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Clarification: by "piecewise", I mean a finite number of pieces.

I'm sure this must be true, but my search for a citation was in vain (although I did learn the new term "polyconvex").

Thanks!

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2 Answers 2

up vote 10 down vote accepted

I don't think this is true. Suppose one of the sets is essentially $\{(x,y):y\geq x^2\}$ in the plane (cut off in some smooth way at the top, to make it compact). And suppose the other one is the same except that the parabolic lower boundary has been replaced by the graph of something like $y=x^2+e^{-1/x^2}\sin (1/x)$. In other words add a fierce oscillation but damped so strongly that the region above the curve is still convex (i.e., $d^2y/dx^2$ remains positive). (I haven't done the arithmetic to make sure my $e^{-1/x^2}$ damping is sufficient; if it isn't, then replace it by a more vigorous damping.) The union of the two convex sets will have infinitely many corners, where $\sin (1/x)$ is 0.

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Thanks! I thought assuming convexity would get rid of the $e^{-1/x^2} \sin(1/x)$ example but I guess not. I don't suppose you know if one can salvage it by assuming the boundaries of $K$ and $L$ are analytic, do you? –  Ryan O'Donnell Nov 10 '10 at 17:35
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Analyticity should salvage it. It certainly eliminates examples like mine, because two analytic curves that intersect infinitely often in a bounded region must coincide (a non-zero analytic function can't ave an accumulation point of zeros). –  Andreas Blass Nov 10 '10 at 22:49

A basic framework in which results like this hold is furnished by o-minimality. This condition implies that the topology is tame (finiteness of number of connected components, triangulations and cell decompositions, etc.). This is applicable to semi-algebraic sets (Tarski), and to many other, comprising subanalytic sets (Denef, van den Dries)...

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