Subject: spacetime Date: Wed, 29 Nov 2000 22:45:39 -0500 (EST) From: Tom Goodwillie OK, I can't resist. (This seems to be my week for fooling around rather than doing the work I'm paid to do.) That spacetime problem has a solution: Normalize the speed of light to be 1. Let X be the "space" in spacetime. It can be any metric space for this purpose. We are seeking a spacelike hypersurface, meaning a real function F on X satisfying F(x)-F(x') < d(x,x') when x and x' are distinct. We are given a function f defined only on a finite subset A of X and satisfying f(a)-f(a') < d(a,a') . Choose K < 1 such that f satisfies the Lipschitz condition (*) f(a)-f(a')\leq Kd(a,a') for all a and a'. For the rest, we don't even need the set A to be finite. Lemma: Let X be a metric space and A a subset. Any function f on A satisfying (*) can be extended to a function on all of X, still satisfying (*). Proof: We'll denote the extension by F. The extension we construct will be the largest possible (pointwise). Note that any extension must satisfy F(x)\leq f(a)+Kd(x,a) for all a in A. Write F_a(x) = f(a)+Kd(x,a). The function F_a satisfies (*), by the triangle inequality. Define F(x) to be the infimum of F_a(x) over all a in A. Then F(a)=f(a) for any a in A, and F will satisfy (*) because the the inequality F_a(x)\leq F_a(x')+Kd(x,x'), valid for all a, carries over to the inf (for any fixed x and x'). We must also make sure that the inf exists for all x. That means finding a lower bound for F_a(x) independent of a. For this, fix some a_0 and observe F_{a_0}(x) - F_a(x) = f(a_0) - f(a) +Kd(x,a_0) - Kd(x,a) \leq K(d(a_0,a)+d(x,a_0)-d(x,a)) \leq 2Kd(x,a_0) Tom Goodwillie