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Cauchy's Theorem in complex analysis Rudin in proof. Why $g$ is a uniformly continuous function?

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    $\begingroup$ Becasue $h(z_n)\to h(z)$ whenever $z_n\to z$ was shown immediately before stating this. $\endgroup$
    – Hagen von Eitzen
    Nov 16, 2014 at 21:46
  • $\begingroup$ Do you know complete proof? For example, why the function g is uniformly continuous. $\endgroup$
    – reza
    Nov 16, 2014 at 22:08

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Any continuous function on a compact set is uniformly continuous.

We can prove this immediately. Fix an $\epsilon > 0$. Then for each point $x$ in the compact set, there is an associated $\delta_x$ as in the $\epsilon$-$\delta$ formulation of continuity. Associate to each point $x$ the ball of radius $\delta_x$ centered at $x$. These balls cover the compact set, and therefore a finite subset of them cover the set. Choosing the minimum radius of these covering balls gives the uniform $\delta$ for the $\epsilon$-$\delta$ formulation of uniform continuity.

As $g$ is continuous on a compact set, it is uniformly continuous.

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