0
$\begingroup$

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

enter image description here

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ Becasue $h(z_n)\to h(z)$ whenever $z_n\to z$ was shown immediately before stating this. $\endgroup$ 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

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.