1. ## Uniformly continuous

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

2. 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

3. Oh, triangle inequality. Didn't think of that.