On proving uniform continuity

Question

I am proving the uniform continuity of certain functions on closed intervals (more like finding the corresponding value of $\delta = \delta(\varepsilon) \; \forall \varepsilon > 0$, that is what the textbook asks me of) and I want to understand the following: The definition of uniform continuity of $f(x)$ on an interval $[a,b]$ states that $f(x)$ is uniformly continuous on $[a,b]$ if and only if $$ \forall \varepsilon > 0 \; \exists \delta > 0 \; \forall x_1, x_2 \in [a,b] \subset D_f: \; |x_1 - x_2| < \delta \implies |f(x_1) - f(x_2)| < \varepsilon. $$ Can I let $x_1$ or $x_2$ be the min or max of $\{x_1,x_2\}$?

Answer 1

A function is continuous at $x=a$ if when given $\varepsilon > 0, $ you can find $\; \delta > 0 $ such that $|a - x_1| < \delta \implies |f(a) - f(x_1)| < \varepsilon.$

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New concept: A function is uniformly continuous if, when given $\varepsilon > 0, \; \exists\ \delta > 0 $ such that $|x_1 - x_2| < \delta \implies |f(x_1) - f(x_2)| < \varepsilon,\ $ i.e. for each $\ \varepsilon > 0,\ $ you can find $\delta>0$ that works for all $\ x_1,\ x_2\ $ in the domain. So in order to prove a function is uniformly continuous, you suppose you are given $\ \varepsilon>0,\ $ and your task is to find such a $\ \delta.$

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Somewhat obviously, a function being uniformly continuous implies it is continuous. However, the converse is only true if the domain is compact i.e. closed and bounded.

Edit: yes, of course you can take let $x_1$ or $x_2$ be the min or max of $\{x_1,x_2\}$ without loss of generality...

For example, you could say, If $D$ is the domain, then let $\ x_1\in D\ $ and let $\ x_2 \geq x_1.$ Although, this will only work if the domain is an ordered set like a subset of $\ \mathbb{R}.$ Which is what it sound like you have to do: "I am proving the uniform continuity of certain functions on closed intervals" (of $\ \mathbb{R}$).