in Spane?) remaining firmly in 2-space:
Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane, C avoids R if and only if C avoids the boundary of R , and there is no point (x, y) in R where df/dx = f = 0 .
Taking f to be x^2 + y^2 = 1, so C is the unit circle, and the only critical point is the origin, and negating both sides of your conjectured equivalence, since I find "intersects" easier to think about than "avoids", your conjecture in this case reduces to R intersects the unit circle if and only if The boundary of R intersects the unit circle or R contains the origin. One direction of this is true: If R intersects the unit circle, but the boundary of R does not, then since the unit circle is connected, R contains the entire unit circle, and since it is simply connected, it contains the entire unit disc. But for the converse, a small disc around the origin is an obvious counterexample. This is the second conjecture you've posted about all plane algebraic curves where the answer is "well, it's trivially false for the circle". Since you're presumably not just spewing out random conjectures, presumably you have some intuition about plane algebraic curves that you are trying to capture and failing. Maybe it would work better for you to post what this intuition is informally, rather than making failed attempts to formalize it and having math-fun play the "well, this is obviously false, let's try to guess what Fred meant to say" game. Andy