Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A.
f_n converges uniformly so |f_n(x)-f(x)|<epsilon.
f_n is uniformly continuous so |x-y|<delta implies |f_n(x)-f_n(y)|<epsilon
it's a standard trick (also known as "3-epsilon trick"):
write |f(x)-f(y)| < |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)| < epsilon+epsilon+epsilon
just be careful with the order of all those epsilons, deltas and n's
Oh, triangle inequality. Didn't think of that.