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I was reading some slides and I stumbled upon this definition of uniform continuity in an interval

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I am unsure on how to trace this back to the definition of uniform continuity that I know: A function $f(x): R \rightarrow R$ is uniformly continuous if $\forall \epsilon > 0$ there exist $\delta > 0$ s.t. $\forall x_1, x_2$ $|x_1 -x_2|< \delta \ \implies |f(x_1) - f(x_2)| < \epsilon$.

In particular what is the use of the supremum here and where is the implication in the definition?

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The supremum replaces the $\forall$ quantifier: essentially, here it expresses that "for all $x,y$ such that $\lvert x-y\rvert \leq h$, $\lvert f(x)-f(y)\rvert \leq \sup (\dots) \leq \delta$". Compare to the implication: "for all $x,y$, whenever $\lvert x-y\rvert \leq h$ then $\lvert f(x)-f(y)\rvert \leq \delta$".

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  • $\begingroup$ why $ \lvert f(x)-f(y)\rvert \leq \sup (\dots) \leq \delta$? sorry for being dense. $\endgroup$
    – Monolite
    Apr 21, 2015 at 22:58
  • $\begingroup$ The $\sup$ is taken over the set of all $(x,y)$ satisfying $\lvert x-y\rvert \leq h$; so in particular, for any $x,y$ in this set, the quantity is at most the $\sup$. $\endgroup$
    – Clement C.
    Apr 21, 2015 at 23:08

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