Suppose f maps a closed and bounded set D to the reals is continuous. Then the...

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) - f(b_n)| > epsilon. Give comment as to why this must 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 that they are in fact equal. Explain why this is leads to a contradiction. Finish the proof.

Expert Solution

since every bounded sequence has a convergent subsequence and a_n and b_n are sequene in D so both are bounded hence we can consider the convergent subsequece. since D is closed so limit also lies in D.

here bounded set is K so please replace D by K.

Colby Messinger answered 2 years ago

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