I found this example of uniformly continuous but had a question regarding a step.
Let $S=R$ and $f(x)=3x+7$. Then $f$ is uniformly continuous on $S$.
Proof: Choose $\epsilon > 0$. Let $\delta= \frac{\epsilon}{3}.$Choose $x_o \in R.$ Choose $x\in R$ Assume $|x-x_o| < \delta$. Then
$|f(x)-f(x_o)|=|(3x+7)-(3x_o+7)|=3|x-x_o| < 3\delta= \epsilon$
My question is how do we get $\delta = \frac{\epsilon}{3}$? Is this unique? Or what thought process must I have to have a delta?
Best Answer
By reverse-engineering. Let $S(e)$ be the sentence $$\forall x,y\;(\;|x-y|<d\implies |f(x)-f(y)<e\;).$$ See what $S(e)$ implies about $d,$ for a given $e>0.$ We find that it implies $d\le e/3.$ If the implication is 2-way ("$\iff$") then we will have $d\le e/3\implies S(e),$ and we're done. If we're not sure that it's 2-way, then we know that $|x-y|<e/3$ is necessary, and more work is needed to see whether it is also sufficient.
In this Q we have $$S(e)\iff [\,\forall x,y\;(|x-y|<d\implies 3|x-y|<e)\,] \iff d\le e/3.$$
Given $e>0,$ this value $e/3$ is the largest possible $d$ that makes $S(e)$ true. Some students get mired in details in problems like this by trying to find the largest possible $d,$ when it doesn't matter. $S(e)$ is also true if $d=e/1000000000 .$