Suppose f maps a closed and bounded set D to the reals is
continuous. Then the Uniform Continuity Theorem says f is uniformly
continuous on D. To prove this, we will suppose it is not true, and
arrive at a contradiction.
So, suppose f is not uniformly continuous on D. If f is not
uniformly continuous on D, then there exists epsilon greater than
zero and sequences (a_n) and (b_n) in D for which |a_n -b_n| <
1/n and |f(a_n)...