Suppose f maps a closed and bounded set D to the reals iscontinuous. Then the Uniform Continuity Theorem says f is uniformlycontinuous on D. To prove this, we will suppose it is not true, andarrive at a contradiction.
So, suppose f is not uniformly continuous on D. If f is notuniformly continuous on D, then there exists epsilon greater thanzero and sequences (a_n) and (b_n) in D for which |a_n -b_n| <1/n and |f(a_n) - f(b_n)| > epsilon. Give comment as to why thismust be true if f is not uniformly continuous on D.
Nothing says the sequences (a_n) and (b_n) converge. But,because f is continuous on D, there exists subsequences of (a_n)and (b_n), say (a_nk) and (b_nk) respectively, that do converge.Give comment as to why this would be true.
Show that both limits of the subsequences must lie in D and thatthey are in fact equal. Explain why this is leads to acontradiction. Finish the proof.
