So the current repaired conjecture is, or anyhow I presume should be, this: CONJECTURE: Given a plane curve C defined by f = 0 with f(x, y) polynomial, and a simply-connected compact region R of the plane with connected interior C avoids R provided 1. C avoids the boundary of R, and 2. f(x,y) has no critical points (meaning grad f = 0) within R. HERE'S MY PROOF: The only way C can fail to avoid R is (i) C hits boundary(R). But that case is already handled in the theorem statement. (ii) A connected component CC of C lies entirely within R's interior.
From now on we only consider case (ii) since that is all we need to consider. Note that any self-crossing points, or isolated points, or end-points of the curve automatically are critical points. (Actually I do not think end-points are possible, but their possibility or impossibility will not matter.) Hence wlog we need only consider the case where CC has no self-crossing or isolated or end points. Since CC is by assumption entirely contained in a compact region with no endpoints and no crossing points (and it s not just an isolated point) it necessarily either closes on itself, or spirals forever, but the latter option is impossible because f is a finite-degree polynomial. So CC is a closed compact curve of the same topological type as a circle.
Now such a CC necessarily contains a critical point of f(x,y) inside it, because f(x,y)=0 on the closed circle-like curve CC and is (wlog) >0 everytwhere that is infinitesimally inside it, and is continuous and differentiable, hence must have a maximum on the compact region inside CC, which necessarily is >0, and which necessarily is a critical point. Q.E.D. GENERALIZATIONS: It seems to me this conjecture ought to be true not only for algebraic curves in the xy plane, but also for algebraic (D-1)-dimensional surfaces in D-dimensional space. Anyhow when D=3. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)