The impression grows on me that this topic may have become mysteriously contentious for reasons unconnected to mathematics. So I'll risk one more version of this proof, expanded to fill gaps which have emerged meantime, and invite anyone still interested to email me off-list. Theorem: Given a continuously differentiable function f(x, y) , and a compact subset R of the Cartesian plane with f nonzero on the boundary of R : f has a zero in R if and only if there is some point (x, y) in R where df/dx = f = 0 . Proof: Denote by S the curve defined implicitly by f(x, y) = 0 . If f has a zero in R , some point lies both on S and in R ; and the entire connected component of S containing it lies in R by continuity, f being nonzero on the boundary of R . Let y' be an endpoint of the interval to which that component of S projects on the y-axis; y' is finite since it lies within the projection of the compact set R . By the implicit function theorem, constraint f(x, y) = 0 defines a function x = g(y) , single-valued in any interval of y where df/dx != 0 . Since g ceases to be single-valued at y' , we have df/dx = f = 0 at (x', y') where x' = g(y') ; also (x', y') in R . The converse is trivial. QED. Please note that the annulus & circle counterexample was retired -- ooh -- at least three days ago. While the worked example concerned is available on request, prospective consumers are warned that it does depend on the ability to sort the set {1,3,2} correctly into ascending order. Further extensions: non-compact R ; higher dimensional R and S . Fred Lunnon On 6/3/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 5/31/14, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, May 30, 2014 at 4:28 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In answer to Andy: it shouldn't matter which compactification is used --- so long as we don't forget about the necessity. "Or some other ..." --- trick question? There aren't any others!
Not true; the Stone-Cech compactification is much, much, larger than RP2.
Or without using such heavy machinery, consider the sequences (0,n) and (1, n). In a compactification, they each have to have a convergent subsequence. But there's no reason that these two convergent subsequences have to converge to the same point, as they do in the two compactifications you mention.
Andy
Compactness is all very well, but practical applications are liable to involve "regions" extending to infinity. Having earlier casually postponed consideration on the grounds that any old completion should solve the problem, I must now admit that it's not so simple. For a start, as Andy pointed out, there are quite a few choices.
A principle which I elucidated early in life --- I believe it was on the occasion I was first exposed to the serial trumpet abuse of Miles Davis --- is that the potential for an activity does not automatically imply a recommendation to engage in it. Numerous other instances since observed include: dropping a 5lb hammer on one's foot; credit default swaps; and Stone–Čech c*mp*ct*f*c*t**n.
Indeed, it's not clear that any of these alternatives has much to offer from the standpoint of computational geometry, besides the classical duo that in my innocence I was relying on. However, Andy's suggestion that (0, n) and (1, n) might be assigned different limits proves prescient, as this example involving a rectangular hyperbola demonstrates (it was time to get away from unit-circle counterexamples) ---
Consider region R = { (x, y) | x >= 0 & y >= 0 } and functions (A) f = x*y + 1 ; or (B) f = x*y - 1 . In case (A) there are no zeros in R , in case (B) there are plenty; however neither projective nor complex boundary points at infinity can distinguish them. [Other examples might involve some complicated singularity at infinity.]
However, I have an elementary workaround which looks intuitively plausible, at least in simple non-compact situations. Apply the (relaxed) intrinsic criterion with respect to THREE different coordinate frames, say: (x, y) ; (y, x) ; (x+y, x-y) . Courtesy of its asymptotes, an interior hyperbola may evade detection in two frames; but it cannot then also hide in the third.
What constraints on a non-compact region & boundary would stand this up? And just how does one go about proving something like this anyway?
Fred Lunnon