# Prove C[0,1] is closed in B[0,1] (sup norm)

by Oster
Tags: norm, prove
 Mentor P: 18,042 Hi Oster! Take |x-y|r, and you'll need to find something for |f(x)-f(y)| See if you can do something with this: $$r<|h(x)-h(y)|=|(h(x)-f(x))+(f(x)-f(y))+(f(y)-h(y))|$$
 HW Helper P: 3,352 Well it can't hurt to see the other way, maybe it'll get you more used to it. One of the most useful characterizations of a closed set is : A set W is closed if and only if every convergent sequence in W has it's limit in W. In R^n with the usual Euclidean metric, convergence you've seen before, it's the normal idea. In a general metric space (X,d) we say a sequence $f_n$ converges to $f$ if $\displaystyle\lim_{n\to\infty} d(f_n, f) = 0$. So in this case, we have to show that every convergent sequence of continuous functions on [0,1] has it's limit in C[0,1]. The norm we have here through is the supremum norm, so convergence in the supremum norm is just the concept of uniform convergence. It is a well known fact (and easy to prove on your own) from real analysis that if a sequence of continuous functions converges uniformly, then its limit is also a continuous function, which suffices to show C[0,1] is closed with respect to this supremum metric.